Class 9

You can Download Real Numbers Questions and Answers, Activity, Notes, Kerala Syllabus 9th Standard Maths Solutions Chapter 10 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Maths Solutions Chapter 10 Real Numbers

Real Numbers Textual Questions and Answers

Textbook Page No. 160

Hss Live Guru 9th Maths Kerala Syllabus Question 1.
Find the distance between the two points on the number line, denoted by each pair of numbers given below: .

Answer:

Real Numbers Class 9 Kerala State Board Question 2.
Find the midpoint of each pair of points in the first problem.
Answer:



Real Numbers Class 9 State Board Kerala Syllabus Question 3.
The part of the number line between the points denoted by the numbers 1/3 and 1/2 is divided into four equal parts. Find the numbers de-noting the ends of each such part.
Answer:

So the portion between 1/3 and 1/2 is divided into 4 equal parts the points are

Textbook Page No. 164

Hsslive Guru 9th Maths Kerala Syllabus Question 1.
Find those x satisfying each of the equations below:
i. |x – 1| = |x – 3|
ii. |x – 3| = |x – 4|
iii. |x + 2| = |x – 5|
iv. |x| = |x + 1|
Answer:
i. \(|x-1|=|x-3|\)
\(|x-1|\) means distance between x and 1
\(|x-3|\) means distance between x and 3. Therefore the distance from x to 1 and 3 are equal.
x is in between 1 and 3 , that is x is the midpoint of 1 and 3.
\(x=\frac{1}{2} \times(1+3)=\frac{1}{2} \times 4=2\)

ii. \(|x-3|=|x-4|\)
The distance from x to 3 and 4 are equal.
∴ x is in between 3 and 4 , that is x is the midpoint of 3 and 4.

iii. \(\begin{array}{l}{|x+2|=|x-5|} \\ {|x+2|=|x-(-2)|}\end{array}\)
The distance from x to -2 and 5 are equal.
∴ x is in between -2 and 5 , that is x is the midpoint of -2 and 5.

iv. |x| = |x + 1|

|x + l] = ]x – (-l)|
The distance from x to -1 and 0 are equal.
∴ x is in between -1 and 0, that is x is the midpoint of -1 and 0.

9th Std Kerala Syllabus Maths Solutions  Question 2.
Prove that if 1 < x < 4 and 1 < y < 4, then |x – y| < 3.
Answer:
1 < x < 4 , possible values of x = 2, 3 1 possible values of y = 2, 3
|x – y| = |2 – 2| = 0 < 3
|x – y| = |3 – 2| – 1 < 3
|x – y| = |2 – 3| = |-l| = l < 3
|x – y| = |3 – 3| = 0 < 3
From this if 1 < x < 4, 1 < y < 4, then |x – y| < 3.

Hss Live Guru Maths 9 Kerala Syllabus  Question 3.
Prove that if x < 3 and y > 7, then |x – y) > 4 .
Answer:
Since x < 3, the maximum value of x is less than 3. Since y > 7 the minimum value of y is greater than 7.
Then the difference between the minimum values of y and the maximum value of x is greater than 7 – 3 = 4.
i.e., |x – y| > 4

9th Standard Maths Notes Kerala Syllabus  Question 4.
Find two numbers x, y such that
\(|x+y|=|x|+|y|\)
Answer:
1. If x = 3, y =7, then
| x + y | =  |13 + 71| = 3 + 7 = 10
|x| + |y| = |3| + |7| = 3 + 7 = 10
|x + y| = |x| + |y|

2. If x = -6, y =-9 , then
|x + y| = |-6 + -9|
= l-15| = 15

Class 9 Maths Chapter 10 Kerala Syllabus Question 5.
Are there numbers x,y such that |x + y| < |x| + |y| ?
Answer:
x = 5, y = -3

|x+y| = |5 + -3| = |5 – 3| = |2| = 2

|x| = |5| = 5

|y| = |-3| = 3 |x| + |y| = 5 + 3 = 8

Among x, y, if one is a positive number and the other is a negative number, then
| x + y | < |x| + |y|. Question 6. Are there numbers x, y such that |x + y| > |x| + |y| ?
Answer:
No

Real Numbers 9th Standard Kerala Syllabus Question 7.
What are the numbers x, for which
|x – 2| + |x – 8| = 6
Answer:
\(|x-2|\) means distance between x and 2.
x can be to the right or left side of the number 2.
\(|x-8|\) means distance between x and 8
x can be to the right or left side of the number 8.
If we add the distance between x and 2 with x and 8 we get 6.
When x is to the left side of 2, when we add the distance of x to 2 and 8 we get a number which is greater than 6.
Distance between 2 and 8
= |2 – 8| = | -6| = 6
The difference between 2 and 8 is 6, so the position of x is between 2 and 8.
That is the value of x is in between 2 and 8 including 2 and 8.
i. e., 2 ≤ x ≤ 8

Std 9 Maths Kerala Syllabus Question 8.
What are the numbers x, for which
|x – 2| + |x – 8| = 10 ?
Taking x as different numbers, what all numbers do we get as
|x – 2| + |x – 8| ?
Answer:
|x – 2| + |x – 8| = 10, |x – 2| – |x – 8| = 10
x – 2 + x – 8 = 10
x = 10
( x – 2) – (x – 8) = 10
There is no number x which satisfies this.
– ( x- 2) + (x – 8) = 10
There is no number x which satisf¬ies this.
-(x – 2) – (x – 8) =10
-x + 2 – x + 8=10
-2x = 0
= x = 0
i.e., x = 0, 10

If x = 1
|x – 2| + |x – 8| = |1 – 2| + |1 – 8|
= | -1 | + | -7| = 1 + 7 = 8

If x = 2
|x – 2| + |x – 8| = |2 – 2| + |2 – 8|
= |0| + | -6| = 0 + 6 = 6

If x = 3
|x – 2| + |x – 8| = |3 – 2| + |3 – 8|
= | -1| + | -5| = 1 + 5 = 6

If x = 4
|x – 2| + |x – 8| = |4 – 2| + |4 – 8|
= | -2| + | -4| = 2 + 4 = 6

If x = 5
|x – 2| + |x – 8| = [5 – 2| +15 – 8|
= | -3| + | -3| = 3 + 3 = 6

If x=6
|x – 2| + |x – 8| = |6 – 2| + |6 – 8|
= | -4|+ | -2| = 4 + 2 = 6

If x = 7
|x – 2| + |x – 8| = |7 – 2| + |17 – 8|
= |-5| + |-1| = 5 + 1 = 6
i.e., |x – 2| + |x – 8| = 6 when 2 < x < 8

If x=9
|x – 2| + |x – 8| = |9 – 2| + |9 – 8|
= |7| + |1| = 7 + 1 = 8

If x= 11
|x – 2| + |x – 8| = |11 – 2| + |11 – 8|
= |9| + |3|= 9 + 3 = 12
…………. etc
Taking x as different numbers, different numbers get as |x – 2| + |x – 8|.

Real Numbers Exam Oriented Questions and Answer

Hss Live Guru Class 9 Maths Kerala Syllabus Question 1.
a. Which number indicates the point located 7 units to left of 5 on the number line?
b. Write down the numbers located 2 units apart from these points.
Answer:
a. The point which is 7 units to the left of 5 = 5 – 7 = -2
b. There is one point 2 unit apart to the left and right side of -2 .
The point which is 2 units to the left of -2 = -4
The point which is 2 units to the right of -2 = 0

9th Standard Maths Textbook Kerala Syllabus Question 2.
How many numbers are on the number line which are 11 units apart from 5. What are they ? Calculate the distance between the numbers.
Answer:
There is one point 11 unit apart to the left and right side of 5.
Point on the left side = 5 + -11 = -6
Point on the right side = 5 + 11 = 16
Distance between them = 16 – (-6) = 16 + 6 = 22 unit

Chapter 10 Maths Class 9 Kerala Syllabus Question 3.
Find the distance between the num-bers given below on the number line.
i. 4, 6
ii. 3, -2
iii. -5, -8
iv. 3/5, 5/6
Answer:
i. Distance = |6 – 4| = |2| = 2

ii. Distance = |3 – (-2)| = |3 + 2| = |5| = 5

iii. Distance = |-5 -(-8)| =|-5 + 8|
= |+3| = 3

9th Std Maths Notes Kerala Syllabus Question 4.
The endpoints of one side of an equilateral triangle are -2 and 4 on the number line. Calculate the area and perimeter of the triangle.
Answer:
The distance between -2 and 4 = 4 – (-2) = 6
One side of a equilateral triangle = 6 unit Perimeter of an equilateral triangle
= 3 × 6 = 18 unit Area of an equilateral triangle

Class 9 Maths Kerala Syllabus Question 5.
The number x is on the number line, then find x if |x – 10| = – |x – 4|.
Answer:
The distance between x to 10 and 4 are same. Therefore x is the midpoint of 4 and 10.

Real Numbers Class 10 State Syllabus Kerala Syllabus Question 6.
The quadrilateral ABCD is a square. A and B the points denote -2 and 3 on the number line.
a. Draw the square ABCD on the number line
b. Calculate the perimeter of the square ABCD.
Answer:

b. AB = |-2 – 3| = 5 unit
Perimeter = 5 × 4 = 20 unit

Question 7.
If |a + 1| = |a + 5 |, |b – 2| = |b – 6|,
|a – x| = |b -x|, then find x.
Answer:
|a + 1| = |a + 5|
Since the point, ‘a’ represented on the num¬ber line is at a the equal distance from -1 and -5.
That is ‘a’ is midpoint of -1 and -5.

|b – 2| = |b – 6|
Since the point, ‘a’ represented on the number line is at a equal distance from 2 and 6.
That is ‘a’ is midpoint of 2 and 6

|x – a| = |x – b| therefore point x is the
midpoint of a = -3 and b = 4

Question 8.
Find the value of x.
a. |x – 2| = |x – 10|
b. |x + 3| = |x – 7|
Answer:
a. The distance between x to 2 and 10 are same. Therefore x is the midpoint of 2 and 10.

b. |x + 3| = |x – (-3)|
The distance between x to -3 and 7 are same. Therefore x is the midpoint of -3 and 7.

Question 9.
If the difference between two numbers is 4. If one number is 8. What values will the other number have?
Answer:
Consider other number as x, then
|x – 8| = 4
i.e., x – 8 = 4 or 8 – x = 4
x = 12 or x = 4
Other number as 12 or 4.

Question 10.
Find the distance between the pairs of numbers given below.
a. 2, 8
b. -2, 8
c. -2, -8
Answer:
a. Distance between 2 and 8
= |x – y| = |2 – 8| |-6| = 6

b. Distance between -2 and 8
= |-2-8| = |-10| = 10

c. Distance between -2 and -8
= |-2 – (-8)| = |-2 + 8| = |6| = 6

Question 11.
Draw the number line. Mark the position of numbers given below.

