Class 10

You can Download Periodic Table and Electronic Configuration Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Chemsitry Solutions Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration

Periodic Table and Electronic Configuration Text Book Questions and Answers

Kerala Syllabus 10th Standard Chemistry Chapter 1 Text Book Page No: 7

→ What is the basis of classification of elements in the periodic table?
Answer:
Atomic Number

Sslc Chemistry Chapter 1 Questions And Answers Text Book Page No: 8

→ Atomic number of sodium is 11 Electronic configuration – 2,8,1
GroupNumber — …………..
Period number — …………
Answer:
Group Number — 1
Period number — 3

→ Is the group 1 element a metal or a nonmetal?
Answer:
Metal

→ Write the electronic configuration of sodium and argon and complete the Table.

Answer:

This article mainly deals with the chemical formula of lithium oxide, the structural lithium oxide formula with its properties and uses.

Periodic Table And Electronic Configuration Class 10 Notes Text Book Page No: 9

→ How many electrons are present in the M shell, the outermost shell of argon?
Answer:
8

→ What is the maximum number of electrons that can be accommodated in the M Shell?
Answer:
18

→ The ‘K’ shell, which is the first shell, has 1 subshell. The next ‘L’ shell has 2, and so on. What will be the number of subshells in the ‘M’ shell and ‘N’
M = ……………… , N = ……………….
Answer:
M = 3, N = 4

→ Which subshell is common to all shells?
Answer:
S

Sslc Chemistry Chapter 1 Notes Kerala Syllabus Text Book Page No: 10

→ Complete the Table 1.3

Answer:

→ Complete the Table 1.4

Answer:

→ What is the maximum number of electronics that can be accommodated in the ‘s’?
Answer:
2

→ What may be the maximum number of electrons to be filled in the ‘p’ subshell?
Answer:
6

Periodic Table And Electronic Configuration Class 10 Text Book Page No: 11.

The atomic number of hydrogen is 1(₁H)

→ How many electrons are present?
Answer:
1

→ In which shell is the electron filled?
Answer:
‘K’ shell

→ In which subshell?
Answer:
S

→ How many electrons are present in helium (₂He)?
Answer:
2

Follow along with the alkene reactions cheat sheet.

Chemistry Class 10 Chapter 1 Kerala Syllabus Text Book Page No: 12

→ Complete the subshell electronic configuration?
Answer:
1s²

→ Write the electronic configuration of Lithium (3Li)
Answer:
1s² 2s¹

→ Complete the electronic configuration of beryllium?
Answer:
Be[Z=4] -1s² 2s²

→ Write the electronic configuration of Boron
Answer:
B[Z=5] -1s¹ 2s² 2p¹

→ Write the electronic configuration of Carbon
Answer:
C[Z=6] – 1s² 2s² 2p²

→ Complete the Table 1.6

Answer:

Class 10 Chemistry Chapter 1 Periodic Table And Electronic Configuration Text Book Page No: 13

→ How was the shell wise electronic configuration of potassium written?
Answer:
2, 8, 8, 1

→ Compare the energies of Is and 2s subshells. Which one has lower energy?
Answer:
1s < 2s

Sslc Chemistry Notes Chapter 1 Kerala Syllabus Question 14.
Among the 3s & 3p subshells which has higher energy?
Answer:
3s < 3p

→Among the 3d & 4s subshells which has higher energy?
Answer:
4s < 3d

→ Write down the subshells in the increasing order of their energies.
Answer:
1s <2s <2p <3s <3p <4s <3d <4p

→ Write the subshell wise electronic configu-ration of potassium.
Answer:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s²

→The electronic configuration of scandium (2lSc) is
Answer:
s² 2s² 2p⁶ 3s² 3p⁶ 3d¹ 4s²

Periodic Table And Electronic Configuration Kerala Syllabus Text Book Page No: 14

→ Write the electronic configu ration of 22Ti, 23V, the two elements after Sc.
Answer:
₂₂Ti — 1s² 2s² 2p⁶ 3s² 3p⁶ 3d² 4s²
₂₃V — 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s²

→Which is the noble gas preceding sodium (11Na)?
Answer:
Neon(Ne)

→ Write its subshell electronic configuration.
Answer:
₁₀Ne – 1s² 2s² 2p⁶

→ Subshell electronic configuration of sodium?
Answer:
₁₁Na – 1s² 2s² 2p⁶ 3s¹

Periodic Table And Electronic Configuration Class 10 Important Questions Text Book Page No: 15

→ Using the symbol of neon, write the subshell electronic configuration of sodium?
Answer:
[Ne] 3s¹

→ Complete the Table 1.7

Answer:

→ Write the subshell electronic configuration of 24Cr
Answer:
₂₄Cr – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹

→ On the basis of this, identify the correct electronic configuration of 29Cu from those given below:
Answer:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹ 4s² – False
1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ – True

Periodic Table And Electronic Configuration Class 10 Question Paper Text Book Page No: 16

If the subshell wise electronic configuration of an atom is 1s² 2s² 2p⁶ 3s², find answers to the following:

→ How many shells are present in this atom?
Answer:
3

→Which are the subshells of each shell?
Answer:
K — Is, L — 2s, 2p, M — 3s

→Which is the subshell to which the last electron was added?
Answer:
3s

→ What is the total number of electrons in the atom?
Answer:
12

→ What is its atomic number?
Answer:
12

→ How can the subshell electronic configuration be written in a short form?
Answer:
[Ne]3s²

Periodic Table And Electronic Configuration Class 10 Questions And Answers Text Book Page No: 17

→ Complete the Table 1.8

Answer:

→ Which is the subshell of lithium to which the last electron was added?
Answer:
S

→ What about the subshell to which the last electron of nitrogen was added
Answer:
p

→ What is the relation between the subshell to which the last electron was added and the block to which the element belongs?
Answer:
The subshell in which the last electron enters represent the block in which the element belongs.

→ Write the subshell electronic configuration of the following elements and find the blocks to which they belong.
a. 4Be: ………………..
b. 26Fe……………..
c. 18Ar: ……………
Answer:
a. 4Be : 1s² 2s² — s block
b. 26Fe : Is² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² — d block
c. 18Ar : 1s² 2s² 2p⁶ 3s² 3p⁶ — p block

Text Book Page No: 18

→ Complete the Table 1.9

Answer:

→ Complete the Table 1.10

Answer:

→ What is the relation between number of electrons present in the last ‘s’ subshell and their group number?
Answer:
The number of electrons in the outermost ‘s’
subshell = The group number

Text Book Page No: 20

→ When the s block elements react, do they donate or accept electrons?
Answer:
They donate electrons.

→ Which type of chemical bond is usually formed?
Answer:
Ionic bonds

→ How many electrons are donated by the first group elements in chemical reactions ?-
Answer:
One

→ How many electrons are donated by the second group elements in chemical reaction?
Answer:
Two

→ Complete the table 1.11

Answer:

→ ‘s’ block elements are present at the extreme left side of the periodic table. Relating to their position, what other characteristics can be listed out?
Answer:

• More metallic character s
• Less ionization energy
• Less electronegativity
• Lose of electrons in chemical reaction
• Compounds are mostly ionic
• Oxides and hydroxides are basic in nature

Text Book Page No: 21

→ Which are the group included in the p block
Answer:
13, 14, 15, 16, 17, 18

→ In which subshell did the filling of the last electron take place?
Answer:
p subshell

→ Complete the table 1.13.

Answer:

Text Book Page No: 22

The outermost subshell wise electronic configuration of an element Y (Symbol is hot real) is 3s2 3p4.

→ To which period and group does this element belongs to?
Answer:
Period = 3, Group = 16

→ Write down the outermost subshell electronic configuration of the element coming just below it in the same group?
Answer:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁴

→ Find out examples of elements in such different states with the help of the periodic tables?
Answer:
Solid – Li, Be, B, C, Na, Mg, Al, Si
Liquid – Br
Gas – H, He, N, O, F, Ne

→ Which element has the highest ionization energy in each period?
Answer:
Group 18 elements.

Text Book Page No: 23

→ The elements having the highest electronegativity is in the p block. Find its name and position?
Answer:
Fluorine F, Period – 2, p block, Group 17

→ Analyze the general characteristics of the p block elements and prepare a note on this?
Answer:

• The outermost p subshell of the p block elements contains 1 to 6 electrons.
• Elements showing positive oxidation state and negative oxidation state are members of this block.
• There are metals and nonmetals in these blocks.
• Elements in the solid, liquid and gaseous states are present in p block.

→ Complete the table 1.14

Answer:

→ Which element has a valency 1?
Answer:
Y

→Which element shows metallic character?
Answer:
X

→ Which element has the highest ionization energy?
Answer:
Y

→ Write the chemical formula of the compound formed by the combination of X and Y and label the oxidation states?
Answer:
Compound: X Y₂
Oxidation state: X²⁺, Y^1-

→ Where is the position of d block elements in the periodic table?
Answer:
3rd Group to 12th Group

→ From which period onwards does the d block begin?
Answer:
4

Text Book Page No: 24

→ Put a tick mark ✓’ against the statements below, which are applicable to d block elements.
Answer:
1. ‘✓’ These are metals.
2. ‘✓’ The last electron is filled in the penultimate shell.
3. ‘✗’ In the case of these elements in the 4th period, the last electron is filled in 4s.
4. ‘✓’ These are found in groups 3 to 12 of the periodic table.

Text Book Page No: 25

→ Complete the table 1.16

Answer:

→ How does Fe change to Fe²⁺?
Answer:
By losing 2 electrons from 4s valence subshell.

Text Book Page No: 26

→ Write down the subshell electronic configuration of Fe21.
Answer:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶

There is only a small difference of energy between the outermost s subshell and the penultimate d subshell of transition elements.

→ If so, which will be the subshell from which iron loses the third electron?
Answer:
From 3d sub-shell

→ Write the electronic configuration of Fe3+ on the basis of this.
Answer:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵

→ Write the subshell electronic configuration of Manganese (Mn).
Answer:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s²

→ Complete the table 1.17

Answer:

Text Book Page No: 27

→ Examine these compounds available. Find more colored compounds and extend the list.
Answer:

• Copper sulfate CuSO₄.5H₂O – blue,
• Copper nitrate Cu(NO₃)₂.6H₂O – pink.
• Potassium permanganate KMnO₄ – violet.
• Ferrous sulfate FeSO₄.7H₂O – Green,
• Ferrous nitrate (Fe(NO₃)₂.6H₂O) – light green

Text Book Page No: 28

→ List out the dements of the s block?
Answer:
A, B

→ Which elements shows+2 oxidation state?
Answer:
B, C, D

→ Which elements contains 5 electrons in the outermost shell?
Answer:
E

→ Which is the element that has 5 p electrons in the outermost shell?
Answer:
G

→ Which are the elements in which the last electron enters the d subshell?
Answer:
C, D

→ Which element has the highest ionization energy?
Answer:
H

→ Which is the highly reactive nonmetal?
Answer:
G

→ Which elements show -2 oxidation state?
Answer:
F

Text Book Page No: 29

The outermost electron configuration of an element in this is 2s² 2p⁶

→ Which is the element?
Answer:
H

→ Write down the complete subshell electronic configuration?
Answer:
Is² 2s² 2p⁶

→ Write any two characteristics of this element?
Answer:

• Noble element / gases.
• The outermost shell is completely filled

→ Write the chemical number of questions, the answer of which is an element in the table
Answer:
A G

Periodic Table and Electronic Configuration Let Us Assess

Question 1.
Based on the hints given, find out the atomic number and write down the subshell electronic configuration of elements (Symbols used are not real).
i. A – period 3 group 17
ii. B – period 4 group 6
Answer:
A₁₇ — 1s² 2s² 2p⁶ 3s² 3p⁵
B₂₄ — 1S² 2s² 2p⁶ 3S² 3p⁶ 3d⁵ 4S¹

Question 2.
When the last electron of an atom was filled in the 3d subshell, the subshell electronic configuration was recorded as 3d8 Answer the questions related to this atom.
1. Complete subshell electronic configuration
2. Atomic number
3. Block
4. Period number
5. Group number
Answer:
1. Complete subshell electronic configuration:
1 s², 2s², 2p⁶, 3s², 3p⁶, 3d⁸, 4s²
2. Atomic number: 28
3. Block : d
4. Period number: 4
5. Groupnumber : 8 + 2 = 10

Question 3.
Pick out the wrong ones from the subshell electronic configuration given below.
a. 1s² 2s² 2p⁷
b. 1s² 2s² 2p²
c. 1s² 2s² 2p⁵ 3s²
d. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d² 4s²
e. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d² 4s²
Answer:
Wrong electronic configuration
a. 1s², 2s², 2p⁷
(2p maximum 6 electrons only)

c. 1s², 2s², 2p⁵, 3s¹ (electrons are filled in 3s only after filling 6 electrons in 2p)

d. 1s², 2s², 2p⁶, 3s², 3p⁶, 3d², 4s¹ (electrons are filled in 3d only after filling 2 electrons in 4s)

Question 4.
The element X in group 17 has 3 shells. If so,
a. Write the subshell electronic configuration of the element.
b. Write the period number,
c. What will be the chemical formula of the compound formed if the element X reacts with element Y of the third period which contains one electron in the p subshell?
Answer:
a. Three shells are K, L, M. The subshells are 1s, 2s, 2p, 3s, 3p, 3d
Group number: 17
Electrons in last shell: 7
Shell electronic configuration: 2,8,7
Sub-shell electronic configuration 1s², 2s², 2p⁶, 3s², 3p⁵

b. Period-3

c. Y – Third period
∴ shells – 3
1 electron in p – subshell
Total electrons in valence shell 2+1=3 (2 electrons in s + 1 electron in p)
Valency of x – 1(1 electron is recieved – electro negative atom)
Valency of y – 3 (3 electrons are lost – electro positive atom)
Therefore they combine to form compounds with chemical formula YX3
(Symbol of electropositive element first followed by electro negative element).

Question 5.
The element Cu with atomic number 29 undergoes chemical reaction to formation with oxidation number +2.
a. Write down the subshell electronic configuration of this ion.
b. Can this element show variable valency? Why?
c. Write down the chemical formula of one compound formed when this element reacts with chlorine (17CI).
Answer:
a. ₂₉Cu — 1s², 2s², 2p⁶, 3s², 3p⁶, 3d¹⁰, 4s¹
Cu²⁺ — 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁹

b. Yes. One electron can be lost from 4s subshell and can’exist as Cu⁺ ion, It is a d-block element.

c. Copper react with chlorine to form two compounds Cu⁺, Cu²⁺ ions react with chlorine to form CuCl and CuCl2 respectively.

Question 6.
Certain subshells of an atom are given below. 2s, 2d, 3f, 3d, 5s, 3p
a. Which are the subshells that are not possible?
b. Give the reason.
Answer:
a. Not possible sub-shells are 2d, 3f

b. d – subshell is not possible in 2^nd shell
f – subshell is not possible in 3^rd shell

Periodic Table and Electronic Configuration Extended Activities

Question 1.
Prepare the comprehensive table which indicates the name, symbol, electron configuration, subshell configuration of elements having atomic number 1 to 36?
Answer:

Question 2.
Some information related to the elements of the p bllock in the 17th group of the periodic table are given in the table below. Complete the table and analyze the following questions?
Answer:

a. What is the family names of elements belonging to the 17^th group?
Answer:
Halogen

b. What is their common valency?
Answer:
1

c. Which element has the highest electro negativity ?
Answer:
F

d. Which element has the highest ionization energy?
Answer:
F

e. List out the name and chemical formula of the compounds formed by these elements with block elements?
Answer:

• sodium chloride – NaCl
• potassium chloride – KCl
• magnesium chloride- MgCl₂
• calcium chloride- CaCl₂
• magnesium fluoride- MgF₂
• calcium fluoride – Ca F₂
• sodium iodide – Nal
• potassium iodide – KI
• potassium bromide – KBr
• potassium fluoride – KF

Periodic Table and Electronic Configuration Orukkam Questions and Answers

Question 1.
Complete the table of details about shells and subshells.

a. No of electrons in KLMN shell.
b. No of electrons in each shell.

c. Which subshell is common to all sub-shells?
d. Write names of subshells in accordance with increasing energy level,
e. Identify the incorrect subshell electronic configuration.
– 1s²
– ls² 2p⁶
– ls² 2s² 2p⁶
– 1s² 2s² 2p⁶ 3s² 3p²
Answer:

a, K – 2 ; L – 8 ; M – 18 ; N – 32

c. s- Subshell

d. 1s <2s <2p <3s <3p <4s <3d <4p <5s <4d <5p <6s <4f <5s

e. 1s² 2p⁶

Question 2.
Atomic number of iron is 26. It exhibits Fe²⁺, Fe³⁺ oxidation state. Write the subshell electronic configuration.

Answer:

Question 3.
Manganese, a d-block element exhibits I different oxidation state. Why?
a. Include chemical formulae of more compounds of manganese in the table, write their ; oxidation state and subshell electronic configuration.

b. Write the oxidation number and subshell electronic configuration K, Cl and O.
Answer:
Manganese shows different oxidation states because in manganese 4s and 3d subshell electrons take part in chemical reactions.

Question 4.
Find out atomic number, group, block period using subshell electronic configuration and then complete the table.

Answer:

Question 5.
Write down the characteristics of s,d,p, f block elements
Answer:
s-block elements:
Elements in which last electron enters into s-subshell are called s-block elements. It contains group I elements (Alkali metals) and group II elements (Alkaline earth metals).

1st group elements lose one electron during chemical combination. Therefore its oxidation state is +1.

2nd group elements lose two electrons from valence shell during chemical combination and their oxidation state is +2.

The highest shell number in a sub-shell electronic configuration is the period number of that element.

1. Group number characteristics = no.of electrons in valence sub-shell.
2. s block ionization energy & electro negativity decreases downwards.
3. Metallic character & reactivity increases downwards.
4. Lose electrons during chemical combination j and they form ionic compounds.
Their oxides and hydroxides are basic.
Their atomic radii are high in a period.

p-block elements:

• Last electron enters into p-subshell.
• Group 13 -18 elements.
• Highly reactive elements are non-metals – group 17,
• These are elements with positive and negative oxidation state.

Group number of p-block elements = electrons in last p-subshell + 12

d-block elements:

• Last electron enters into penultimate d-subshell
• Known as transition elements.
• Metals
• Shows similarity in group and period.
• Variable oxidation states.
• Form coloured compounds.

Group = electrons in ‘d’-subshdl + electrons in s-subshell.

f-block elements:

• Last electron enters into antepenultimate f sub-shell.
• Contains Lanthanoids and Actinoids.
• Variable oxidation state.
• Most of the Actinoids are radioactive.
• Most of the elements are artificial.
• U, Th, Pu are used in nuclear reactors.
• Some elements are used as catalyst in pet-roleum industry.

Periodic Table and Electronic Configuration Evaluation Questions

Question 1.
Write down subshell electronic configuration of Cu¹⁺ and Cu²⁺
Answer:
Cu¹⁺ – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰
Cu²⁺ – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹

Question 2.
How many ‘s’ subshell electrons are in 1s², 2s², 2p⁶, 3s², 3p²
Answer:
6 Electrons

Question 3.
11,17,10 are the atomic number of elements X, Y, and Z.
a. Write down their subshell electronic configuration, group, block, period,
b. Write the molecular formulae of the compound formed when any two of the above elements are combined.
c. Write down the oxidation numbers of the elements in those compounds. Write the subshell electronic configuration of both ions.
Answer:

b. X Y

Question 4.
Element ‘X’ is having atomic number 28, it gives two electrons to element ‘Y’.
a Write down the electronic configuration of ‘X’ and its ion
b. In which block ‘X’ belongs?
c. Write down the characteristics of that block
Answer:
a. X₂₈ – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s²
X²⁺– 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸

b. d block Compound

c. 1. It exhibits variable oxidation states
2. Forms colored compounds
3. Last electron enters d subshell

Question 5.

a. Write down the group and period of each element.
b. What are the use of writing electronic configuration this fashion?
Answer:

b. Group and period of the element can be identified easily. In the Same way long electron configuration can be avoided.

Question 6.
₂₄Cr – [Ar] 3d⁵ 4s¹
₂₉Cu – [Ar] 3d¹⁰ 4s¹
Why chromium and copper exhibits such electronic configuration ?
Answer:
Half filled and completely filled subshells are most stable. Change in the electronic configuration of ₂₄Cr &, ₂₉Cu is due to .this. The electrons in these elements are arranged in such away to give these elements stability.

Periodic Table and Electronic Configuration SCERT Questions and Answers

Question 1.
The electronic configuration of the elements A, B, C, Dare given below.
A – 1s² 2s² 2p⁶ 3s² 3p⁴
B – 1s² 2s² 2p⁶ 3s²
C – 1s² 2s² 2p⁶ 3s² 3p⁵
D – 1s² 2s² 2p⁶ 3s¹
a. Which of these elements show +2 oxidation state?
b. Which metal belongs to 17^th group?
c. Which is the period number of the element A ? What is the basis of your findings?
d. Which of these elements can form basic Oxides?
Answer:
a. B

b. C

c. Period number: 3, Period number = No.of shells

d. B, D

Question 2.
Two compounds of iron are jpven below.
FeSO₄ Fe₂(SO₄)₃
(The oxidation state of sulfate radical is-2)
a. Which ofthese compounds show +2 oxidation state for Fe?
b. Which compounds has Fe³⁺ ion?
c. Write the subshell electronic configuration of Fe³⁺ ion.
d. Why do transition elements show variable oxidation states?
Answer:
a., FeSO₄
b. Fe₂(SO₄)₃
c. Fe³⁺ – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵

d. The energy difference between the outer most ‘s’ subshell and penultinate ‘d’ subshell is very small. Hence under suitable conditions, the electrons in ‘d’ subshell also take part in chemical reaction.

Question 3.
Identify the incorrect electronic configurations and correct them.
i) 1s² 2s² 2p³
ii) 1s² 2s² 2p⁶ 3s¹
iii) 1s² 2s² 2p⁶ 2d⁷
iv) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴
Answer:
iii). 1s² 2s² 2p⁶ 3s² 3p⁵ .
iv). 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d²

Question 4.
Complete the table.

Answer:
a. – 2
b. 1
c. 17,
d. – 1
e. 12
f. +12

Question 5.
a. Two compounds XY₂, XZ₄ are given. The oxidation state of Z is 1. What will be the oxidation state of Y ?
b. Write the molecular formula of the compound formed by Y when it combines with aluminum (Al) having oxidation state +3.
Answer:
a. Y= – 2 (oxidation state of X is +4)
b. Al₂ Y₃

Question 6.
Pick out the statements which suit to f-block elements.
a. All of them are naturally occurring elements.
b. Uranium and Thorium are f block elements.
c. Last electrons is filled in the shell pre-ceding the outermost shell.
d. last electrons are filled up in the antepenultimate shell.
e. Includes some radioactive elements.
f. Some of them are used as catalyst in petroleum industry.
Answer:
b, d, e, f

Question 7.
The atomic number of four elements are given below. (The symbols ore not real)
A – 8
B – 10
C – 12
D – 18
a. Write the sub-shell electronic configuration of the elements,
b. Which of them are inert gases?
c. Write the chemical formula of the compound formed by two elements other than inert gases.
Answer:
a. A – 1s² 2s² 2p⁴
B – 1s² 2s² 2p⁶
C – 1s² 2s² 2p⁶ 3s²
D – 1s² 2s² 2p⁶ 3s² 3p⁶

b. B, D

c. CA, (C₂ A₂ is simplified and written as CA)

Question 8.
The subshell electronic configuration of two elements ends as follows. (Symbols are not real)
P – 3s² Q – 3p⁴
a. Write the complete subshell electronic configuration.
b. Find out the oxidation state of each element.
c. The chemical formula of the compound formed by these elements is PQ. Is this statement correct? Justify your answer.
Answer:
a. P – 1s² 2s² 2p⁶ 3s²
Q – 1s² 2s² 2p⁶ 3s² 3p⁴

b. P = +2, Q = – 2 :

c. Right, valency of both P and Q is ‘2’

Question 9.
Match the following.

Answer:

Question 10.
The atomic number of two elements are given below.
Si – 14 Ni – 28
a. Write the subshell electronic configu-ration of these elements.
b. Find out the group and period of each element.
Answer:
a. Si – 1s² 2s² 2p⁶ 3s² 3p²
Ni – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s²
b. Si – Period Number – 3, Group number – 14 Ni – Period Number – 4, Group number – 10

Question 11.
The element ‘X’ has 4 shells and its 3d subshell has 6 electrons. (Symbol is not real)
a. Write the complete electronic configu-ration of the element.
b. What is its group number? Which is the block?
c. Write any two characteristics of the block to which element X belongs to.
d. From which subshell the electrons are lost when the element X shows +2 oxidation state.
Answer:
a. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s²
b. Group number – 8, Block – d
c. All of them are metals
d – block elements are placed in group 3 to group 12
d. s – Sub shell

Question 12.
The outermost electronic configuration of the element A is 2s² 2p². (Symbol is not real)
a. Find out the group number and block of the element.
b. Write the chemical formula of the compound formed by A when it combines with chlorine.
c. Write the complete electronic configuretion of the element just below ‘A’ in the j periodic table.
Answer:
a. Group number – 14, Block – P
b. ACl₄
c. 1s² 2s² 2p⁶ 3s² 3p²

Question 13.
The figure of an incomplete periodic table is given below.

a. Which one of these elements shows -2 oxidation state?
b. Which of these elements have 3 electrons in their outermost p subshell?
c. Which element has the highest atomic radius? Which one has the least?
d. Which of these elements show variable oxidation state?
e. Which of these elements has the highest ionization energy?
Answer:
a. G

b.F

c. The element having highest atomic radius – A
The element having lowest Atomic radius – H

d. D, C

e. H

Question 14.
Examine the given electronic configurations.
A – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s²
B – 1s² 2s² 2p⁶ 3s¹
C – 1s² 2s² 2p¹ 3s² 3p⁶
D – 1s² 2s² 2p⁶ 3s²
E – 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
a. Which of these elements belongs to 4th period?
b. Which elements belongs to the same group ?
c. Which element doesn’t participate in chemical reactions generally ?
d. Which element has highest metallic character ?
Answer:
a. A, E
b. B, E
c. C
d. E

Question 15.
The atomic number of the elements X and Y are 20, 26 respectively. When these elements combine with chlorine, three compounds XCl₂, YCl₂, YCl₃ are formed.
a. What is the specialty of the oxidation number of Y, compared to that of X?
b. Explain the reason for this, on the basis of the subshell based electronic configuration.
Answer:
a. Element X has constant oxidation state. Y shows variable oxidation states.
b. X₂₀ = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s²
X₂₆ = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d⁶
Y is a transitional element. In chemical reactions only two elections in ‘ s’ subshell or besides ‘s’ subshell electrons ‘d’ sub shell electrons also take part.

Periodic Table and Electronic Configuration Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 1.
Arrange the following sub-shells in the in-creasing order of energy 5p, 2s, 4f, 3s, 4s, 3d, 6s
Answer:
2s < 3s < 4s < 3d < 5p < 6s < 4f

Question 2.
Last electron in f-block elements goes to
a. Which shell? Outer shell/Penultimate shell /Antepenultimate shell
b. Which sub-shell? Outer f-subshell Penultimate f-subshell/Antepenulti mate f-subshell.
Answer:
a. Antepenultimate shell
b. Antepenultimate f-sub-shell

Question 3.
Sub-shell electronic configuration of X is given below.
1s², 2s², 2p⁵
a. The element Y is coming just below the element in same group. Then write the sub-shell electronic configuration of Y.
b. Write the sub-shell electronic configuration of the element next to X in same period.
Answer:
a. 1s², 2s², 2p⁶, 3s², 3p²
b. Is², 2s², 2p⁶

Question 4.
A compound of vanadium pentoxide (V20;) is used as catalyst.
a. What is the oxidation state of vanadium in this compound?
b. How vanadium ion is represented?
c. Write the sub-shell electronic configuration of this ion (V – 23)
Answer:
a. +5
b. V⁵⁺
c. 1s², 2s², 2p⁶, 3s², 3p⁶

Short Answer Type Questions (Score 2)

Question 5.
Find the wrong electronic configurations from the following. What is wrong in these?
a. 1s²,2s²,2p⁶,3s²,3p⁶,3d⁹,4s²
b. 1s²
c. 1s², 2s¹, 2p⁶
d. 1s², 2s², 2p⁶, 3s², 3p²
Answer:
(a) and (c) are wrong electronic con figurations.

In (a) one electron from 4s is to be trans-ferred to 3d since completely filled configu-rations are more stable. So the correct electronic configuration is 1s², 2s², 2p⁶, 3s², 3p⁶, 3d¹⁰, 4s¹

In (c) electrons are filled in 2p only after filling electrons in 2s.

Question 6.
Group and period number of two elements are given.
P – group 17, period – 3
Q – group 2, period – 3
a. Write the sub-shell electronic configuration of each.
b Write the chemical formula of the compound formed by their combination.
Answer:
Answer:
a. P – 1s², 2s², 2p⁶, 3s², 3p⁵
Q – 1s², 2s², 2p⁶, 3s²

b. Q is electropositive. P is electro negative;
∴Chemical formula QP2

Question 7.
Write the reason for the statement given
below.
a. d-block elements in the same period show similarity.
b. Transition elements show variable oxidation state.
Answer:
a. Valence shell electrons of d-block elements in same periods are almost same. Valence shell electrons are entering in chemical reaction. Therefore they shows similarity.

b. Energy of electrons in s-subshell and inner d- subshells are almost same. Therefore s- electrons or s and d electrons take part in chemical reaction and show variable oxidation state.

Short Answer Type Questions (Score 3)

Question 8.
Write the sub-shell electronic configuration of following elements. Predict the block, group and period. (Symbols are not real)
a. M – 27 b. N – 19 c. P – 15
Answer:
a. 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁷, 4s²
block – d; group – 9; period – 4.
b. 1s², 2s², 2p⁶, 3s², 3p⁶, 4s¹
block – s; group – 1; period – 4
c. 1s², 2s², 2p⁶, 3s², 3p³
block – p; group – 15; period – 3

Question 9.
Observe the model of periodic table.

a. Which element is having S electrons in valence shell?
b. Which elements are having 2 electrons in valence sub-shell?
c. Which element is having last electron in3p?
d. Which element ends with electronic configuration 4d⁵, 5s¹ ?
Answer:
a. B;
b.A, C;
c. C, D;
d. E

Question 10.
Calculate oxidation state of transition elements in the following compounds.
Answer:
a. KMnO4 – Mn – 7⁺
b. Cr₂ O₃ – Cr – 3⁺
c. K₂Cr₂O₇ – Cr – 6⁺

Question 11.
Atomic number of some elements are given. A – 15, B – 8, C – 11, D – 18, E – 20, F – 34, G – 10
a. Which are the elements in same period?
b. Which are the elements in same group?
Answer:
A -1s², 2s², 2p⁶, 3s², 3p³
(group -15 period – 3)
B – 1s², 2s², 2p⁴ (group –16 period – 2)
C – 1s², 2s², 2p⁶, 31 (group – 1 period – 3)
D – 1s², 2s², 2p⁶, 3s², 3p⁶ (group – 18 period – 3)
E – 1s², 2s², 2p⁶, 3s², 3p⁶, 4s².(group – 2 period – 4)
F – 1s², 2s², 2p⁶, 3s², 3p⁶, 3d¹⁰, 4s², 4p⁴ (group – 16 period – 4)
G – 1s², 2s², 2p⁶ (group – 18 period – 2)
a. A, D same period ;
B, G same period 1
b. B, F same group ;
D, G same group

Long Answer Type Questions (Score 4)

Question 12.
Electronic configuration of some elements are given. Write answers to the following questions.
i. [Ne] 3s²
ii. [Ar ] 3d²,4s²
iii. [Xe] 6s²
iv. [Ne]3s²
v [Ne] 3s²,3p⁵
a. Which metal is having high reactivity?
b. Which is having possibility of formation of colored compounds?
c. Which is the non-metal?
d Which element shows the possibility of +2 oxidation state?
Answer:
a. [Xe] 6s¹
b. [Ar ] 3d², 4s²
c. [Ne] 3s², 3p⁵
d. [Ne] 3s²,[Ar] 3d², 4s²

Question 13.
Pick the wrong statement from the following.
a. Elements with atomic number 5 belong to group 15.
b. Electronic configuration of scandium (Atomic number 21) is 2,8,8,3.
c. d-block elements are known as transition elements.
d. All s-block elements are metals.
Answer:
a. Wrong. It belongs to group 13.
b. Wrong. Electronic configuration
2,8,9,2 (1s², 2s², 2p⁶, 3s², 3p⁶,3d¹, 4s²)
c. Correct .
d. Correct

Question 14.
Look at the Bohr model of X-atom.

a. Write the sub-shell electronic configuration of this atom.
b. Mention the compounds in which d-subshell electrons are taking part in chemical reaction during their formation.
XCl₂, XO₂, X₂O₇
c. Write the electronic configuration of X ions in the above three compounds.
Answer:
a. 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁵, 4s²

b. XCl₂ – ion X²⁺ (electrons in 4s only)
XO₂ – ion X+⁴ (2 electrons in 4s and 2 electrons in 3d)
X₂O₇ – ion X⁷⁺ (2 electrons in 4s and 5 electrons in 3d)
XO₂, X₂O₇ d – subshell electrons are taking part in chemical reaction during the formation of X207

c. X²⁺ – 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁵
X⁴⁺ – 1s², 2s², 2p⁶, 3s², 3p⁶, 3d³
X⁷⁺ – 1s², 2s², 2p⁶, 3s², 3p⁶

Question 15.
Select the suitable one from the following columns A, B, C.

Answer:

You can Download Gas Laws Mole Concept Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Chemsitry Solutions Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept

Gas Laws Mole Concept Text Book Questions and Answers

Gas Laws And Mole Concept Questions And Answers Kerala Syllabus 10th Text Book Page No: 33

→ Complete the table 2.1

Answer:

→ If a gas which is kept in a cylinder having a volume of 1 liter, is completely transferred to another 5-liter cylinder then what will be the volume of the gas?
Answer:
5 liter

→ Press the piston after closing the nozzle of the syringe. What will happen to the volume of air inside the syringe?
Answer:
Volume decreases

→ Explain it on the basis of the distance between the molecules of gas and their freedom of movement?
Answer:
Gases molecules are separated from each other by a large distance. As a result, there will be a lot of vacant spaces. So as the piston is pressed, the molecUles come closer and this volume decreases

Use the Molarity Calculator Chemistry to calculate the mass, volume or concentration required to prepare a solution of compound of known molecular weight.

Sslc Chemistry Chapter 2 Kerala Syllabus  Text Book Page No: 34

→ What is the specialty of the movement of the molecules?
Answer:
Molecules move in all possible directions

→ What assumption can be made regarding the possibility of collision between gas molecules?
Answer:
The molecules collide each other.

→ Which energy gained due to the movement of molecules? Potential energy/Kinetic, energy.
Answer:
Kinetic energy

→ When a gas is heated, temperature is increased. What happens to the movement of molecules if the temperature of the gas is increased?
Answer:
Speed of motion increased

→ Asa result, what happens to the energy of the molecules?
Answer:
Energy of molecules increase

→ Volume
Answer:
The space needed for a substance to occupy is its volume. The volume of solids and liquids are definite. But the volume of a gas is the volume of its container in which it is present.

→ Pressure
Answer:
The force exerted at unit area is pressure, Therefore, force at unit area/ pressure
\(=\frac{\text {Force exerted at the surface}}{\text {Area of the surface}}\)

→ Temperature
Answer:
The average Kinetic energy of all the molecules in a substance is its temperature

Sslc Chemistry Chapter 2 Questions And Answers Text Book Page No: 35

→ Is there any change in the number of molecules?
Answer:
No

→ What happens to the pressure when the volume is decreased?
Answer:
Pressure increases

→ What is the specialty of the movement of the molecules?
Answer:
Molecules move in all possible directions

Mass to Moles Calculator — The quantity of substance n in moles is equal to the mass m in grams divided by the molar mass M in g/mol.

→ What changes can you observe in the volume of the gas inside the syringe?
Answer:
Volume is formed to be decreasing.

→ What about decreasing the pressure?
Answer:
Volume increasing

→ What relation do you arrive at between pressure and volume of the gas?
Answer:
As pressure increases volume decreases when pressure is reduced volume increased.

Gas Laws And Mole Concept Kerala Syllabus 10th Text Book Page No: 36

→ The size of the air bubbles rising from the bottom of an aquarium increases. Can you explain the reason?
Answer:
As the bubbles move upward, the pressure on them decreases. This causes increase in volume. So as the bubbles move upward, their size increases.

→ What do you observe?
Answer:
Ink rises through the tube

→ What is the reason for the rising of the ink upwards?
Answer:
When the bottle is placed in hot water, air inside the bottle becomes hot. This causes expansion of air. This pushes ink in the tube. So ink rises through the tube.

→ What did you observe on cooling the bottle after taking it out? Why?
Answer:
Ink comes down. Because as air becomes cool, Its volume decreases.

→ What can you infer about the relation between the volume and temperature of a gas?
Answer:
When temperature is increased volume of gas increases. Similarly, when temperature is decreased, The volume decreases.

→ Complete the table 2.2

Answer:

Sslc Chemistry Chapter 2 Notes Kerala Syllabus Text Book Page No: 37

→ In which unit is the temperature stated?
Answer:
Kelvin (K)

→ What happens to the volume when the temperature is increased?
Answer:
Volume increases.

→ If an inflated ballon is kept in sunlight, it will burst. What may be the reason for this?
Answer:
When the ballon placed in sun light, temperature increases. so volume of air inside the ballon increases. Thus ballon expands and finally bursts.

→ What happens to the volume of the gas when its pressure is decreased or temperature is increased. volume increased/decreased.
Answer:
volume increased

→ If the temperature and pressure are kept constant how can we increase the volume?
Answer:
Fill some more gas.

→ Fill the cylinder with a little more gas. Does the number of molecules increase or decrease now?
Answer:
Then number of molecules increases

→ What is the relation between the volume and number of molecules?
Answer:
When the number of molecules increases, volume increases.

→ According to Avagadro’s law when the temperature and pressure remain constant on which factor does the volume of gas depend?
Answer:
Depends on the number of molecules.

Kerala Syllabus 10th Standard Chemistry Chapter 2 Text Book Page No: 38

→ If the mass of a coin 5g, then what will be the mass of thousand coins?
Answer:
5 x 1000 = 5000g

→ If the mass of coins in a bag is 50,000 g, then how many coins will be there?
Answer:
\(\frac { 50000 }{ 5 }\) = 10000

→ Like this we can calculate the number of coins on the basis of mass. Can‘t we?
Answer:
Yes, it becomes easy.

→ Is their any relation between the mass and the number, if the particles are of the same mass.
Answer:
Yes

→ What may be the method of stating the mass of atoms?
Answer:
The mass of an atom is compared to the mass, of another atom and expressed as a number which shows how many times it is heavier than the other atom. The atomic mass of elements are expressed by considering 1/12 mass of an atom of carbon-12 as one unit.

→ What do you understand from the statement that the atomic mass of Helium is 4?
Answer:
Atomic mass of Helium is 4. That is mass of one atom of Helium is 4 times of 1/12^th mass of carbon atom.

Enter the formula and press “calculate” to work out the molar mass calculator with steps, the number of moles in 1 g and the percentage by mass of each element.

Chemistry Class 10 Chapter 2 Kerala Syllabus Text Book Page No: 39

→ How many oxygen atoms combine with one carbon atom?
Answer:
2 Oxygen atoms

→ How many oxygen atoms combine with 1000. carbon atoms?
Answer:
2000 oxygen atoms.

→ How many atoms are present in 12g carbon?
Answer:
6.022 x 10B carb Answer:
2 × 6.022 × 10²³ oxygen atoms

→ What will be the mass of these tabs ?
Answer:
2 × 16 = 32g

Gas Laws And Mole Concept Extra Questions 10th Text Book Page No: 40

→ Complete the table 2.5

Answer:

→ 1GAM sodium means 23g sodium. This contains 6.022 × 1023 atoms. If so, how many GAM is present in 69 g sodium? How many atoms are present in it?
Answer:
\(\frac { 69 }{ 23 }\) = 3 GAM,
3 × 6.22 × 10²³ sodium atoms.

Gas Laws And Mole Concept Notes Pdf Kerala Syllabus 10th Text Book Page No: 41

→ How many GAMs are present in each the samples given below? Calculate the, number of atoms present in each of die sample ? (Atomic mass N= 14, O= 16)
1. 42g Nitrogen,
2. 80g Oxygen
Answer:
1. 42g Nitrogen:
No.of GAM = \(\frac { 42 }{ 14 }\) = 3GAM
No.of atoms = 3 × 6.022 × 10²³

2. 80g Oxygen:
No.of GAM = \(\frac { 80 }{ 16 }\) = 5GAM
No.of atoms = 5 × 6.022 × 10²³

→ Complete the table 2.6

Answer:

→ Calculate the molecular mass of glucose (C₆ H₁₂ O₆) and sulphuric acid (H₂ SO₄)
Answer:
Molecular mass of glucose
= 6 × 12 + 12 × 1 + 6 × 16 = 72 + 12 + 96 = 180 g
Molecular mass of sulphuric acid
2 × 1+ 1 × 32 + 4 × 16 = 2 + 32 + 64 = 98 g

Gas Laws And Mole Concept Questions And Answers Pdf 10th Text Book Page No: 42

→ Complete the table 2.7

Answer:

→ What is the molecular mass of oxygen?
Answer:
32 g

→ How many GMM is present in 32g oxygen?
Answer:
1 GMM

→ How many molecules are present in it?
Answer:
6.022 × 10²³ oxygen molecules

→ How many GMM is present in 28 gm nitrogen?
Answer:
1 GMM

→ How many molecules are present in N₂?
Answer:
6.022 × 10²³ N₂, molecules

→ How many GMM is present in 18 gm water?
Answer:
1 GMM

→ How many H₂O molecules arf present in it?
Answer:
6.022 × 10²³ H₂O Molecules

→ Calculate the number of GMM present in 96g oxygen?
Answer:
\(\frac { 96 }{ 32 }\) = 3GMM

Sslc Chemistry Chapter 2 Notes Pdf Kerala Syllabus Text Book Page No: 43

How many GMM are present in each of the given samples? Calculate the number of molecules present in each sample ?

→ 360 g glucose (Molecular mass = 180)
Answer:
No.of GMM = \(\frac { 360 }{ 180 }\) = 2 GMM
No. of molecules = 2 × 6.022 × 10²³

→ 90g water (Molecular mass = 18)
Answer:
No.of GMM = \(\frac { 90}{ 18 }\) = 5 GMM
No. of molecules = 5 × 6.022 × 10²³

→ How many molecules of water are present . in one mole of water ?
Answer:
6.022 × 10²³ water molecules

→ What is its mass?
Answer:
18 g

→How many GMM is present in it?
Answer:
1 GMM

Sslc Chemistry 2nd Chapter Notes Kerala Syllabus Text Book Page No: 44

22.4 L of a gas at STP = 1 mole
44.8 Lofa gas at STP = = 2 mole
224 L of a gas at STP = \(\frac { 224 }{ 22.4 }\) = 10 mole

Gas Law And Mole Concept Kerala Syllabus 10th Text Book Page No: 45

Complete the flow chart given below, related to one mole of substance.

Answer:

Gas Laws Mole Concept Let Us Assess

Gas Laws And Mole Concept Notes Kerala Syllabus 10th Question 1.
Examine the date given in the table (Temperature and number of molecules of the gas are kept constant).
a. Calculate P × V
b. Which is the gas law related to this?
Answer:
a. 8L atm
b. Boyle’s law

Gas Laws And Mole Concept Pdf Kerala Syllabus 10th Question 2.
Analyse the situations given below and explain the gas law associated with it.
a. When an inflated balloon is immersed in water, its size decreases.
b. A balloon is being inflated
Answer:
a. Avogadro’s law
b. Boyle’s law

Gas Laws And Mole Concept Class 10 Kerala Syllabus Question 3.
Certain data regarding various gases kept under the same conditions of temperature and pressure are given below.

a. Complete the table?
b. Which gas law is applicable here?
Answer:
a.

b. Avogadro’s law

Hss Live Guru 10th Chemistry Kerala Syllabus Question 4.
a. Calculate the mass of 112 L CO₂ gas kept at STP (molecular mass = 44)
b. How many molecules of CO₂ are present in it?
Answer:

Sslc Chemistry Chapter 2 Gas Laws And Mole Concept Question 5.
Calculate the volume of 170g of ammonia at STP ? (Molecular mass 17)
Answer:
Number of moles = \(\frac { Given mass }{ GMM }\) = \(\frac { 170 }{ 17 }\) = 10 moles
Volume at STP = mole × 22.4 L = 10 × 22.4 L = 224 L

Hsslive Chemistry 10th Kerala Syllabus Question 6.
Find out the number of moles of molecules present in the samples given below (GMM-N₂=28g, H₂O= 18g)
a. 56g N₂
b. 90g H₂O
Answer:
a. Number of mol molecules = \(\frac { Mass }{ GMM }\) = \(\frac { 56 }{ 28 }\) = 2

b. Number of mol molecules = \(\frac { 90 }{ 18 }\) = 5

10th Chemistry 2nd Chapter Kerala Syllabus Question 7.
The molecular mass of ammonia is 17.
a. How much is the GMM of ammonia?
b. Find out the number of moles of molecules present in 170g of ammonia.
c. Calculate the number of ammonia molecules present in the above sample of ammonia?
Answer:
a. GMM of ammonia (NH3) = 14 + 3 × 1 = 17g = 1GMM

b. Number of mole molecules = \(\frac { Mass }{ GMM }\) = \(\frac { 170 }{ 17 }\) = 10

c. Number Of molecules = Mole × 6.022 × 10²³
= 10 × 6.022 × 10²³

Class 10 Chemistry Chapter 2 Kerala Syllabus Question 8.
The molecule’s mass of oxygen is 32.
a. What is the GMM of O₂
b. How many moles of molecules are there in 64g of oxygen? How many molecules are there in it?
c. Calculate the number of oxygen atoms present in 64g of oxygen?
Answer:
a. GMM of O₂ =2 × 16 = 32 g

b. No. of mole molecules = \(\frac { Mass }{ GMM }\) = \(\frac { 64 }{ 32 }\) = 2
Number of molecules = mole × 6.022 × 10²³
= 2 × 6.022 × 10²³
c. c. Number of atoms = Number of molecules × number of atoms in one molecules Number of atoms in one molecules of oxygen (O₂) =2
∴ total number of atoms = 2 × 6.022 × 10²³ × 2
=4 × 6.022 × 10²³

Gas Laws Mole Concept Extended activities

Hss Live Guru Chemistry 10 Kerala Syllabus Question 1.
How many grams of carbon and oxygen are required to get the same number of atoms as in one gram of Helium?
Answer:
GAM of Helium = 4 g
Number of mole atoms in 4 g of Helium = 6.022 × 10²³
Number of atoms in 1 gofHelium = \(\frac { 1 }{ 4 }\) × 6.022 × 10²³
GAM ofCarbon = 12 g
∴ Number of atoms in 12gofCarbon=6.022 x 10²³
∴ Mass required for \(\frac { 1 }{ 4 }\) × 6.022 × 10²³
Carbon atoms = \(3 \mathrm{g}\left(12 \times \frac{1}{4}\right)\)
GAM of Oxygen = 16 g.
∴ Number of atoms in 16 g of Oxygen = 6.022 × 10²³
∴ Mass required for \(\frac { 1 }{ 4 }\) × 6.022 × 10²³
Oxygen atoms = \(4 \mathrm{g}\left(16 \times \frac{1}{4}\right)\)

Kerala Syllabus 10th Standard Chemistry Guide Question 2.
Examine the samples given:
a. 20 g of He
b. 44.8 L of NH₃ at STP
c. 67.2 L of N₂ at STP
d. 1 mol of H₂SO₄
e. 180 g of water.
i. Arrange the samples in increasing order of the number of molecules in each.
ii. What will be the ascending order of the total number of atoms?
iii. What will be the masses of samples b, c, and d?
Answer:

Mole Concept Class 11 Question 3.
In 90 grams of water.
a. How many molecules are present?
b.What will be the total number of atoms?
c. What will be the total number of electrons in this sample?
Answer:
a. GMM= 18 g
∴ Number of molecules
\(\frac { 90 }{ 18 }\) × 6.022 × 10²³ = 5 × 6.022 × 10²³ 18

b. ∴ Number of atoms = 3 × 5 × 6.022 × 10²³ = 15 × 6.022 × 10²³

c. Electrons in 1 H atom = 1
Electrons in 1 O atom = 8
Total electrons in H₂O molecule? 10
Total electrons in 90 g H₂O = 10 × 5 × 6.022 × 10²³ = 50 × 6.022 × 10²³

Gas Laws Mole Concept Orukkam Questions and Answers

Question 1.
a. Complete the table based on the data given in the box.

b. Express atomic weight and molecular weight in grams. How many moles is this? Find out the number of Atoms or molecules in it?

Answer:
a.

b.

Question 2.
Complete the table based on the molecules given in the first column and then answer the question given below.

a. 10 Mole of water = …………. g …………… Molecules
5 mole of CaO = ……….. g ………… Molecules
2 Mole of H₂SO₄ = …………. g ………… Molecules
1/2 Mole of Al₂O₃= …………. g ………….. Molecules
Answer:

a. 10 Mole of water = 10 × 18 = 180g, 10 × 6.022 × 10²³ Molecules
5 mole of CaO = 280g, 5 × 6.022 × 10²³ Mol-ecules
2 Mole of H₂SO₄ = 196g, 2 × 6.022 × 10²³ Mol-ecules
2- Mole of AlO = 100g, 1/2 × 6.022 × 10²³

Question 3.
Complete the table.

Answer:

Question 4.
Complete the data.

Answer:

Question 5.
Based on the reaction given below, write the answers for the questions.
N₂ + 3H₂ → 2NH₃ ;
a. Write the ratio of reactant molecules and product molecules.
b. How many moles of Ammonia forms when we take 2 moles of Nitrogen and six moles of Hydrogen?
c. Two moles of Nitrogen and three moles of hydrogen are taken in jar? Will they react together?
d. How many moles of Nitrogen and Hydrogen is needed for rearing 20 moles of Ammonia?
Answer:
a. 1:3:2
b. 2 Mole
c. No
d. 10 Mole Nitrogen 30 Mole Hydrogen

Question 6.
Balance the given equation and then write down the answers for the questions given below.
CH₄+O₂ → CO₂ + H₂O
a. How many moles of C02 formed when 20 moles of Methane burns in air?
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
b. Based on the equation above, How many moles of CO₂ is formed when 10 moles of Ethane is burned in air ?
Answer:
a. When 1 mole of methane bum in air 1 mole of CO₂ gas is formed.
When 20 moles methane burn 20
moles of CO₂ are formed.
The molecular weight of 20 moles = 20 × 44 = 880 g

b. When 2 moles of ethane is burned 4 moles of CO2 is formed.
The number of moles of CO₂ when 1 mole of ethane is burned = 4/2 = 2 mole.
The number of moles when 10 moles of ethane bums = 2 x 10 = 20 mole.
Weight of 20 moles = 20 × 44 = 880 g

Question 7.
Based on the given equation write down the answers.
2H₂+ O₂ → H₂O
a. How much Oxygen and Hydrogen is needed for making 1800g of water vapor?
b. How many moles of Oxygen is needed for the reaction with one mole of Hydrogen?
Answer:
a. 36 g of water vapour can be made using 4 g hydrogen.
The amount of hydrogen required to make 1 g water vapour = 4/36
The amount of hydrogen required to produce 1800 g watervapour= 4/26 × 1800 = 200 g
Mass of oxygen = 1800 – 200 = 1600 g

b. 0.5 Mole

Gas Laws Mole Concept Evaluation Questions

Question 1.
Find out the number of moles of hydrogen and Oxygen atoms present in 10 moles of HCI.
Answer:
One mole HC₁ contains 1 mole of hydrogen and 1 mole of chlorine.
Hydrogen contained in 10 moles of HC₁ = 1 × 10 =10 mol,
Clatom = 10 × 1 = 10 mol

Question 2.
Find out the mass of Hydrogen atom and chlorine atom in 10 moles of HCI.
Answer:
Mass of 10 mole hydrogen atom = 10 × 1 = 10g
Mass of 10 mole chlorine atom =10 × 35.5 = 355 g.

Question 3.
a Find out the mass of one mole of CaCO3. How many moles of calcium present in 1000g CaCO₃?
b. How many moles of Oxygen present in 1000gms of CaCO₃?
Answer:
a. Mass of 1 mole of CaCO₃ = 40 + 12 + 48 = 100 gram.
No of moles ofCa in 1000 g CaCO₃ = \(\frac { 1000 }{ 100 }\) = 10 mol

b. No of moles of Ca in 10 moles CaCO₃ =10 × 1 = 10 mol
No of moles of oxygen in 10 moles of CaCO₃ = 10 × 3 = 30mol.

This Combined Gas Law Calculator can help you estimate either the pressure, temperature or the volume of gas.

Question 4.
Find out number of moles of water formed when 4gms of Hydrogen and 32 gms of Oxygen combined together. What is the result when 5 gms of Hydrogen and 32 gms of Oxygen combined together?
Answer:
Ans. When 4 g of hydrogen and 32g of oxygen are combined 36 g of water \(\frac { 37 }{ 18 }\) = 2 mol
5g H + 32 gO → 37g H₂O
No of moles in 3 7 g of water = \(\frac { 37 }{ 18 }\) = 2.055 mol

Question 5.
a How much grams of NaCl is needed for making 2 molar solution (NaCl – 58.5). Wh-at is the amount of water needed for this?
b. How will you change a two molar solution of Sodium Chloride into 5 major?
Answer:
a. Mass of 2 moles of NaCl = 2 × 58.5 = 117 g
1 liter water is required for this.

b. When 2 moles of NaCl is dissolved in 4 liter of water 5 molar solution is obtained.

Finally, you encounter how to find molar concentration step-by-step manually, and if your preference indulges with instant calculations.

Question 6.
How many moles of Cl₂ present in 11.2 L of same in STP? Find out the mass of this?
Answer:
No.of moles present in 11.2 litre of chlorine = \(\frac { 11.2 }{ 22.4 }\) = 0.5 mol
Mass of 0.5 moles of chlorine = 0.5 × 35.5 = 17.759 g.

Question 7.
Find out the mass of Oxygen atom in 44.8L of CO₂ in STP.
Answer:
No of moles in 44.8 litre of CO₂ \(\frac { 44.8 }{ 22.4 }\) 2 mol,
1 Mole of CO₂ contains 1 mole of C and 1 mole of O₂
∴ 2 Mole of CO₂ contains 2 mole of O₂ or 4 mole of oxygen atom.
Man of oxygen atom 4 × 16 = 64 g

Question 8.
Find out the amount of CO₂ formed when the burning of one mole of Ethane.
Answer:
2 moles of CO₂ is formed when 1 mole of ethane burns.
Mass of 2 moles of CO₂ = 2 × 44 = 88g

Question 9.
Why are atomic mass of some elements are infractions ?
Answer:
The atomic masses of some elements are infractions because they exist as a mixture of isotopes of different masses. The fractional atomic masses arise because of this mixture.
Average mass = \(\frac { Total mass of all atoms }{ numbers of atoms }\)

Gas Laws Mole Concept SCERT Questions and Answers

Question 1.
One GAM substance contains Avogadro number of particles in it.
a. How many particles are there in Avogadro number ?
b. Write the number of atoms present in each of the following.
i. 32g Sulphur
ii. 32g Oxygen
iii. 32g Carbon
(Atomic mass S = 32, O = 16, C = 12)
Answer:
a. 6.022 × 10²³
b. i.6.022 × 10²³
ii. 2 x 6.022 × 10²³
iii \(\frac { 32 }{ 12 }\) × 6.022 × 10²³

Question 2.
a. Group the following into pairs having same number of atoms.
A. 2g Hydrogen
B. 16g Oxygen
C. 14g Nitrogen
D. 8g Helium (Atomic mass H=1, O= 16, N =14, He=4)
b. How many atoms are present in each pair?
Answer:
a. A, D2g Hydrogen, 8g Helium
B, C 16g Oxygen, 14g Nitrogen

b. A, D – 2 × 6.022 × 10²³
B,C – 6.022 × 10²³

Question 3.
N₂ + 3H₂ → 2NH₃
a. What is the ratio between the reactant molecules in the above reaction?
b. Complete the following table.

Answer:
a. 1:3
b. a – 2 NH₃,
b – l H₂,
c – 12H₂,
d – 4NH₃

Question 4.
2H₂ + O₂ → 2H₂O
a. What is the ratio between the reactant molecules in the above reaction?
b How many O₂ molecules are required to react 100 H₂ molecules completely?
c. How many water molecules are formed when 1000 H₂ molecules are reacted completely
Answer:
a. 2:1
b. 50 O₂ molecules
c. 1000 H₂O molecules

Question 5.
Complete the following table. (All the elements given are diatomic. Atomic mass O=16, N=14, CI=35.5)

Answer:
a. 6.022 × 10²³
b. 6.022 × 10²³
c. 71 g
d. 14 g

Question 6.
A sample of substances are given.

Hint: Molecular mass NH3 = 17, N2 = 28, H2SO4 = 98, O2=32
a. Which of these samples have same number of molecules?
b. Which of these samples has least number of molecules?
Answer:
a. 68g NH₃, 128gO₃
b. 49 g H₂SO₄

Question7.
Pick out the correct statements from the following. Also, correct the incorrect statements.
a The number of molecules present in 1 mol hydrogen and 1 mol oxygen are same,
b. 2 mol chlorine contains 4 x 6.022 x 10²³ chlorine molecules.
c. The mass of 1/2 mol nitrogen gas is 14 g.
d. 0.5 mol water has the mass 9g. There are 6.022 × 10²³ H20 molecules in it. (Atomic mass H = 1, O = 16, CI= 35.5, N= 14)
Answer:
Correct statements – a, c
No. of molecules in 2 mol chlorine is 2 × 6.022 × 10²³
Mass of 0.5 mol water is 9g. So it contains 0.5 × 6.022 × 10²³ H₂O molecules.

Question 8.
Complete the following.

Answer:
a. 2 × 6.022 × 10²³
b. 1GMM
c. 6.022 × 10²³

Question 9.
67.2 L of Carbon dioxide gas is filled in a cylinder at STP.
a. Calculate the mass of CO₂ present in it. (Atomic mass- C = 12, O = 16)
b. Calculate the number of molecules present in the cylinder.
Answer:
a. Molecular mass of CO₂=12 × 1+16 × 2 = 12 + 32 = 44
No. of moles in 67.2L CO₂ at STP = \(\frac { 67.2L }{ 22.4L }\) = 3
b. 3 × 6.022 × 10²³

Question 10.

Answer:
a. 2
b. 2 × 6.022 × 10²³
c. 17 g
d. 51 g
e. 3
f. 3 × 22.4 L

Question 11.
CH₄ + 2O₂ → CO₂ + 2H₂O
The equation describes the combustion of methane in air.
a. How many moles of oxygen is required for the complete combustion of 16g CH₄?
b. Calculate the amount of CO₂ formed when 100g of CH₄ is completely burnt?
Answer:
a. 2 mol
b. Amount of CO₂ produced by the combustion of 16g CH4 = 44g
Amount of CO₂ produced by the combustion of 1 gm CH4 = \(\frac { 44 }{ 16 }\) g
Amount of CO₂ produced by the combustion of 100g CH4 = \(\frac { 44 }{ 16 }\) × 100g

Question 12.
45 g glucose is taken in a beaker and made into 1 L (MM = 180).
a. Calculate the molarity of the solution,
b. Above solution is made up to 2 L by adding more water. What will be the molarity of the resultant solution?
c. How will you prepare IM solution of glucose with the same quantity (45 g) of glucose?
Answer:
a. 0.25
b. M = \(\frac { n }{ v }\) = \(\frac { 0.25 }{ 2 }\)
c. Add 250 ml water in 45 g glucose

Question 13.
Two gases occupy equal volume at STP are shown below.
(Atomic mass S = 32, O =16, N = 14

a. Find the mass of the gas in B.
b. Calculate the number of molecules present in B.
Answer:
a. No. of moles in 320g SO₂ = \(\frac { 320 }{ 64 }\) = 5
Mass of 5 mol NO₂ = 5 × 46 = 230 g
b. No.of molecules present in B = 5 × 6.022 × 10²³

Question 14.
The balanced chemical equation of a reaction (at STP) is given below.
2H₂(g) + O₂(g) → 2H₂O(g)
a. Calculate the volume of oxygen required to combine completely with 224 L of the hydrogen at STP.
b. Calculate the mass of water formed as a result of the reaction (a).
Answer:
a. 112 L
b. Volume of water obtained when 224 L hydrogen completely reacts with oxygen = 224 L
No. of moles in 224 L water = \(\frac { 224 }{ 22.4 }\) = 10
Massof 10 mol water = 10 × 18 = 180 g

Question 15.
Complete the table.