Answer:

Question 12.
The number x is located to the right side of 3 on the number line, if
|x – 3|= 0, then
a. What is the value of x ?
b. Calculate |x – 3| + |x – 9|.
Answer:
a. |x – 3| = 0
-(x – 3) = 0 or x – 3 = 0
-x = -3 or x = 3
x=3

b. |x – 3| + |x – 9|
= |3 – 3| + |3 – 9|
= |0| + |-6 | = 6

Question 13.
If |x – 5| = 10, then find the value of x .
Answer:
The distance between x and 5 is 10.
If x is to the right side of 5 , then x = 5 + 10 = 10
If x is to the left side of 5 , then
x = 5 – 10 = -5

Question 14.
If |x + 2| = |x – 8|, then find the value of x ?
Answer:
|x + 2| = |x – (-2)|
The distance of x to -2 and 8 are equal.
Therefore position of x is between -2 and 8.

Question 15.
If |x – 1| =|x – 3|, then find the value of x.
Answer:
Distance from 1 and 3 to x are same.
Therefore position of x is between 1 and 3.
Position of x = \(\frac { 1 + 3 }{ 2 }\) = \(\frac { 4 }{ 2 }\) = 2

Question 16.
Using the common form of rational numbers, prove that the sum, difference, product and quotient of any two rational numbers is again a rational number.
Answer:
i. If a, b, c and d are natural numbers,
\(\frac { a }{ b }\), \(\frac { c }{ d }\) are rational numbers.

Sum and product of all-natural numbers are always natural numbers. So here the numerator and denominator are natural numbers.
That is if x and y are natural numbers
then this can be expressed as x/y.
Therefore the sum is rational

ii. Difference:
a, b, c and d are natural numbers

Hence it is rational.

iii. Product:
a, b, c and d are natural numbers

Hence it is rational.

iv. Quotient:
a, b, c and d are natural numbers

Hence it is rational.

Question 17.
Prove that the product of any irrational number and non – zero rational number is an irrational number.
Answer:
Let a be the irrational number and b the rational number and a × b = c, that is to prove that c is irrational.
Consider c is a rational number, a × b = c
From this we get b = c/a
That is b = \(\frac { One rational }{ other rational }\) = One rational
If we take b as irrational here we get b as a rational number. This is a wrong decision. This is due to our wrong assumption that is c is a rational number. The assumption that c is a rational number is a wrong one.
∴ c is an irrational number.
That is the product of any irrational number and non – zero rational number is an irrational number.

Question 18.
Give an example of two different irrational numbers whose product is a rational number.
Answer:
3√2 and 5√2 are irrational numbers
3√2 × 5√2 = 3 × 5 × √2 × √2 = 3 × 5 × 2 = 30 is a rational nummber. More examples
1. √3 × 7√3 = 4 × 7 × 3 = 84
2. 2√5 × 5√5 = 2 × 5 × 5 = 50
3. √20 × √5 = √20 × √5 = √100 = 10

You can Download Protectors of Biosphere Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Biology Solutions Part 1 Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Biology Solutions Chapter 1 Protectors of Biosphere

Protectors of Biosphere Textual Questions and Answers

Kerala Syllabus 9th Standard Biology Notes Chapter 1 Global Warming

Global warming is the increase of earth’s average surface temperature due to effect of greenhouse gases. Deforestation, atmospheric pollution, burning of fossil fuels, etc.
1) At least do not ruin the life of those trees, depending on whom we live”.
2) ‘Plant together let’s make the world greener.’
3) It is our duty to save environment duties.
4) Plant trees: for our future.
5) Plant trees: It is the only way to prevent Global warming.

Kerala Syllabus 9th Standard Biology Notes Question 1.
………… is the result of deforestation, atmospheric pollution, etc.?
Answer:
Global warming.

Protectors Of Biosphere Class 9 Notes Kerala Syllabus Question 2.
What is Global warming?
Answer:
Global warming is the increase of earth’s average surface temperature due to effect of greenhouse gases.

Class 9 Biology Notes Kerala Syllabus  Question 3.
What are the causes of Global warming?
Answer:
Deforestation, atmospheric pollution, burning of fossil fuels, etc.

9th Biology Notes Kerala Syllabus Question 4.
“Trees are essential for protecting our nature and life.” Prepare slogans highlighting this idea to protect our environment?
Answer:

• At least do not ruin the life of those trees, depending on whom we live”.
• ‘Plant together let’s make the world greener.’
• It is our duty to save environment duties.
• Plant trees: for our future.
• Plant trees: It is the only way to prevent Global warming.

Pigments In Leaves

9th Class Biology Notes Kerala Syllabus Question 5.
Which of the pigments given below is the main pigment that performs photosynthesis?
a) Chlorophyll a
b) Chlorophyll b
c) Xanthophyll
d) Carotene
Answer:
a) Chlorophyll a

Kerala Syllabus 9th Standard Biology Notes Malayalam Medium Chapter 1 Question 6.
Complete the equation related to photosynthesis

Answer:

Protectors Of Biosphere Class 9 Kerala Syllabus Question 7.
Complete the word relations.
a) Light reaction: Grana
Dark reaction: ……………
b) Green plants: land
……………..: Ocean
c) …………..: bluish green
Carotene: yellowish-orange
Answer:
a) Stroma
b) Algae
c) Chlorophyll a

9th Class Biology Notes Chapter 1 Kerala Syllabus Question 8.
………. is the main source of energy on the earth
Answer:
Sunlight

Kerala Syllabus 9th Standard Biology Notes Pdf Question 9.
Which group of living things can absorb solar energy directly?
Answer:
Plants

9th Class Biology Chapter 1 Kerala Syllabus Question 10.

Have you noticed Veena’s doubt’? Write down your inference?
Answer:
More number of chloroplasts are seen on the upper part of leaf than on the lower part. That is why the lower part of the leafless green in color.

9th Class Biology Chapter 1 Notes Kerala Syllabus Question 11.
Odd one out, give reason Carotene, xanthophyll, chlorophyll a
Answer:
Chlorophyll a. Others accessory pigments.

Hsslive Guru Class 9 Biology Kerala Syllabus Question 12.
………. part of leaf are seen chloroplasts more in number
Answer:
Upper part

Kerala Syllabus 9th Standard Biology Notes Malayalam Medium Question 13.
Photosynthesis in green plants taken mainly in
Answer:
leaves

Hss Live Guru 9 Biology Kerala Syllabus Question 14.
How many chloroplasts will be there in one square millimeter of a leaf?
Answer:
At an average 5 lakh chloroplasts

9th Class Biology First Chapter Kerala Syllabus Question 15.
Identify the missing parts in the picture.

Answer:
a) Grana
b) Stroma

Question 16.
The fluid part of the chloroplast is
Answer:
Stroma

Question 17.
What are the pigments seen in the grana?
Answer:
Chlorophyll a, Chlorophyll b, Carotene, and Xanthophyll are seen in the grana.

Question 18.
Describe the structure of chloroplast?
Answer:
The chloroplast is an organelle with a double-layered membrane. The fluid part of the chloroplast is the stroma. The membranous sacs arranged on above other is the grana. The membrane-bound structures joining the grana are the stroma lamellae, Pigments that can absorb sunlight are seen in the grana.

Question 19.
What is the function of accessory pigments?
Answer:
Chlorophyll b, Carotene, and Xanthophyll are accessory pigments. They are seen in the grana. They can absorb light and transfer it to chlorophyll a.

Question 20.
How do carbon dioxide and water reach the leaves for photosynthesis?

Answer:

The Chemistry Of Photosynthesis

Question 21.
Differentiate between Light reaction and Dark reaction; Prepare a table?
Answer:

Question 22.
Find out whether the statements given below are true or false?
a) Dark reaction occurs in the grana.
b) Light reaction stops when availability of light decreases.
Answer:
a) False
b) True

Question 23.
Light reaction occurs in the ………….
Answer:
grana of chloroplast

Question 24.
Dark reaction occurs in the …………..
Answer:
Stroma of chloroplast.

Question 25.
Complete the flow chart related to photosynthesis.

Answer:
a) Light is required
b) Glucose is produced
c) Takes place in the grana

Question 26.
During light reaction, light energy is converted to ……… and stored as ………..
Answer:
Chemical energy, ATP molecules

Question 27.
Water splits into & in the light reaction
Answer:
Hydrogen & Oxygen

Question 28.
Define Dark reaction?
Answer:
Sunlight is not utilized in this phase that occurs in the stroma of chloroplast. In this process, hydrogen is added to C02 to form glucose using the energy of ATP molecules.

Question 29.
Illustrate an experiment to prove that plants released oxygen during the time of photosynthesis.
Answer:

Question 30.
Dark reaction is also known as …………
Answer:
Calvin cycle

Question 31.
……….. is known as the energy currencies of the cell.
Answer:
ATP

Question 32.
Expansion of ATP =
Answer:
Adenosine Tri Phosphate

Question 33.
About 70-80% of oxygen in the atmospheric air is contributed by ………..
Answer:
algae in the ocean

Question 34.
What is the contribution of Melvin Calvin?
Answer:
Melvin Calvin explained the various stages of the formation of glucose during photosynthesis.

Question 35.
Prepare the equation showing the process of photosynthesis?
Answer:

After Photosynthesis

Question 36.

Isn’t doubt genuine? What happens to the glucose formed as a result of photosynthesis? Explain your opinion?
Answer:
Glucose is easily soluble in water. It can’t be stored in plant body. Therefore plants store glucose in the form of insoluble starch in leaves. Plants utilize starch as a source of energy for life activities and to prepare substances required for growth. Starch is converted to sucrose and it transported through phloem to various plant parts and to store in different forms such as starch, protein, fat, fructose and sucrose.

Question 37.
Prepare a flow chart showing the chemical changes of glucose after its formation
Answer:

Question 38.
In which form is starch transported to various parts of the plant body for storage?
Answer:
Sucrose

Question 39.
Complete the table?

Answer:

Question 40.
Plants store glucose in the form of ………….. in leaves.
Answer:
Insoluble starch.

Question 41.
Write the examples of economically important plant products.
Answer:
Coffee, Rubber, Pepper, Cocoa

Ocean At Par With Land

Question 42.
‘Ocean – an ecosystem’ – Prepare a flow chart.
Answer:
Sunlight + CO₂ → Algae & other aquatic Organisms → Photosynthesis
Fish and other aquatic organisms → Birds & Terrestrial organisms

Question 43.
How does pollution in the ocean affect the organisms?
Answer:
Ocean is polluted due to sewage, toxic chemicals from industries, land runoff, large scale oil spills, ocean mining, Littering, etc. As a result of ocean pollution, there are several negative impacts such as
1) Effect of toxic wastes on marine animals.
2) Disruption to the cycle of coral reefs
3) Depletes oxygen content in water
4) Failure in the reproductive system of sea animals.
5) Effects on food chain.
6) It affects human health.
In this way ocean pollution adversely affects the organisms.

Plants – Earth’S Wealth

Question 44.
Do we obtain food only from producers?
Answer:
No plants are the cheapest and effective natural mechanism for the purification of air. It offers great service to the biological world by absorbing CO₂ from the
atmosphere and releasing oxygen.

Question 45.
The service rendered by plants for the sustenance of the living world is unique comment on this statement.
Answer:
Plants serve as the cheapest, effective and natural means for the purification of air. Plants also have a . major role in the mitigation of natural disasters. Mangrove forests help in controlling Tsunami to some extend. Bamboo forests, reed, vetiver, lemongrass, etc protect the river banks from collapsing during flood. Trees and bushes in mountains and hills prevent soil erosion and landslides.

Protectors of Biosphere Additional Question and Answers

Question 46.
“Trees are essential for protecting our nature and life.” Prepare slogans highlighting this idea to protect our environment?
Answer:

• At least do not ruin the life of those trees, depending on whom we live”.
• Plant together let’s make the world greener.’
• It is our duty to save environment duties.
• Plant trees: for our future.
• Plant trees: It is the only way to prevent Global warming.

Question 47
What is photosynthesis?
Answer:
Photosynthesis is a process used by plants and other organisms to convert light energy into chemical energy that can be later released to fuel the organisms’ activities.

Question 48.
What do you mean by greenhouse gases? Give examples?
Answer:
Greenhouse gases are gases that contribute to the greenhouse effect by absorbing infrared radiation. Carbon dioxide and chloroform carbons are examples of greenhouse gases.

Question 49.
“They kill Good Trees To put out Bad Newspapers” -comment.
Answer:
There are wide range of cutting trees all over the world. Many such instances are for making newspapers. In the same newspapers, we proclaim that trees should not be cut down. It is really a controversy. Similarly, bad news and wrong messages are passed onto people & generations by spoiling the life-supporting trees. Therefore
protection of trees is more important.

Question 50.
“Sale of AC’ has increased in Kerala due to severe heat” – Comment on this pews paper statement?
Answer:
The temperature in Kerala has gone up to the record figure this year. This has happened due to the practice of cutting down trees and pollution. To escape from the heat people started purchasing AC instead of planting trees. The excessive use of AC will again increase heat in the atmosphere by increased emission of chloroform carbon. This will finally intensify the magnitude of Global warming.

Question 51.
Plants give out approximately ………. of oxygen when they use one tonne CO₂
Answer:
118 kg

Question 52.
Why the plants are called the lungs of the earth? Explain.
Answer:
Plants are the cheapest and effective natural mechanism for the purification of air. They absorb carbon dioxide from the atmosphere & release oxygen. It is estimated that plants give out approximately 118 kilograms of oxygen when they use one tonne CO₂. As the plant cover on the earth decreases, this recycling mechanism stops and air pollution becomes severe. That is why the plants are called the lungs of the earth.

You can Download Statistics Questions and Answers, Activity, Notes, Kerala Syllabus 9th Standard Maths Solutions Chapter 13 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Maths Solutions Chapter 13 Statistics

Statistics Textual Questions and Answers

Textbook Page No. 193

Statistics Class 9 Kerala Syllabus Question 1.
The weight of 6 players in a volleyball team are all different and the average weight is 60 kilograms.
i. Prove that the team has at least one player weighing more than 60 kilograms.
ii. Prove that the team has at least one player weighing less than 60 kilograms.
Answer:
i. Total weight of 6 players = 60 × 6 = 360 kg
The team contains players having weights 60, less than 60 or greater than 60. If the weights of all the players are less than 60, the average will also be less than 60. This is not possible. Therefore there will be at least one player having weight greater than 60.

ii. If the weight of all the players are greater than 60, the average will also be greater than 60. Therefore there will be at least one player having weight less than 60.

Hss Live Guru 9th Maths Kerala Syllabus Question 2.
Find two sets of 6 numbers with average 60, satisfying each of the conditions below:
i. 4 of the numbers are less than 60 and 2 of them greater than 60.
ii. 4 of the numbers are greater than 60 and 2 of them less than 60.
Answer:
Total sum = 60 × 6 = 360
i. 20, 30, 40, 50, 100, 120

ii. 5, 15, 70, 80, 90, 100
Other ways are also possible.

Kerala Syllabus 9th Standard Maths Notes Question 3.
The table shows the children in a class, sorted according to the marks they got for a math test.

Calculate the average marks of the class.
Answer:
Total number of children is 45. Repeated addition can be written as multiplication.

Average mark = \(\frac { Total }{ Number }\) = \(\frac { 297 }{ 45 }\) = 6.6

Class 9 Maths Solutions Kerala Syllabus Question 4.
The table below shows the days in a month sorted according to the amount of rainfall in a locality

What is the average rainfall per day during this month?
Answer:

The average of the rain fall per day during that month = \(\frac { Total rain fall }{ Number of days }\)
= \(\frac { 1545 }{ 30 }\) = 51.5mm

Hsslive Guru 9th Maths Kerala Syllabus Question 5.
The details of rubber sheets a farmer got during a month are given below.

i. How many kilograms of rubber did he get a day on average in this month?
ii. The price of rubber is 120 rupees per kilogram. What is his average income per day this month from selling rubber?
Answer:


i. Average Quantity of rubber per day = \(\frac { 381 }{ 30 }\) = 12.77kg
ii. . If the price is Rs. 120 per kg, then average incomeperday = 12.7 × 120 = Rs.1524

Textbook Page No. 197

Kerala Syllabus 9th Standard Maths Solutions Question 1.
Find different sets of 6 different numbers between 10 and 30 with each number given below as mean:
i. 20
ii. 15
iii. 25
Answer:
i. The mean of 6 numbers is 20.
ie; sum = 6 x 20= 120 (Write 3 pairs with sum 40)
i.e., 15, 25, 18, 22, 19, 21

ii. The mean of 6 numbers is 15
i.e; sum =6 x 15=90
(Write 30 pairs with sum 3)
12, 18, 13, 17, 14, 16

iii. Mean is 25
sum = 25 x 6 = 150
(Write 50 pairs with sum 3)
22, 28, 23, 27, 24, 26

Hss Live Guru Class 9 Maths Kerala Syllabus Question 2.
The table below shows the children in a class, sorted according to their heights.

What is the mean height of a child in this class?
Answer:

Mean height = \(\frac { Total height }{ No of children }\)
= \(\frac { 7892 }{ 50 }\) = 157.84 cm

Hss Live Class 9 Maths Kerala Syllabus Question 3.
The teachers in a university are sorted according to their ages, as shown below.

What is the mean age of a teacher in this university?
Answer:

Mean age = \(\frac { Total age }{ No of persons }\)
= \(\frac { 2775 }{ 70 }\) = 39.64

Kerala Syllabus 9th Standard Maths Question 4.
The table below shows children in a class sorted according to their weights.

The mean weight is calculated as 26 kilograms. How many children have weights between 23 and 25 kilograms?
Answer:
Let’s prepare the table for finding the mean by considering the number of children in the group 23 to 25, as ‘x’.

Mean weight = 26 kg

i.e; the number of children having weight between 23 to 25 is 7.

Statistics Exam oriented Questions and Answers

Kerala Syllabus Std 9 Maths Solutions Question 1.
The details of rubber sheets got for a month by a farmer are given in the table.

During this month he got an average of 10 sheets per day. If so in how many days did he get 11kg per day ?
Answer:

Let ‘x’ be the number of days in which he got 11 kg rubber sheet.

∴ He got 11 kgs of sheets for 5 days.

Kerala Syllabus 9th Standard Maths Notes Malayalam Medium Question 2.
In a factory, there are workers belonging to four categories. The average income and the number of workers in each category are given. What is the mean income when all the workers in the four categories are combined?

Answer:

Mean income = \(\frac { 320000 }{ 40 }\)
= Rs. 8000

Kerala Syllabus 9th Standard Maths Solution Question 3.
The daily wages of 10 workers in a factory are given below.
400, 350, 450, 500, 400, 500, 350, 500, 350, 450
If one more person is joined, the mean becomes Rs. 450. What is the daily wage of the new person?
Answer:
Total wages of 10 workers = 4250 Total wages of 11 workers= 11 × 450 = Rs. 4950
Wage of the 11th person = 4950 – 4250

Kerala State Class 9 Maths Solutions Question 4.
Find 10 different numbers between 10 and 30 whose mean is 20.
Answer:
Given mean is 20
Sum 20 × 10 = 200.
We have to find 10 different numbers whose sum is 200 (for this find 5 pairs of sum 40)
(15, 25) (16, 24) (17, 23) (18, 22) (19, 21)
The numbers are 15, 16, 17, 18, 19, 21, 22, 23, 24, 25

9th Maths Notes Kerala Syllabus Question 5.
A table categorizing the workers in an office on the basis of their salary is given below.

Find the mean of salary.
Answer:

Mean income = \(\frac { 349500 }{ 15 }\)
= R.s 23300

Kerala Syllabus 9th Standard Notes Maths Question 6.
i. Find the mean of natural numbers from 1 to 100.
ii. What is the mean of even numbers from 1 to 100? What is the mean of odd numbers?
iii. What is the difference between the means of the first 100 even numbers and odd numbers?
iv. What is the difference between the means of the first 200 even numbers and 200 odd numbers?
Answer:
Sum of the natrural numbers from 1 to n = \(\frac n{ n + 1 }{ 2 }\)


Sum of the first 100 odd numbers
= 1002 = 100 × 100

In general, the difference between the means of n even numbers and n odd numbers is always 1.

Question 7.
A table tabulating the players in a cricket team on the basis of their age is given below.

Calculate the mean age of the players?
Answer:

Mean age of players = 306/12 = 25.5

You can Download अकाल में सारस Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Hindi Solutions Unit 4 Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Hindi Solutions Unit 4 Chapter 1 अकाल में सारस (कविता)

अकाल में सारस Textual Questions and Answers

अकाल में सारस विश्लेषणात्मक प्रश्न

अकाल में सारस कविता का आशय Kerala Syllabus 9th प्रश्ना 1.
धान की सूखी पत्तियों की गंध’ –से किसका आभास होता है?

उत्तर:
धान के पौधे गर्मियों में सूखते हैं। धान की सूखी पत्तियों की गंध से गर्मियों का आभास मिलता है।

Akal Mem Saras Hindi Poem Kerala Syllabus 9th प्रश्ना 2.
सारसों ने जल भर कटोरे को क्यों न देखा होगा?

उत्तर:
सारसों के लिए कटोरे भर का जल पर्याप्त नहीं है। वे ताल-तलैयों की तलाश कर रहे थे। उनके लिए प्राकृतिक जल स्रोतों की ज़रूरत थी। इसलिए वे जल भर कटोरे को न देखा होगा।

अकाल में सारस Summary Kerala Syllabus 9th प्रश्ना 3.
सारसों ने जाते-जाते शहर की ओर क्यों मुड़कर देखा होगा?

उत्तर:
शायद सारसों को वह जगह पसंदीदा थी। लेकिन उनकी आशा के विरुद्ध वहाँ कोई जलस्रोत न देखा। वे निराश लौटने लगे। वहाँ के लोगों के प्रति उनके मन में दया या घृणा की भावना थी। दया इसलिए कि ये लोग कितने अभागे हैं। घृणा शायद इसलिए कि इन्होंने प्राकृतिक जलस्रोतों का ह्रास किया है।

अकाल में सारस Additional Questions and Answers

अकाल में सारस आशयग्रहण के प्रश्न

Akaal Mein Saras Summary In Hindi Kerala Syllabus 9th प्रश्ना 1.
वे देर तक करते रहे शहर की परिक्रमा’ -सारस किसकी तलाश में शहर की परिक्रमा करते होंगे?