Hint: (MM – CO₂ = 44, CH₄ = 16, SO₂ = 64)
Answer:
a. 67.2 l
b. 132 g
c. 1/4.
d. 4g
e. 11.2 L
f. 1/2.

Question 16.
Analyse the following equation
2NO(g) + O₂(g) → 2NO₂(g)
a. Calculate the number of the moles of NO required to combine completely with 112 L of Oxygen at STP.
b. Calculate the mass of NO₂ formed when 112L of oxygen reacts completely?
Answer:
a. 10 mol
b. 2NO(g) + O₂(g) → 2NO₂(g) (2 : 1: 2)
No. of moles in 112L O₂ = 5 mol
According to equation no. of moles of NO₂ obtained by reacting oxygen completely with nitric oxide = 2
No. of moles of NO₂ obtained by reacting 5 mol oxygen completely = 10
Massof 10 mol NO₂ = 10 × 46 = 460g

Question 17.
The chemical equation of the decomposition of calcium carbonate is given below.
CaCO₃→ CaO + CO₂
(HintMM: CaCO₃ – 100, CaO – 56, CO₂ – 44)
a. Calculate the mass of CaCO3 required to get 224 g of CaO?
b. Calculate the number of CO2 molecules fronted when 224g of CaO is obtained?
Answer:
a. CaCO₃ → CaO + CO₂
100g 56g 44g
I I : I
Amount of CaCO₃ required to get 56g of CaO = 100g
Amount of CaCO₃ required to get 1 g of CaO = \(\frac { 100 }{ 56 }\)
Amount of CaCO₃ required to get 224 g of CaO = \(\frac { 100 }{ 56 }\) × 224 = 400g
b. 4 × 6.022 × 10²³

Question 18.
You are requested to make 20 moles of NaCl into packets of 100g each. (Hint^Molecular mass of NaCl is 58.5)
a. How many packets of NaCl can be prepared?
b. Is there any NaCl remaining? If so, how much?
Answer:
a. Mass of 20 mol NaCl = 20 × 58.5 = 1170g
1170 g NaCl can be made into 11 packets with 100g each.
b. Remaining NaCl = 1170 – 1100 = 70g

Gas Laws Mole Concept Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 1.
GAM of Hydrogen is 1 g.
a. How many number of atoms are there in 1 g of Hydrogen?
b. Find the mass of 1 atom of hydrogen.
Answer:
a. 1g hydrogen = 6.022 × 10²³ atoms
b. Mass of 1 atom of hydrogen
= \(\frac{1 g}{6.022 \times 10^{23}}\) = 1.66 × 10-24 g

Question 2.
Number of molecules of substance is 3.011 × 10²⁴.
a. What is the number of molecules of 1 mole of any substance?
b. Find the number of moles of 3.011 × 10²⁴ molecules.
Answer:
a. 6.022 ×10²³
b. Number of moles of molecules = \(\frac { Number of molecules }{ N_A }\)
= \(\frac{3.011 \times 10^{24}}{6.022 \times 10^{23}}=5\)

Short Answer Type Questions (Score 2)

Question 3.
Fill the patterns.

Answer:
a. GMM
b. 22.4
c. Number of molecules
d. Number of molecules
e. Volume in litres
f. Mass

Question 4.
Identify the incorrect statements from those given with respect to the arrangements of molecules in gases.
a. The minute molecules are present without any freedom of movement
b. Collision take place between the molecules,
c. Increasing the number of molecules at constant volume causes the decrease in number of collisions.
d. The energy of molecules are comparatively high.
Answer:
a. The statements (a) and (c) are not correct

Question 5.
The pressure of 20L of a gas kept at 300 K is found to be 2 atoms. If the pressure is increased to 3 atom at the same temperature, what will be the new volume?
Answer:
According to Boyles law, P_V = a constant
Therefore, P₁ V₁ = P₂ V₂
Here, P₁ = 2atm V₁ = 20L P₂ = 3atm V₂=?
∴ 2 × 20 = 3 × V₂
Thus, V₂ = \(\frac { 2 × 20 }{ 3 }\) = 13.3 L

Charles Law Calculator is a free online tool that displays the volume of gas that tends to expand when heated.

Question 6.
If the temperature of 5L of a gas at atmospheric pressure is changed from 200K to 50 K, what will be the volume?
Answer:
According to Charles law, \(\frac { V }{ T }\) = a constant

Question 7.
What will be mass of 89.6 L of ammonia (NH3) gas at STP?
Answer:
89.6 L of NH3 gas at STP = \(\frac { 89.6 }{ 22.4 }\) = 4 mol
GMM of NH3 = 14 + 3 = 17g
Mass = mole × GMM = 4 × 17 = 68g

Question 8.
8. a. What is molar volume?
b. What is the molar volueofa gas at STP?
Answer:
a. The volume of one mole of a gas is called molar volume,
b. 22.4 L

Question 9.
Look at the balanced equation given.
2NaOH + CO₂ → Na₂ CO₃ + H₂0
a. Find out the mass of NaOH needed for 264 g CO₂ to react completely.
b. Find out the total number of moles of water molecules when CO₂ reacts.
Answer:
a. GMM of CO₂ = 44 g
∴Number of moles in the molecule of 264 g CO₂ = \(\frac { 264 }{ 44 }\) = 6
According to the equation NaOH needed for the reaction of 1 mole CO₂= 2 moles
∴ NaOH needed for the reaction of 6 moles CO₂ = 2 × 6 = 12 moles
GMM of NaOH = 23 + 16 + 1 = 40 g
Total mass of NaOH = 12 × 40 = 480 g

b. H₂O formed when lmole CO₂ reacts = 1 mole
∴ Total number of moles of water molecules when 6 mole CO₂ reacts = 6 moles

Question 10.
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
This is the equation of ignition of cooking gas butane.
Calculate the volume of CO₂ in STP during the complete ignition of 14 kg of cooking gas.
Answer:
Mass of Butane (C₄H₁₀) = 14 kg = 1400g
GMM of C₄H₁₀ = 4 × 12 + 10 × 1 = 58 g
∴ Number of moles in molecules = \(\frac { 1400 }{ 58 }\) = 241.38
Amount of CO₂when 2 moles of C₄ H₁₀ ignites = 8 moles of C₄H₁₀ ignites = \(\frac { 8 }{ 2 }\) × 965.52 moles
∴ Volume of CO₂ formed in STP
= 965.52 × 22.4 L = 21627.65 L

Question 11.
Write down the preparation of 100 ml NaOH solution of 0.1 M.
Answer:
GMM of NaOH = 40 g
Molarity = \(\frac{\text { Number of moles of solute }}{\text { Volume of solution in litres }}=\frac{\mathbf{n}}{\mathbf{v}}\)
M = 0.1 V=100ml = 0.1 L
0.1 = n/0.1
∴ n = 0.1 × 0.1 = 0.01
Mass needed to prepare 100 ml NaOH in 0.1M = 0.01 × 40 = 0.4g
Take 0.4 g NaOH in a beaker. Dissolve it Hilly by adding a little amount of water. Then, again add water to make it 100 ml.

Question 12.
The molarity of250 ml of Na₂CO₃ solution is 0.5 M. Find the mass of Na₂CO₃.
Answer:

Question 13.
63 g HNO₃ is in the dilute solution of 200 ml HNO₃ (Nitric acid). Find the molarity.
Answer:
a.

Short Answer Type Questions (Score 3)

Question 14.
Some equations related to gas laws are given below.
i. V α P
ii. \(\frac { V }{ T }\) = a constant
iii. V α n
iv. P_v = a constant
a. Which of these are correct?
b. Write the gas law to which it is related for the correct equations.
Answer:
a. Equations (ii), (iii) and (iv) are correct

b. (ii) Charles law
(iii) Avogadro’s law
(iv) Boyles law

Question 15.
In 100 g of CaCO₃
a. Find out the number of moles of each element and atom.
b. Find out the total number of atoms of each element.
Answer:

Question 16.
Fill the blanks in the given table.

Answer:

Long Answer Type Questions (Spore 4)

Question 17.
See CO₂ gas is taken in a cylinder provided with a piston. The cylinder is dipped in hot water.
a. What happens to the movement of CO₂ molecules?
b. What change do you expect in the position of the piston?
c. What is the relation between temperature and the volume of a gas?
d State this gas law.
Answer:
a. As the temperature increase, the energy of molecules increases. This increases the speed of the motion of molecules.
b. Piston is pulled in the upward direction. So piston moves upward.
c. As temperature increases, volume increases.
d. At constant pressure, the volume of a definite mass of gas is directly proportional to the temperature in Kelvin scale. (Charles law)

Question 18.
The molecular formula of ammonium sulfate is (NH₄)₂SO₄.
a. Find the gram molecular mass (GMM) of ammonium sulfate.
b. Calculate the number of molecules and atoms in 1.32g of ammonium sulfate.
Answer:
a. GMM of (NH₄)₂SO₄
= (14+4) × 2 + 32 + 4 × 16 = 36 + 32 + 64
= 132 g

Question 19.
Fill in the blanks of the table given below.

Answer:

Question 20.
See the diagram given below.

Answer:
a. 196g
b. 2 × 6.022 × 10²³
c. 2 GMM
d. 2 × 6.022 × 10²³

Question 21.
Write in pairs, equal number of atoms from those given below,
a. 2g Hydrogen
b. 16 g Oxygen
c. 14 g Nitrogen
d. 8 g Helium
Answer:

Question 22.
Certain compounds and its masses are given,
i) 68 g NH₃
ii) 28 g N₂
iii) 9 g H₂O
iv) 128 g O₂
a. Which of these compounds have equal number of molecules?
b. How many molecules are there?
c. How many atoms are there in 9 g of water?
Answer:

Question 23.
368 g NO ₂gas is given. Find the answers of each one given below,
a. GMM of NO₂
b. Number of moles of molecules of 368 g NO₂
c. Number of molecules
d. Number of atoms
e. Volume in STP
Answer:
a. GMM of NO₂
= 14 + 2 × 16 = 46 g

b. Number of mole = \(\frac { 368 }{ 46 }\) = 8

c. Number of molecules = Number of moles × NA
= 8 × 6.022 × 10²³

d. Number of atoms = Number of atoms in one molecule × Number of molecules
= 3 × 8 × 6.022 × 10²³ = 24 × 6.022 × 10²³

e. Volume in STP = Number of moles × 22.4
L = 8 × 22.4 L = 179.2 L

Question 24.
Find out the GMMofthe following. Also find out the total number of atoms.
a. 20 g Nitrogen (H₂)
b. 88.75 g Chlorine (Cl₂)
c. 4 g Calcium (Ca₂) ,
d. 7.75 g phosphorus (p₄)
(H = 1, Cl =35.5, Ca =40, P=31)
Answer:
a. 20 g of hydrogen (H2): GMM = \(\frac { 20 }{ 2 }\) = 10
Number of molecules = 10 × 6.022 × 10²³
Total number of atoms = 2 × 10 × 6.022 × 10²³
= 20 × 6.022 × 10²³

b. 88.75g of chlorine
Number of GMM = \(\frac { 88.75 }{ 71 }\) = 1.25
Number of molecules = 1.25 × 6.022 × 10²³
Total number of atoms = 2 × 1.25 × 6.022 × 10²³
= 2.5 × 6.022 × 10²³

c. 4 g Calcium (Ca)
Number of GMM = \(\frac { 4 }{ 40 }\) = \(\frac { 1 }{ 10 }\) = 0.1
Number of molecules = 0.1 × 6.022 × 10²³
Total number of atoms = 1 × 0.1 × 6.022 × 10²³
= 0.1 × 6.022 × 10²³

d. 7.75 g Phosphorus Number of GMM = \(\frac { 7.75 }{ 31 }\) = \(\frac { 1 }{ 4 }\) = 0.25
Number of molecules = 0.25 × 6.022 × 10²³

You can Download Moving Images Questions and Answers, Kerala SSLC 10th IT Theory Questions and Answers Chapter 9 help you to revise complete Syllabus and score more marks in your examinations.

Kerala Syllabus SSLC IT Theory Questions and Answers Chapter 9 Moving Images

Moving Images Questions and Answers

Section -1

Question 1.
The animation software in IT @ School GNU/Linux is
a) Synfig studio
b) Tupi
c) Gimp
d) Paint
Answer:
a) Synfig studio

Question 2.
Which type of animation software is synfig studio.
a) Proprietary
b) not able to take copy
c) pay and use
d) free software
Answer:
d) free software

Question 3.
To prepare a film, How many times images appear one after the other continuously in front of our eyes in one second?
a) 48
b) 12
c) 24
d) 36
Answer:
c) 24

Question 4.
Say the important stage of an animation film?
a) character designing
b) background
c) characters
d) action
Answer:
a) character designing

Question 5.
Construction of a storyboard is a preparation of
a) Drawing a picture
b) finding time zones
c) producing an animation
d) character designing
Answer:
c) Producing an animation

Question 6.
Synfig Studio is a free-animation software
a) Drawing pictures
b) Studio
c) Three dimensional
d) Two dimensional
Answer:
d) Two dimensional

Question 7.
Synfig studio software is designed by
a) Stall men
b) Robert B Quattlebaum
c) Leslie Lamport
d) Donald Knuth
Answer:
b) Robert B Quattlebaum

Question 8.
What versions of animation software can run in GNU/Linux and Microsoft Windows?
a) Grass
b) Notepad
c) Synfig studio
d) Pencil
Answer:
c) Synfig studio

Question 9.
Which tool is used to fill colors to the objects in synfig studio?

Answer:

Question 10.
Which tool is used to draw rectangular objects in synfig studio?

Answer:

Question 11.
Which tool is used to move the objects in synfig studio?

Answer:

Question 12.
Expand FPS
a) Frames per system
b) Frames per second
c) File properties settings
d) Frames per hour
Answer:
b) Frames per second

Question 13.
Animation is done by fast and continuous movements of images in a two-dimensional canvas. The images are known as
a) characters
b) FPS
c) Frames
d) Scene
Answer:
c) Frames

Question 14
Which menu is used to studio
a) File
b) view
c) window canvas
d)canvas
Answer:
d)canvas

Question 15.
To change the FPS in Synfig studio, by clicking
a) canvas → properties → time
b) view → pause
c) window → toolbox
d) canvas properties → image
Answer:
a) canvas → properties → time

Question 16.
We can import Images into synfig studio and use them
a) odf
b) vector
c) Pdf
d) Bitmap
Answer:
d) Bitmap

Question 17.
FPS = 24, Time =5, for an animation. Find the total number of frames in that animation
a) 24,
b) 48
c) 120
d) 920
Answer:
c) 120

Question 18.
The frames that represent important positions are known as
a) Tweening
b) current time
c) key
d)keyframe
Answer:
d) keyframe

Question 19.
What happens when you press

play button
a) Animation works
b) Animation stops
c) no change
d) to save
Answer:
a) Animation works

Question 20.
The software fills up the frames in between two keyframes is called.
a) Interpolation
b) keyframe
c) Tweening
d) Edit
Answer:
c) Tweening

Question 21.
Rey frame utility is set in one of the panels in synfig audio. Give the names of that panel
a) Time track panel
b) layers panel
c) parameters panel
d) panel
Answer:
c) parameters panel

Question 22.
Name the extension of project file in synfig studio?
a). svg
b) .pdf
c) .sifz
d) .ods
Answer:
c) .sifz

Question 23.
In which panel to give the number of frames in current time?
a) Time track panel
b) layers panel
c) meters panel
d) panel
Answer:
a) Time track panel

Question 24.
What is the current time to animation from first frame
a) 0 f
b) 60 f
c) 120f
d) 121 f
Answer:
a) 0 f

Question 25.
What is the name for the frame with current time is ‘o’ f
a) first keyframe
b) last keyframe
c) middle keyframe
d) keyframe
Answer:
a) first keyframe

Question 26.
Give the order of activity to export a project file on synfig studio
a) File → Render
b) File → Save
c) File → Export
d) File → Save as
Answer:
a) File → Render

Question 25.
Which is not a member related to synfig studio?
a) parameters panel
b) time track panel
c) panel
d) layers panel
Answer:
c) panel

Question 28.
The utility to animate the flap its wings of a bird is
a) joining wings
b) adding time loop layer
c) copy of the first wing
d) adding the layer
Answer:
b) adding time loop layer

Question 29.
Write the activity to include an image to synfig studio
a) file → open
b) file → new
c) file → import
d) file → save
Answer:
c) file → import

Question 29.
in which menu import facility is available”?
a) Edit
b) file
c) canvas
d) windows
Answer:
b) file

Question 30.
Select ofie not group
a) . dv
b) .flv
c) . mpeg
d). svg
Answer:
d) .svg

Question 32.
What is the maximum value of current time?
a) 20 f
b) 60 f
c) 120 f
d) 1 f ,
Answer:
c) 120 f

Question 33.
What is the least value of current time?
a) 60 f
b) 120 f
c) 40 f
d) Of
Answer:
d)0f ‘

Question 34.
Keyframe are constructed in different type or same type
a) same type
b) different type
c) not constructing a keyframe
d) Equal type
Answer:
b) different type

Question 35.
Select any two free animation software’s
a) Anim studio
b) pencil
c) synfig studio
d) adobe flash
Answer:
b) pencil
and
c) synfig studio

Question 36.
An important stage in preparing animation is character designing. What is the meaning of character designing?
a) characters
b) Draw the characters
c) bringing character to life with humanity & personality
d) Giving life to the story
Answer:
b) Draw the characters
and
c) bringing character to life with humanity & personality

Question 37.
What are the uses of multi-colored buttons on the handle of a image?
a) To switch on the Animate editing mode
b) To adjust size
c) To take copy
d) To rotate if needed.
Answer:
b) To adjust size
and
d) To rotate if needed

Question 38.
Give the two activities done before starting animation from first keyframe
a) Current time is 60 f
b) Current time is 0 f
c) Before editing the motion, animate the edit mode button is active
d) press the play button to see the animation
Answer:
b) Current time is 0 f
and
c) Before editing the motion, animate the edit mode button is active

Question 39.
Sky and one star are drawn in synfig studio canvas. In the layers panel we can see a rectangle 001, and star 001. What are the tools use dot draw them
a) Circle tool
b) Fill tool
c) Star tool
d) Rectangle tool
Answer:
c) Star tool
and
d) Rectangle tool

Question 40.
Animation became much easier in film industry, with the arrival of new technologies. What are they?
a) computers
b) Radio
c) Animation software’s
d) office software.
Answer:
a) computers
and
c) Animation software’s

Question 41.
Create animations is one of the stages in animation film. There are two other stages before creating an animation. What are they?
a) To save animation project
b) Character designing
c) Playing an animation
d) Preparation of storyboards
Answer:
b) Character designing
and
d) Preparation of storyboards

Question 42.
Utilities of synfig studio windows are given below. Select one of each set and make a list.

Set-I

a) Database panel
b) Work area
c) Layers Panel
d) Task area
Answer:
c) Layers Panel

Set -II

a) Audio track
b) Video track
c) Time track panel
d) Fill
Answer:
c) Time track panel

Set-III

a) Parameters panel
b) Table
c) Storyboard
d) Path
Answer:
a) Parameters panel

Set -IV

a) Database window
b) tooIX
c) Editor window
d) Window
Answer:
b) tooIX

pH Calculator is a free online tool that displays the pH value for the given chemical solution.

Question 43
Select one video formats from each set

Set I

a) PNG
b) gif
c) xcf
d) sifz
Answer:
b) gif

Set II

a) dv
b) ods
c) odf
d) odp
Answer:
a) dv

Set III

a) jpg
b) flv
c) jpg
d) py
Answer:
b) flv

Set -IV

a) Html
b) ph
c) htm
d) mpeg
Answer:
d) mpeg

Question 44.
There are a lots of job opportunities related to animation in Government/public sectors of India and abroad. A list of job opportunities are given below. Select one from each set.

Set-I

a) Cinema production
b) Still photograph
c) Newspaper
d) Shops
Answer:
a) Cinema production

Set-II

a) Schools
b) advertising agencies
c) building construction
b) advertising agencies
Answer:
d) Library

Set-III

a) Radio
b) FM radio
c) TV
d) Tape recorder
c) TV

Set-IV

a) mobile
b) computer
c) calculator
d) computer games
Answer:
d) computer games

Question 45.
Images are included in synfig studio through import menu what are the uses of handles on the image? Select one form each set.

Set-1

a) To join the parts of an image
b) To arrange its position
c) To animation
d) Frame
Answer:
b) To arrange its position

Set – II

a) To arrange its size
b) To reduce its size
c) To increase its size
d) Can’t change its size
Answer:
a) To arrange its size

Set-II

a) To rotate
b) Can’t rotate
c) To remove the image
d) To include and image
Answer:
a) To rotate

Set-IV

a) To add a background image
b) Creating a scene
c) To adjust the views of the images
d) To change its background-color
Answer:
c) To adjust the views of the images

Question 46.
The layers are seen in the layers panel of a synfig studio some activities related to the layers are given in sets. Select correct activities from each set.

Set-1

a) Can’t change its order
b) can change its order
c) Arrange the order before including the layer
d) Layers are not visible
Answer;
b) can change its order

Set -II

a) Can’t group together
b) Objects
c) Can group together
d) Can’t change its color
Answer:
c) Can group together

Set-III

a) To take a copy of a layer
b) Layer is fixed
c) Can edit animation
d) Can’t delete the layer
Answer:
a) To take a copy of a layer

Set-IV

a) Removing the layer may effect he other
b) Can delete a layer
c) Can edit animation
d) Can’t delete the layers
Answer:
b) Can delete a layer

You can Download Publishing Questions and Answers, Kerala SSLC 10th IT Theory Questions and Answers Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

Kerala Syllabus SSLC IT Theory Questions and Answers Chapter 2 Publishing

Publishing Questions & Answers

Question 1.
Clickthrough Insert → fields → more fields in writer software. We get a field window contains utility.
a) File
b) Edit
c) Database
d) None
Answer:
c) Database

Question 2.
Formattings of a text is copped to another text is using tool
a) Clone formatting
b) style
c) Paragraph formatting
d) styles and formatting
Answer:
a) Clone formatting

Question 3.
Which is the Clone formatting tool?

Answer:

Question 4.
A word can define headings, subheadings and para-graph separately is called
a) Style
b) Insert
c) Formatting
d) Clone
Answer:
a) Style

Question 5.
In which menu lies the Index and Tables
a) File
b) Insert
c) Edit
d) Data
Answer:
b) Insert

Question 6.
Where does contain the table of contents in your text?
a) Middle of the text
b) Not entered
c) At the beginning
d) At the end
Answer:
c) At the beginning

Question 7
The Software developed by Leslie Lamport is
a) Linux
b) Windows
c) Android
d) LaTex
Answer:
d) LaTex

Question 8.
The program for type setting technical documents is developed by Donald Knuth is
a) LaTex
b) Tex.
c) Linux
d) Windows
Answer:
b) Tex

Question 9.
If you click at the word preface on the table of contents in writer. What message appears?
a) Ctrl-click to follow link
b) Click to follow link
c) Follow link
d) Click to follow link to close the window
Answer:
a) Ctrl-click to follow link

Question 10.
Which menu lies the field’s utility?
a) Edit
b) Insert
c) File
d) formatting
Answer:
b) Insert

Question 11.
Which software having the mail merge facility?
a) Libre office writer
b) Python
c) Firefox
d) Calculator
Answer:
a) Libre office writer

The Full form of DSLR is Digital Single Lens Reflex Cameras.

Question 12.
To include the names and address one by one in a table in the letters using a feature called
a) Style
b) Mail merge
c) Database
d) Indexable
Answer:
b) Mail merge

Question 13.
Which style is used to the headings more attractive?
a) Character style
b) Page style
c) Paragraph style
d) List style
Answer:
c) Paragraph style

Question 14.
Anu was prepared a style named main head. Then he selects the all headings in a report and clicks on the style named main head. What happens for the headings?
a) All headings are changed to same type
b) No change to the headings
c) Headings are changed to different sizes
d) Headings are seen in different colors
Answer:
a) AI1 headings are changed to same type

Question 15.
The use of organizer tab in styles and formattings window for creating a new style is?
a) No use at all
b) To applying the style
c) Giving the name for the style
d) Don’t give a name for the style
Answer:
c) Giving the name for the style

Questions 16.
Which software includes the styles and formattings utility
a) Cale
b) Python
c) Writer
d) Sun clock
Answer:
c) Writer

Question 17.
Pick out the correct statement related to a frame
a) Can’t able to type inside the frame
b) Frame can move at any place on the page
c) Frame can’t move on the page
d) Can’t able to insert an image in a frame
Answer:
b) Frame can move at any place on the page

Question 18.
Where did you get the fields window?
a) Insert →Fields → More fields
b) Insert → Media → Fields
c) Format → More fields
d) Tools → More fields
Answer:
a) Insert → Fields → More fields

Question 19.
A correct statement related to clone formatting is
a) Use clone formatting for a large report
b) Formattings of headings can be changed by changing the formatting of each heading separately
c) Formattings of headings can be changed by changing the formatting of one heading
d) Formattings of heading cannot be copped to another
b) Formattings of headings can be changed by changing the formatting of each heading separately

Section – II

Question 20.
Right-click on the heading (styles name) on a paragraph style, we get the utilities
a) Modify
b) Colour
c) New
d) Pattern
Answer:
a) Modify

Question 21.
Libre office writer file prepared in Malayalam fonts and English fonts, on which fonts are changed to change the styles of the files?
a) If the file is in Malayalam fonts, then change the CTL font
b) If the file is in English fonts, then change the CTL font
c) If the file is in Malayalam fonts, then change western text font
d) If the file is in English fonts, then change the Western text font
Answer:
a and b

Question 22.
Select the correct statements
a) The styles in a writer are changed and use them
b) The styles in a writer cannot be changed
c) New styles cannot be created
d) New styles can prepared in a writer.
Answer:
a and d

Question 23.
Select two correct statements related to a frame
a) Can’t move from one part of a page
b) To place text or images within a document separated from the main contents
c) To place anywhere in the page
d) Not separated from the main contents
Answer:
b and c

Question 24.
Participant’s cards are prepared by mail merge utility and select print from file menu. Then you get a window to select two utilities to save the output file what are they?
a) Save as single document
b) Save as image
c) Save as individual documents
d) Save picture as
Answer:
a and c

Question 25.
Using mail merge to preparing the letters for the parents about the Youth Festival. What are the preparations we make for this?
a) Making a list of parents address in Libre Office writer
b) Making a list of parents addresses in Libre Office Calc
c) Preparing the letters to the parents in Libre office writer
d) Preparing the letters to the parents in Libre office calc
Answer:
b and c

Question 26.
Why did scientific articles and research reports are prepared by LaTex Software?
a) Facility to use English fonts
b) Facility to use Malayalam fonts
c) Several features for typesetting Symbols
d) Several features for typesetting equations.
Answer:
c and d

Question 27.
What are the difficulties to prepare a text /document using clone formatting?
a) Separate formattings the contents of a text is difficult
b) Formattings of heading can be changed by changing the formattings of each heading separately
c) Formattings of headings/ paragraph can be changed by changing one
d) One by one formatting is very easy
Answer:
a and b

Question 28.
The order of preparing a table of contents to the school report are following. Select one from each set.

Set -1

a) Close the report
b) Type the school report
c) Correct the report
d) Open the prepared school report
Answer:
d) Open the prepared school report

Set – II

a) Click on the last page of the report
b) Click on the place where the table of contents to be inserted
c) Insert the table of contents on the cover page
d) Click on the blank space of the report
Answer:
b) Click on the place where the table of contents to be inserted

Set – III

a) Select Index and Tables from Insert Menu
b) Select fields from Insert menu
c) Select modify
d) Select styles and formatting from insert menu
Answer:
a) Select Index and Tables from Insert Menu

Set – IV

a) Close the window
b) A new window appears, give a heading and background color for the table of contents and click OK.
c) Click cancel on the new window
d) Leave the window without any change
Answer:
b) A new window appears, give a heading and background color for the table of contents and click OK.

Question 29.
What are the styles included in styles and formatting window?

Set -1

a) Paragraph style
b) Western-style
c) Formatting style
d) More style
Answer;
a) Paragraph style

Set – II

a) Apply style
b) Character styles
c) Style Box
d) Clone formatting
Answer:
b) Character styles

Set – III

a) Heading style
b) Report style
c) Frame style
d) Life style
Answer:
c) Frame style

Set-IV

a) Page style
b) Heading 1
c) Head
d) Address
Answer:
a) Page style

Question 30.
The uses of mail merge utility in LibreOffice writer.

Set-I

a) Preparing Notice
b) Preparing the letter for parents about youth festival
c) Writing the mark list
d) To prepare address
Answer:
b) Preparing the letter for parents about youth festival

Set – II

a) Creating a style
b) Taking a print
c) To taking a copy
d) To prepare participant’s card
Answer:
d)To prepare participant’s card

Set – III

a) To prepare certificate
b) To prepare a letter to a friend
c) To prepare a document
d) To prepare a study report
Answer:
a) To prepare certificate

Set – IV

a) To calculate the achievement
b) To prepare electricity bill
c) To apply style
d) To prepare the table of contents
Answer:
b) To prepare electricity bill

Question 31.
Which are the styles we are using from paragraph style? Make a list using from each set

Set – I

a) Table
b) Report
c) Heading
d) Contents
Answer:
c) Heading

Set – II

a) Caption
b) Font
c) Footer
d) Color
Answer:
a) Caption

Set-III

a) Find
b) Link
c) Object
d) Index
Answer:
d) Index

Set-IV

a) Body
b) Media
c) Text body
d) Frame
Answer:
c) Text body

You can Download Solids Questions and Answers, Activity, Notes, Kerala Syllabus 10th Standard Maths Solutions Chapter 8 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Maths Solutions Chapter 8 Solids

Solids Textbook Questions & Answers

Textbook Page No. 191

Solids Class 10 Kerala Syllabus Kerala Syllabus Chapter 8 Question 1.
A square of side 5 centimetres, and four isosceles triangles!of base 5 centimetres and height 8 centimetres, are to be put together to make a square pyramid. How many square centimetres of paper is needed?
Answer:
Area of base = 5 × 5 = 25 cm² Area of one triangle 1/2 × 5 × 8 = 20 cm² Curved surface area = 4 × 20 5cm = 80 cm²
Paper is needed to make a square pyramid = 25 + 80 = 105 cm²

Solids in Maths SSLC Question 2.
A toy is in the shape of a square pyramid of base edge 16 centimetres and slant height 10 centimetres. What is the total cost of painting 500 such toys, at 80 rupees per square metre?
Answer:
Surface Area of the toy
= 16 × 16 + 4 × 1/2 × 16 × 10
=256 + 320 = 576 cm²
Surface Area of 500 toys = 500 × 576 = 288000 cm²

Kerala Syllabus 10th Standard Maths Question 3.
The lateral faces of a square pyramid are equilateral triangles and the length of a base edge is 30 centimetres. What is its surface area?
Answer:
Lateral faces are equilateral surfaces Surface area
= Base Area + Curved surface area

= 900 + 900 √3
=900 + 1558.8 = 2458.8 cm² = 2459 cm²

The Chebyshev’s theorem calculator counts the probability of an event being far from its expected value.