उत्तर:
सारस देश-देशांतर में जानेवाले पक्षी हैं। वे ऋतुओं के बदलने के अनुसार एक देश से दूसरे में चले जाते हैं। यहाँ सारस पानी की तलाश कर रहे हैं।

अकाल में सारस कविता की व्याख्या Kerala Syllabus 9th प्रश्ना 2.
बुढ़िया ने क्या सोचकर पानी लाकर रखा होगा?

उत्तर:
सारसों को देखकर बुढ़िया ने सोचा कि वे पानी की तलाश में हैं। इसलिए बुढ़िया ने कटोरे में पानी लाकर रखा।

अकाल में सारस Grammar

अकाल में सारस व्याकरण के प्रश्न

अकाल में सारस Summary In Hindi Kerala Syllabus 9th प्रश्ना 1.
तुलना करें :

i. तीन बजे दिन में / आ गए वे

ii. वे दिन में तीन बजे आ गए।

iii. कवितांश की संरचना और वाक्य की संरचना का अंतर पहचानें।

iv. कविता को गद्य में बदलने पर शब्दों के क्रम में क्या परिवर्तन आया है?

उत्तर:
वाक्य की संरचना कर्ता, कर्म और क्रिया का पालन किया है। लेकिन कविता की संरचना में यह क्रम नहीं है। कविता की सुंदरता को बढ़ाने के लिए कभी-कुभी शब्दों के क्रम में आवश्यक परिवर्तन करते हैं।

Hss Live Guru 9th Hindi Kerala Syllabus प्रश्ना 2.
इन पंक्तियों को गद्य की संरचना में बदलकर लिखें।

उत्तर:

9th Class Hindi Notes Kerala Syllabus प्रश्ना 3.
लेख लिखें :

1. प्रकृति के जलस्रोतों का नाश होता जा रहा है।
2. साथ ही पर्यावरण का संतुलन भी बिगड़ रहा है।

उत्तर:
अकाल में सारस’ कविता मनुष्यों के बुरे व्यवहार की संकेत करनेवाली कविता है। प्राकृतिक संसाधन केवल मनुष्यों के लिए नहीं है। पशु-पक्षी, मछलियाँ और दूसरे प्राणी भी इनके हकदार है। दरअसल ऋतु बदलने पर सारस पक्षी पानी आदि की तलाश में एक देश से दूसरे देश में चले जाते हैं। लेकिन यहाँ कहीं पानी नहीं है। सारसों को तो पता तक नहीं था कि लोग उन्हें सारस कहते हैं। यहाँ तो पानी और धान की पत्तियाँ सूख गई हैं। उनकी गंध हवा में हैं। यही गंध सारसों के डैनों से नीचे झर रही है। एक बुढ़िया अपने आँगन में जलभरा कटोरा रख देती है।

लेकिन सारसों ने न तो बुढ़िया को देखा। क्योंकि वे जलस्रोतों की तलाश में आए थे। उनकी निगाहों में दया थी या घृणा, हम समझ नहीं पाते। दया शायद इसलिए कि सारस तो उड़कर कहीं और चले जाएँगे, जहाँ पानी होगा। किंतु ये दयनीय लोग जो अपना पानी तक नहीं बचा पाए, अपने खेत-खलिहान और घर-बार छोड़कर, कहाँ और कैसे जाएँगे? और घृणा इसलिए कि अपना पानी तो नष्ट किया ही हमारा भी नष्ट कर दिया। क्योंकि पर्यावरण नष्ट करने के ज़िम्मेदार तो मनुष्य ही हैं।

अकाल में सारस Summary in Malayalam and Translation

Akal प्रश्ना 1.
बरसों बीते
बादलों को इधर
बरसे नहीं।
ये पंक्तियाँ किस हालत की ओर ।
संकेत करती है?

उत्तर:
पानी की कमी एक बड़ी समस्था हो.
पर्यावरण में हुए बदलाव से बारिश
की मात्रा कम होती जा रही है। इस
हालत की ओर ये पंक्तियाँ संकेत देती है 

अकाल में सारस शब्दार्थ

You can Download The World of Carbon Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Chemistry Solutions Part 2 Chapter 7 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Chemistry Solutions Chapter 7 The World of Carbon

The World of Carbon Textual Questions and Answers

Activities In The Text

Kerala Syllabus 9th Standard Chemistry Notes Question 1.
Find the position of carbon in the periodic table and the complete the table.

Answer:

Hss Live Guru 9th Chemistry Kerala Syllabus Question 2.
What are allotropes?
Answer:
Different forms of the same element having different physical properties but with same chemical properties are known as Allotropes and this phenomenon is called Allotropy.

9th Class Chemistry Chapter 7 Notes Kerala Syllabus Question 3.
What are the characteristics of diamond.
Answer:

• Very hard
• Transparent
• Not a conductor of electricity
• High thermal conductivity
• High refractive index

9th Chemistry Notes Kerala Syllabus  Question 4.
Write the uses of diamond
Answer:

• It is used to make ornaments.
• It is used for cutting glass.

9th Standard Chemistry Kerala Syllabus Question 5.
Explain the structure of diamond with Figure?
Answer:
In diamond each carbon atom is linked by covalent bond with four other carbon atom surrounding it. This strong bonding is responsible for the hardness of diamond. Due to the absence of free electron in this crystal structure, it does not conduct electricity

Chemistry 9th Class Notes Kerala Syllabus Question 6.
Write the important properties of Graphite?
Answer:

• It is Gray in color
• It is good conductor of electricity
• It is a smooth solid
• It does not vapourize
• It is Lustrous
• It is nonvolatile

Hss Live Guru Chemistry 9th Kerala Syllabus Question 7.
Give reason Graphite is used as a lubricant to reduce friction in machine parts?
Answer:
It is a smooth solid. It does not vapourize. It also has high melting point

Hsslive Guru Chemistry Class 9 Kerala Syllabus  Question 8.
Explain the structure of Graphite?
Answer:

In Graphite each carbon atoms is united with three surrounding carbon atom through covalent bond and forms a sheet-like structure. These sheets or layers are stacked one above the other to form three dimensional structure.

Each layer is made up of hexagons there is no covalent bonding between the layers. These layers are held together by weak Vander Waal’s physical forces. Hence these layers can slide over one another.

Hsslive Guru 9 Chemistry Kerala Syllabus Question 9.
What are amorphous carbon?
Answer:
Cock, coal, charcoal, bone charcoal, etc. are non-crystalline allotropes of carbon. These are commonly called amorphous carbon.

Kerala Syllabus 9th Chemistry Notes Question 10.
Which carbon compound is present in the atmosphere?
Answer:
Carbon dioxide

Hss Live 9th Chemistry Kerala Syllabus Question 11.
Which carbon compound is produced by the combustion of fuels?
Answer:
Carbon dioxide

9th Class Chemistry Notes Kerala Syllabus Question 12.
How carbon dioxide is prepared of in the laboratory?
Answer:
Carbon dioxide is prepared in the laboratory by the action of marble piece or calcium carbonate with dil.
HCl CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Hsslive Guru Std 9 Chemistry Kerala Syllabus Question 13.
Which are the reactants are used to prepare carbon dioxide (CO₂) in the laboratory?
Answer:
CaCO₃ and HCl

Hsslive Guru 9th Chemistry Kerala Syllabus Question 14.
Complete the equation of the reaction
Answer:
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Chemistry Solutions Class 9 Kerala Syllabus Question 15.
How can we identify that the gas formed here is C02?
Answer:

1. Show a burning splinter into the gas. It gets extinguished.
2. lt turns lime water in milky

Hsslive Guru Chemistry 9 Kerala Syllabus Question 16.
Which properties of CO₂ are familiar to you?
Answer:
Extinguishes fire

Hsslive 9th Chemistry Kerala Syllabus Question 17.
What are the properties of carbon dioxide known to you? Tick the correct ones given below.
Answer:

• Coloured / Colourless✓
• Supports combustion / Does not support combustion✓
• Denser than air / Density higher than air✓
• Has a characteristic odor/ Odourless ✓

Kerala Syllabus 9th Standard Chemistry Question 18.
Whether the aqueous solution of CO₂ is acidic or alkaline?
Answer:
Acidic (carbonic acid, H₂CO₃)

Hsslive Class 9 Chemistry Kerala Syllabus Question 19.
Write the chemical formulae and uses of some carbonates.
Answer:
Na₂CO₃ – Soda ash
CaCO₂ – limestone, marble.

9th Std Chemistry Notes Kerala Syllabus Question 20.
How carbonates can be identified.
Answer:
Add some dilute HC; to the given salt. If a colorless gas that turns lime-water milky is formed, that salt will be a carbonate. The gas formed is CO₂.

Question 21.
The variety of carbon compounds is essential for the existence of life on the earth. Figure shows the ways in which carbon dioxide is exchanged over the earth. This is known as carbon cycle.

Name the process by which carbon dioxide is utilized by plants?
Answer:
Photosynthesis

Question 22.
What are the activities that increase the amount of carbon dioxide in air?
Answer:
Combustion, Respiration, Decomposition

Question 23.
Is the tremendous increase in the amount of CO₂ in atmosphere advantages?
Answer:
No, It causes greenhouse effect. Greenhouse effect is the reason for global warming.

Question 24.
What is green house effect?
Answer:
The process of increasing atmospheric temperature due to the increase in the amount of carbon dioxide in the atmosphere is called green house effect.

Question 25.
What is global warming?
Answer:
As a result of the green house effect, the average temperature of the earth and atmosphere increases. This is known as global warming.

Question 26.
Discuss the consequences of global warming in the following.
1. In ice layers
2. In ocean islands
3. In the field of agriculture
4. In the climate
Answer:

1. Ice layers melt and rivers will be flooded
2. The ocean islands will be submerged.
3. Fields of agriculture will be submerged
4. Causes rise in atmospheric temperature.

Question 27.
Suggest some measures to resist global warming effectively.
Answer:

1. Limit the use of fossil fuels
2. Plant more trees

Question 28.
Give the uses of carbon dioxide?
Answer:

• It is used in fire extinguisher
• It is used for the preparation of soda water/soft drink
• It is used for the preparation of washing soda, baking soda, etc.
• It is used for the preparation of fertilizers like urea
• Used for the preparation of carbogen which is used for artificial breathing. Carbogen is 95% 02 and 5% C02
• Dry ice is solid carbon dioxide, which is used in stage shows for creating special effects resembling clouds.

Question 29.
How carbon monoxide is formed?
Answer:
Carbon dioxide is the gas formed when carbon reacts with oxygen.
However, if the relative amount of carbon increases or that of oxygen decreases the reaction takes place as given below.
2C + O₂ → 2CO

• The gas thus formed is carbon monoxide. It is a poisonous gas.
• It is formed by the incomplete combustion of carbon in a limited supply of air.