Sslc Maths Chapter 8 Kerala Syllabus Question 4.
The perimeter of the base of square^»yra- mid is 40 centimetres and the total length of all its edges is 92 centimetres. Calculate its surface area.
Answer:
Base perimeter = 4a = 40
a=10cm
Total length of edges = 92 cm
Length of total laterals edge = 92 – 40 = 52
Length of one laterals edge = \(\frac { 52 }{ 4 }\) = 13 cm
Surface area of pyramid = a 2 + 2 al
= 102 + 2 × 10 × 13
= 100 + 260 = 360 cm2

KBPS full form, KBPS stands for, meaning, what is KBPS, description, example, explanation, acronym for, abbreviation, definitions, full name.

Surface Area of Kerala Question 5.
Can we make a square pyramid with the lateral surface area equal to the base area?
Answer:
Curved surface area = 2al
Area of base = a2
a2=2al
a = 21 ⇒ 1 = a/2
For making a square pyramid first we must determine its base, one side of the lateral will be the base. Other two sides make half of base by reducing the angle. That is angle at apex will be less than 90°.

Textbook Page No. 193

Sslc Maths Solids Kerala Syllabus Chapter 8 Question 1.
Using a square and four triangles with dimensions as specified in the picture, a pyramid is made.

What is the height of this pyramid?
What if the square and triangles are like this?

Answer:

Sslc Maths Solids Equations Kerala Syllabus Chapter 8 Question 2.
A square pyramid of base edge 10 centimetres and height 12 centimetres is to be made of paper. What should be the dimensions of the triangles?
Answer:
Base edge of a square pyramid = 10 cm
Let h be the height
Base edge a = 10 cm
Height h= 12 cm
Slant height =

Sslc Maths Chapter 8 Solids Kerala Syllabus Question 3.
Prove that in any square pyramid, the squares of the height, slant height and lateral edge are in arithmetic sequence.
Answer:
Height = h,
Slant height = l,
Lateral edge = e

Solids in Maths Question 4. A square pyramid is to be made with the triangles shown here as a lateral face. What I would be its height? What if the base edge is 40 centimetres instead of 30 centimetres?

Answer:

It is impossible to make a square pyramid of base edge 40cm.

Textbook Page No. 195

Solids Chapter Class 10 Kerala Syllabus Question 1.
What is the volume of a square pyramid of base edge 10 centimetres and slant height 15 centimetres?
Answer:
a = 10, l = 15
Volume of pyramid =

Sslc Maths Chapter Solids Kerala Syllabus Chapter 8 Question 2.
Two square pyramids have the same volume. The base edge of one is half that of the other. How many times the height of the second pyramid is the height of the first?
Answer:

The height of the second pyramid is 4 times the height of the first pyramid.

Solids Class 10 Kerala Syllabus Chapter 8 Question 3.
The base edges of two square pyramids are in the ratio 1:2 and their heights in the ratio 1:3. The volume of the first is 180 cubic centimetres. What is the volume of the second?
Answer:

Solids In Maths Sslc Kerala Syllabus Chapter 8 Question 4.
All edges of a square pyramid are 18 centimetres. What is its volume?
Answer:
Length of base edge a = 18 cm

Class 10 Maths Solids Kerala Syllabus Chapter 8 Question 5.
The slant height of a square pyramid is 25 centimetres and its surface area is 896 square centimetres. What is its volume?
Answer:
l = 25 cm
Surface area = 896 cm
a2 + 2al = 896
a2 + 2a × 25 = 896
a2 + 50a – 896 = 0

Sslc Maths Chapter 8 Solutions Kerala Syllabus Question 6.
All edges of a square pyramid are of the same length and its height is 12 centimetres. What is its volume?
Answer:

Sslc Maths Solids Questions Kerala Syllabus Chapter 8 Question 7.
What is the surface area of a square pyramid of base perimeter 64 centimetres and volume 1280 cubic centimetres?
Answer:
Base perimeter 4a = 64
a= 16 cm

Surface area = a2 + 2al
= 162 + 2 × 16 × 17 = 256 + 544 = 800 cm²

Textbook Page No. 198

Maths Solids Class 10 Kerala Syllabus Chapter 8 Question 1.
What are the radius of the base and slant height of a cone made by rolling up a sector of central angle 60° cut out from a circle of radius 10 centimetres?
Answer:
R = Radius of the circle

Radius of cone = 1.66 cm
Radius of circular part = slant height of cone = 10 cm.

Sslc Solids Solutions Kerala Syllabus Chapter 8 Question 2.
What is the central angle of the sector to be used to make a cone of base radius 10 centimetres and slant height 25 centimetres?
Answer:
Central anglejof the sector
(x) = \(\frac { r }{ l }\) × 360,
r = 10 cm, l = 25 cm
\(=\frac{10}{25} \times 360\)
= 144°

Solids Maths Questions Kerala Syllabus Chapter 8 Question 3.
What is the ratio of the base-radius and slant height of a cone made by rolling up a semicircle?
Answer:
Radius of bigger circle = R
Radius of smaller circle = r
Radius of circular base of the pyramid = r = \(\frac { R }{ 2 }\)
Ratio between radius and slant height

Textbook Page No. 199

Sslc Maths Solutions Kerala Syllabus Chapter 8 Question 1.
What is the area of the curved surface of a cone of base radius 12 centimetres and slant height 25 centimetres?
Answer:

r = 12cm l = 25cm
Curved surface area = πrl
π × 12 × 25 = 300π = π × 12 × 25
= 300 × 3.14 = 314 × 3
= 942 cm2

Kerala Syllabus 10 Maths Solutions Kerala Syllabus Chapter 8 Question 2.
What is the surface area of a cone of base diameter 30 centimetres and height 40 centimetres?
Answer:
Radius = \(\frac { 30 }{ 2 }\)= 15 cm =r
Height = h = 40 cm
Total surface area = Base area + Curved surface
area = πr2 + πrl

= 865 πr = 2718.6 m²

Std 10 Kerala Syllabus Maths Solutions Chapter 8 Question 3.
A Conical firework is of. base diameter 10 centimetres and height 12 centimetres, 10000 such fireworks are to be wrapped in colour paper. The price of the colour paper is 2 rupees per square metre. What is the total cost?
Answer:
r = 5 cm
h =12 cm
t = 13 cm
Total surface area of one firework = Base area + Curved surface area

Question 4.
Prove that for a cone made by rolling up a semicircle, the area of the curved surface is twice the base area.
Answer:
Perimeter of base of a cone is equal to half of perimeter of large cone.
Radius of pyramid = R/2
Perimeter of base

= 2 × Base area , that is twice

Textbook Page No. 200

Question 1.
The base radius and height of a cylindrical block of wood are 15 centimetres and 40 centimetres. What is the volume of the largest cone that can be carved out of this?
Answer:
r = 15cm, h = 40cm
Volume of cone = \(\frac { 1 }{ 3 }\) πr² h

= \(\frac { 1 }{ 3 }\) π × 15 × 15 × 40
= 3000π = 3000 × 3.14
= 314 × 30
= 9420 cm³

Question 2.
The base radius and height of a solid metal cylinder are 12 centimetres and 20 centimetres. By melting it and recasting, how many cones of base radius 4 centimetres and height 5 centimetres can be made?
Answer:

No. of cones = Volume of cylinder/Volume of cone Volume of cylinder
= π × 12 × 12 × 20
Volume of cone = \(\frac { 1 }{ 3 }\) π × 4 × 4 × 5

Question 3.
A sector of central angle 216° is cut out from a circle of radius 25 centimetres and is rolled up into a cone. What are the base radius and height of the cone? What is its volume?
Answer:

Question 4.
The base radii of two cones are in the ratio 3:5 and their heights are in the ratio 2 : 3. What is the ratio of their volumes?
Answer:

Question 5.
Two cones have the same volume and their base radii are in the ratio 4:5. What is the ratio of their heights?
Answer:

Textbook Page No. 203

Question 1.
The surface area of a solid sphere is 120 square centimetres. If it is cut into two halves, what would be the surface area 0f each hemisphere?
Answer:
Surface area of the solid sphere

Surface area of one hemisphere = 3 πr²
3 πr² = 3 π × \(\frac { 30 }{ π }\) = 90 cm²

Question 2.
The volumes of two spheres are in the ratio 27 : 64. What is the ratio of their radii? And the ratio of their surface areas?
Answer:
Ratio of volumes

Question 3.
base radius and length of a metalder are 4 centimetres and 10 centimetres, If it is melted and recast into spheres of radius 2 centimetres each, how many spheres can be made?
Answer:

Question 4.
A metal sphere of radius 12 centimetres is melted and recast into 27 small spheres. What is the radius of each sphere?
Answer:
Radius = 12cm
Volume of bigger sphere
\(=\frac{4}{3} \pi \mathrm{R}^{3}=\frac{4}{3} \pi \times 12^{3}\)
If the radius of smaller sphere is ‘r’
Volume of 27 smaller spheres = Volume of the bigger sphere

Question 5.
From a solid sphere of radius 10 centimetres, a cone of height 16 centimetres is carved out What fraction of the volume of the sphere is the volume of the cone?
Answer:
Radius of sphere = 10 cm
Radius of cone

Question 6.
The picture shows the dimensions of a petrol tank.

How many litres of petrol can it hold?
Answer:
Length of circular cylinder = 4 m
Height = 4 m
Radius = 1m
Volume of circular cylinder = π × 12 × 4 = 4π
= 4 × 3.14 = 12.56 cm³
Volume of two hemisphere = Volume of a sphere

Litres of petrol the tank can hold = 12.56 + 4.19 = 16.73 m³ = 16750 litre

Question 7.
A solid sphere is cut into two hemispheres. From one, a square pyramid and from the other a cone, each of maximum possible size are carved out. What is the ratio of their volumes?
Answer:

Solids Orukkam Questions & Answers

Worksheet 1

Question 1.
The base edge of a square pyramid is Stem, height 3cm. Calculate slant height and lateral edge
Answer:
Base edge = 8 cm, height = 3 cm

Question 2.
Slant height of a square pyramid is 10cm, height 6cm .Calculate total length of the edges.
Answer:
Slant height =10 cm, height = 6 cm

Question 3.
The slant height of a square pyramid is 12 cm, lateral edge 13 cm. Calculate height
Answer:
Slant height = 12 cm
lateral edge = 13 cm

Question 4.
The length of base edge is 24 cm, slant height 13 cm. Find height and lateral edge
Answer:
Base edge = 24 cm
Slant height = 13 cm
Height \(=\sqrt{(13)^{2}-(12)^{2}}=\sqrt{25}=5 \mathrm{cm}\)
Length of lateral edge = \(=\sqrt{(13)^{2}-(12)^{2}}=\sqrt{313} \mathrm{cm}\)

Worksheet 2

Question 5.
A sector is folded in such a way as to get a cone. Radius of the sector is 12 cm, central angle 120°.Calculate radius and slant height
Answer:
Slant height of cone = radius of sector = 12 cm
Radius of cone = \(\frac { 120 }{ 360 }\) of radius of sector = 12 × \(\frac { 120 }{ 360 }\) = 4 cm

Question 6.
The central angle of a sector is 90°, radius 16cm, calculate slant height and radius
Answer:
slant height of cone = 16 cm

Question 7.
Slant height of a cone is 20 cm, radius 10 cm. What should be the radius and central angle of the sector?
Answer:
Radius of sector = slant height of cone = 20 cm
Central angle of the sector = radius of cone \(\times \frac{360}{R}=10 \times \frac{360}{20}=18^{\circ}\)

Question 8.
Radius of a cone is 4cm, slant height is 5/2 times radius. Calculate the radius and central angle of the sector.
Answer:
Radius of sector = slant height of cone =

Worksheet 3

Question 9.
The base edge of a square pyramid is 6cm, height 4cm, calculate slant height and total surface area.
Answer:
Length of base edge = 6 cm, height = 4 cm,
slant height = \(\sqrt{(3)^{2}+(4)^{2}}=\) \(\sqrt{9+16}=\sqrt{25}=5 \mathrm{cm}\)
Total surface area = base area + curved surface area=(6)2 + 2 × 6 × 5 = 36 + 60 = 96 cm².

Question 10.
The height of a square pyramid is 12cm, slant height 15cm , calculate total surface area and volume
Answer:
height =12 cm. slant height = 15 cm
\(a=2 \sqrt{15^{2}-12^{2}}\) = 2 × 9 = 18 cm²
Total surface area = base- area + curved surface area
= (18)2 + 2 × 18 × 15 = 324 + 540 = 864 cm²
Volume = \(\frac{1}{3}(18)^{2} \times 12=1296 \mathrm{cm}^{3}\)

Question 11.
The base perimeter of a square pyramid is 48cm. Slant height is 10cm. Calculate lateral surface area and volume.
Answer:
Base perimeter = 48cm
baseedge = \(\frac { 48 }{ 4 }\) = 12 cm, slant height = 10cm
height = \(\sqrt{(10)^{2}-(6)^{2}}=\sqrt{64}=8 \mathrm{cm}\)
Curved siuface area= 2 × 12 × 10 = 240 cm²
Volume = \(\frac { 1 }{ 3 }\) (12)2 × 8 = 384 cm³

Question 12.
The height of a square pyramid is 15cm, volume 1620cm. Calculate the total surface area.
Answer:
Volume of square pyramid =

Worksheet 4

Question 13.
The base area of a cone is 25 π cm, curved surface area 165 π. Calculate total surface area.
Answer:
Total surface area of cone = base area + curved surface area
= 25 π +165 π = 190 cm²

Question 14.
Base area of a cone is 81 π, height 12 cm. Calculate volume
Answer:
Volume of cone = \(\frac { 1 }{ 3 }\) × base perimeter ×
height = \(\frac { 1 }{ 3 }\) × 81 π × 12 = 324 π cm³

Question 15.
The height of a cone is 4cm, slant height 5cm. Calculate total surface area
Answer:
Height of cone = 4 cm
Slant height = 5 cm

Surface area = π (3)² + π x 3 x 5 = 9 π + 15 π =24 π cm²

Question 16.
Radius of a cone is 10cm, volume 3140 cubic centimeter. Calculate total surface area
Answer:
Radius of cone = 10 cm

Question 17.
Calculate the surface area and volume of a sphere of radius 3 cm.
Answer:
Surface area of sphere = 4 π (3)² = 36 π cm²
Volume = \(\frac{4}{3} \pi r^{3}=\frac{4}{3} \times \pi \times(3)^{3}=36 \pi \mathrm{cm}^{3}\)

Workshee 5

Question 18.
Calculate the volume of a sphere of surface area 144 π square centimetre.
Answer:

Solids SCERT Questions & Answers

Question 19.
The measurements of the lateral surface of a square pyramid are shown in the figure. Calculate die base edge and slant height of die pyramid. [Score: 2, Time: 3 minute]

Answer:
Base edge = 10cm (1)
The given figure can be divided into n two right-angled triangles their angles are 30°, 60°, 90°, so ratio of their side will be 1: √3: 2.
2x = 10, x = 5
Slant height = 5 √3 cm (1)³

Question 20.
Is It possible to construct a pyramid of base edge 24 cm and lateral edge 13 cm? Justify [Score: 2, Time: 3 minute]
Answer:
Since slant height is 5 cm, such a pyramid can’t be constructed. (1)
Slant height should be greater than half of the base edge. \(\sqrt{13^{2}-12^{2}}=5 \mathrm{cm}\) (1)

Question 21.
Lateral surface of a square pyramid is shown in the Figure. All angles are equal

Find the total length of all edges of the square: pyramid. Find the slant height What is the |atio between height and slant height [Score: 4, Time: 5 minutes]
Answer:
Sum of edges = 8 x 8 = 64 cm (1)
Slant height = 4√3 cm (1)

Question 22.
Devikamade a square pyramid having base edge 40cii and height 15cm. Unfortunately, one lateral face got separated from the pyramid. Check which figure given below shows the isosceles triangle that got separated. [Score: 3, Time: 5 minute]
Answer:

Square pyramid can’t be constructed since slant height 1 should be greater than half the base edge. (1)

If base edge = 40 and slant leight = 35, then height can’t be 15 (1)

Here height of pyramid is 15, so this is the isosceles triangle that got separated (1)

Question 23.
A tent constructed in the form of a square pyramid of base perimeter 80 metres and lateral edge 26 metres
a. Calculate the slant height of the tent
b. Calculate the area of tarpaulin sheet. required to cover the lateral faces of the tent. [Score: 3, Time: 5 minute]
Answer:
Base perimeter = 80 m, Base edge= 20 m Lateral edge = 26 m

= 960 sq.metre (2)

Question 24.
The triangle given in the figure is one lateral face of a square pyramid.
a. Calculate the slant height.
b. Find the lateral surface area of the pyramid, [Score: 3, Time: 4 minutes]

Answer:
Slant height \(=\sqrt{17^{2}-8^{2}}\)
\(=\sqrt{275}=15 \mathrm{cm}\) (2)

Question 25.
A square pyramid is made from a solid cube having edge 30cm. Calculate the surface area. [Score: 3, Time: 5 minute]
Answer:
Base edge of the square pyramid = 30 cm
Height = 30cm
Slant height = \(\sqrt{30^{2}+15^{2}}\)
\(=\sqrt{900+225}=\sqrt{1125}=15 \sqrt{5}\) (1)
Lateral surface area = \(=4 \times \frac{1}{2} \times 15 \sqrt{2} \times 30\)
= 60 x 15√5 = 900√5 sq.cm (1)
Total surface area = 900 + 900√5
= 900(1 + √5) sq.cm (1)

Question 26.
The lateral faces of a square pyramid are equilateral triangles Lateral, edge = 20 cm
a. Calculate the slant height
b. Find its surface area.
C. Find its volume. [Score: 5, Time: 6 minute]
Answer:
a. Slant height = 10 √3 cm (1)
b. Lateral surface area

Question 27.
Prove that the ratio between the base edge, slant height and height of a square pyrarmid having equal edges is 2: √3 : √2. [Score: 4, Time: 5 minute]
Answer:
Slant height = √3 a
Height = \(\begin{array}{l}{=\sqrt{(\sqrt{3} a)^{2}-a^{2}}} \\ {=\sqrt{2} a}\end{array}\)
Base edge: Slant height : Height
= 2a: √3a : √2 a = 2: √3 : √2 (1)

Question 28.
The ratio between the base edges of two square pyramids is 1: 2. The heights are also in the same ratio. If the volume of the first pyramid islO cubic centimeters, calculate the volume of the sec ond one. [Score: 3, Time: 4 minute]
Answer:
v₁ : v₂ = 1 : 8
v₁ = \(\frac { 1 }{ 3 }\) a₂ h
v₂ = \(\frac{1}{3}(2 a)^{2} \times 2 h, v_{1}: v_{2}=1: 8\) (1)
Volume of the second pyramid = 800 cm² (1)

Question 29.
Meera constructed a square pyramid of base edge 10cm and height 6cm Manu made a square pyramid having base edge 5cm and height 4cm. Find the volume of the pyramids and compare the measurements. [Score: 3, Time: 4 minutes]
Answer:
Volume of Meera’s pyramid = \(\frac { 1 }{ 3 }\) x Baste edge x height 3
= \(\frac { 1 }{ 3 }\) x 102 x 6 =200 cm3 (1)
Volume of Manu’s pyramid = \(\frac { 1 }{ 3 }\) x 52 x 24 = 200cm³
Volumes are equal (1)

Question 30.
The central angle of a sector is 288° If this sector is rolled up to make a cone, find the ratio between the radius and slant height of the cone. [Score :: 4, Time: 5 minutes]
Answer:
360 x \(\frac { 4 }{ 5 }\) = 288
∴ Radius of the cone 4 = \(\frac { 4 }{ 5 }\) x radius of the big circle (1)
∴ If r is the radius of the circle Radius of the
Radius of the cone = \(\frac { 4 }{ 5 }\) r (1)
But radius of the circle = slant height of cone
i.e., 1 = r (1)
∴Ratio between the radius of the cone and slant height
\(=\frac{4}{5} r: r=\frac{4}{5}: 1=4: 5\) (1)

Question 31.
The ratio between the radius and slant height of a cone is 2 : 3. Find the central angle of the sector to make the cone. [Score: 3, Time: 4 minutes]
Answer:
Ratio between the radius and slant height 2 : 3 (1)
Area length of the sector is equal to \(\frac { 2 }{ 3 }\) part of the circle perimeter. (1)
Central angle of the sector = 360 x \(\frac { 2 }{ 3 }\) = 240° (1)

Question 32.
The central angle of a circle is divided in the ratio 2 : 3 to form two sectors. Two cones are made by rolling up the two Rectors.
a. Find out the ratio between the base perimeters of the cones.
b. What is the ratio between the curved surface areas. [Score: 3, Time 6 minutes]
Answer:
a. Central angles are in the ratio 2 : 3 , so, let the perimeter of the two clones which made by rolling up this two sectors be \(\frac { 2 }{ 5 }\)
part and \(\frac { 3 }{ 5 }\) part of the perimeter of the circle (1)
That is perimeter of each sector be \(2 \pi r \times \frac{2}{5} \text { and } 2 \pi r \times \frac{3}{5}\)
Ratio between base perimeters of cone = \(2 \pi r \times \frac{2}{5}: 2 \pi r \times \frac{3}{5}=2: 3\)
b. Base perimeter of the cones will be \(\frac { 2 }{ 5 }\) and \(\frac { 3 }{ 5 }\) parts of the circumference of the 5 circle. (1)
Ratio between the perimeters of die cones = \(\pi r^{2} \times \frac{2}{5}: \pi r^{2} \times \frac{3}{5}=2 ; 3\) (1)

Question 33.
Find the ratio between the radius and slant height of a cone by roiling up a sector with central angle 120°. If the curved surface area is 108 π, find the radius and slant height of the cone. [Score:5,Time:7minutesv]
Answer:
Area of the sector with central angle 120° is one-third of the area of the circle. (1)
Curved surface area = 108 π
Ares of die; sector = 108 π,
which is one-third of the area of the circle Area of the circle = 108 π × 3
πr² = 324 π (1)
Radius of the circle r = 18 cm
Slant height = 18 cm (1)
Radius of the cone = 6 cm (1)

Question 34.
A wooden cone is has radius 30crn and height 40crn. Find its slant height. Calculate the cost to paint the face of 10 such cones at the rate of Rs.50/- per square metre. [Score: 5, Time: 7 minutes]
Answer:
Base Radius = 30 cm
Height =40 cm
Slant height = \(\sqrt{40^{2}+30^{2}}=50\) (1)
Surface area of the cone = πr² + πrl = π × 302 + π × 30 × 50 (1)
= 900π + 1500π = 2400π (2)
Total cost to paint 10 cones \(=\frac{2400 \pi \times 10 \times 50}{10000}\)
= \(=\frac{2400 \times 3.14 \times 10 \times 50}{10000}=377 \mathrm{Rs}\) (1)

Question 35.
Two cones are made using two sectors of central angles 60° and 120° of a circle. If the radius of the smaller cone is 5cm
a. Calculate the radius and base area of the smaller cone.
b. Find the surface area of the bigger cone. [Score: 5, Time: 8 minutes]
Answer:
a. Central angle of the small sector = 60°
\(\frac { 1 }{ 6 }\) part of the area of the circle Base radius of cone formed from above sector = 5 cm (1)
Radius of the circle = 5 × 6 = 30 (1)
Similarly, area of the sector of central angle 120° = \(\frac { 1 }{ 3 }\) of the area of the circle Base radius ofthe bigger cone = 30 × \(\frac { 1 }{ 3 }\) = 10 (1)
Base area of the bigger cone = π × 102 = 100× (1)
b. Curved surface area ofthe bigger cone = π × 10 × 30 = 300π (1)
Surface area = 100π + 300π = 400π cm² (1)

Question 36.
Three solids a square pyramid, a cone and a sphere have been carved out from three solid cubes of the same size. Find the volume of each solid. [Score: 5, Time: 7 minutes]
Answer:
Volume of square pyramid = \(\frac { 1 }{ 3 }\) a³ (1)
Volume of cone

Question 37.
A metal sphere is melted and recasted into a cone. Both have same radii
a. Find the relationship between the height of the cone and the radius of the sphere.
b. Which solid has greater surface area? Justify. [Score: 5, Time: 7 minutes]
Answer:
a. If r is the radius of the sphere, then its volume = \(\frac { 4 }{ 3 }\) πr³ (1)
If h is the height of the cone, then its volume = \(\frac{1}{3} \pi r^{2} h=\frac{4}{3} \pi r^{3}\)
h = 4r (1)
Height of pyramid is four times the radius of sphere.
Surface area of the sphere = 4πr²
Slant height of the cone = \(\sqrt{(4 r)^{2}+r^{2}}=\sqrt{17 r^{2}}=\sqrt{17} r\) (1)
Surface area of the cone = \(\pi r^{2}+\pi r \sqrt{17} r=\pi r^{2}(1+\sqrt{17})\)
b. .Surface area of the cone is greater (1)

Question 38.
A hemisphere and a cone with same radii are attached to get a solid as given in the figure. Radius of the hemisphere is 9 cm. The height of the two solids together is 21 cm.
a. Find the height of the cone.
b. Find the volume of the cone
c. Find the volume of the solids [Score: 4, Time: 6 minutes]
Answer:
Height of the cone 21 – 9 = 12 cm (1)

Solids Exam Oriented Questi0ns& Answers

Short Answer Type Questions (Score 2)

Questi0n 39.
Total surface area of a solid hemisphere is 675 π sqcm. Find the curved surface area of the solid hemisphere.
Answer:
3 πr² = 675π cm²
r2 = 225
The CSA of the solid hemisphere,
CSA = 2πr² = 2π × 225 = 450π cm²

Questi0n 40.
The volume of a solid right circular cone is 4928 cm3. If its height is 24 cm, them find the radius of the cone.
Answer:
V = 4928 cm³ and h = 24 cm

Questi0n 41.
The figure given below has the total length 20cm height and common diameter 6cm. Find the volume of the figure.

Answer:
volume of solid

Questi0n 42.
The circular plate of radius 12cm is cut out into six sectors having same size. Calculate the slant height and radius of circular cone used to make one sector.
Answer:
slant height = 12cm
centre angle = \(\frac { 360 }{ 60 }\) = 60
radius of square pyramid = 12 × \(\frac { 60 }{ 360 }\) = 2 cm

Questi0n 43.
A circus tent is in the shape of a square pyramid. The area of the base is 1600m² and its height is 375m then, how much canvas would be needed for this and also find its perimeter?
Answer:

Questi0n 44.
Height of a cone is 40cm. Slant height is 41cm.
a. Find diameter of its base,
b. Find volume
Answer:

Questi0n 45.
Radius and slant height of a solid right circular cone are 35cm and 37 cm respectively. Find the curved surface area and total surface area of the cone.
Answer:
r= 35 cm
l = 37 cm
CSA = πrI = π(35 x 37)=4070 cm²
TSA = πr (1 + r )
\(=\frac{22}{7} \times 35 \times(37+35)=7920 \mathrm{cm}^{2}\)

Questi0n 46.
Surface area of a wooden sphere is 40 cm2. It is cut into two identical hemispheres. Find
a. The area of the plane surface of one of the hemispheres,
b. Its surface area.
Answer:
a. Surface area of the sphere = 4 πr²
Here, 4 πr² = 40cm²
πr² = 10cm²
∴ Area of plane surface = πr² = 10cm²
b. Surface area of one piece (hemisphere)
= 3 πr² = 3 x 10 = 30cm²

Short Answer Type Questions (Score 3)

Questi0n 47.
A toy in the shape of a square pyramid has base edge 16 cm and slant height 10 cm. 500 of these are to be painted and the cost is 80 rupees per square meter. What would be the total cost?
Answer:
Surface area = Curved surface area + base area
4 × 1/2 × 16 × 10+ 162 = 320 + 256 = 576cm
Total surface area of 500 square prism is = 500 × 576 = 288000 cm²

Questi0n 48.
The base radius of a circular cone is 9cm and its height is 12cm. What is the radius and central angle of die sector used to make it?
Answer:
base radius r =9 cm
height h =12cm
slant height l = \(\begin{array}{l}{=\sqrt{12^{2}+9^{2}}=\sqrt{144+81}} \\ {=\sqrt{225}=15 \mathrm{cm}}\end{array}\)
radius of the sector = 15cm
centre angle of sector = 360 × \(\frac { 9 }{ 15 }\) = 216°

Questi0n 49.
The central angle of a sector is 120°. What is the ratio of radius and slant height of a circular cone made by it? What is the radius and slant height of a cone if its curved surface area is 108 π cm².
Answer:
The ratio of radius and slant height = \(\frac { 120 }{ 360 }\) = \(\frac { 1 }{ 3 }\) = 1 : 3
curved surface area (πrl) =108π
rl=108; r × 3 r = 108
3 × r² = 108; r² = 36
radius (r) = 6 cm; slant height (l) = 3r = 3 × 6 = 18cm

Questi0n 50.
For constructing a square pyramid, Rabiya cut of four triangles and a square. Figure given below shows the measures of these triangles and square. Can you make a square pyramid by using these measures? Explain the reason.

Answer:
Base edge = 42 cm
Slant edge = 29 cm
Slant height = \(\sqrt{29^{2}-\left(\frac{42}{2}\right)^{2}}=\sqrt{29^{2}-21^{2}}\)
\(\sqrt{841-441}=\sqrt{400}=20 \mathrm{cm}\)
Slant height is less than the half of the base edge 21cm so not possible for making a square pyramid

Long Answer Type Questions (Score 4)

Questi0n 51.
A sector shown in the figure is rolled up and made a cone. Find its
a. Slant height
b. Base radius
c. Volume

Answer:
a. Slant height = radius of the
sector = 30cm


\(\frac{2000 \sqrt{2 x}}{3} \mathrm{cm}^{3}\)

Questi0n 52.
Paddy is filled in a cylindrical shaped vessel. Then it has the following shape. How many litres of paddy does it contain?

Answer:

Questi0n 53.
A petrol tank is in the shape of a cylinder with hemisphere of the same radius as the base of the cylinder attached to both ends. If the total length of the tank is 5 meters and the base radius of the cylinder is 1 metre, how many litres of petrol can it hold?
Answer:
Volume of cylinder = πr²h = 3πcm³
r=1, h=3
Volume of hemisphere = 2/3 πr³

Long Answer Type Questions (Score 5)

Questi0n 54.
A toy is in the shape of a hemisphere attached to one end of the cone Total height of the toy is 14.5cm, and common diameter is 7cm.
a. Draw a rough figure based on this fact,
b. Find the volume of the toy.
Answer:

Questi0n 55.
Prove that:
i. ∠BAT = ∠BPA
ii. ∠BAS = ∠AQB

Answer:

Solids Memory Map

You can Download Public Administration Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Social Science Solutions Part 1 Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Social Science Solutions Part 1  Chapter 3 Public Administration

Public Administration Text Book Questions and Answers

Public Administration Class 10 Kerala Syllabus Question 1.
Some of government institutions and their functions are given below. Expand the table by writing more.
Answer:

Institutions	Functions
Primary health center	Provide treatment facilities
Krishi bhavan	Promotes agriculture
Police station	Maintains law and order
Schools	Provide learning opportunities
Post offices	Provide communications
Courts	Protecting liberties and rights

Sslc History Chapter 3 Notes Kerala Syllabus Question 2.
Discuss and list out the changes in the objectives of public administration in monarchy and democracy.
Answer:

Monarchy	Democracy
Completely under king.	Values on the human rights, liberty and democracy
Law is the king	Ensures man’s liberty
King is last and cannot be questioned	Humans are given complete control
Ruler may be from selected families	All are given chances to vote and elect

State Syllabus Class 10 Social Science Notes Pdf  Question 3.
Whom do you wish to get the services of public administration? Discuss in your class and write your conclusion.
Answer:
The main aim of public administration is to ensure equality and justice to all is, especially, the deprived society. Government has taken measures for the social safety and enlistment. Crores are spent for the purpose. But corruption and political intervention are a hindrance.

• Justice for the deserved.
• Education, employment and treatment facilities for the deprived ones.
• Control the cost of commodities.
• Strict laws for the upliftment of women.
• I believe these to be the main aims of public administration.

Public Administration Malayalam Notes Kerala Syllabus Question 4.
Visit a nearby Government. Office and prepare a report on the features of bureaucracy there.
Answer:
Bureaucracy is the main weapon of the government. They play various responsibilities. Employee may be highly experienced as they serve for long years. Ministers are the heads. Employees influence the decisions of the government. Though the ruling party changes, bureaucracy remains the same. Employees help the ministers in order to execute plans.

Kerala Syllabus 10th Standard Social Science Notes Malayalam Medium Question 5.
Discuss and prepare a note on the changes to be brought in the administrative system.
Answer:

• Ensure justice and equal rights to all sections of the society.
• Create corruption free administration and responsibility among the workers.
• Constitution must be to save the victims and punish culprits.
• Government service must be reached at fixed time.
• Protect the rights of women and ensure their safety.
• Ensure the welfare of all.

Use our free online relative standard deviation RSD calculator to know the standard deviation and %RSD for the given mean of data.

Social Science Class 10 Kerala Syllabus Question 6.How far the Right to Information Act make the general administration system efficient. Evaluate.
Answer:
In 2005, RTI Act was passed by the parliament. To get information is the fundamental right of all citizens. The main objectives of this Act are to prevent corruption, create responsibility and make the functioning of the government transparent. The citizens will get copies of public documents if they apply for them.

Sslc History Chapter 1 Notes Pdf Kerala Syllabus Question 7.
What are the situations in our society where the Right to Information Act can be positively used? Discuss and prepare a note.
Answer:
In 2005, RTI Act was passed by the parliament. To get information is the fundamental right of all citizens. RTI Act is helpful socially, for all the subjects, quickly in our society.
Example: Waiting for the results after University Exam if delayed, RTI commission could be approached :

• To get information about certain legal affairs.
• Any problems concerned with government.

Kerala Syllabus 10th Standard Social Science Notes Question 8.
Discuss whether the Right to Service Act is helpful for the people to get the service they should obtain from Government offices.
Answer:
The responsible employer would have to pay the penalty if he fails in his duty. So, the people can get the service from Government offices at the right time.

Hsslive Guru Social Science Kerala Syllabus Question 9.
What is the use of Ombudsman to the public? Prepare a note.
Answer:
Elected representatives and bureaucrats are part of administration. Complaints are filed against their corruptions to Ombudsman. People can directly approach Ombudsman with complaints.

Social Science 10th Kerala Syllabus Kerala Syllabus Question 10.
Can we make Government services transparent and corruption free through the above mentioned system? Conduct a debate.
Answer:
For:

• Need not wait in Government offices for services.
• Receives Govt, service with less expense and quickly.
• Increases the efficiency and excellence of services.
• Receives feedback quickly.

Against:

• Is completely hidden in files.
• Life is free of corruption.
• Govt, services are obtained based on the financial status.

Public Administration Let Us Assess

Social 10th Class Notes State Syllabus Kerala Syllabus Question 1.
Explain the need of public administration in a country?
Answer:
Public administration is related to governmental administration. It is the effective way of utilizing men and materials for implementing the existing law policies and developmental projects. Primary Health Centers, Police station Krishi Bhavan, Village office, Corporation office, Election Commission, and other government organisations are all the parts of public administration.

The responsibilities of these institutions must be strictly executed for the public welfare.There must be an accurate method of public administration for all the sectors of people in our country. The following is a list of a few public administration.

Centers and their duties are given:
1. Vanitha Commission: Ensures the rights and welfare of women.

2. Election Commission: Relates with the election procedures.

3. Human Rights Commission: Protects the rights of men.

4. Village office: Ensures the primary needs of people.

10th Standard Social Science Notes Kerala Syllabus Question 2.
How are employees selected in a public administration system?
Answer:
Certain methods are adopted in the selection of employees to the public administrative system in India. The first step is the notification through Public Service Commission of India.Through this, the vacancies in various sections of governments are made known to the public. Later, through the exam conducted on the date notified or on the basis of interviews, candidates are selected and appointed.

Civil Service began during the British rule. It came to be the Civil Service of India after Independence. All the employees who work under the central and state governments and the employees under public sector undertakings are a part of India’s Civil Service. It is divided into All India Service, Central Service and State Service. The selection procedures of employees to these are given below.

All India Service:

• Recruits at national level.
• Appoints in Central Service or State Service.
  eg: IAS, IPS.

Central Service:

•  Recruits at national level.
•  Appoints in central government department only.
  eg: Indian Foreign Service, Indian Railway Service.

State Service:

• Recruits at State level
• Appoints in state government department
  eg: Sales Tax officer.

Candidates to All India Services and Central Services are recruited by the Union Public Service Commission (UPSC). The chairman and members of this commission are selected by the President of India. The UPSC has elaborate mechanisms for the selection based on qualifications.

In the state level, candidates are recruited by the Public Service Commission (PSC) of the state. The governor appoints the Chairman and the member of the State Public Service Commission. UPSC and state PSCs are on the basis of constitutional provisions and can be called constitutional institutions.

Std 10 Social Science Notes Kerala Syllabus Question 3.
What are the features of bureaucracy ?
Answer:
Hierarchical organisation:
Bureaucracy is organised in such a way that there is one employee at the top and the number increases, when it reaches the lower levels. This is known as hierarchical organisation.

Permanence:
Persons appointed will continue in service till the age of retirement.

Appointment based on qualification:
Employees are recruited and appointed on the basis of educational qualification.

Political Neutrality:
Bureaucrats are liable to implement the policies of which ever party comes to power. Party interests should not reflect in their work.

Professionalism:
Every government employee must be skilled in their work.

Sslc Social Science Notes Malayalam Medium Pdf 2021 Kerala Syllabus Question 4.
Classify the bureaucracy in India and explain ?
Answer:
Kerala Public Service Commission notifies for the recruitment of employees to the government service. Then candidates are selected on the basis of competitive examinations and interviews and are appointed in different government sectors. All those appointed in this way become a part of the civil service of India.

The aim of it is to bring welfare programmers speedily to all. Now there are specific services at the central and state levels. All the employees who work under public sector undertakings are the part of India’s civil service. There are All India services, central services and state services.

10th Social Science Notes Pdf State Syllabus Question 5.
What are the measures taken for the administrative reforms in India? Prepare a note.
Answer:
Government has taken a number of steps to increase the efficiency of services and to provide service to people within a time limit. This is called as administrative reforms. It makes the administration friendly and effective. Administrative reform commissions are made at national and state levels. Here are some steps taken for the reformation in our country.

E-Governance:
The single window system for Higher Secondary education, online applications for several scholarships etc are example of E-Governance. The use of electronic technology has helped to obtain the services of government effectively in a faster way.

Right to Information:
Every citizen has the right to collect the information from any government office about its working. This is under the Right to Information Act in 2005. The efforts of Mazdoor Kiran Shakthi Samghathan of Rajasthan led to this Act. The interventions of several organisations and social activities helped in passing this Act in 2005. The Right to Information ensures the right of all citizens of India to receive information.

Information Commission:
Files documents, circulars, memos, advice or orders, agreements statistics, reports, log books, press notes, samples, models, information in the form of electronic data, e-mail etc. related to government offices belong to public departments. Information commission helps to know everything.To perform the functions under the Right to information Act, Information Commissions are constituted at National and state levels.

A chief Information Commissioner and members not more than ten are in the Information Commission. If the information given is wrong and unsatisfactory we have the right to approach the Information Commission. If the commission is convinced a fine of Rs. 250 can be imposed on the employee.

Right to Service:
This ensures service to the people. This law determines the time limit for every service given by a government office. If the deserved service is not given within the time limit the responsible employee should pay fine. As per the Right to Service, an officer is appointed to give proper guidance and help to the applicants.

Lokpal and Lokayuktha:
These are the institutions to prevent corruption at administrative, bureaucratic and political levels. Lokpal has the power to register cases on issues of corruption against public workers and then suggest necessary actions. Lokayuktha hears the corruption cases at the state level. Both follow judicial procedures.

Central Vigilance Commission:
This is constituted at the national level to prevent corruption. It came in to force in 1964. It was formed to prevent corruption at central government offices.

Ombudsman:
Complaints of corruption against the elected representatives and bureaucrats who are a part of public administration are filed to ombudsman. A retired Judge of the High Court is appointed as Ombudsman. People can directly approach the Ombudsman with complaints.

He has the power to summon anyone on receiving the complaints, can order inquiry and recommend actions. Ombudsman began in the banking sector to hear the complaints of clients and rectify them.These actions recommended for the welfare of people as a part of administration maintains the public administration more trans parent.

Question 6.
How is E-Governance helpful to the public?
Answer:

• Need not wait in the government offices for services.
• Can receive service with the help of information technology.
• It services offers fast and at less expense.
• Enhances efficiency of offices and quality of the service.

Question 7.
Right to Information and Right to Service help to make public administration popular. Substantiate this statement.
Answer:
We can collect information from any government office about its working. People got this opportunity under the right to information act, 2005. This ensures the right of all citizens of India to receive information. The main objectives of this act are to prevent corruption, create responsibility and make the functioning of government transparent.

The citizens will get copies of public documents if they apply for them. Right to Service act is a law which ensures services to the public. As per the Right to Service Act, an officer is appointed in every government office to give guidance and proper help to the applicants. That is Right to Information and Right to Service helps to make public administration popular.

Question 8.
List out the steps taken to prevent corruption in India.
Answer:
1. Lokpal and Lokayuktha
2. Central vigilance commission
3. Ombudsman.

Public Administration Extended Activities

Question 1.
Prepare an application for getting information from an office under the Right to Information Act.
Answer:
From
Madhavan
Mahal Nivas
Ollavanna

To
Secretary
Grama Panchayath
Ollavanna

Based on RTI Act of 2005
Sir,
Sub: Amount spent for the electrification of houses for schedule tribes in the year 2015-16.
Could you kindly give details about how many applicants there were and how many money was spent in previous years. Expecting the reply within 15 days.
Ollavanna — Madhava
25/08/19  —  sd/-

Question 2.
Prepare a chart o the importance of All India Services.
Answer:
All the employees who work under the central and the state governments and the employees under the public sector under takings are part of India’s civil service. India’s civil service classification as All India Service, Central Service ans State Service.

All India Service.
Recruits at National level.
Appoints in the central or state service.
The UPSC has elaborate mechanisms for the recruitment of candidates based on qualification.
e.g., Indian Administrative Service.
Indian Foreign Service.
Indian Police Service.

Question 3.
Prepare a questionnaire to conduct an interview with the District Collector.
Answer:
Interview with kozhikode District Collector.

Sir, could you please share with me the inspiration behind your selection this position?

What were the changes you have thought to bring in Kozhikode before becoming the District Collector?

Sir, even in the presence of Lokpal, Lokayuktha, Central Vigilance Commission, Ombudsman being launched against corruption still corruption exists in Kerala. Don’t you long for a change ? How do you react as a district collector?

Sir, hope you have noted that vegetables from Tamil Nadu were sold less here in the Onam season. Do you expect the same for Vishu? As a District Collector, what all steps would you take to improve the agriculture sector of Kerala?

Though Kerala has developed in the educational field, the number of suicidal attempts have also increased. A change in our educational system is unavoidable. Sir, what all changes could be made possible?

Sir, M. C. Noufal was arrested here in Kozhikode for raping a woman from Bangladesh. Though many were arrested they were refused saved. Let me ask you sir with due respect, don’t you long for a change in the system of punishment here ?

Public Administration Orukkam Questions and Answers

Question 1.
List out the changes and objectives of public Administrative in monarchy and democracy.
Monarchy – The interest of the Monarch were the basis of public administration.
Democracy –
Answer:

Democracy – Importance is given to the interests of the people.

Question 2.
Complete the diagram showing the importance of public administration
Answer:

• Formulate government policies.
• Ensure welfare of the people.
• Find out solutions to public issues.
• Provides goods and services.

Question 3.
Some features of public administration are given in column A. Find out the definitions of each one of them in column B.
i. Hierarchical Organisation – one employee at the top and the number increases when it reaches the lower levels.
ii. Permanence- i
iii.Appointment on the Basic of qualification- iii
iv. Political Neutrality – iii.
x Professionalism – iv
Answer:
i. Permanence – Persons appointed will continue in service till the age of retirement.
ii. Appointment on the basis of qualification- Employees are recruited and appointed on the basis of educational qualification.
iii. Political Neutrality – Bureaucrats are liable to implement the policies of which ever party comes to power. Party inters ts should not reflect in their work. They should act neutrally.
iv. Professionalism – Every government employee must be skilled in their work.

Question 4.
Find out the functions of the following constitutional institutions.
1. State public service commission.
2. Union Public Service Commission.
Answer:
1. State Public Service Commission – At the state level, candidates are recruited by the public service commission of the state.

2. Union Public Service Commission- Candidates to all India services and central services are recruited by the Indian Public service commission.

Question 5.
Complete the short showing the classification of India’s Civil Service.
All India Service
Central Service
State Service
Answer:
All India Service: Recruits at national level, Appoints in the central or state service.
Eg: Indian Administrative Service, Indian Police.

Central Service : Recruits at national level, Appoints in central government departments only.
Eg: Indian foreign service, Indian Railway Service.

State Service: Recruits at state level, Appoints in state government departments only.
Eg: Sales Tax Officer

Question 6.
Find out and list the benefits of E – Governance to Public.
Need not to wait in government offices for services.
Answer:

• Can receive service with the help of information technology.
• Government services offered speedily and with less expense.
• Efficiency of the offices and quality of the service get enhanced.

Question 7.
Prepare a sample applications of Right to Information Act 2005
Answer:
From
Smitha Vijayan
DeviVihar
Arppookara

To
Secretary
Grama panchayath
Arpookara

Sir,
I request you to furnish information regarding the following questions under the right to information act 2005.

• In the academic year 2015-2016 what is the amount of money taken from the fund for the constriction of houses for the backward castes in arpookara grama panchayath.
• What was the amount of money spent in the past years for this purpose.
• Expecting reply within 15 days.

Question 8.
Lokpal, and Lokayukta are institutions constituted to prevent corruption at administrative, bureacrative and political level, find out the features and complete the chart.

Answer:

Lokpal	Lokayukta
The institution constituted at national level to prevent corruption is called Lokpal.	Lokayukta is the institution constituted at the state level to hear the corruption cases.
Lokpal has the power to register cases on issues of corruption against employees and pub lie workers and can suggest necessary actions.	Follow judicial Procedures.

Question 9.
Right to service Act is a law which ensures services to the people. Find out and list the treasures of Right to service Act. This law determines the time limit for every service given by a government office.
Answer:

• If the deserved service is not given within this time limit, the responsible employee should pay a fine.
• An officer is appointed in every government office to give guidance and proper help to the applicants.

Question 10.
Compare and the list out central vigilance commission and state vigilance commission.
Answer:
Central Vigilance Commission :
The Central Vigilance Commission constituted at national level to prevent Corruption, Central Vigilance Commission, it came into effect into 1964, Formed to prevent corruption in central government offices, in every department there will be a chief vigilance officer.

State Vigilance Commission :
Constituted at State level to prevent corruption, Inquires in to corruption in the state government offices.

Question 11.
Complete the diagram of administrative reform measures adopted by the government.

Answer:

• Right to information.
• Information commission.
• Lokpal and lokayuktha.

12. Complete the concept map given below.

Answer:

• Create responsibility
• The citizen will get copies of public documents if they apply for them.

Public Administration Evaluation Questions

Question 1.
‘Public Administration is about Governmental Administration’ on the basis of Gladden’s definition examine the features of public administration.
Answer:
From the definition of Gladden we can under stand that the important area of public administration is governmental administration. Public administration is the effective utilization of men and materials for the implementation of existing laws, governmental policies, programmer and developmental projects. Government try to find the solutions to various problems and ensure the welfare of the people through public administration.
Significance of Public administration.

• Formulate government policies.
• Provide goods and services.
• Ensure welfare the people.
• Find out solutions to public issues.

Question 2.
Explain Gandhiji’s vision on public administration.
Answer:
Gandhiji expected the protection of the interests of all through public administration. But there are a number of persons in our society who require special consideration and protection. Fie opined that public administration should consider them specially and protect them. Gandhiji’s concept of Grama Swaraj influenced India’s outlook of public administration to a great extent.

Question 3.
Define Bureaucracy. Examine the features of Bureaucracy?
Answer:
The employees who work under public administrative system are together known as Bureaucracy.

Features of Bureaucracy:
Hierarchical Organisation:
Bureaucracy is organised in such a way that there is one employee at the top and the number increases when it reaches the lower levels. This is known as Hierarchical organisation.

Permanance:
Persons appointed will continue in service till the age of retirement. Appointment on the basis of qualification Employees are recruited and appointed on the basis of educational qualification.

Political Neutrality:
Bureaucrats are liable to implement the policies of whichever party comes to power. Party interests should not reflect in their work.They should act neutrally.

Professionalism:
Every Government employee must be skilled in their work.

Question 4.
Name of the feature of bureaucracy which intercepts quick decisions.
a. Political Neutrality.
b.Professionalism.
c. Permanence.
d. Hierarchical organisation.
Answer:.
Hierarchical Organisation.

Question 5.
The Rajasthan based organisation paved the way for the legislation of right to Information Act.
a. Narmada Bachao Andolan.
b. Swabhimana Prasthanam
c. Mazdoor Kisan sakthi Sangathan.
d. Bharatiya Kissan Union.
Answer:
Mazdoor Kisan Shakti Sangathan.

Question 6.
Write a short note on the functions of central and state information commission.
Answer:
To perform the functions under the right to information act, Information Commissions are constituted at national and state levels.

Question 7.
State level organisation to prevent corruption at administrative, bureaucratize and political levels
a. Lokpal b. Lokayuktha
c. Central Vigilance Commission d Child rights commission
Answer:
Lokyukta

Question 8.
Match the items of column A with B

A	B
1. All India Service	i. Sales Tax Officer
2.Central Service	ii. Indian Police Service
3.State Service	iii.Indian Railway Service

Answer:
1-ii,
2-iii,
3-i.

Question 9.
The Institution constituted at the national level to prevent corruption in 1964
a. NitiAyog.
b. Administrative Tribunal.
c. Central Information Commission.
d. Central Vigilance Commission.
Answer:
Central Vigilance Commission.

Question 10.
UPSC and PSC are caused constitutional institutions why?
Answer:
UPSC( Union Public Service Commission) and state PSC are constituted on the basic of constitutional provisions.So they are called constitutional institutions.

Question 11.
Define E-Governance and write down two examples of E – Governance.
Answer:
E- governance is the use of electronic technology in administration . This help to obtain government services easily in a speedy manner. The single window system for admission to higher secondary courses, Online applications for various scholarships etc are exam pies for E- governance.

Question 12.
Explain different administrative reforms adopted by government for increasing the affiance of service.
Answer:
information Commission:
To perform the functions under the right to information act, In formation Commissions are constituted at the national and state levels. There will be a chief Information commission and not more than ten members in the Information Com mission.

Central Vigilance Commission:
The Central Vigilance Commission is the institution constituted at the national level to prevent corruption. It came into effect in 1964. It is formed to prevent corruption in the central government offices.The Central Vigilance Commissioner is the head of the Central Vigilance Commission.

Lokpal and Lokayuktha:
Lokpal and lokayuktha are institutions constituted to prevent corruption at administrative, bureaucratic and political levels, the institution constituted at the national level to prevent corruption is lokpal.

Lokpal has the power to register cases on issues of corruption against employees and public workers and can suggest necessary actions. Lokayukta is the institution constituted at the state level to hear the corruption cases.

Ombudsman:
Elected representatives and bureaucrats are part of public administration. Complaints can be filed against their corruption,nepotism or financial misappropriation or negligence of duties. Ombudsman is constituted for this purpose.

Public Administration SCERT Questions and Answers

Question 1.
Pick out any two public administration institutions and write about their functions.
Answer:

Institution	Functions
a. Krishi Bhavan	Promotes agriculture
b. Police Station	Maintains law and order
c. Primary health center	Provides treatment for illness

Question 2.
What are the differences found in the public administration under monarchy and democracy?
Answer:

• In monarchy, the interest of the monarch is the basis of public administration,
• In democracy importance is given to the interests of the people.

Question 3.
Explain the importance of public administration.
Answer:

• Ensures welfare of people
• Formulates government policies
• Provides goods and services
• Finds out solution for public grievances

Question 4.
Explain the importance of bureaucracy in public administration.
Answer:

• Makes the public administration system dynamic,
• Services of the government made accessible to the people ,
• Performs the day- to-day administration of the country.

Question 5.
What is meant by hierarchical organisation and permanence of bureaucracy.
Answer:

• Organization made up of one employee at the top and more towards the bottom.
• Persons once appointed as employee will continue in service till the age of retirement. This is permanency.

Question 6.
Prepare a note on civil service in India.
Answer:

• All India service
• Central services
• Stateservice. Explain hints

Question 7.
Why PSC and UPSC are called as constitutional institutions?
Answer:
UPSC and PSC are constituted on the basis of constitutional provisions. So they are cal led as constitutional institutions.

Question 8.
Find out the factors that adversely affect the efficiency of public administration.
Answer:

• Inefficiency of bureaucracy,
• Corruption
• Shortage of employees

Question 9.
Write a brief note on e-Governance implemented as part of administrative reforms?
Answer:

• e-Governance is the use of electronic technology in administration. This helps people to obtain government services quickly and easily .
• Information technology is used in the field of public service.
• Government service is made available less expensively expensively

Question 10.
A road constructed before six months in your place is damaged now. You became aware that there is some corruption, Prepare an application under Right to Information Act addressing panchayat secretary to get the details of it.
Answer:
To Prepare an application under the RTI Act.

Question 11.
What are the benefits attained by society as a result of formulation of the Right to Information Act.
Answer:

• Controlled corruption,
• Increased the responsibility of bureaucrats,
• Functioning of government became transparent.

Question 12.
Explain the structure of Information Commission?
Answer:

• Central Information Commission,
• Chief Information Commissioner and not more than ten members,
• State Information Commission.

Question 13.
How the Right to Service Act is helpful to people?
Answer:

• Ensures government services to people.
• Gets service within time limit.
• Employee should pay fine in case of delay.
• Appoints an officer in charge as per Right to Service Act in all offices to give service.

Question 14.
Differentiate the functioning of Lokpal and Lokayuktha?
Answer:
Lokpal:

• Institution constituted to prevent corruption charges at national level
• Has the power to register cases of corruption against employees and public workers.

Lokayuktha :

• Constituted at state level to hear corruption cases.
• Follow Judicial procedures

Question 15.
From the following select the statement appropriate to central vigilance commission and state vigilance commission?
a. Esquires about corruption in state government offices.
b. Institution to prevent corruption at national level.
Answer:
a. State vigilance commission .
b. Central vigilance commission

Question 16.
How the functioning of Ombudsman helps the public to prevent corruption.
Answer:

• Complaints against corruption among elected representatives and bureaucrats can be filed in Ombudsman,
• People can directly approach Ombudsman to give complaints.
• Ombudsman can enquirer into such complaints and recommend actions.

Question 17.
Compare and list the different levels of civil services in India.
Answer:

• All India Service
• Central Service
• State Service

Question 18.
What is considered as corruption by modern society?
Answer:

• Delayed service, Denial of right to service is corruption.
• Making service as a favor.

Public Administration Exam Oriented Questions and Answers

Question 1.
The chairman and the members of this commission are appointed by
Answer:
The president of India

NSUI full form stands for National Students’ Union of India that is a student wing of famous politician Indian National Congress party.

Question 2.
What is the full form of U. P. S.C ?
Answer:
Union Public Service Commission.

Question 3.
In which year Central Vigilance Commission come into effect ?
Answer:
1964

Question 4.
Write a note on Akshaya Center and E-literacy.
Answer:
For the benefit of people, Akshaya centers have been constituted to make use of Government service delivered through E – governance. It also aims at making people E-literate. E – literacy is the awareness about basic information about technology.

Question 5.
Prepare a seminar report on the importance of public administration.
Answer:
Without public administration, the government cannot operate and manage activities effectively and efficiently. The administration plays a vital role for delivering and distributing the public services to all comers of the country.

Administration spreads all over the country for supplying the governmental and public goods and services up to the villages and door to door. The administration is not operating and managing the activities properly and smoothly in developing countries. The scopes of administration shows the importance of public administration.

Following are the importance of Public Administration:
Management of Public Service, Distribution Social Change, Disaster Management, Population Control, Preservation of Human Right, Management of Industrial Relationship, Internal and External attached Economic Development. These points show that the administration is used all over the sections of the country.

Question 6.
Write a note on Administrative Tribunal.
Answer:
Actions are taken by various government departments against the government officials. The Administrative Tribunal is the institution where the employees can lodge their complaints against such actions. ”

Question 7.
The famous western administrative thinker Gladden says “ Public administration is concerned with an administration of the government”. On the basis of this definition, explain the relationship between Public administration and the government.
Answer:
From this definition we can understand that the important area of public administration is governmental administration. Public administration is the effective utilization of men and materials for the implementation of existing laws, governmental policies, programmers and developmental projects and the government has constituted a number of institutions for this purpose. All these governmental institutions are part of public administration. They function for the welfare of the people.

An administrative system is needed for governments to exist and function. The history of public administration begins with the formation of state. Based on differences in the form of government we can find differences in public administration also. In a monarchy, the interests of the monarch was the basis of public administration. But in a democratic system, importance is given to the interests of the people. Democratic administration becomes more effective and efficient through public administration.

You can Download Second Degree Equations Questions and Answers, Activity, Notes, Kerala Syllabus 10th Standard Maths Solutions Chapter 4 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Maths Solutions Chapter 4 Second Degree Equations

Second Degree Equations Text Book Questions and Answers

Textbook Page No. 81

Second Degree Equation Class 10 Kerala Syllabus Question 1.
When each side of a square was reduced by 2 metres, the area became 49 square metres. What was the length of a side of the original square?
Answer:
Let the length of each side of the original square be x, then

The length of each side of the original square = 9 cm

Second Degree Equation Class 10 Questions And Answers Kerala Syllabus Question 2.
A square ground has a 2 metre wide path all around it. The total area of the ground and the path is 1225 square metres. What is the area of the ground alone?
Answer:
’Each side of the ground = x
Each side of ground+path = x + 4
(x + 4)² = 1225 = 35²;
x + 4 = 35
x= 35 – 4 = 31

Area of the ground alone = 31² = 961 m²

Second Degree Equation Class 10 Extra Questions Kerala Syllabus Question 3.
The square of a term in the arithmetic sequence 2, 5, 8, ……., is 2500. What is its position?
Answer:
Let 2500 is the square of the n^th term of the arithmetic sequence 2, 5, 8, …………

It is the 17^th term.

To solve a system of linear equations with steps, use the system of linear equations calculator.