Question 30.
How carbon monoxide becomes fatal?
Answer:
When carbon monoxide is inhaled, it reacts with the hemoglobin in the blood and forms carboxy hemoglobin. As a result, the oxygen-carrying capacity of blood decreases leading even to death.

Question 31.
What measures can be taken to avoid situations that produce carbon monoxide?
Answer:

• Avoid incomplete combustion
• Ensure the supply of oxygen.
• Service motor vehicles regularly

Question 32.
Explain the use of carbon monoxide?
Answer:

• Used as a gaseous fuel
• Industrially important gases like water gas (A mixture of CO and H₂) producer gas (A mixture of CO and N₂)
• Used as reducing agent in metallurgy

Question 33.
Write the chemical formulae of washing soda baking soda and marble.
Answer;
Washing Soda (Na₂CO₃.10H₂O)
Baking soda (NaHCO₃)
Marble (CaCO₃)

Question 34.
What is organic compounds?
Answer:
Organic compounds are carbon compounds except the inorganic compounds like CO,CO₂, carbonates, bicarbonates etc.

Question 35.
How many electrons are there in the outermost shell of carbon?
Answer:
Four

Question 36.
What is the valency of carbon?
Answer:
4

Question 37.
Complete the table given below.

Answer:

Question 38.
What are hydrocarbon?
Answer:
Hydrocarbons are compounds containing only car¬bon and hydrogen.

Question 39.
What is catenation?
Answer:
Catenation is the ability of the atoms of an element to combine among themselves. In comparison to other elements, the ability for catenation is very high for carbon.

Question 40.
What are characteristics responsible for the increase in number of carbon compounds?
Answer:

• Valency of carbon is 4
• Ability of catenation is high
• Single, double and triple bonds are possible be-tween carbon atoms.
• Carbon atoms combine together to form many straight-chain, ring or branched-chain compounds.

Let’S Assess

Question 1.
The names of some allotropes of carbon, their properties and uses are given in the table, but not in the correct order Match them suitably.

Answer:
Diamond:

• Manufacture of ornaments
• Transparent
• High refractive index

Graphite:

• Electric conductor
• Smooth
• Lubricant

Question 2.
Some statements related to carbon monoxide and carbon dioxide are given. Classify them correctly.
a) formed as a result of the incomplete combustion of carbon compounds’
b) aqueous solution shows acidic nature.
c) poisonous gas
d) used in fire extinguishers
e) an be used as a fuel
f) formed as a result of the complete combustion of carbon compounds.
g) can be prepared from carbonates and bicarbonates.
h) is a component of producer gas and water gas.
Answer:
a) carbon monoxide
b) Carbon dioxide
c) Carbon monoxide
d) Carbon dioxide
e) Carbon monoxide
f) Carbon dioxide
g) Carbon dioxide
h) Carbon monoxide

Question 3.
a) Write the chemical formula of calcium carbonate.
b) Which gas is formed when calcium carbonate reacts with acids?
c) What is the name of an aqueous solution of this gas?
Answer:
a) CaCO₃
b) Carbon dioxide
c) Soda water (carbonic acid)

Question 4.
Graphite, which is an allotrope of carbon, is a con-ductor of electricity. But diamond, another allotrope is not a conductor of electricity. Why?
Answer:
diamond each carbon atom is linked by covalent bond with four other carbon atom surrounding it. This strong bonding is responsible for the hardness of diamonds. Due to the absence of free electron in this crystal structure, it does not conduct electricity

Question 5.
Write the structure of a straight-chain and a a ring hydrocarbon having four carbon atoms.
Answer:

Extended Activities

Question 1.
Arrange the objects as shown in the figure and conduct the experiment. Based on your observations, what is the conclusion that you reach at?

Answer:
The bulb glows because pencil led or graphite is a good conductor of Electricity.

Question 2.
Lighted candles of different lengths are arranged in a trough as shown in the figure. Pour a saturated solution of sodium bicarbonate (baking soda) into the trough. Add a little vinegar to the solution. What do you observe? Give reasons for the observation.

Answer:
Candle with smallest height is extinguished first and the candle with highest height is extinguished last. The reason is the density of carbon dioxide is greater than that of air. Carbon dioxide is produced when baking soda react with vinegar.

Question 3.
Let’s make a fire extinguisher Arrange the apparatus as shown in the figure (a). Add the vinegar contained in a test tube to the sodium bicarbonate (baking soda) solution (figure b) by tilting the wash bottle. Introduce the resultant gas to a candle flame. Record your observation. What is your inference?

Answer:
The candle flame gets extinguished. The reason is when Baking soda reacts with vinegar to produce carbon dioxide.

You can Download The Biology of Movement Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Biology Solutions Part 2 Chapter 6 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Biology Solutions Chapter 6 The Biology of Movement

The Biology of Movement Textual Questions and Answers

The Biology Of Movement 9th Chapter 6 Question 1.
Are exercise and games necessary?
Answer:
On physical strength increases as we involve in interesting exercises such as games. Exercise reduces mental stress and helps us to work energetically.

9th Class Biology Notes Kerala Syllabus Chapter 6 Question 2.
Prepare a note on how exercise is beneficial to the body?
Biology Answer:
Exercise helps us in many ways.
It increases blood circulation all over the body. Cardiac muscles become strong. More capillaries are formed in muscles. Increases the efficiency of muscles. Stored fat is broken down thereby reduces obesity. Sweats more and so more waste is eliminated through sweat. Exchange of respiratory gases becomes more effective. Vital capacity increases.

Kerala Syllabus 9th Standard Biology Notes Chapter 6 Question 3.
‘Exercise helps our respiratory system more healthy.’ Do you agree with this statement? Substantiate your answer?
Answer:
Exercise increases our vital capacity and exchange of respiratory gases becomes more effective,

Involuntary Movements

Kerala Syllabus 9th Class Biology Notes Chapter 6 Question 4.
Prepare a table about voluntary movements and involuntary movements?

Answer:

Hss Live Guru 9th Biology Chapter 6 Question 5.
What do you mean by voluntary movements?
Answer:
The movements which occur according to our will is called voluntary movements.

9th Biology Notes Kerala Syllabus Chapter 6 Question 6.
Define involuntary movements?
Answer:
The movements which are not controlled by our will is called involuntary movements.

Types Of Muscles

9th Class Biology Chapter 6 Notes  Question 7.
Which muscle make voluntary movements possible?
Answer:
Skeletal muscle

Hss Live Guru Biology 9 Chapter 6 Question 8.
Striated muscle have cells
Answer:
Cylindrical

9th Standard Biology Notes Chapter 6 Question 9.
Where do you find smooth muscles in human body?
Answer:
Smooth muscles are seen in internal organs like the stomach, small intestine and in blood vessels.

Class 9 Biology Notes Kerala Syllabus Chapter 6 Question 10.
Smooth muscles are also known as
Answer:
Nonstriated muscles

Kerala Syllabus 9th Standard Biology Notes Pdf Chapter 6 Question 11.
Shape of smooth muscle is
Answer:
Spindle

Biology Notes For Class 9 Kerala Syllabus Chapter 6 Question 12.
Cardiac muscles are seen on the
Answer:
Walls of the heart

Kerala Syllabus 9th Standard Biology Solutions Chapter 6 Question 13.
…….. & ……… makes involuntary movements possible.
Answer:
Smooth muscle and cardiac muscle

Kerala Syllabus 9th Standard Biology Notes Malayalam Medium Question 14.
Skeletal muscle: Cylindrical shape
…………………..: Spindle shape
Answer:
Smooth muscle

Kerala Syllabus 9th Std Biology Solutions Chapter 6 Question 15.
Skeletal muscle: striated muscle
………………….: nonstriated muscle
Answer:
Smooth muscle

9th Standard Biology Question 16.
Different types of muscles and their characteristics. Prepare a table.
Answer:

Muscles Fatigue

Question 17.
What do you mean by muscle fatigue?
Answer:
When we are engaged in continuous and strenuous exercises, lactic acid accumulates in the muscles due to anaerobic respiration. This increases acidity in muscles and slows down the action of many enzymes associated with muscle contraction. As a result, muscles get exhausted and temporarily lose their power of contraction. This condition is called muscle fatigue.

Bones and Movement

Question 18.
The human skeleton system consists of bones.
Answer:
206

Question 19.
Based on the position, the human skeleton can be divided into ………. & …………
Answer:
Axial skeleton and appendicular skeleton.

Question 20.
Number of bones in the human skull is
Answer:
29

Question 21.
How many bones are there in human ribs?
Answer:
12 × 2 = 24

Question 22.
Hind limbs: 60 bones
………………: 26 bones
Answer:
Vertebral column

Question 23.
There are bones in the pelvic girdle of human beings.
Answer:
1 × 2 = 2

Question 24.
Muscles which contracts on folding the forelimb?
Answer:
Flexor muscle

Question 25.
Complete the illustration given below

Answer:

Question 26.
Muscle which contracts on extending the forelimb?
Answer:
Extensor muscle

Question 27.
Muscle which relaxes on folding the forelimb?
Answer:
Extensor muscle

Question 28.
Muscle which relaxes on extending the forelimb?
Answer:
Flexor muscle

Question 29.
What is antagonistic muscle?
Answer;
A movement is effective and complete when muscles work in unison with bones. In forelimb, when one muscle contracts the other muscle relaxes. These types of muscles which are opposite in action are called antagonistic muscles.

Question 30.
The basis of almost all the movements of the body is the proper functioning’ of ………….
Answer:
Antagonistic muscles

Joints And Movements

Question 31.
Complete the table of skeletal joints. Which shows its position and peculiarities.

Answer:

Structure Of Joint

Question 32.
…………. are the meeting place of two bones
Answer:
Joints

Question 33.
Explain the function of joints
Answer:
joints help in the movement of bones. Joints give more flexibility to bones to move. The nature of movements varies with the nature of joints.

Question 34.
…………. secretes synovial fluid
Answer:
Synovial membrane

Question 35.
…………. covers and protects the joints
Answer:
Capsule

Question 36.
…………. reduces friction between the bones
Answer:
Cartilage

Question 37.
What is the function of synovial fluid?
Answer:
Synovial fluid functions as a lubricant between the bones.

Question 38.
What is the function of ligaments?
Answer:
Ligaments ensure that bones are not displaced and holds them in position.

Question 39.
What are the functions of the skeletal system?
Answer:
Skeletal system facilitating movements, maintains posture, helps in hearing, protects our internal organs from damage, produces blood cells and maintains the mineral homeostasis.

Skeletal And Muscular Disorders

Rheumatic Arthritis:

• Caused by infection in joints, injuries, degenerative changes due to old age.
• Damage to cartilage
• Severe pain, incapable of moving joints

Dislocation:

• Displacement of bones in joints
• Damage to ligaments
• Severe pain oedema and difficulty in movements

Sprain:

• The stretching or breaking of ligaments
• Severe pain and oedema

Osteoporosis:

• A condition in which bones become brittle and cause fracture
• This may be due to the deficiency of calcium, defects in metabolic activities and deficiency of Vitamin D

Muscular dystrophy:

• A condition that leads to degeneration of muscles due to various reasons.
• Muscles become weak
• Generally, affect boys.