Sslc Maths Second Degree Equations Kerala Syllabus Question 4.
2000 rupees was deposited in a scheme in which interest is compounded annually. After two years the amount in the account was 2205 rupees. What is the rate of interest?
Answer:
Amount (P) = 2000
Compound interest (r %)
Year (n)

Textbook Page No. 86

Sslc Second Degree Equation Questions Kerala Syllabus Question 1.
1 added to the product of two consecutive even numbers gives 289. What are the numbers?
Answer:
Let the two consecutive even numbers be x, x+2
x, x+2
x(x+2) + 1 = 289
x² + 2x = 288
x²+ 2x – 288 = 0
x²+ 2x = 288
(x+1)² = 288 + 1
(x+1)² = 289
x+1 = ± 17
x = 16, – 18
The numbers are 16, 18

Second Degree Equation Questions And Answers Kerala Syllabus Question 2.
9 added to the product of two consecutive multiples of 6 gives 729. What are the numbers?
Answer:
Let the two consecutive multiples of 6 be x, x+6
x, x+6
x(x + 6)+ 9 = 729
x² + 6 x + 9 = 729
x² + 6x = 720
x²+ 6x – 720 = O
x² +6x = 720
(x+3)²= 720+9
(x+3)² = 729
x+3 = ± 27
x = 24, – 30
The numbers are 24, – 30

10th Maths Second Degree Equation Kerala Syllabus Question 3.
How many terms of the arithmetic sequence 5, 7, 9, …, must be added to get 140?
Answer:
First term f= 5,
Common difference = 2

10 terms should be added to get 140

Second Degree Equation Class 10 Model Question Paper Kerala Syllabus Question 4.
16 added to the sum of the first few terms of the arithmetic sequence 9, 11, 13, gave 256. How many terms were added?
Answer:
9, 11, 13,…………
First term f= 9,
Common difference d = 2
Sum of first n terms

Second Degree Equation Class 10 Notes Kerala Syllabus Question 5.
An isosceles triangle has to be made like this

The height should be 2 meters less than the base. The area of the triangle should be 12 square meters. What should be the length of its sides?
Answer:
Base = AB = x
Height= CD = x – 2

Sslc Maths Chapter 4 Questions And Answers Kerala Syllabus Question 6.
A 2.6-meter long rod leans against a wall, its foot 1 meter from the wall. When the foot is moved a little away from the wall, its upper end slides the same length down. How much farther is the foot moved?

Answer:

Textbook Page No. 91

Sslc Maths Chapter 4 Solutions Kerala Syllabus Question 1.
The product of a number and 2 more than that is 168, what are the numbers?
Answer:
Let the number be x
x(x+2) = 168
x² + 2x = I68
x² + 2x + 1 = I68 + 1
(x+1)² = 169
x + 1 = ± 13
x + 1 = 13
x + 1= – 13
x = 13 – 1 = 12
x = -13 – 1 = – 14
The number is 12 and 14 or – 12 and – 14

Maths Chapter 4 Class 10 Kerala Syllabus Question 2.
Find two numbers with sum 4 and product 2.
Answer:
Sum of the numbers = 4
If, First number = x
Then second number = 4 – x

Class 10 Maths Second Degree Equation Kerala Syllabus Question 3.
How many terms of the arithmetic sequence 99, 97, 95, … must be added to get 900?
Answer:

The spectrally extended signal produced by nonlinear function calculator 520 is likely to have a pronounced dropoff in amplitude as frequency increases.

Class 10 Maths Chapter 4 Kerala Syllabus Kerala Syllabus Question 4.
A rod 28 centimeters long is to be bent to make a rectangle.
i. Can a rectangle of diagonal 8 centimeters be made?
ii. Can a rectangle of diagonal 10 centimeters be made?
iii. How about a rectangle of diagonal 14 centimeters?
Calculate the lengths of the sides of the rectangles that can be made.
Answer:




∴ Cannot make a rectangle of diagonal 14 centimeters.

Textbook Page No. 97

Maths Second Degree Equation Kerala Syllabus Question 1.
The perimeter of a rectangle is 42 meters and its diagonal is 15 meters. What are the lengths of its sides?
Answer:

Sslc Second Degree Equation Questions And Answers Kerala Syllabus Question 2.
How many consecutive natural numbers starting from 1 should be added to get 300?
Answer:
Sum of natural numbers starting from 1:

Second Degree Equation Class 10 Kerala Syllabus Kerala Syllabus Question 3.
What number added to 1 gives its own square?
Answer:
Let the number be x

Second Degree Equation Kerala Syllabus Question 4.
In writing the equation to construct a rectangle of specified perimeter and area, the perimeter was wrongly written as 24 instead of 42. The length of a side was then computed as 10 meters. What is the area in the problem? What are the lengths of the sides of the rectangle in the correct problem?
Answer:
If length is 10 and perimter is 24
2 (l + b) = 24
2(10 + b) = 24
10 + 5 = 12
b = 2
Width= 2
Area= lb =10 x 2 = 20 sq.cm
Area in the correct problem =20
lb = 20
2(l + b) = 42
l + b = 21
l + \(\frac { 20 }{ l }\) = 21

Kerala Syllabus 10th Standard Maths Chapter 4 Kerala Syllabus Question 5.
In copying a second-degree equation to solve it, the term without x was written as 24 instead of – 24. The answers found were 4 and 6. What are the answers of the correct problem?
Answer:

Second Degree Equations Orukkam Questions and Answers

Worksheet 1

Second Degree Equations Exercises Kerala Syllabus Question 1.
Write two numbers whose square is 25?
Answer:
Let x be the number
x² = 25
x= √25 = ±5
numbers are –5, +5

Second Degree Equation Class 10 Formulas Kerala Syllabus Question 2.
When the square of a number is added to the number we get 30. What are the numbers?
Answer:
Let x be the number ,
x² + x = 30
x² + x – 30 = 0, (x + 6) (x – 5) = 0
x = – 6, 5
numbers are – 6, +5

Second Degree Equations Class 10 Kerala Syllabus Question 3.
Find the side of the square whose area and perimeter are numerically equal.
Answer:
Let x be the length of side.
Perimeter = Area
x² = 4x
x = 4
Length of side = 4

Question 4.
How many odd numbers from 1 makes the sum 961?
Answer:
Sum of continuous n odd numbers = n²
n² = 961
n = √961 = ± 31
Sum of continuous 31 odd numbers

Question 5.
A man’s age after 15 years will be the square of his age 15 years ago. What is his present age?
Answer:

Worksheet 2
Form the equation

Question 6.
The sum of a number and its square is ten times that number
Answer:
Let x be the number
x² + x = 10 x
x² – 9x = 0

Question 7.
The sum of a number and its square root is 6.
Answer:
Let x be the number
x + √x = 6
x – 6 = √x
(x – 6)² =(–√x)2
x² – 12x + 36 = x
x² – 13 x + 36 = 0

Question 8.
The sum of first n natural numbers is 210.
Answer:
\(\frac { n(n+1) }{ 2 }\) = 210
n² + n = 420
n² + n – 420 = 0

Question 9.
The area of a rectangle whose length is 5 more than its width
Answer:
Let x be the length of side.
other side = x + 5
x (x + 5) = 150
x² + 5x – 150 = 0

Question 10.
The sum of a number and its reciprocal is \(\frac { 5 }{ 2 }\)
Answer:
Let x be the number

Question 11.
The sum of even numbers from 2 in an order is 240
Answer:
2 + 4 + 6 + + 2n = 420
2(1 + 2 + 3 + ….+ n) = 420

Question 12.
A man’s age after 15 years will be the square of his age 15 years ago.
Answer:
Let x be the present age
(x – 15)² = x + 15
x² – 30 x + 225 = x +15
x² – 31x + 210 = 0

Worksheet 3

Question 13.
When 8 times a number is added to its square we get 8. Find the number by making the equation properly.
Answer:
x² + 2 x = 8
x² + 2x + 1 = 8 + 1
(x + 1)² =9
x + 1 =3
x = 2
Number = 2

Question 14.
Which term in the sequence 2, 5, 8 …….. gives its square 2500?
Answer:
nth term = 3n + (2 – 3) = 3n – 1
(3n – 1)² = 2500,
3n – 1 = 50,
n = 17
square of 17^th term is 2500

Question 15..
A man’s age after 15 years will be the square of his age 15 years ago. Find the age
Answer:
Let x be the age
x + 15 = (x – 15)²
x + 15 = x² – 30 x + 225
x² – 31 x = – 210

Question 16.
The length of a rectangle is 2 more than its width. The area of the rectangle is 80. Find length and breadth.
Answer:
Let x be the width, then length = x + 2
x (x + 2) = 80
x² + 2x = 80
x² + 2x + 1 = 80 + 1
(x + 1)² = 81
x + 1 =9
x = 8
length = 10
width = 8

Question 17.
The sum of a number and its reciprocal is \(\frac { 5 }{ 2 }\) Find the number.
Answer:
Let x be the number

Question 18.
The sum of some even numbers starting from 2 is 420. Find the number of even numbers added.
Answer:

Worksheet 4

Question 19.
Sum of the squares of three consecutive natural numbers is 110.
Answer:
If the numbers are x, x + 1, x + 2, then
x² + ( x + 1)² + (x + 2)² = 110
x² + x² + 2x + 1 + x² + 4x + 4 = 110
3x² + 6x – 105 = 0

Question 20.
The product of the digits of a two-digit number is 12.When 36 is added to the number we get a two-digit number in which the digits are reversed. Find the two-digit number .
Answer:
Let the two digit number be 10 x + y, then
x y = 12
y = \(\frac { 12 }{ x }\)
If we change the positions of digits 10 y + x
10 y + x = 36+ 10 x + y
Let’s take x be the digits in the position of 10.
Let’s take \(\frac { 12 }{ x }\) as the digits in the position of unit.
Number which digits are reversed \(\frac { 120 }{ x }\) +x

Question 21.
Serena and Johan had 45 diamond stones. They sold 5 stones. The product of the remaining stones is 124. Find the number of stones each had
Answer:
Let x be the stones Serena has and 45 – x be the stone Johan has. After selling 5
(x – 5)(45 – x – 5) = 124
(x – 5)(40 – x) = 124
x² – 45x + 200 = – 124
x² – 45x + 324 = 0


Number of diamond stones Serena has = 36
Number of diamond stones Johan has = 9

Question 22.
The sum of a number and its reciprocal is \(1 \frac{1}{2}\) . Find the number.
Answer:
Let x be the number

Question 23.
The sum of two numbers is 15. Sum of its reciprocals is \(\frac { 3 }{ 10 }\). Find the numbers.
Answer:
Let x, 15 – x be the numbers

Question 24.
A two-digit number is four times sum of its digits. The number is three times product of the digits. Find the number
Answer:
Let x, y be the numbers
10 x + y = 4 (x + y)
6 x = 3 y

Worksheet 5

Question 25.
A train travels a distance of 300km constant speed. If the speed of the train is increased by 5 km, the journey would have taken 2 hours less. Find the original speed of the train
Answer:
Let x km/hr be the speed.
Let the time required to travel in the same speed be \(\frac { 300 }{ x }\) hour. If the speed is increased by 5 km/ hr the time will decreased by 2 hours.

Question 26.
An express train takes 3 hours less than a passenger train for a journey of 600 km. If the speed of the passenger train is 10 less than the speed of the express train find the speeds of both trains (Use Pythagoras theorem in distance, not in speeds)
Answer:
Let x km/hr be the speed of the express.

Speed of the express 50 km/hr
Speed of the passenger 50 km/hr

Worksheet 6

Question 27.
One year ago a man’s age is eight times the age of his son. At present man’s age is the square of son’s age. Find the present age.
Answer:
Let x be the present age of son

Question 28.
A man’s age after 15 years will be the square of his age 15 years ago. Find the present age by forming a second degree equation
Answer:
Let x be the present age .
x + 15 = (x – 15)2
x + 15 = x² – 30x + 225
x² – 31x + 210 = 0

Present age = 21

Question 29.
The product of Layas’s age before 5 years and after 8 years is 30. Find the present age.
Answer:
Let x be the present age of Laya
(x – 5)(x + 8) = 30
x² – 3x – 40 = 30,
x² – 3x – 70 = 0

Present age of Laya = 7

Worksheet 7

Question 30.
Sravani teacher asked the students to construct a rectangle having area 5 square unit and perimeter 8. Jeevan, a wise student of the class, after making some calculations told that it is not possible to construct such a rectangle. Can you agree with him. Justify reasonably
Answer:
Let x be the length, then the area will be 5 cm², So the width will be \(\frac { 5 }{ x }\) cm. When the perimeter is 8 cm, 4 – x will be the width. The width which was obtained first will be \(\frac { 5 }{ x }\) and the width obtained now i.e., 4 – x will be equal.

– 4 < 0, x will not be obtained. Hence it is not possible to draw a rectangle with the given values.

Question 31.
The perimeter of a rectangle is 34 cm, area 60 square centimeter. Find the sides
Answer:
Let x be the length
Perimeter 34 cm, therefore width will be 17-x cm
Perimeter = 60 cm, therefore width be \(\frac { 60 }{ x }\) cm

x = 5 cm
length = 12 cm therefore width = 17 – 12 = 5 cm
length = 12 cm ,
width = 5 cm

Question 32.
The length of the rectangle is 4 more than its breadth. Area of the rectangle is 140 square centimeter. Calculate length and breadth
Answer:
Let x be the width of rectangle
length = x + 4 cm
x(x + 4) = 140
x² – 4 x – 140 = 0

width =10 cm
length = 14 cm

Question 33.
When the sides of a square are increased by 4, area become 256. Find the length of the first square.
Answer:
Let x be the side of square

x will not -20
The side of the first square = 12 cm

Question 34.
The area of a right-angled triangle is 60 square unit. The one of the perpendicular sides is 10 more than other. Find the sides of the triangle.
Answer:
Let x cm, x + 10 cm be the perpendicular sides

Question 35.
The area of an isosceles triangle is 60 square meters. One of the equal sides is is 13 cm. Find the third side. Take base x then h = \(\sqrt{13^{2}-x^{2}}\)
Answer:

Second Degree Equations SCERT Question Pool Questions and Answers

Question 36.
When the sides of a square are increased by 8 cm each, its area becomes 1225 sq. cm Frame an equation using the above data by taking the side of the smaller square as x cm. Find the sides of both the squares. [Score: 3, Time: 5 Minutes]
Answer:
One side of smaller square = x
One side of bigger square = x + 8
Area = (x + 8)² = 1225 (1)
x + 8 = 35
x = 35 – 8 = 27 (1)
Side of smaller square = 27 cm
Side of bigger square = 35 cm (1)

Question 37.
The difference of two positive numbers is 6. Their product is 216. Find the numbers. [Score: 3, Time: 4 Minutes]
Answer:
Let the numbers be x, x + 6
Products : x (x+6) = 216 (1)
x² + 6x = 216
x² + 6x + 9 = 216 + 9 = 225
(x + 3)² = 225 (1)
x + 3 = 15
x = 15 – 3 = 12
Numbers : 12, 18 (1)

Question 38.
In a right triangle one of the perpendicular sides is one less than 2 times the smaller side. Hypotenuse is one more than 2 times the same smaller side. By taking the smaller side as x cm, write the algebraic expression for the other two sides. Compute all the 3 sides of the right triangle. [Score: 4, Time: 5 Minutes]
Answer:
Smaller side = x
Perpendicular side = 2x – 1
Hypotenus = 2x + 1 (1)
x² + (2x – 1)² = (2x + 1)² (1)
x² + 4x² – 4x + 1 = 4x² + 4x + 1
x² – 8x = 0 (1)
x (x – 8) = 0
x = 0 or x = 8
Side=8 cm, 15 cm, 17 cm (1)

Question 39.
Find the sides of a rectangle whose perimeter is 100 metres and area 600 sq. metres. [Score : 4, Time : 5 Minutes]
Answer:
Perimeter = 100 m .
Length+Breadth = 50 m.
Length = 25 + x.
Breadth = 25 – x
Area = (25 + x)(25 – x) = 600 (1)
25² – x² = 600 (1)
x² = 625 – 600 = 25, x = 5 (1)
Sides, 25 + 5 = 30 m
25 – 5 = 20 m (1)

Question 40.
The one’s place of a two-digit number is 4. The product of the number and digit sum is 238.
a. If ten’s place digit is taken as x, Write the number.
b. Frame a second-degree equation and find the number. [Score : 4, Time : 6 Minutes]
Answer:
Digit in the ten’s place = x
Number = 10 x + 4 (1)
The product of the number and digit sum
= (x + 4) (10 x + 4) = 238 (1)

Question 41.
How many consecutive natural numbers from 1 should be added to get 465? [Score : 4, Time : 5 Minutes]
Answer:
Sum of first n natural numbers

Question 42.
The product of the digits of a two-digit number is 12. When 36 is added to this number, got a new number with digits reversed. Find the number. [Score : 5, Time : 7 Minutes]
Answer:

Question 43.
For a two-digit number, one’s place is 3 more than its ten’s place. The product of this number and its digit sum is square of double the digit sum. What is the number? [Score : 4, Time : 6 Minutes]
Answer:

Question 44.
A pavement of Width 2 metres is built around a square shaped garden. The area of the pavement alone is 116 square metres. Find one side of the garden.
[Score : 3, Time: 4 Minutes]
Answer:
Side of garden = x
Area of the pavement Area of the pavement Area of the pavement Side of garden including pavement = x + 4 (1)
Area of the pavement = (x + 4)² – x² = 116
x² + 8x + 16 – x² = 116 (1)
8x + 16 = 116,
\(x=\frac{116-16}{8}=12.5 \mathrm{m}\)
Side of garden = 12.5m (1)

Question 45.
A rectangle of Width 8 centimetres is cut off from a square sheet along its side. The remaining rectangular portion has an area 84 sq. metres. Calculate the side of the square. [Score : 4, Time : 6 Minutes]
Answer:
Let the side of square = x
Sides of rectangle = x, x – 8 (1)
Area = x(x – 8) = 84
x² – 8x = 84 (1)
x² – 8x + 16 = 84 + 16
(x – 4)²= 100, (1)
x – 4= 10, x= 14
Side of square = 14 centimetre

Question 46.
The lengh of a rectangle is 3 metre more than 3 times its breadth. Its diagonal is 1 metre more than the length. Find the lengh and breadth of the rectangle. [Score: 4, Time: 7 Minutes]
Answer:
Sidc of rectangle = x
Length =3x +3, Diagonal = 3x + 4 (1)
(3x + 4)² = x² + (3x + 3)² (1)
9x² + 24 x + 16 = x² + 9x² + 18x + 9
x² – 6x = 7, x² – 6x = 7 (1)
x² – 6x + 9 = 16, (x – 3)² = 16
x – 3 = 4, x = 4 + 3 = 7 (1)
length of rectangle = 3 × 7 + 3 = 24 metre breadth of rectangle = 7 metre

Question 47.
In the figure AB is the diameter of the circle. CD = 10 cm. BC is 15 cm less than Ac. Find AB?

Answer:

Question 48.
The chords AB and CD of a circle intersect at. If MA = 6 cm, MB = 8 cm and CD = 16 cm. Find MC and MD.

Answer:
CD = 16 centimetre MC = 8 – x, MD = 8 + x
then, MA x MB = MC x MD (1)
6 x 8 = (8 – x)(8 + x)
48 = 64 – x², x² = 16, x = 4 (1)
MC = 8 – 4 = 4 centimetre (1)
MD = 8 + 4 = 12 centimetre (1)

Question 49.
In the figure, AD is drawn perpendicular to the side opposite to the right angled vertex A. BC-13 cm and AD=6 cm.
a. Take BD = x and express DC in terms of x.
b. Frame a second-degree equation and find the lengths BD and DC. [Score : 4, Time: 5 Minutes]

Answer:

Question 50.
Can the sum of a number and it’s reciprocal be 2/3? [Score: 4, Time: 5 Minutes]
Answer:
Number = x , then its reciprocal = 1/x

Since the discriminant is negative. We do not get a solution.
∴ The sum of a numbrer and it’s reciprocal never gives 2/3. (1)

Question 51.
The sum of a number and its reciprocal is \(\frac { 13 }{ 6 }\) What is the number ? [Score: 4, Time: 5 Minutes]
Answer:
Let number = x, then its reciprocal = 1/x

Question 52.
The sum of a number is 12 and sum of its reciprocal is \(\frac { 3 }{ 8 }\) Find the numbers. [Score: 4, Time: 5 Minutes]
Answer:
If we take numbers as 6 + x, 6 – x (1)

Question 53.
Consider an arithmetic sequence with common difference 20. If the sum of reciprocals of two consecutive terms of this sequence is 1/24, find the first term of the arithmetic sequence. [Score: 4,Time: 6 Minutes]
Answer:
’Terms x – 10, x + 10 (1)

Question 54.
Prove that the difference of a number and its reciprocal will be always positive. [Score: 3, Time : 5 Minutes]
Answer:
Number = x, its reciprocal = 1/x (1)
then, \(x-\frac{1}{x}=k\) (1)

Since k being a positive number, k² + 4 also negative. (1)

Question 55.
In copying a second-degree equation, the number without x was written as -30 in¬stead of 30. The answers found were 15 and -2. What are the answers of the correct problem? [Score: 5, Time: 8 Minutes]
Answer:

Second Degree Equations Exam oriented Questions and Answers

Short Answer Type Questions (Score 2)

Question 56.
Find two numbers whose sum is 6 and product is 9.
Answer:
’First number = x
Second number = 6 – x
x ( 6 – x ) = 9
6x – x² = 9
0 = 9 – 6x + x²
x² – 6x + 9 = 0
(x – 3)² = 0
(x – 3 ) (x – 3) = 0
x = 3
Numbers are 3, 3.

Question 57.
Ammu is 7 years younger than Divya. If 2 is added to the product of their ages, we get 200. Find their ages.
Answer:
Age of Ammu = x
Age of Divya = x – 7
x(x – 7)+2 = 200
x² – 7x + 2 = 200
x² – 7x – 198 = 0 (x – 18) (x + 11) = 0
x= 18, x = – 11
(age will not be a negative number)
Age of Ammu = 18,
Age of Divya =18 – 7 = 11

Question 58.
In order to solve a quadratic equation, Arun did the following steps. Find the values of x by completing each step.

Answer:

Question 59.
The perimeter of a rectangle is 42 cm and its diagonal is 15 cm. Find the dimensions of the rectangle.
Answer:
Let breadth = x; perimeter = 2 (length + breadth) length + breadth = 42/2 = 21
∴ length = 21 – x
x² + (21 – x) = 152
x² + 441 – 42x + x² = 225
2x² + 216 – 42x =0
x² – 21x + 108 = 0
(x – 12) (x – 9) = 0
x = 12 or x = 9
breadth = 9 unit
length = 12 unit

Question 60.
If the equation 4x² – 5 x + k = 0 has two equal solutions. Find the value of k.
Answer:
If there is one solution, the discriminant = 0

Short Answer Type Questions (Score 3)

Question 61.
The product of a number and the number 8 more than it is 105.
a. What is the least number to be added to make the product a perfect square?
b. What are the numbers in the problem?
Answer:
a. Let the number be x, then second number is x+ 8
x(x + 8) = 105
x² + 8x = 105
The number to be added to get a perfect square is 16

Question 62.
On Arts day, 180 sweets were distributed equally among the students. Tasting one sweet Deepa said, “It is a pity that 9 of our friends are absent today.”
Hearing this Deepu said, “Because of that we got one sweet more.”
a. Find out the total number of students,
b. How many students were present on that day?
Answer:
a. Let the number of students = n
Number of sweets given to one student = \(\frac { 180 }{ n }\)
Since 9 students were absent = \(\frac { 180 }{ n-9 }\)

b. Total present = 45 – 9 = 36

Question 63.
One of the perpendicular side of a right angled triangle has length 4 cm more than twice of the other side. It has surface area 80 cm². Find out the lengths of the perpendicular sides.
Answer:
Let one of the perpendicular side = x
Second side = 2x + 4, Area = 80
1/2 × (2x + 4) = 80
2x² + 4x = 160
x² + 2x = 80
x² + 2x + 12 = 80 +12
(x + 1)² = 81
x +1 = ± 9
x + 1 = 9
x = 8
x = – 10 cannot be accepted as – ve sign cannot exist for the length of the side
∴ One side = 8cm
Second side = 2 × 8 + 4 = 20 cm

Long Answer Type Questions (Score 4)

Question 64.
Reji’s father bought several note books of the same price. Total price Rs. 360. If the price of each book were less by 2 rupees, he would get 2 books more.
a. Find the number of books if the price of one book is x.
b. Find the number of books if the price of one books is less by 2 rupees.
c. Form a quadratic equation.
d. Find the number of books bought.
Answer:
(a) Price of each book be x rupees,
No. of Notebooks = \(\frac { 360 }{ x }\)

(b) If the cost of each book is Rs (x – 2)
No. of books brought for Rs. 360 = \(\frac { 360 }{ x – 2 }\)

Question 65.
1235 Sacks of rice is to be brought from Trivandrum to Kottayam. For this a minitruck takes 6 trips more than that of an ordinary truck. An ordinary truck can carry 30 sacks more than that of a mini truck. Find the capacity of each truck.
Answer:
Let the capacity of mini truck be x sacks and ordinary truck be x + 30 sacks

Long Answer Type Questions (Score 5)

Question 66.
a. The sum of a number and it’s reciprocal is \(\frac { 25 }{ 12 }\) What is the number?
b. Prove that the sum of a positive number and it’s reciprocal is always greater than or equal to 2.
Answer:

Free Limit using Substitution Calculator – Find limits using the substitution method step-by-step.

Question 67.
Sum of the area of two squares is 500 m². If the difference of their perimeters is 40m, find the sides of the two squares.
Answer:
Let the side of the squares be x and y meters.
According to the condition,
x² + y² = 500 (1)
4x – 4y = 40
(x – y) = 10
y = x – 10
Substituting the value of y in (1), we get
x² + (x – 10)² – 500
2x² – 20x – 400 = 0
x² – 10x – 200 = 0
x = 20 or x = – 10
As the side cannot be negative, x = 20
Hence, side of the first square, x = 20 m
Side of the second square, y = 20 – 10 = 10 m

Question 68.
a By increasing the speed of a bus by 10 km/hr, it takes one and half hours less to cover a journey of 450km. Find the original speed of the bus.
b. 250 Rupees is divided equally among a certain number of children. If there were 25 children more, each would have received 50 paise less. Find the number of children.
Answer:
Let speed of the bus be x km/hr

Second Degree Equations Memory Map

The value of x in the second-degree equations can be found out by mainly 3 ways.
1. Factorization
2. Completing the square
3. By using equation

You can Download Statistics and Algebra Questions and Answers, Activity, Notes, Kerala Syllabus 10th Standard Maths Solutions Chapter 11 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Maths Solutions Chapter 11 Statistics

Statistics Textbook Questions & Answers

Textbook Page No. 245

Statistics Class 10 Kerala Syllabus Questions 1.
The distance covered by an athlete in long jump practice are
6.10, 6.20, 6.18, 6.20, 6.25, 6.21, 6.15, 6.10
in meters. Find the mean and median. Why is it that there is not much difference between these?
Answer:
Mean = \(\begin{array}{l}{=\frac{6.10+6.20+6.18+6.20+6.25+6.21+6.15+6.10}{8}} \\ {=\frac{49.39}{8}=6.17}\end{array}\)
If distances are written in ascending order
6.10, 6.15, 6.18, 6.20, 6.21, 6.25
Median = \(\frac{6.18+6.20}{2}=6.19 \mathrm{m}\)
Mean and median are gives the average dis¬tance covered by a person. Hence they will not have much difference..

Polynomial in Descending Order Calculator is a free online tool that determines the descending order of a polynomial in just a few taps.

Sslc Maths Statistics Kerala Syllabus Questions 2.
The table below gives the rainfall during one week of September 2015 in various districts of Kerala.

District	Rainfall (mm)
Kasaragod	66.7
Kannur	56.9
Kozhikode	33-5
Wayanad	20.5
Malappuram	13-5
Palakkad	56.9
Thrissur	53-4
Ernakulam	70.6
Kottayam	50.3
Idukki	30.5
Pathanamthitta	56.4
Alapuzha	45-5
Kollam	56.3
Thiruvananthapuram	89.0

Calculate the mean and median rainfall in Kerala during this week. Why is the mean less than median?
Answer:
Mean = Total amount of rain / No. of districts
= \(\frac { 700 }{ 14 }\) = 50
In ascending order:
13.5, 20.5, 30.5, 33.5, 45.5, 50.3, 53.4, 56.3, 56.4, 56.9, 56.9, 66.7, 70.6, 89
Median \(=\frac{53.4+56.3}{2}=54.85\)
The mean is less than median because the number contains are far small and large numbers than mean.

Use this number sequence calculator to easily calculate the n-th term of an arithmetic, geometric or fibonacci sequence.

Sslc Statistics Solutions Kerala Syllabus Questions 3.
Prove that for a set of numbers arithmetic sequence, the mean and median are equal.
Answer:
Let a, a+d, a+3d, a+4d are the numbers of an arithmetic sequence, then median = \(\frac{a+a+4 d}{2}=\frac{2 a+4 d}{2}=a+2 d\)
Median will be the term which is at center = a + 2d
∴ A set of numbers in arithmetic sequence, the mean and median are equal.

Textbook Page No. 248

Sslc Maths Statistics Notes Kerala Syllabus  Questions 1.
35 households in a neighborhood are sorted according to their monthly income in the table below.

Monthly income (Rs)	Number of households
4000	3
5000	7
6000	8
7000	5
8000	5
9000	4
10000	3

Calculate the median income.
Answer:

Monthly income (Rs)	Number of households
up to 4000	3
up to5000	10
up to 6000	18
up to 7000	23
up to 8000	28
up to 9000	32
up to10000	35

In the table monthly income up to 18th place be 6000 rupees.
That is 18^th place family also includes in the middle of total number of families.
∴ Median of the income = Rs. 6000

Sslc Maths Chapter 11 Statistics Kerala Syllabus Questions 2.
The table below shows the workers in a factory sorted according to their daily wages.

Daily wages (Rs)	Number of workers
400	2
500	4
600	5
700	7
800	5
900	4
1000	3

Calculate the median daily wage.
Answer:

Daily wages (Rs)	Number of workers
up to 400	2
up to 500	6
up to 600	11
up to 700	18
up to 800	23
up to 900	27
up to 1000	30

Total number of workers = 30
Half= 15
So median is the wage of 15th worker.
The daily wages between the place 11 and 18 = 700 rupees
Median of daily wages = 700 rupees

Sslc Maths Chapter 11 Kerala Syllabus Questions 3.
The table below gives the number of babies born in a hospital during a week, sorted according to their birth weight.

Weight (kg)	Number of babies
2.500	4
2.600	6
2.750	8
2.800	10
3.000	12
3-i5o	10
3-250	8
3-300	7
3-5°o	5

Calculate the median birth-weight
Answer:

Weight (kg)	Number of babies
up to 2.500	4
up to 2.600	10
up to 2.750	18
up to 2.800	28
up to 3.000	40
up to 3.150	50
up to 3.250	58
up to 3.300	65
up to 3.500	70

Total number of babies = 70
Half =35
So median is the weight of 35^th baby.
The weight of 35 111 child be in between 29 and 40 placed child, its weight will be 3 kg.
∴ Median of the weight = 3 kg.

Textbook Page No. 254

Sslc Statistics Notes Kerala Syllabus Questions 1.
The table shows some households sorted according to their usage of electricity:

Electricity usage (units)	Number of households
80 – 90	3
90 – 100	6
100 – 110	7
110 – 120	10
120 – 130	9
130 – 140	4

Calculate the median usage of electricity.
Answer:

Usage of electricity (units)	Number of households
less than 90	3
less than 100	9
less than 110	16
less than 120	26
less than 130	35
less than 140	39

Half of the number of houses = 20
We have to find the electricity usage of the 20th house. According to this, we can divide

Statistics Class 10 State Syllabus Question 2.
Answer:

Weight	Number
less than 45.5	5
less than 50.5	12
less than 55.5	22
less than 60.5	30
less than 65.5	34

Total number of children = 34
It is an even number, so we will take the half of sum of weight of the children, those are in the 17 and 18 positions. According to this child between 13 and 22 have weight between 50.5 and 55.5. Our required children (between and 17 and 18) are in these positions. Divide 5 years from 50.5 to 55.5 into 10 equal, parts. Let consider each part have one child.

Weight of each part = \(\frac { 5 }{ 10 }\) = \(\frac { 1 }{ 2 }\).
Hence weight of a child in 13th place is in middle of 50.5 and 51. That is 50.75. Further, each student’s weight is increased by 5.
Hence weight of the man in the 17th position

Median \(=\frac{52.75+53.25}{2}=53\)

Statistics 10th Standard State Syllabus Question 3.
The workers of a company are arranged as given below. Calculate median

Income (Rs)	Number of workers
450	2
500	3
550	5
600	8
650	6
700	5
750	1

Answer:

Income (Rs)	Number of workers
up to 450	2
up to 500	5
up to 550	10
up to 600	18
up to 650	24
up to 700	29
up to 750	30

Total number of workers = 30
Half = 15
So median is the wage of 15th worker.
The daily wage between the place 10 and 18 = 600 rupees
Median of daily wage = 600 rupees

Statistics SCERT Questions & Answers

Scert Class 10 Maths Statistics Kerala Syllabus Question 5.
10 households in a neighborhood are sorted according to their monthly income are given below
16500, 21700, 18600, 21050, 19500, 17000, 21000, 18000, 22000, 75000
a. What is the mean income of these 10 families?
b. How many families have monthly incomes less than the mean income? Prove that in such situation this average is suitable or not?
Answer:
a. Mean = \(\frac { sum }{ number }\) = \(\frac { 248000 }{ 10 }\) = 24800
b. 9 families have monthly income less than the mean income. So in this situation this is not a suitable average.

Sslc Maths Statistics Questions And Answers Kerala Syllabus Question 6.
Number of members in 10 families, collected by mathematics club survey are given. Calculate mean; median and explain. Which is the suitable average?.
4, 2, 3, 5, 4, 3, 2, 20, 4, 3
Answer:
a. Mean = 5
b. Median = 3.5
Suitable average median = 3.5

Sslc Statistics Questions Kerala Syllabus Question 7.
Weekly Wages of 9 persons Working in a factory are given. Find the median 2100; 3500, 2100, 2500, 2800, 4900, 2300, 2200, 3300
Answer:
Write the number in order.
2100, 2100, 2200, 2300, 2500, 2800, 3300, 3300, 3500 (1)

Median = 2500

Statistics Class 10 State Board Kerala Syllabus Question 8.
The table shows the workers doing different jobs in a factory according to their daily wages.

Daily wages(Rs)	Number of workers
225	4
250	7
270	9
300	5
350	3
400	2

Calculate median of daily wages.

Answer:

Daily wages(Rs)	Number of workers
up to 225	4
up to 250	11
up to 270	20
up to 300	25
up to 350	28
upto400	30

The worker in the 12th position to 20th position has daily wage 270 ie, Median

Sslc Maths Chapter 11 Solutions Kerala Syllabus Question 9.
The table below shows the 60 children in a class sorted according to their heights

Height (cm)	Number of children’s
140-145	5
145-150	8
150-155	12
155-160	16
160-165	11
165-170	5
170-175	3

Find the median height?
Answer:

Height (cm)	Number of children’s
Belowl45	5
Below 150	13
Below 155	25
Below 160	41
Below 165	52
Below 170	57
Below 175	60 1

Sslc Maths Chapter Statistics Kerala Syllabus  Question 10.
Answer:
a. 100
b. 25
c.

Mid value (x)	No.of Workers (y)	x, y
130	16	2080
150	11	1650
170	20	3400
190	28	5320
210	18	3780
230	7	1610

Mean = \(\frac { 17840 }{ 100 }\) = 178.4

Hss Live Guru 10th Maths Kerala Syllabus Question 11.
The mean of the frequency table given below is 50. Then find out the values of a and b.

Answer:

mean = 50
∴ \(\frac{3480+30 a+70 b}{120}=50\)
30a + 70b = 6000 – 3480
30a + 70b = 2520 (1)
17 + 32 + 19 + a + b = 120
68 + a + b = 120,
a + b = 52 (2)
(2) x 30 => 30a + 30b = 1560
30a + 70b = 2520,
30a + 30b = 1560,
40b = 960,
b = \(\frac { 960 }{ 40 }\) = 24,
a + 24 = 52
a = 52 – 24 = 28,
∴ a = 28, b = 24

Long Answer Type Questions (Score 5)

10th Standard Maths Statistics Kerala Syllabus Question 21.
The table below shows groups of children in a class according to their heights:

Height (cm)	Number of children
135-140	5
140-145	8
145 – 150	10
150-155	9
155-160	6
160-165	3

a. If the children are lined up according to their heights, the median is the height of the child in which position?
b. According to the table, the height of this child is between what limits?
c. What are the assumptions used to compute the median?
d. What is the median height according to these assumptions?
Answer:

Height (cm)	Number of children
Below 140	5
Below 145	13
Below 150	23
Below 155	32
Below 160	38
Below 165	41

a. Height of the 21st child is the median height.
b. Height of the 21st child is between 145 cm and 150 cm.
c. Methods to find the median are.
1. Divide 5cm in between 145 cm and 150 cm into 10 equal sections.
2. Consider that the height of each sub-group is exactly on the midpoint of the subgroup.
Height of the 14th child is in between 145 cm and 145 \(\frac { 5 }{ 10 }\) cm.
i.e., 145 \(\frac { 5 }{ 20 }\) cm.
Similarly, the height of the 15th student is in between 145 \(\frac { 5 }{ 10 }\) cm and

Hence height of each child can be increased by \(\frac { 5 }{ 10 }\) cm.
There are 7 children to reach the 21st child from 14th child.
i.e., 14 th term is 145 \(\frac { 5 }{ 20 }\) and common difference is \(\frac { 5 }{ 10 }\).
Mean is the 21 st term of the arithmetic sequence.

Use our simple online Limit Of Sequence Calculator to find the Limit with step-by-step explanation.

Statistics Menton Map

Arithmetic mean is the sum divided by the number of terms.

Mean = \(\frac { sum of terms }{ Number of terms }\)

When the numbers are arranged in a ascending order, then the middle term is the median.
i.e., half of the total frequency will give the median.

You can Download Arithmetic Sequences Questions and Answers, Activity, Notes, Kerala Syllabus 10th Standard Maths Solutions Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Maths Solutions Chapter 1 Arithmetic Sequences

Arithmetic Sequences Text Book Questions and Answers

Textbook Page No. 10

Arithmetic Sequence Questions And Answers 10th Standard Question 1.
Make the following number sequences, from the sequence of equilateral triangles, squares, regular pentagons and so on, of regular polygons:
Number of sides 3, 4, 5, ………
Sum of inner angles
Sum of outer angles
One inner angle
One outer angle
Arithmetic Sequence worksheet with Answer:
No. of triangles of a regular polygon having no. of sides are 3, 4, 5 ………… n respectively is given as 1, 2, 3, 4. Sum of interior angles of a triangle is 180°.

Then the sequence of Sum of interior angles
= 180, 180 × 2, 180 × 3 …. = 180, 360, 540, ……………….
Sum of exterior angles of any geometric structure having any number of sides is always 360°.
Sum of exterior angles = 360, 360, …………..
One interior angle
= \(\frac { 360 }{ 3 }\), \(\frac { 360 }{ 4 }\), \(\frac { 360 }{ 5 }\), ……………. = 120, 90, 72, ………….

Find the sequence calculator can determine the terms (as well as the sum of all terms) of the arithmetic, geometric, or Fibonacci sequence.

Arithmetic Sequence Class 10 Kerala Syllabus Question 2.
Look at these triangles made with dots. How many dots are there in each ?

Compute the number of dots needed to make the next three triangles.
Arithmetic Sequence worksheet with Answer:
3, 6, 10
The number of dots needed to make the next three triangles will be:
10 + 5 = 15
15 + 6 = 21
21 + 7 = 28
15, 21, 28 Dots

Arithmetic Sequence Question 3.
Write down the sequence of natural numbers leaving remainder 1 on division by 3 and the sequence of natural numbers leaving remainder 2 on division by 3.
Answer:
The numbers that leave 1 as remainder when divided by 3 are 1, 4, 7, 10, 13, ………..
(Sequence each with difference of 3 and starting from 1)
The numbers that leave 2 as remainder when divided by 3 are 2, 5, 8, 11, 14, ………….
( Sequence each with difference of 3 and starting from 2)

Kerala Syllabus 10th Standard Maths Chapter 1 Question 4.
Write down the sequence of natural numbers ending in 1 or 6 and describe it in two other ways.
Answer:
1, 6, 11, 16, 21, ……………..
Numbers, each with difference of 5 and starting from 1.
Numbers, when divided by 5, leaves 1 as remainder.

SSLC Maths Arithmetic Sequence Question 5.
A tank contains 1000 litres of water and it flows out at the rate of 5 litres per second. How much water is there in the tank after each second? Write their numbers as a sequence.
Answer:
Water in the tank initially = 1000 litre
Water in the tank after first second
= 1000 – 5 = 995 litre
Water in the tank after next second
= 995 – 5 = 990 litre
Water in the tank after third second
= 990 – 5 = 9.85 litre
Sequence 1000, 995, 990, 985, 980, ………………

Textbook Page No. 15

Sslc Maths Chapter 1 Question Answer Question 1.
Write the algebraic expression for each of the sequences below:
i. Sequence of odd numbers
ii. Sequence of natural numbers which leave remainder 1 on division by 3.
iii. The sequence of natural numbers ending in 1.
iv. The sequence of natural numbers ending in 1 or 6.
Answer:
i. Sequence of odd numbers = 1, 3, 5, 7 …………
That is 2 × 1 – 1,2 × 2 – 1,
Common difference a = 2,
First term, a + b=1, b = 1
Algebraic expression xn = an + b = 2n – 1
n=1, 2, 3 ………..

ii. Sequence of natural numbers which leave remainder 1 on division by 3.
1, 4, 7, 10, 13, 16
x₁ = 1
x₂ = 1 + 3(1) = 4
x₃ = 1 + 3(2) = 7
………………………
……………………..
x_n= 1 + 3(n – l) = 3n – 2
Algebraic expression xn = 3n – 2
n = 1, 2, 3 …………..

iii. The sequence of natural numbers ending in 1 is 1, 11, 21, 31……….
That is 10 × 1 – 9, 10 × 2 – 9, ……………….
Algebraic expression x_n = 10n – 9
n = 1,2,3 ………….

iv. The sequence of natural numbers ending in 1 or 6 is 1, 6, 11, 16, 21 ………….
That is 5 × 1 – 4, 5 × 2 – 4,
Algebraic expression xn = 5n – 4
n=1,2,3 ……………….

An arithmetic common difference calculator is an online free tool to find the arithmetic sequence with the given common difference.

Arithmetic Sequence Pictures Question 2.
For the sequence of regular polygons starting with an equilateral triangle, write the algebraic expressions for the sequence of the sums of inner angles, the sums of the outer angles, the measures of an inner angle, and the measure of an outer angle.
Answer:
Let n be the number of sides
Sum of interior angles: 180°, 360°, 540°, 720°, ……………
Algebraic expression x_n = 180n
Sum of exterior angles 360°, 360°, 360°,………..
Algebraic expression xn = 360°
Sequence of one interior angle:
\(\frac { 180° }{ 3 }\), \(\frac { 360° }{ 4 }\), \(\frac { 540° }{ 5 }\) ………………
Algebraic expression xn = \(\frac{180^{0} n}{n+2}\)
Sequence of one exterior angle:
\(\frac { 360° }{ 3 }\), \(\frac { 360° }{ 4 }\), \(\frac { 360° }{ 5 }\), …………..
Algebraic expression xn = \(=\frac{360^{0}}{n+2}\)

Class 10 Maths Chapter 1 Kerala Syllabus Question 3.
Look at these pictures:

The first picture is got by removing the small triangle formed by joining the midpoints of an equilateral triangle. The second picture is got by removing such a middle triangle from each of the red triangles of the first picture. The third picture shows the same thing done on the second.
i. How many red triangles are there in each picture?
ii. Taking the area of the original uncut triangle as 1, compute the area of a small triangle in each picture.
iii. What is the total area of all the red triangles in each picture?
iv. Write the algebraic expressions for these three sequences obtained by continuing this process.
Answer:
i. The red triangles in the first picture = 3
Red triangles in the second picture = 9
Red triangles in the third picture = 27

ii. Area of the original uncut triangle = 1
Area of the first triangle = \(\frac { 1 }{ 4 }\)
Area of red triangles in the second picture = \(\frac { 1 }{ 16 }\)
Area of red triangles in the third picture = \(\frac { 1 }{ 64 }\)

iii. Total area of red triangle in the first picture = = \(3 \times \frac{1}{4}=\frac{3}{4}\)
Total area of red triangles in the second picture = \(9 \times \frac{1}{16}=\frac{9}{16}\)
Total area of red triangles in the third picture = \(27 \times \frac{1}{64}=\frac{27}{64}\)

This free Recursive sequence formula calculator can determine the terms (as well as the sum of all terms) of the arithmetic, geometric, or Fibonacci sequence.

Textbook Page No. 18

Arithmetic Sequence Class 10 SCERT Question 1.
Check whether each of the sequences given below is an arithmetic sequence. Give reasons. For the arithmetic sequences, write the common difference also.
i. Sequence of odd numbers.
ii. Sequence of even numbers.
iii. Sequence of fractions got as half the odd numbers.
iv. Sequence of powers of 2.
V. Sequence of reciprocals of natural numbers.
Answer:
i. Sequence of odd numbers 1, 3, 5, 7,…….
Difference of adjacent numbers = 2
Common difference = 2

ii. Sequence of even numbers 2, 4, 6, 8,…….
Difference of adjacent numbers = 2 Common difference = 2
∴ This is an arithmetic sequence.

iii. Sequence of half of the odd numbers

Difference of adjacent numbers = 1
Common difference = \(\frac{3}{2}-\frac{1}{2}=1=\frac{5}{2}-\frac{3}{2}\)
∴ This is an arithmetic sequence.

iv. Sequence of powers of 2 21, 22, 23, 24,
2, 4, 8,16, ………………
Difference of adjacent numbers = 4 – 2 ≠ 8 – 4
It does not have a common difference
∴ This is not an arithmetic sequence

v. Sequence of reciprocals of natural numbers

Difference of adjacent numbers

It does not have a common difference
∴ This is not an arithmetic sequence

SSLC Maths Chapter 1 Question 2.
Look at these pictures:

If the pattern is continued, do the numbers of coloured squares form an arithmetic sequence? Give reasons.
Arithmetic Sequence Questions and Answer:
Numbers of coloured squares in each picture 8, 12, 16, ……………
Common difference =12 – 8 = 4 = 16 – 12
∴ This is an arithmetic sequence

10th Class Maths Chapter 1 Arithmetic Sequence Question 3.
See the pictures below:

i. How many small squares are there in each rectangle?
ii. How many large squares?
iii. How many squares in all?
Continuing this pattern, is each such sequence of numbers, an arithmetic sequence?
Answer:
i. No., of small squares in the first picture = 2
No. of small squares in the second picture = 4
No. of small squares in the third picture = 6
No. of small squares in the fourth picture = 8

ii. No. of big squares in first picture = 0
No. of big squares in second picture = 1
No. of big squares in third picture = 2
No. of big squares in fourth picture = 3

iii. Total squares in first picture = 2
Total squares in second picture = 5
Total squares in third picture = 8
Total squares in fourth picture = 11
Sequence of no. of small squares 2, 4, 6, 8
This is an arithmetic sequence with common difference = 2
Sequence of no. of big squares, 0, 1, 2, 3
This is an arithmetic sequence with common difference = 1
Sequence of Total no. of squares 2, 5, 8, 11
This is an arithmetic sequence with common difference = 3

Question 4.
In the staircase shown here the height of the first step is 10 centimetres and the height of each step after it is 17.5 centimetres.
i. How high from the ground would be some-one climbing up, after each step?
ii. Write these numbers as a sequence

Answer:
i. High from the ground climbing up first step = 10 cm
High from the ground climbing up sec-ond step = 10 + 17.5 = 27.5cm
High from the ground climbing up third step = 27. 5+ 17.5 = 45 cm
High from the ground climbing up fourth step = 45 + 17.5 = 62.5 cm
High from the ground climbing up fourth step = 62.5 + 17.5 = 80 cm
High from the ground climbing up fourth ‘ step = 80 + 17.5 = 97.5cm

ii. Sequence of height
10, 27.5, 45, 62.5, 80, 97.5 ……………

Question 5.
In this picture, the perpendiculars to the bottom line are equally spaced. Prove that, continuing like this, the lengths of perpendiculars form an arithmetic sequence.

Answer:

In figure ΔOAB, ΔOCD are similar right-angled triangles. That is if one side and its included angles of a triangle are equal to the one side and included angles of another triangle, then the triangles are similar.
∴ sides are proportional.
\(\frac{O A}{O C}=\frac{x}{2 x}=\frac{A B}{C D} \quad \Rightarrow C D=2 A B\) similarly
EF = 3AB. Sequence of length of perpendicular AB, 2AB, 3AB,………..
That is AB is the common difference, so perpendicular lengths are in arithmetic sequence

Question 6.
The algebraic expression of a sequence is x_n = n³ – 6n² + 13n – 7
Is it an arithmetic sequence?
Answer:
Algebraic expression = n³ – 6n² + 13n – 7
n = 1 ⇒ =1 – 6 + 13 – 7 = 1
n = 2 ⇒ = 8 – 24 + 26 – 7 = 3
n = 3 ⇒ =27 – 6 × 9 + 39 – 7
= 27 – 54 + 32 = 5
n = 4 ⇒ = 64 – 96 + 52 – 7 = 13
1, 3, 5, 13,………. This is not an arithmetic sequence.lt does not have a common difference

Textbook Page No. 21

Question 1.
In each of the arithmetic sequences below, some terms are missing and their positions are marked with . Find them.

Answer:

Common difference = 42 – 24 = 18
Numbers to be found out = 42 + 18 = 60
60 + 18 = 78
Arithmetic sequence 24, 42, 60, 78, ………….

Common difference = 42 – 24 = 18
Numbers to be found out = 24 – 18 = 6
42 + 18 = 60
24 – 18, 24, 42, 42 + 18
Arithmetic sequence 6, 24, 42, 60, ………….

Common difference = 18
Numbers to be found out = 24 – 18 = 6
6 – 18 = – 12 6 – 18, 24 – 18, 24, 42
Arithmetic sequence – 12, 6, 24, 42, ………

Common difference = 9
[(42 – 24)/2 = 18/2 = 9)
Numbers to be found out =
24 + 9 = 33, 33 + 9 = 42, 42 + 9 = 51
Arithmetic sequence 24, 33, 42, 51 ……..

Common difference = (42 + 24)/2 = 18/2 = 9
Numbers to be found out = 24 – 9, 24, 24 +9, 42
Arithmetic sequence 15, 24, 33, 42

Let numbers to be found out be x and y If common differences are equal = x – 24 = 42 – y

Question 2.
The terms in two positions of some arithmetic sequences are given below.
Write the first five terms of each:
i. 3^rd term 34
6^th term 67
ii. 3^rd term 43
6^th term 76
iii. 3^rd term 2
5^th term 3
iv. 4^th term 2
7^th term 3
v. 2^nd term 5
5^th term 2
Answer:
i. Third term =34
Sixth term =67
We get the 6th term from the 3 rd term, we must add the common difference (6 – 3 = 3)3 time.
Three times of common difference 67 – 34 = 33
Common difference = \(\frac { 33 }{ 3 }\) = 11
Second term = 34 – 11 = 23
First term = 23 – 11 = 12
First five terms 12, 23, 34, 45, 56

ii. Third term =43
Sixth term = 76
Thrice the common difference = 76 – 43 = 33
Common difference = \(\frac { 33 }{ 3 }\) = 11
Second term = 43 – 11 = 32
First term =32 – 11 =21
First five terms 21, 32, 43, 54, 65.

iii. Third term = 2
Fifth term = 3
Twice the common difference= 3 – 2 = 1
Common difference = 1/2
Second term =2 – 1/2 = 1 1/2
First term = 1 1/2 – 1/2 = 1

iv. Fourth term =2
Seventh term = 3
Thrice the common difference = 3 – 2 = 1
Common difference = 1/3

v. Second term = 5
Fifth term = 2
Thrice the common difference = 2 – 5 = -3
Common difference = –3/3 = –1
First term = 5 – –1 = 6
First five terms 6, 5, 4, 3, 2

Question 3.
The 5th term of an arithmetic sequence is 38 and the 9th term is 66. What is its 25th term?
Fifth term = 38
Ninth term = 66
Term difference = 66 – 38 = 28
Position difference = 9 – 5 = 4 – 2 8
Common difference = 28/4 = 7
25th term = Fifth term = 20 × Common difference
25th term = Fifth term + 20 × Common difference = 38 + 20 × 7 = 38 + 140 = 178

Question 4.
Is 101 a term of the arithmetic sequence 13, 24, 35, …? What about 1001?
Answer:
13, 24, 35,………..
Common difference = 24 – 13 = 11
Difference between 101 and 13
= 101 – 13 = 88 = 8 × 11
That is we can obtained 101 by adding 8 times of common difference with 13.
So, ‘101’ is the 9th term of this arithmetic sequence
101 is the 9th term of this sequence
Difference between 1001 and 13
= 1001 – 13 = 988
This is not a multiple of 11.
Therefore ‘1oo1’ is not a term of this arithmetic sequence.

Question 5.
How many three-digit numbers are there, which leave a remainder 3 on division by 7?
Answer:
101, 108, …………………………. 997
First-term = 101
Common difference = 7
Last term = 997
Term difference = 997 – 101 = 896 = 128 × 7
Last term = First term + 128 × common difference
i. e., 997 is the 129th term.
129 three-digit numbers are there which leave remainder 3 on division by 7

Question 6.
Fill up the empty cells of the given square such that the numbers in each row and column form arithmetic sequences:

What if we use some other numbers instead of 1, 4, 28 and 7?
Answer:
1. In the first row differences between 1st and 4th terrm =3
Position difference = 4 – 1 = 3
Common difference = 3/3 = 1
Arithmetic sequence 1, 2, 3,4

2. In the first coloumn Term difference = 7 – 1 = 6
Position difference = 4 – 1 = 3
Common difference = 6/3 = 2
Arithmetic sequence 1, 3, 5, 7……….

3. In the fourth row
Termdifference = 28 – 7 = 21
Position difference = 4 – 1=3
Common difference = 21/3 = 7
Arithmetic sequence 7, 14, 21, 28

4. In the second coloumn 2, –,– 14
Term difference = 14 – 2 = 12
Position difference = 4 – 1=3
Common difference = \(\frac { 14 – 2 }{ 3 }\) = 4
Arithmetic sequence 2, 6, 10, 14

5. In the third coloumn 3, –, –, 21
Term difference = 21 – 3 = 18
Position difference = 4 – 1=3
Common difference = \(\frac { 21 – 3 }{ 3 }\) = 6
Arithmetic sequence 3, 9,15, 21

6. In the fourth coloumn 4, –, –, 28
Term difference = 28 – 4 = 24
Position difference = 4 – 1=3
Common difference = \(\frac { 28 – 4 }{ 3 }\) = 8
Arithmetic sequence 4,12, 20, 28

Question 7.
In the table below, some arithmetic sequences are given with two numbers against each. Check whether each belongs to the sequence or not.

Answer:
i. Sequence 11, 22, 33, …………..
Difference = 22 – 11 = 11
Difference between 123 and 11.
=123 – 11 = 112
112 is not a multiple of 11, so 123 is not a term of this arithmetic series.

Difference between 132 and 11
= 132 – 11 = 121
121 is a multiple of 11, so 132 is a term of this arithmetic series.

ii. Sequence 12, 23, 34,
Common difference = 23 – 12 = 11 Difference between 100 and 12
= 100 – 12 = 88
88 is a multiple of 11, so 100 is a term of this arithmetic series.

Difference between 1000 and 12.
= 1000 – 12 = 988
988 is not a multiple of 11, so 1000 is not a term of this arithmetic series.

iii. Sequence 21, 32, 43, ……………
Common difference = 32 – 21 = 11
Difference between 100 and 21.
= 100 – 21 = 79
79 is not a multiple of 11, so 1000 is not a term of this arithmetic series.

Difference between 1000 and 21
=1000 – 21 = 979
979 is a multiple of 11, so 1000 is a term of this arithmetic series.

Textbook Page No. 24

Question 1.
The 8th term of an arithmetic sequence is 12 and its 12th term is 8. What is the algebraic expression for this sequence?
Answer:
Difference between terms = 8 – 12 = -4
To get the 12th term from the 8td term, we must add the common difference (9 – 5 = 4) 4 times.
Common difference = – 4 / 4 = -1
Algebraic expression for arithmetic series
x_n = an + b
12 = – 1 × 8 + b ⇒ 12 = – 8 + b
b = 12 + 8 = 20
Algebraic expression for arithmetic series =
xn = –1 × n + 20 = 20 – n

Question 2.
The Bird problem in Class 8 (The lesson, Equations) can be slightly changed as follows.
One bird said:
“We and we again, together with half of us and half of that, and one more is a natural number”
Write all the possible number of birds starting from the least. For each of these, write the sum told by the bird also.
Find the algebraic expression for these two sequences.
Answer:
If we take x as the number of birds

∴ 11x + 4 is a multiple of 4.
⇒ 11x is a multiple of 4.
x is a multiple of 4.
No. of birds 4, 8, 12, 16 ……………
(a= 4, a+b = 4, b = 0)
Algebraic expression of the series = an + b = 4n No. of sums
\(\left(\frac{11 x}{4}+1, x=4,8,12 \ldots .\right)=12,23,34,45, \ldots \ldots\)
Algebraic expression for the series = 12 + (n -1)11 = 11 n + 1

Question 3.
Prove that the arithmetic sequence with first term 1/3 and common difference 1/6 contains all natural numbers.
Answer:
f = \(\frac { 1 }{ 3 }\)
d = \(\frac { 1 }{ 6 }\)
\(x_{n}=d n+(f-d)=\frac{1}{6} n+\frac{1}{6}=\frac{1}{6}(n+1)\)
As n = 5, 11, 17,………….
∴ All natural numbers will occur in this arithmetic sequence

Question 4.
Prove that the arithmetic sequence with first term 1/3 and common difference 2/3 contains all odd numbers, but no even number.
Answer:

Value of (2n -1) is always an odd number, Therefore there is no even number.

Question 5.
Prove that the squares of all the terms of the arithmetic sequence 4,7,10,…. belong to the sequence.
Answer:

This is in the form 3k + 1, so squares of all terms are in this series. (When we divide 3k + 1 by 3 get 1 as remainder)

Question 6.
Prove that the arithmetic sequence 5,8,11,……. contains no perfect squares.
Answer:
f = 5
d = 3
\(x_{n}=d n+(f-d)=3 n+(5-3)=3 n+2\)
That is when we divide a perfect square with 3 we get remainder as 1 or 0. Here we can divide 3k + 2 by 3 to get 2 as remainder and it doesn’t contain any perfect square.

Question 7.
Write the whole numbers in the arithmetic sequence \(\frac { 11 }{ 8 }\), \(\frac { 14 }{ 8 }\), \(\frac { 17 }{ 8 }\), ………..
they form an arithmetic sequence?
Answer:

If ‘n’ is a multiple of 8 then 4, 7, 10, … is a sequence having whole numbers
Common difference = 3,This is an arithmetic sequence.

Textbook Page No. 28

Question 1.
Write three arithmetic sequences with 30 as the sum of the first five terms.
Answer:
If the five terms are
x, x + 1, x + 2, x +3, x + 4
x + x+ 1 + x + 2 + x + 3 + x + 4 = 30
5x + 1o = 30
5x = 20
x = 4
The arithmetic sequence is 4, 5, 6, 7, 8

If the five terms are
x, x + 2, x + 4, x +6, x + 8
x + x + 2 + x + 4 + x + 6 + x + 8 = 30
5x + 20 = 30
5x = 30 – 20 = 10
x = 2
The arithmetic sequence is 2, 4, 6, 8, 10

If the five terms are
x, x + 3, x + 6, x + 9, x + 12
x + x + 3+ x+ 6 + x + 9+ x + 12 = 30
5X + 30 = 30
5x = 0
x = 0
The arithmetic sequence is 0, 3, 6, 9,12
The arithmetic sequences whose sum of first five terms give 30 are:
4, 5, 6, 7, 8, …………..
2, 4, 6, 8, 10, …………..
0, 3, 6, 9, 12, ………….

Arithmetic Sequence Question 2.
The first term of an arithmetic sequence is 1 and the sum of the first four terms is 100.
Find the first four terms.
Answer:
First term f = 1
If first four terms are
f, f+d, f+2d, f+3d
f + f + d + f + 2d + f + 3d = 100
4f + 6d = 100
= 2f + 3d = 50
2+3d = 50
= 3d = 50 – 2 = 48
d = 16
The arithmetic sequence is 1, 17, 33, 49, ………….