Skeleton Outside the Muscles

Locomotion Without Skeleton

Question 40.
Different types of movement in organisms which move without skeleton.
Answer:

Locomotion And Movement

Movement is the displacement occurring in any part of the body. Displacement of the entire body is called locomotion

The diversity of locomotion in animal world:

Do plants move?

Question 41.
Plants exhibit movements in response to various
Answer:
Stimuli

Question 42.
What are the various stimuli which cause movements in plants?
Answer:
Light, gravity, water, touch, chemicals, etc. are the various stimuli which cause movements in plants.

Question 43.
Complete the table relating to the plant movement

Answer:

Question 44.
Identify the type of movement, roots grow towards water
Answer:
Hydrotropism

Question 45.
What do you mean by tropic movement?
Answer:
If the direction of plant movements is in accordance with the direction of stimulus it is called tropic movements.

Question 46.
Is there any relation between stimulus and the direction of movement in mimosa?
Answer:
No. Hence it is nastic movement.

Question 47.
What do you mean by nastic movement?
Answer:
If the direction of plant movement is not in accordance with the stimulus, it is called nastic movement.

Question 48.
Write some examples for nasty plant movements from your surroundings.
Answer:
Movements of Mimosa pudica, prayer plant, venus flytrap, etc.

Let Us Assess

Question 1.
What is the reason for muscle fatigue?
a) Lack of glucose in muscle cells
b) Lack of oxygen in muscle cells
c) Increase in the level of carbon dioxide in muscle cells
d) cellular respiration ceases
Answer:
b) Lack of oxygen in muscle cells

Question 2.
Observe the figure and answer the following questions. What changes do you observe in the growth of root and stem in a plant, if it is kept stationary as shown in the figure for a few days? Why?

Answer:
Roots grow towards gravity and the stem grows against gravity.

Question 3.
Identify the odd one giving reason
a) Coconut trees near a river bend towards the river
b) Root of trees near a well grows towards the well
c) Leaves of touch-me-not fold when we touch it
d) Roots of plants grows towards gravity
Answer:
Leaves of touch-me-not fold when we touch it because it is a nastic movement.

You can Download Prisms Questions and Answers, Activity, Notes, Kerala Syllabus 9th Standard Maths Solutions Chapter 11 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Maths Solutions Chapter 11 Prisms

Prisms Textual Questions and Answers

Textbook Page No. 171

Kerala 9th Maths Solutions Chapter 11 Question 1.
The base of a prism is an equilateral triangle of perimeter 15 centimetres and its height is 5 centimetres. Calculate its volume.
Answer:
Base perimeter of an equilateral triangle = 15 cm.
Base edge = 15/3 = 5 cm
Height = 5 cm
Volume = Base area x Height

Kerala Syllabus 9th Standard Maths Notes Chapter 11 Question 2.
A hexagonal hole of each side 2 metres is dug in the school ground to collect rainwater. It is 3 metres deep. It now has water one metre deep. How much litres of water is in it?
Answer:
One side of the hexagon = 2 m
It is given that the depth of the pit is 3 metre but the water level is only in 1 metre. So we take the height as 1 metre.
Volume of water = Volume of the hexagonal prism = Area of the hexagon × height

Area of the regular hexagon is equal to six times the area of equilateral triangle.

Kerala Syllabus 9th Standard Maths Solutions Chapter 11 Question 3.
A hollow prism of base a square of side 16 centimetres contains water 10 centimetres high. If a solid cube of side 8 centimetres is immersed in it, by how much would the water level rise?
Answer:
When a solid cube is immersed into water, the volume of water is raised.
Sum of the volume of the water at first time and volume of the solid cube immersed equally to the product of base area and height of water level now.
Volume of the water at first = Base area × height
= 16 × 16 × 10 = 2560 cm3
Volume of the solid cube when edges are 8 cm
= 8 × 8 × 8 = 512 cm3
Height at first = 10 cm
Water level raised = 12 -10 = 2 cm

Textbook Page No. 174

Class 9 Maths Solution Kerala Syllabus Chapter 11 Question 1.
The base of a prism is an equilateral triangle of perimeter 12 centimetres and its height is 5 centimetres. What is its total surface area?
Answer:
Base perimeter of equilateral triangular prism = 12 cm
Base side = 12/3 = 4 cm
Lateral surface area of equilateral trian-gular prism = Base perimeter × Height
=12 × 5 = 60 cm²
Base area of equilateral triangular prism = \(\frac { √3 }{ 4 }\) × 4² = 4√3 = 6.92 cm²
Total surface area of equilateral triangular prism =60 + 2 × 6.92 = 73.84 cm²

Kerala Syllabus 9th Standard Notes Maths Chapter 11 Question 2.
Two identical prisms with right tri-angles as base are joined to form a rectangular prism as shown below:

What is total surface area?
Answer:
The sides of the rectangular prism are
Length = 12 cm
Breadth = 5 cm and
Height = 15 cm.
Base area = 2 × perimeter of rectangle
2 × 12 × 5 = 120 cm²
Lateral surface area = base perimeter × height
= 2 (length + breadth) × height = 2(12 + 5) × 15
= 2 × 17 × 15 = 510 cm²
Total surface area = 120 + 510 = 630 cm²

Class 9 Maths Notes Kerala Syllabus Chapter 11 Question 3.
A water trough in the shape of a prism has trapezoidal faces. The dimensions of a base are shown in this picture:

The length of the trough is 80 centimetres. It is to be painted inside and outside. How much would be the cost at 100 rupees per square metre?
Answer:
Now we add the area of two edge faces and three lateral faces ( 1 on bottom and 2 on sides).
In figure, AB = 50 cm, CD = 70 cm

EC = \(\frac { 70 – 50 }{ 2 }\) = 10 cm
BE = 24 cm
BC² = BE² + EC2 = 24² + 10²
= 576 + 100 = 676
BC = √676 = 26 cm.
Area of the trapezoidal faces of the water

= 12 × 120 = 1440 cm² Area of two trapezoidal faces
= 2 × 1440 = 2880 cm² Area of two rectangular faces on sides
= 2 × 80 × 26=4160 cm²
Area of rectangular face at bottom
=50 × 80 = 4000 cm²
Total Area= 2880 + 4160 + 4000 = 11040 cm² = 1.1 m²
Total area to be painted
= 2 × 1.1 =2.2 m²
Total cost = 2.2 × 100 = 220
= Rs. 220

Textbook Page No. 176

9th Standard Maths Notes Kerala Syllabus Chapter 11 Question 1.
The base radius of an iron cylinder is 15 centimetres and its height is 32 centimetres. It is melted and re-cast into a cylinder of base radius 20 centimetres. What is the height of this cylinder?
Answer:
Volume of the first cylinder
= π R2 H = π (15)² x 32 = 7200 π
Volume of the melted and recast cylinder = π r2h = π (20)² x h = 400 π h
Volume remains constant when melted.
400 π h=7200π
h = \(\frac { 7200π }{ 400π }\) =18
Height of the second cylinder = 18 cm

9th Class Maths Notes Malayalam Medium Chapter 11 Question 2.
The base radii of two cylinders of the same height are in the ratio 3:4. What is the ratio of their volumes?
Answer:
Let r₁ r₂ be the radii of two cylinders,
then r₁ : r₂ = 3 : 4,

Kerala Syllabus 9 Standard Maths Chapter 11 Question 3.
The base radii of two cylinders are in the ratio 2:3 and their heights are in the ratio 5:4.
i. What is the ratio of their volu-mes?
ii. The volume of the first cylinder is 720 cubic centimetres. What is the volume of the second?
Answer:
i. If the base radius of the first cylinder is 2r then base radius of the second one is 3r.
Height of the first cylinder is 5h and second cylinder is 4h.
Volume of the first cylinder
= Base area × height .
= π × 2r × 2r × 5h = 20 π r² h.
Volume of the second cylinder
= π × 3r × 3r × 4h=36 πr²h.
The ratio between the volumes
= 20πr²h : 36πr²h = 20 : 36.= 5 : 9

ii. The volume of the first cylinder is 720 cm3 is given. Let consider the volume of the second cylinder be x, then the ratio between the volumes is 5: 9
5 : 9 = 720 : x
5 × x = 9 × 720
x = 1296
Volume of the second cylinder = 1296 cm³

Textbook Page No. 178

Question 1.
The inner diameter of a well is 2.5 metres and it is 8 metres deep. What would be the cost of cementing its inside at 350 rupees per square metre?
Answer:
Area of the part cemented = Base perimeter × height

= 20π = 20 × 3.14 = 62.8cm
Total cost of cementing = 62.8 × 350 = Rs. 21980

Question 2.
The diameter of a road roller is 80 centimetres and it is 1.20 metres long:

What is the area of levelled surface, when it rolls once?
Answer:
Radius of roller = 40 cm
Length of roller (height) = 1.20 m = 120 cm
Area of leveled surface, when it rolls once
= Curved surface area of the roller
= 2 × π × Radius × Height
= 2 × 3.14 × 40 × 120 = 30144 cm² = 3.0144 m²

Question 3.
The base area and the curved surface area of a cylinder are equal. What is the ratio of the base radius and height?
Answer:
Curved surface area of the cylinder = Base perimeter x height = 2 πrh
Base perimeter = πr²
If it is equal, 2 π rh = πr²
2rh = r × r, 2h = r
i.e., The radius is twice the height.

Prisms Exam oriented Questions and Answers

Question 1.
A cylinder has base radius 4 cm and height 10 cm. Then find its lateral surface area.
Answer:
Perimeter of a circle with radius 4 cm = 2 × π × 4 = 8π cm.
Curved surface area of the cylinder = Base perimeter × height = 8 π × 10= 80 π cm²

Question 2.
The volume and base area of a square prism are 3600 cubic centimetre and 144 cm² respectively. What is its total surface area ?
Answer:
Volume of the prism = 3600 cm³
Base area × height = 3600 cm³
144 × height = 3600
height = 3600/144 = 25 cm
Surface area = = Base area + Lateral surface area
Lateral surface area = Base perimeter × height
= 12 × 4 × 25 = 1200 cm²
(Base area = 144, One side = √144 =12 cm)
Total surface area = 144 + 1200 = 1344 cm²

Question 3.
A cylinder has height 20 cm and base radius 4 cm. Then find its vol-ume.
Answer:
When the square of the radius is multiplied by π we get the area of the circle.
Base area of the cylinder
= π × 4² = 16 π cm²
The height of the cylinder is 20cm Volume = 16 π × 20 = 320 π cm³

Question 4.
The base edge of a square prism is 15 cm. The total surface area is 1950 m².
i. What is its height ?
ii. Calculate the volume.
Answer:
i .Base edge = 15 cm
Total surface area = Base area + Lateral surface area
= 2a² + 4ah (base egde is ‘a’)
1950 = 2 × 15² + 4 × 15 × h
1950 = 450 + 60h
60h = 1950 – 450 = 1500
h = 1500/60 = 25 cm

ii. Volume = a² × h = 15 × 15 × 25
= 5625 cm³

Question 5.
A square-shaped plot has length 24 m. A pond of 4 m length, 3 m breadth and 1.5 m height is dug here. If the sand dug out is levelled equally in the remaining area of the plot. Find the height of the levelled sand.
Answer:
Area of the land
= 32 × 24
= 768 m²