Question 3.
Prove that for any four consecutive terms of an arithmetic sequence, the sum of the two terms on the two ends and the sum of the two terms in the middle are the same.
Answer:
First four terms of an arithmetic sequence is x,
x+ d, x+ 2d, x+ 3d
Sum of the two terms on the two ends = 2x + 3d
Sum of the two terms in the middle
= (x + d) +(x + 2d) = 2x + 3d
They both are equal

Question 4.
Write four arithmetic sequences with 100 as the sum of the first four terms.
Answer:
If the first four terms are
x – 3d, x – d, x + d, x + 3d
x – 3d + x – d + x + d + x + 3d = 100
4x = 100
x = 25
i. If d= 1
The arithmetic sequence is 22, 24, 26, 28

ii. If d= 2
The arithmetic sequence is 19, 23, 27, 31

iii. If d = 3
The arithmetic sequence is 16, 22, 28, 34

iv. If d = 4
The arithmetic sequence is 13, 21, 29, 37

Question 5.
Write the first three terms of each of the arithmetic sequences described below:
i. First-term 30; the sum of the first three terms is 300.
ii. First-term 30; the sum of the first four terms is 300.
iii. First-term 30; the sum of the first five terms is 300.
iv. First-term 30; the sum of the first six terms is 300.
Answer:
i. First-term = 30
Sum of first three terms = 300
In the arithmetic sequence sum of any three consecutive natural numbers is thrice the middle number
∴ Second term = \(\frac { 300 }{ 3 }\) = 100
∴ Common difference = 300 – 30 = 270
∴ Third term =100 + 70 = 170
Sequence 30, 100, 170, ……….

ii. First-term = 30
Sum of first four terms = 300
Four consecutive terms of an arithmetic sequence, the sums of the first and the last is equal to the sum of the second and the third.
First term + Fourth term = Second term + Third term = \(\frac { 300 }{ 2 }\) = 150
∴ Fourth term = 150 – 30 = 120
30, ………, ………., 120
Term difference = 120 – 30 = 90
Position difference = 4 – 1 = 3
Common difference = 90/3 = 30
∴ Sequence = 30, 60, 90, 120,………….

iii. First term = 30
Sum of first five terms = 300
Sum of the five consecutive terms of arithmetic sequence is five times of its middle term.
Third term = 300/5 = 60
Common difference = 15
∴ Sequence = 30, 45, 60, 75, 90, ……….

iv. First term =30
Sum of first six terms =300
Fist term + Sixth term = Second term + Fifth term = Third term + Fourth term = 300/3 = 100
Sixth term = 100 – First term = 100 – 30 = 70
Term difference = 70 – 30 = 40
Position difference =6 – 1 = 5
Common difference = \(\frac { 40 }{ 5 }\) = 8
Sequence = 30, 38, 46, 54, 62, 70,

Question 6.
The sum of the first five terms of an arithmetic sequence is 150 and the sum of the first ten terms is 550.
i. What is the third term of the sequence?
ii. What is the eighth term?
iii. What are the first three terms of the sequence?
Answer:
i. Sum of first five terms = 150
Sum of the five consecutive terms of arithmetic sequence is five times of its middle term.
Third term = \(\frac { 150 }{ 5 }\) = 30

ii. First term + Tenth term = Second term + Nineth term = Third term + Eighth term = Fourth term + Seventh term = Fifth term + Sixth term = \(\frac { 550 }{ 5 }\) = 110
Third term + Eighth term = 110
Eighth term =110 – Third term
= 110 – 30 = 80

iii. Third term = 30
Eighth term = 80
Term difference = 80 – 30 = 50
Position difference = 8 – 3 = 5
Common difference = \(\frac { 50 }{ 5 }\) = 10
∴ Sequence = 10, 20, 30,…………….

Question 7.
The angles of a pentagon are in arithmetic sequence. Prove that its smallest angle is greater than 36°.
Answer:
Let the smallest angle in the pentagon be x
x + x + d + x + 2d + x + 3d + x + 4d = 540
5x + 10d = 540,
x + 2d = 108
x = 36°, then
d = 36,
angles are 36, 72, 108, 144, 180.
180° will not an angle of a pentagon (Sum of exterior and interior angle is 180°).
Therefore the smallest angle in a pentagon will always be greater than 36°

Textbook Page No. 35

Question 1.
Find the sum of the first 25 terms of each of the arithmetic sequences below.
i. 11, 22, 33,….
ii. 12, 23, 34,…
iii. 21, 32, 43,….
iv. 19, 28, 37,…
vi. 1, 6, 11,….
Answer:

Question 2.
What is the difference between the sum of the first 20 terms and the next 20 terms of the arithmetic sequence 6, 10, 14,…?
Answer:

Differences = 2480 – 880 = 1600

Question 3.
Calculate the difference between the sums of the first 20 terms of the arithmetic sequences 6, 10, 14,… and 15, 19, 23,……
Answer:
6, 10, 14,
Algebraic expression of the above arithmetic sequence = 4n + 2

15, 19, 23, …… Algebraic expression of the arithmetic sequence = 4n +11

Differences = 1060 – 880 = 180

Question 4.
Find the sum of all three-digit numbers, which are multiples of 9.
Answer:
First 3 digit number divisible by 9 = 108
Last 3 digit number divisible by 9 = 999 108, 117,126, …………… 999

Question 5.
The expressions for the sum to n terms of some arithmetic sequences are given below. Find the expression for the nth term of each:
i. n² + 2n
ii. 2n² + n
iii.n² – 2n
iv.2n² – n
v. n² – n
Answer:
i. n² + 2n, First term =12+ 2 × 1 = 3
Sum of first two terms = 22 + 2 × 2 = 8
Second term = 8 – 3 = 5
The arithmetic sequence is 3, 5,…………..
nth term = dn + (f – d) = 2n + (3 – 2)
= 2n + 1

ii. 2n² + n
First term =2 + 1 = 3
Sum of first two terms =2 × 22 + 2
= 8 + 2 = 10
Second term = 10 – 3 =7
The arithmetic sequence is 3, 7, …………

iii. n² – 2n
First term = 1 – 2 = – 1
Sum of first two terms =22 – 4 = o
Second term = 0-(-1) = 0 + 1 = 1
The arithmetic sequence is –1, 1, ……………

iv. 2n² – n
First term = 2 – 1 = 1 Sum of first two terms = 2 – 22 – 2
=8 – 2=6
Second term = 6 – 1 = 5
The arithmetic sequence is 1, 5,……….

v. n² – n
First term = 1 – 1 = 0
Sum of first two terms = 22 – 2 = 2
Second term = 2 – 0 = 2
The arithmetic sequence is 0, 2,……….

Question 6.
Calculate in head, the sums of the following arithmetic sequences.

Question 7.
Prove that the sum of any number of terms of the arithmetic sequence 16, 24, 32,….. starting from the first, added to 9 gives a perfect square
Answer:
16, 24, 32,…………. Algebraic expression of the arithmetic sequence = 8n + 8

Hence it is a perfect square

Question 8.

i. Write the next two lines of the pattern above.
ii. Write the first and the last numbers of the 10th line.
iii. Find the sum of all the numbers in the first ten lines.
Answer:
i. First row contains one element, second row contains two elements, therefore j fifth row will contain five elements and j sixth row will contain six elements.
Consider the first row
1, 2, 4, …………..
1, 1+1, 2 + 2, 4 + 3, …………
Therefore
1, 1+1, 2 + 2, 4 + 3, 7 + 4, 11+5 …………..
1, 2, 4, 7, 11, 16 …………
Generally it written as
1 + 1 (1+2 + 3 + ……….)
Fist term in the fifth row = 1 + 1(1 + 2 + 3 + 4) = 1 + 10 = 11
Fist term in the sixth row= 1 + 1(1 + 2 + 3 + 4+ 5) = 1 + 15 = 16
Common differences in each row = 1
Fifth row 11, 12, 13, 14, 15.
Sixthrow 16, 17, 18, 19, 20, 21.

ii. Fist term in the tenth row = 1 + 1(1 + 2 + 3+ 4 + 5 + 6+ 7 + 8 + 9) = \(=1+\frac{1 \times 9 \times 10}{2}=46\)
Last term in the tenth row = 46 + 1 × 9 = 46 + 9 = 55

iii. Numbers in the first ten lines 1, 2, 3, 4, 5, ………….. 55
Sum of all the numbers in the first ten lines = \(\frac{n(n+1)}{2}=\frac{55 \times 56}{2}=1540\)

Question 9.

Write the next two lines of the pattern above. Calculate the first and last terms of the 20th line.
Answer:
4, 7, 13, 22, 34, 49,…….
(4, 4+3, 7+6, 13+9, 22 + 12, 34+15),
Generally it written as 4 + 3(1 + 2 + 3 + ………..)
First term in the fifth row = 4 + 3(1 +2 + 3+ 4) = 4 + 30 = 34
First term in the sixth row =
4 + 3(1 + 2+ 3 + 4 + 5) = 4 + 45 = 49
Common differences in each row = 3
Fifth row 34, 37, 40, 43, 46,..
Sixth row 49, 52, 55, 58, 61, 64…
4
7 10
13 16 19
22 25 28 31
34 37 40 43 46
49 52 55 58 61 64
………………………….
First term in the 20th row = 4 + 3(1 + 2 + 3 + 4 +……………. +19)

Arithmetic Sequences Orukkam Questions & Answers

Worksheet 1
Answer the following questions

Question 1.
Write the sequence of natural numbers
Answer:
1,2,3,4,

Question 2.
Write the sequence of odd numbers
Answer:
1,3,5,7,

Question 3.
Write the sequence of even numbers
Answer:
2,4,6,8,10,

Question 4.
Write the sequence of multiples of 3.
Answer:
3,6,9,12,15,

Question 5.
Write the sequence of numbers which leaves the remainder 1 on dividing by 4.
Answer:
0 × 4 + 1, 1 × 4 + 1, 2 × 4 + 1, 3 × 4 + 1, 4 × 4 + 1, 5 × 4 + 1, …………
=1, 5, 9, 13, 17, 21,

Question 6.
Write the sequence of prime numbers.
Answer:
2,3,5,7,11,13,17,

Question 7.
Write the sequence of perfect squares.
Answer:
1, 4, 9, 16, 25, 36,

Question 8.
Write the sequence of numbers which leaves the remainder 0 on dividing by 6
Answer:
6, 12, 18, 24, 30, 36, ………….

Question 9.
Write the sequence starting from 1 and is added subsequently
Answer:

Question 10.
Write the sequence starting from 1/2 and 3/4 is added subsequently
Answer:

Question 11.
Write the sequence starting from 60 and 0 is added subsequently
Answer:
60, 60, 60,…………..

Worksheet 2

Question 12.
Write the sequence of the perimeters of the equilateral triangles having sides 1cm, 2cm, 3cm.
a. Write the sequence of area
b. Write the sequence of sum of angles.
Answer:
’If we Increase 1 cm of sides of an equilateral triangle having sides 1cm, 2cm, 3cm.
a. Area =1 cm2, Area after increasing
sides by 1 cm \(\frac{\sqrt{3} \times 2^{2}}{4}=\sqrt{3}\) cm2
Area after increasing thelength of side 2cm by
V3x32 9V3 ,
1cm = \(=\frac{\sqrt{3} \times 3^{2}}{4}=\frac{9 \sqrt{3}}{4}\) cm2
Area after increasing thelength of side 3cm by

c. The sum of angles of a triangle will be 180° always for any measures of sides. Sequence of sum of angles 180, 180, 180,……..

Question 13.
Write the sequence of numbers which leaves the remainder 3 on dividing by 5 and 10.
Answer:
The number which can be divided by 5 and 10 must be divisible by 10.
∴ The sequence of numbers which leaves the remainder 3 on dividing by 5 and 10 is 3, 13, 23, 43,………

Question 14.
Look at the sequence 1 + (1 + 5), 2 + (2 + 5), 3 + (3 + 5) …….
a. Write next two terms
b. Write its algbraic form.
Answer:
a. Next two terms
4 + (4 + 5), 5 + (5 + 5),

b. Algebraic expression
x_n = n + (n + 5) = n + n + 5
x_n = 2n + 5

Question 15.
Write the terms of the sequence 5 × (1+6), 10 × (2+6), 15 × (3+6), 20 × (4+6) in the form :
first term 5 × 1(1 + 6),
second term 5 × 2(2 + 6).
Write its algebraic expression
Answer:
5 x (1+6), 10 x (2 + 6), 15 x (3 + 6), 20 x (4 + 6), … this sequence can be written as 5 x 1(1 + 6), 5 x 2 (2 + 6), 5 x 3 (3 + 6), 5 x 4 (4 + 6). Algebraic expression = 5 x n (n + 6)
i.e. x_n = 5 n² + 30n.

Worksheet 3

Question 16.
Write eighth terms of an arithmetic sequence using the numbers given below. (22, 15, 18, 4, 10, 14, 6, 12)
Answer:
6, 10, 14, 18, 22, ……………

Question 17.
Write the missing terms in the arithmetic sequence given below

Question 18.
The difference between 12th and 8th term of an arithmetic sequence is 20. Find the common difference.
Answer:

Question 19.
The tenth term of an arithmetic sequence is 65 and its 15^th term is 80. Is 200 a term of this sequence?
Answer:
’To get the 15^th term from the 10th term, we must add the common difference 5 times.
5 times of common difference = 80 – 65 = 15
Common difference (d) = 15/5 = 3
Dividing 65 by 3 we get the remainder as 2.
Dividing 200 by 3 we get the remainder as 2.
Therefore 200 is a term of this sequence.

Question 20.
The 20^th term of an arithmetic sequence is 64 and its 21^th term is 70. Can the difference between two terms 46? Why?
Answer:
20^th term = 64
21^th term = 70
Common difference = 70 – 64 = 6 Difference between any two terms of this sequence will be a multiple of 6.
Dividing 46 by 6 we get the remainder as 4, So difference between any two terms of the sequence will not be 46.

Question 21.
The angles of a quadrilateral are in an arithmetic sequence. The largest angle is 150°. Find other angles.
Answer:
Sum of four angles of a quadrilateral = 360° Angles are in arithmetic sequence, so first angle + 3 x common difference (d) = 150°
i.e., f + 3d= 150

The four angles are f, 70, f + 2d, 150 .
\(\mathrm{f}+2 \mathrm{d}=\frac{150+70}{2}=110\)
Common difference = 110 – 70 = 40
First angle = 70 – 40 = 30
So, the four angles are 30, 70, 110, 150

Question 22.
What will be the remainder on dividing a term of the sequence 3n + 7 by its common difference?
Answer:
Dividing 3n + 7 by 3 we get the remainder as 1.

Worksheet 4

Question 23.
Write the algebra of the following sequences and its sum of n terms
1.5, 10, 15, 20, ………..
2. 6, 11, 16, 21, ………
3. 4, 9,14,19, ……….
4. 3, 8, 13, 18, …………..
Answer:
1. 5, 10, 15, 20,
First term f = 5
Common difference = d = 10 – 5 = 5
General form xn = dn + (f – d)
= 5n + (5 – 5) = 5n

2. 6, 11, 16,21, …………..
First term f = 6

3. 4, 9, 14, 19, ………….
First term f = 4

4. 3, 8, 13, 18, …………
First term f = 3

Worksheet 6

Question 24.
Write the sequence of the squares of all odd numbers. What is its algebra?
Answer:
Odd numbers = 1, 3, 5, 7, 9, 11, ………
f= l, d = 3 – 1 =2, f – d= 1 – 2 = – 1
Algebraic form of odd numbers
x_n= dn+ (f – d) = 2n – 1
The sequence of the squares of all odd numbers = 1, 9, 2 5, 49,……….
Algebraic form = (2n -1 )²

Question 25.
Write the sequence formed by the number of diagonals from a vertex of a triangle, a quadrilateral, a pentagon etc. What is its algebra?
Answer:
Diagonal drawn from one vertex of a triangle = 0 = 3 – 3 = 0
Diagonal drawn from one vertex of a quadrilateral = 4 – 3 = 1
Diagonal drawn from one vertex of a pentagon = 5 – 3 = 2
Diagonal drawn from one vertex of a n sided polygon = n – 3
Sequence of number of diagonals 0, 1, 2, 3, 4, ……….
Algebraic expression x_n = n – 3

Question 26.
Write the sequence of the number of diagonals in a quadrilateral, pentagon, hexagon etc. What is its algebra?
Answer:

Question 27.
Can the difference between any two terms of an arithemetic sequence having common difference 6 be 2016? Justify your answer.
Answer:
\(\frac{\text { Differences of terms }}{\text { Common difference }}=\frac{2016}{6}=336\)
It is a natural number, so the difference between any two terms of an arithemetic sequence having common difference 6 be 2016.

Question 28.
Write algebra of the sum of the sequence 6n + 5. Can the sum 2000 ? Why?
Answer:
nth term = 6n + 5
1st term = 6 + 5 = 11
Sum of n terms = n/2 (11 + 6n + 5)
= n/2 (6n + 16) = n(3n + 8) = 3n² + 8n
Algebraic form of the sum = 3n² + 8n
To check wheater the sum is 2ooo
3n² + 8n = 2000
i.e., 3n² + 8n – 2000 = 0

It is not a natural number. So 2000 will not be a sum.

Worksheet 7

Question 29.
Prove that the squares of the sequence 1, 3, 5,……..belongs to that sequence itself.
Answer:
Odd-numbered arithmetic sequence will be 1, 3, 5, …
Here the arithmetic sequence have common difference 2. Their squares are also odd numbers. Therefore the squares of the sequence 1, 3, 5, …belongs to that sequence itself. nth term = 2n – 1.

Question 30.
In an arithmetic sequence having terms natural numbers, prove that if one of the terms is a perfect square, it will have more that this as the perfect square term.
As we know when a definite number of common difference of an arithmetic sequence is added to a term we get another term of the same sequence. If n2 is a perfect square term, add (2n + d) times d to n². n² + (2n + d) x d = (n + d)². This is nothing but a perfect square term.

Question 31.
If the angles of a right triangle are in an arithmetic sequence, find them by making suitable equations.
Answer:
Let angles be f – d, f, f + d f – d + f + f+d = 180
3f = 180, f = 60, f + d = 90, d = 30 Angles are 30, 60 and 90

Question 32.
If ten times tenth term of an arithmetic sequence is equal to fifteen times fifteenth term, find 25^th term. Calculate the product of first 25^th terms.
Answer:
10 × (f + 9d) = 15 × ( f + 14 d)
⇒ 10 f + 90 d = 15 f + 210 d
⇒ 5 f+ 120 d = 0
⇒ f + 24 d=0
25^th term = f + 24 d = 0
25^th term is 0. Product of 25 terms is 0.

Question 33.
Write the algebraic form of 1,4,7,10,… is 100 a term of this sequence. Why? Prove that the square of any term of this sequence belongs to that sequence.
Answer:
1,4, 7, 10,…
f = 1, d = 3
x_n= dn + (f – d) = 3n + (1 – 3) = 3n – 2
When 4 is divided by 3 we get remainder as 1 When 100 is divided by 3 we get remainder as 1 So, 100 is a term of the arithmetic sequence.
Square = (3n – 2)² = 9n² – 12n + 4 (9n² – 12n + 4) + 3
When 9n² – 12n + 4 is divided by 3the remainder is 1.
So square of the term also in the sequence.
∴ The square of any term of this sequence belongs to that sequence.

Question 34.
Write the sequence obtained by adding two adjacent consecutive terms in counting numbers starting from 1.
Write the algebraic expression of this sequence.
[Score: 3, Time: 4 minutes]
Answer:
Counting numbers:
1, 2, 3, 4, 5, … (1)
Sequence obtained by adding:
: 1+2, 2 + 3, 3 + 4, 4 + 5, … (1) Two adjacent consecutive terms
3, 5, 7, 9, …
Algebraic expression of above sequence :
n + (n+l) = 2n + 1 (1)

Question 35.
Consider circles, points on its circumference and chords as shown in the figure. Mark two points on the circle and draw a chord. Mark one more point and draw three chords. Continue this process by adding one more point each time. [Score: 4, Time: 7 minutes]
a. Write the number of chords in each figure as a sequence.
b. Write the algebraic expression of this sequence.
c. Find the number of chords in the 10^th figure.

Answer:
a. No. of chords in figure 1 = 1
No. of chords in figure 2 = 1 + 2 = 3
No. of chords in figure 3 = 1+ 2 + 3 = 6 (1)
Sequence of number of chords
= 1, 3, 6, 10, … (1)

b. No. of chords in figure n
\(=1+2+3+\ldots+n=\frac{n(n+1)}{2}\) (1)

c. No. of chords in the 10^th figure
\(=\frac{10 \times 11}{2}=55\) (1)

Question 36.
A pattern is formed using sticks of equal length as shown below:
[Score: 4, Time: 9 minutes]

a. Write the sequence of number of sticks used in each figure.
b. Write the sequence of number of squares and rectangles in each figure,
c. Write the algebraic expression in the above two sequences.
d. Find the number of sticks and squares in the 10^th figure.
Answer:
a. No of sticks in figure 1 = 1 + 3 = 4
No of sticks in figure 2
= 1 + 3 + 3 = 1+2 × 3 =7 No of sticks in the figure 3
= 1 + 3 × 3 = 10 No of sticks in the figure 4
= 1+4 × 3 = 13 Sequence of number of sticks
= 4, 7, 10, 13, … (i)

b. No of squares and rectangles in the figure 1 = 1
No of squares and rectangles in the figure 2 = 2 + 1 = 3
No of squares and rectangles in the figure 3 =3 + 2+ 1=6
No ofsquares and rectangles in the figure 4 = 4 + 3 + 2 + 1 = 10
Sequence of squares and rectangles = 1, 3, 6, 10, …. (1)

c. No of sticks in the nth figure 1
= 1+ n × 3 = 3n + 1
No of squares and rectangles in the nth figure = 1 + 2+ 3….. +n = \(=\frac{n(n+1)}{2}\) (1)

d. No of sticks in the 10th figure = 3 × 10 + 1 = 31
No of squares and rectangles in the 10th figure = \(\frac{10 \times 11}{2}=55\) = 55 (1)

37. Consider an arithmetic sequence with common difference 6 and 7^th term 52. Find the 15^th term of the arithmetic sequence. Is it possible, to get a difference of 100 between any two terms of this sequence?
[Score: 3, Time: 5 minutes]
Answer:
15^th term can be obtained by adding 8 times the common difference to the 7^th term.
x₁₅ = x₇ +8d (1)
= 52 + 8 × 6 = 100 (1)
The difference between any two terms of an Arithmetic sequence will be a multiple of com-mon difference. 100 can’t be the difference be-tween any two terms of this sequence, since it is not a multiple of 6. (1)

Question 38.
Consider an arithmetic sequence whose 7^th term is 34 and 15^th term is 66.
[Score: 3, Time: 5 minutes]
a. Find the common difference,
b. Find the 20th term.
Answer:
a. 15^th term can be obtained by adding 7^th term ‘ and 8 times the common difference.
x₁₅ = x₇ + 8d (1)
66 = 34 + 8d
8d = 66 – 34 = 32
d = \(\frac { 32 }{ 8 }\) = 4 (1)

d. 20^th term can be obtained by adding 15^th term and 5 times the common difference.
x₂₀ = x₁₅ + 8d (1)
= 66 + 5 × 4 = 86

Question 39.
Consider an arithmetic sequence \(\frac { 17 }{ 7 }\), \(\frac { 20 }{ 7 }\), \(\frac { 23 }{ 7 }\), ……….
a. Write the algebraic expression of the sequence.
b. Write the sequence of counting numbers in the above given sequence. Is the newly obtained sequence an arithmetic sequence. [Score:. 4, Time: 6 minutes]
Answer:
a. Common Difference = \(\frac { 20 }{ 7 }\) – \(\frac { 17 }{ 7 }\) = \(\frac { 3 }{ 7 }\) (1)
Algebraic expression of the sequence

Question 40.
Let the algebraic expression of an arithmetic sequence be 5n + b. If there is no perfect square in this sequence, find the counting number less than 5 that can be the value of ‘b’. [Score:4, Time: 5minutes]
Answer:

Any perfect square when divided by 5 leaves remainder 0, 1,4 So if remainder is 2 and 3 then it will not be a perfect square.
If there is no perfect square in the sequence, the only possibility for value of ‘b’ less than 5 is 2 and 3. (1)

Arithmetic Sequences Exam Oriented Questions & Answers

Short Answer Type Questions (Score 2)

Question 41.
—, 18, —, 28 are four consecutive terms of an arithmetic sequence. Fill in the blanks.
Answer:
18 + 2d = 28
2d = 28 – 18 = 10
d = 5
Sequence 13, 18, 23, 28.

Question 42.
98 is a term of the arithmetic sequence having common difference 7. is 2016 a term of this sequence. Why?
Answer:
If (2016 – 98) is a multiple of common difference 7, then 2016 is a term of the arithmetic sequence.
2016 – 98 = 1918
1918 is a multiple of 7.
(Quotient = 274, Remainder = 0)
∴ 2016 is a term of this sequence.

Question 43.
For the arithmetic sequence 6, 12, 18,………
a. What is the common difference?
b. Find the 10th term?
Answer:
a. Common difference = 6
b. 10th term = f + 9d = 6 + (9 × 6)
= 6 + 54 = 60

Question 44.
The algebraic form of an arithmetic sequence is 3 + 2n.
a What is the first form of the sequence?
b. What will be the remainder if the terms of the sequences are divided by 2?
Answer:
a. Firstterm = 3 + 2 x 1 = 5
b. d = 2 (coefficient of n be the common difference)
The remainder divided by 2 = 1

Short Answer Type Questions (Score 3)

Question 45.
The nth term of an arithmetic sequence is an = 5 – 6n. Finditssumofnterms?
Answer:

Question 46.
Consider the multiples of 7 in between 100 and 500.
a. What are the first and last numbers?
b. How many terms are there in the sequence?
Answer:
a. Firstteim=100 – 2 + 7 = 105
Last term = 500 – 3 = 497

Question 47.
For an arithmetic sequence 22, 26, 30, …………..
a. What is the common difference?
b. Will 50 be a term of this sequence? Why?
c. Can the difference between any two terms of this sequence be 50? Justify your answer?
Answer:
a. Common differenced = 26 – 22 = 4
b. \(\frac{50-22}{4}=7\)
So, 50 is a term of this sequence,
c. 50 is not a multiple of 4. So, 50 is cannot be a difference of two terms.

Question 48.
Find the smallest 3 digit number which is the multiple of 6. Find the sum of all the three-digit numbers which are the multiple of six.
Answer:
Smallest 3 digit number which is the multiple of 6 = 102,
Highest =996
Common difference = 6
Arithmetic series : 102, 108, …………….. 996
Number of three digit numbers

Question 49.

Answer:

Question 50.
Find the sum of first 24 terms of the list of numbers whose nth term is given by a_n = 3 + 2n.
Answer:

Long Answer Type Questions (Score 4)

Question 51.
a. Write the arithmetic sequence with first term 2 and common difference 3.
b. Check whether 100 is a term in this sequence.
c. Check whether the difference of any two terms of this sequence will be 2015.
d. Find the position of the term 125 in this sequence.
Answer:
a. f = 2
d. = 3
Sequence is 2, 5, 8, 11, ……….

b. If (100 – 2) is not a multiple of common difference 3, then 100 is not a term of the arithmetic sequence.
c. 2015 is not a multiple of common difference 3, so 2015 will not be the difference of any two terms of this sequence.

d. x_n= 3n – 1
3n-1 = 125
3n= 126
n = 42

Question 52.
When 60 added to the first term of an A.P, we get its 11^th term. Which number should be added to its first term to get the 19^th term? Can 75 be the difference between any of the two terms of this sequence?
Answer:
Differences between first term and 11th term is = 60
(11 – 1 = 10)
10 times of common difference = 60
Common difference = 60/10 = 6
Differences between first term and 19^th term is = 18 x common difference = 18 x 6 = 108 When 108 is added to the first term, we get 19^th
term. The difference between two terms in an A.P is the multiple of common difference. 75 is not the multiple of common difference 6. So 75 cannot be the difference between two terms in the series.

Question 53.
i. What is the sum of first 20 natural numbers?
ii. The algebraic form of an arithmetic sequence is 6n + 5. Find the sum of first 20 terms of this sequence?
Answer:

Question 54.
23rd term of an arithmetic sequence is 32. 35^th term is 104. Then
a. What is the common difference?
b. Which is the middle term of first 35 terms of this sequence?
c. Find the sum of first 35 terms of this sequence.
Answer:

b. Middle term offirst 35 terms = \(\sqrt { 4 } \)
c. 18 th term = X₁₉ = X₂₃ – 5d = 32 -5 × 6 = 32 – 30 = 2
Sum of 35 terms = 18^th term × 35 = 70

Question 55.
The difference between the 15^th term and the 5^th term of an A.P is 40. Which number is to be added to its 12 Answer:
The difference between the 15^th term and the 5^th term = 40
i.e., ten times an of a common difference = 40
Common difference = 40/10 = 4
When we add 8 x common difference we will get 20^th term.
The difference between the 12^th term and the 20^th term = (20 – 12) × common difference
= 8 x 4 = 32
The difference between the first term and the 21^th term = (21 – 1) × Common difference
= 20 × = 80

Long Answer Type Questions (Score 5)

Question 56.
Consider the arithmetic sequence 171, 167, 163,………..
i) Is ‘0’ is a term of this sequence? Why?
ii) How many positive terms are in this sequence?
Answer:
Arithmetic senes : 171, 167, 163 …………….

∴ 0 will not be a term of the sequence

Question 57.

a. How many numbers are there in the 30th row of this number pyramid?
b. Which is the last number in the 30th row?
c. Which is the first number in the 30th row?
d. What is the sum of all terms in the first 30 rows?
Answer:
a. Total numbers in each sequence can be written as 1, 3, 5, …. xn = 2_n – 1
Numbers in the 30th row = 2 x 30 – 1 = 59

b. Last number in the first row = 12 = 1
Last number in the second row = 22 = 4
Last number in the third row = 32 = 9
Last number in thew 30th row = 302 = 900

c. Number of terms in the 30th row = 59
Last number in the 30th row = 900
First number in the 30th row + 58d = Last term in the 30th row
First number in the 30th row + 58 x 1 = 900 First number in the 30th row = 900 – 58 = 842

d. The sum of all terms in the first 30 rows = \(\begin{array}{l}{\frac{900 \times 901}{2}} \\ {=450 \times 901=405540}\end{array}\)

Arithmetic Sequences Memory Map