Area of the pond = 4 × 3 = 12m²
Area of the remaining places = 768 – 12 = 756 m².
Volume of the sand dug out = 4 × 3 × 1.5
Height of soil = \(\frac { Volume }{ Base area }\)
= \(\frac { 4 × 3 × 1.5 }{ 756 }\) = 0.0238m = 2.38 cm

Question 6.
Two Aluminium sheets of length 10 cm and breadth 6 cm are folded to make two cylindrical vessels. One is made by folding lengthwise and the other breadthwise, which will have maximum volume?
Answer:

Question 7.
A cylinder made of metal has radius 18 cm and height 40 cm. When this melts how many cylinders can be made of radius 2 cm and height 5 cm?
Answer:
Volume of the first c ylinder = π r²H = π (18)² × 40 = 12960 π cm³
Volume of the newly made cylinder = π (2)² × 5 = 20π cm³
Number of the newfy made cylinders
= \(\frac { 12360π }{ 20π }\) = 648 cylindres

Question 8.
If a wooden piec e is in the shape of square prism has base 12 cm and height 70 cm. What is the maximum volume of the cylinder that can be carved out of it?
Answer:
Diameter of the largest cylinder =12cm Radius = 6 cm, Height = 70 cm
Volume of the cylinder
= base area × height
= πr²h = π × 6 × 6 × 70 = 7912.8 cm³

Question 9.
A tin with length 40 cm width 20 cm and height 20 cm, which is in the shape of a quadrangular prism has sugar-filled in it. If the sugar is measured using a cylindrical vessel with radius 4 cm and height 15 cm, then how many times can the sugar be measured using the vessel?
Answer:
Volume of the quadrangular prism = 50 × 40 × 20
Volume of the cylinder = πr²h.
= 3.14 × 4 × 4 × 15
Number of times the sugar can be measured = \(\frac { 50 × 40 × 20 }{ 3.14 × 4 × 4 × 15 }\) = 53 times

Question 10.
A prism is made by cutting cardboard as shown in the figure.

a. What is the name of the prism?
b. What will be the area of the required cardboard for making the rectangle form?
Answer:
a.Triangular prism
b. Lateral area = base perimeter × height = (15 + 13 + 14) × 20 = 42 × 20
= 840 cm²

Question 11.
Two tins in the shape of cylinder has radii 15 cm and 10 cm respectively. The heights are 25 cm and 18 cm respectively.In the both cylinders, the ghee are filled. When it transferred into another cylindrical-shaped tin. If there is ghee in the bigger tin with height 30 cm. Then find its radius.
Answer:
‘Volume of the ghee in the first tin
= πr²h = π × 15 × 15 × 25 = 5625π cm³
Volume of the ghee in the second tin
= πr²h = π × 10 × 10 × 18 = 1800 π cm³
Total volume = 5625 π + 1800 π
=7425 π cm3 Volume of the ghee in the bigger tin
= πr²h = 7425 π
h =30
πr²h = πr² × 30 = 7425 π
r² = \(\frac { 7425 π }{ 30π }\) = 247.5
\(r=\sqrt{247.5}=15.73 \mathrm{cm}\)

Question 12.
The diameter of a water tank in the shape of a cylinder is 3 m and height 4 m. How many litres of water will the tank hold?
Answer:
Radius = 1.5 m, Height = 4m
Volume of the cylinder
= Base area × height = πr²h = 3.14 × 1.5 × 1.5 × 4 = 28.26 m³
= 28.26 × 100 × 100 × 100 cm³
= 28260000 cm³ = 28260000/1000 = 28260 liter

Question 13.
If a box in the shape of a square prism has length 25 cm, 20 cm width and 7-litre volume. What will be its height?
Answer:
Volume = 7 litre = 7000 cm³
lbh = 7000
25 × 20 × h = 7000
h = \(\frac { 7000 }{ 25 × 20 }\) = 14 cm

Question 14.
If the base length, width, height of a quadrangular prism are 37.5 cm, 18 cm, 40 cm respectively. Find the area of cube which has same volume as that of this prism.
Answer:
Volume of quadrangular prism = base area × height = 37.5 × 18× 40 = 27000 cm³
Volume of quadrangular prism = Volume of cube
∴ Volume of cube = 27000 cm³
Let one side of a cube be x, then a³ = 27000 ; a = 30cm
30 × 30 × 30 = 27000, Hence
Surface area of the cube = 6a² = 6 × 30 × 30 = 5400 cm²

Question 15.
The diameters of two-cylinder are in the ratio 2 : 3 and their heights in the ratio 5: 4. If the volume of the first cylinder is 400 cm³, then find the volume of the second cylinder.
Answer:
The diameters of two-cylinder are in the ratio 2 : 3 so the radius of the cylinders are also 2:3.
Assume that radius of first cylinder is 2r and second cylinder be 3r.
Assume that height of first cylinder is 5h and second cylider be 4h.
Volume of the first cylinder = πr2h
= π × 2r × 2r × 5h = 20 πr²
Volume of the second cylinder = πr2h
= π × 3r × 3r × 4h=36 πr²h
Ratio between volumes
= 20πr²h : 36πr²h = 20 : 36 = 5 : 9
Volume of the first cylinder is 400.
If the volume of second cylinder be x, then
5 : 9 = 400: x; 5x = 400 × 9
x = \(\frac { 400 × 9 }{ 5 }\) = 80 × 9 = 720 cm³

You can Download Pairs of Equations Questions and Answers, Activity, Notes, Kerala Syllabus 9th Standard Maths Solutions Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Maths Solutions Chapter 3 Pairs of Equations

Kerala Syllabus 9th Standard Maths Pairs of Equations Text Book Questions and Answers

Textbook Page No. 36

Do each problem below either in your head, or using an equation with one letter, or two equations with two letters:

Pairs Of Equations Class 9 Questions And Answers Kerala Syllabus Question 1.
In a rectangle of perimeter one metre, one side is five centimetres longer than the other. What are the lengths of the sides?
Answer:
Shortest side = x
Longest side = x + 5
Perimeter = 1 m = 100 cm
2(x + x + 5) = 100
2x + 5 = 50; 2x = 45;
x = 22.5
∴ Shortest side = 22.5
Longest side = 22.5 + 5 = 27.5

Pairs Of Equations Questions And Answers Kerala Syllabus Question 2.
A class has 4 more girls than boys. On a day when only 8 boys were absent, the number of girls was twice that of the boys. How many girls and boys are there in the class?
Answer:
Number of boys = x
Number of girls = x + 4
2(x – 8) = x + 4
2x – 16 = x + 4
2x – x = 4 +16; x = 20
∴ Number of boys = 20
Number of girls = 24

Pairs Of Equations Problems Kerala Syllabus Question 3.
A man invested 10000 rupees, split into two schemes, at annual rates of interest 8% and 9%. After one year he got 875 rupees as interest from both. How much did he invest in each?
Answer:
If one part is x then
the remaining part is 10000 – x
\(x\times \frac { 8 }{ 100 } +\left( 10000-x \right) \times \frac { 9 }{ 100 } =100\)
8x + 90000 – 9x = 87500
90000 – 87500 = x
2500 = x
one part = 2500 and
remaining part = 7500

Kerala Syllabus 9th Standard Maths Chapter 3  Question 4.
A three and a half metre long rod is to be cut into two pieces, one piece is to be bent into a square and the other into an equilateral triangle. The length of their sides must be the same. How should it be cut?
Answer:
Total length = 3½ m
Since the sides of a square and equilateral triangle are equal, all these 7 sides are equal.
∴ Length of one side
\( 3\frac { 1 }{ 2 } \div 7=\frac { 7 }{ 2 } \div 7=\frac { 1 }{ 2 } \)m
Length of the rod for the square
\(= 4\times \frac { 1 }{ 2 } \) = 2m
Length of the rod for the equilateral triangle = \(3\times \frac { 1 }{ 2 } \) = \(1\frac { 1 }{ 2 } \)m

Class 9 Maths Chapter 3 Kerala Syllabus Question 5.
The distance travelled in t seconds by an object starting with a speed of u metres/second and moving along a straight line with speed increasing at the rate of a metres/second every second is given by ut + \(\frac { 1 }{ 2 } \) at² metres. An object moving in this manner travels 10 metres in 2 seconds and 28 metres in 4 seconds. With what speed did it start? At what rate does its speed change?
Answer:
If t = 2
ut + \(\frac { 1 }{ 2 } \)at²= 10
2u + 2a= 10
u + a = 5 — (1)
If t = 4
4u + 8a = 28
u + 2a = 7 — (2)
from (1) and (2)
a = 2
∴ u = 3

Textbook Page No. 40

Pairs Of Equations Class 9 Questions And Answers Pdf  Question 1.
Raju bought seven notebooks of two hundred pages and five of hundred pages, for 107 rupees. Joseph bought five notebooks of two hundred pages and seven of hundred pages, for 97 rupees. What is the price of each kind of notebook?
Answer:
Cost of 200 page note book = x
Cost of 100 page note book = y
7x + 5y= 107 …………(1)
5x + 7y = 97 …………(2)
(1) × 5 \(\Rightarrow \) 35x + 25y = 535 …………(3)
(2) × 7 \(\Rightarrow \) 35x + 49y = 679 …………(4)
(4) – (3) \(\Rightarrow \) 24y = 144
y = \(\frac{144}{24} \) = 6
Substitute y = 6 in equation (1)
7x + 30 = 107; 7x = 77
x = \(\frac{77}{7} \) = 11
Price of the 200 pages notebook = Rs. 11
Price of the 100 pages notebook = Rs. 6

Pairs Of Equations Class 9 Extra Questions And Answers Question 2.
Four times a number and three times number added together make 43. Two times the second number, subtracted from three times the first give 11. What are the numbers?
Answer:
Let the first number = x and
the second number = y
4x + 3y = 43 …………(1)
3x – 2y = 11 …………(2)
(1) × 3 \(\Rightarrow \) 12x + 9y= 129 …………(3)
(2) × 4 \(\Rightarrow \) 12x – 8y = 44 …………(4)
(3) -(4) \(\Rightarrow \) 17y = 85; y = \(\frac{85}{17} \) = 5
Substitute y = 5 in equation (1)
4x + 3y = 43
4x + 15 = 43
4x = 43 – 15 = 28
∴ x = \(\frac{28}{4} \) = 7, y = 5
First number = 7
Second number = 5

9th Standard Maths Chapter 3 Kerala Syllabus Question 3.
The sum of the digits of two – digit number is 11. The number got by interchanging the digits is 27 more than the original number. What is the number?
Answer:
If the numbers are x and y
x + y = 11 …………(1)
10x + y + 27 = 10y + x
9x – 9y = -27
X – y = -3 …………(2)
(1) + (2) 2x = 8; x = 4
x + y = 11
4 + y = 11
y = 7
∴ Required number is 47.

Kerala Syllabus 9th Standard Maths Notes Question 4.
Four years ago, Rahim’s age was three times Ramu’s age. After two years, it would just be double. What are their ages now?
Answer:
Ramu’s present age = x
Rahim’s present age = y
4 years back,
Ramu’s age = x – 4
Rahim’s age = y – 4
3(x – 4) = y – 4
3x – 12 = y – 4
3x – y = 8 ……….(1)
After 2 years,
Ramu’s age = x + 2
Rahim’s age = y + 2
2(x + 2) = y + 2
2x + 4 = y + 2
2x – y = -2 ……….(2)
(1) – (2) \(\Rightarrow \) x = 10
3x – y = 8; 30 – y = 8; y = 22
x = 10, y = 22
Ramu’s present age = 10
Rahim’s present age = 22

Pair Of Equations Class 9 Kerala Syllabus Question 5.
If the length of a rectangle is in-creased by 5 metres and breadth decreased by 3 metres, the area would decrease by 5 square metres. If the length is increased by 3 metres and breadth increased by 2 metres, the area would increase by 50 square metres. What are the length and breadth?
Answer:
length = x; breadth = y
(x + 5)(y – 3) = xy – 5
xy – 3x + 5y – 15 = xy – 5
– 3x + 5y = + 10
3x – 5y = -10 ………..(1)
(x + 3)(y + 2) = xy + 50
xy + 2x + 3y + 6 = xy + 50
2x + 3y = 44 ………..(2)
(2) × 1 \(\Rightarrow \) 6x-10y = -20 ……….(3)
(3) × 2 \(\Rightarrow \) 6x + 9y = 132 …………(4)
(3)- (4) \(\Rightarrow \) -19y = -152
y = \(\frac{-152}{-19} \) = 8, 2x + 3y = 44
2x + 24 = 44; 2x = 20; x = 10
∴ x = 10, y = 8
Length of the rectangle = 10 m
Breadth of the rectangle = 8m

Textbook Page No. 42

Hsslive Guru Maths 9th Kerala Syllabus Question 1.
A 10 metre long rope is to be cut into two pieces and a square is to be made using each. The difference in the areas enclosed must be 1\(\frac{1}{4} \) square metres. How should it be cut?
Answer:
Length of one piece = x m
Length of other piece = (10 – x) m

∴ Rope is divided into 6 m and 4 m.

9th Maths Notes Kerala Syllabus Question 2.
The length of a rectangle is 1 metre more than its breadth. Its area is 3\(\frac{3}{4} \) square metres. What are its length and breadth?
Answer:
Length = x
Breadth = y
x = y + 1; x – y = 1
\(x y = 3 \frac{3}{4} =\frac{15}{4}\)
(x + y)² = (x – y)² + 4xy
1² + 4 × \(\frac{15}{4}\) = 1 + 15 = 16
x – y = 1; x + y = 4
2x = 5; x = 5/2 = 2.5
y=1.5
∴ Length = 2.5 m
Breadth = 1.5 m

Hsslive Maths Class 9 Kerala Syllabus Question 3.
The hypotenuse of a right triangle is 6\(\frac{1}{2} \) centimetres and its area is 7\(\frac{1}{2} \) square centimetres. Calculate the lengths of its perpendicular sides.
Answer:
The perpendicular sides are x and y. Given that,

From (3) and (4)
2x = 24/2 = 12
∴ x = 6
6 – y = 7/2
∴ y = 5/2 = 2.5
∴ Perpendicular sides = 6 and 2.5

Kerala Syllabus 9th Standard Maths Pairs of Equations Exam Oriented Text Book Questions and Answers

Kerala Syllabus 9th Standard Notes Maths Question 1.
There are some oranges in a bag. When 10 oranges more added in the bag; the numbers become 3 times of the oranges initially had. Then how many oranges were there in the bag initially.
Let the number of oranges initially taken = x
X + …….. = 3X
3X – X = ……..
2X = ……..
X = ……… /2 = ……..
Answer:
x + 10 = 3x; 3x – x = 10 ; 2x = 10;
x = \(\frac{10}{2} \) = 5

Kerala Syllabus 9th Standard Maths Notes Pdf Question 2.
A box contains some white balls and some black balls. The number of black balls is 8 more than the number of white balls. The total number of balls is 4 times the number of white balls. Find out the number of white balls and the number of black balls.
Number of white balls = x
Number of black balls = ……… + 8
Total number of balls = ……. ‘ x;
i. e. (x) + (x + 8) = ………. x;
2x + 8 = ……. x;
8 = …… x – 2x
= …… x; x = 8/ …….
white balls = ……….
black balls = ……… + 8 = ……..
Answer:
Number of white balls = x
Number of black balls = x + 8
Total number of balls = 4 ‘ x;
(x) + (x + 8) = 4x ; 2x + 8 = 4x;
8 = 4x – 2x = 2x ; x = \(\frac{8}{2} \) = 4
white balls = 4
black balls = 4 + 8 = 12

Pairs Of Equations Questions Kerala Syllabus Question 3.
The sum of two numbers is 36 and the difference is 8. Find the numbers.
Let x, y be the numbers
x + y = 36
x – y = 8
(x + y) + (x – y) = ……. + ……
2x = …..
x = …… /2 = ……..
x – y = 8
……. – y = 8
…….. – 8 = y
Answer:
x + y = 36; x – y = 8
(x + y) + (x – y) = 36 + 8
2x = 44; x = \(\frac{44}{2} \) = 22
x – y = 8; 22 – y = 8;
22 – 8 = y; y = 14
numbers 22, 14

Kerala Syllabus 9th Std Maths Notes  Question 4.
The cost of 2 pencils and 5 pens is Rs 17, two pencils and 3 pens of the same rate is Rs 11. Find out the prices of a pencil and a pen.
Let the price of pencil = x;
Price of pen = y;
∴ 2x + 5y = …….. 2x +….. y = …..
2x + …… y = 11
(2x + 5y) (-2x + ….. y) = -11
…… y = …….
y = \(\frac{…..}{……} \)
2x + 5x ….. = 17;
2x = 17 – ……..; x = ……. /2; = ……..
Answer:
2x + 5y = 17; 2x + 3y = 11;
(2x + 5y) – (2x + 3y) = 17 – 11; 2y = 6
y = \(\frac{6}{2} \) = 3; 2x + 5 × 3 = 17;
2x = 17 – 15;
x = \(\frac{2}{2} \) = 1
Price of pencil = Rs 1
Price of pen = Rs 3

Question 5.
Twice of a number added with thrice of another number gives 23. Four times the first number and 5 times the second number when added gives 41. Find out the numbers.
First number = x
Second number = y
∴ 2x + 3y = ……
4x + 5y = ……
2(2x + 3y) = 2 …….
4x + 6y = …….
(4x + 6y) – (4x + 5y) = ( …… ) – ( ….. )
y = …….; 2x + 3x ……. = …….
2x = ( …… ) – ( ……. ); x = ………./2
Answer:
2x + 3 y = 23
4x + 5y = 41
2(2x + 3y) = 2 × 23; 4x + 6y = 46
(4x + 6y) – (4x + 5y) = 46 – 41;
y = 5
2x + 3 × 5 = 23
2x = 23 – 15;
x = \(\frac{8}{2} \) = 4
Price of pencil = Rs 4
Price of pen = Rs 5

Question 6.
Rama spends Rs 97 to buy 4 two hundred page note books and 5 hundred page note books. Geetha spends Rs 101 to buy 5 two hundred page note book and 4 one hundred page note books. What is the prices of two types of note books?
Let the cost of two hundred page note books = x
The cost of one hundred page note books = y
(1) 4x + 5y = 97
(1) × 5 \(\Rightarrow \) 20x + ……..y = ……..
(2) 5x + 4y = 101
(2) × 4 \(\Rightarrow \) 20x + …….y = …….
……y = ( ….. ) – ( ……. );
y = …….. /……..
4x + 5x( ……. ) = 97
4x = 97 – ( ……. )
x = ……../4
Answer:
(1) 4x + 5y = 97
(2) 5x + 4y = 101
(1) × 5 → 20x + 25y = 485
(2) × 4 → 20x + 16y = 404
9y = 485 – 404
y = \(\frac{81}{9} \) = 9
4x + 5 × 9 = 97
4x = 97 – 45 = 52
Cost of two hundred page note book = Rs 13
Cost of one hundred page note book = Rs 9

Question 7.
6 years back the age of Muneer was 3 times the age of Mujeeb. After 4 years the age of Muneer becomes twice the age of Mujeeb. Find the age of two of them now.
Answer:
Age of Mujeeb 6 years back = x
Age of Muneer 6 years back = 3x
After 4 years
3x + 4 = 2(x + 4)
3x + 4 = 2x + 8
3x – 2x = 8 – 4;
x = 4
Age of Mujeeb 6 years back = 4 + 6 = 10
Age of Muneer 6 years back = 3(4 + 6) = 18 years.

Question 8.
The cost of 4 chairs and 5 tables is Rs 6600 and the cost of 5 chairs and 3 tables is Rs 5000 at the same prices. What are the prices of a table and a chair?
Answer:
Cost of a chair = Rs a
Cost of a table = Rs b
4a + 5b = 6600 ¾ …………(1)
5a + 3b = 5000 ¾ …………(2)
(1) × 5 → 20a + 25b = 33000
(2) × 4 → 20a + 12b = 20000
(20a + 25b) – (20a + 12b) = 33000 – 20000
(1) \(\Rightarrow \) 13b = 13000
b = \(\frac{13000}{13} \) = Rs 1000
4a + 5b = 6600;
4a + 5 × 1000 = 6600
4a = 6600 – 5000 = 1600
a = \(\frac{1600}{4} \) = Rs 400
Cost of a table = Rs 1000
Cost of a chair = Rs 400

You can Download नंगे पैर(Nange pair) Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Hindi Solutions Unit 3 Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Hindi Solutions Unit 3 Chapter 1 नंगे पैर (कहानी)

नंगे पैर Textual Questions and Answers

नंगे पैर विश्लेषणात्मक प्रश्न

Nange Pair Notes Kerala Syllabus 9th प्रश्ना 1.
‘बंद दरवाज़ा खोलने में उसे थोड़ा समय लगा। दरवाज़े से बाहर और हॉल को पारकर आगे आने में देर लगी।’ देर लगने का कारण क्या होगा?

उत्तर:
बेबी एक अपाहिज लड़की है। उसके एक पैर नहीं है। वह बैसाखियों के सहारे चलती है। इस लिए देर लगी।

9th Hindi Notes Kerala Syllabus प्रश्ना 2.
‘दोपहर में बेबी को अकेलापन लगता’ -बेबी बाहर निकल नहीं पाती। जो अकेलापन सहते हैं, वे हमसे क्या-क्या चाहते होंगे?

उत्तर:
जो अकेलापन सहते हैं वे दूसरों की हाज़िरी चाहते हैं। बोलने-बियाने के लिए दोस्तों को चाहते हैं। उनकी आँखें हमेशा दूसरों के आने की प्रतीक्षा करती रहती हैं।

Kerala Syllabus 9th Standard Hindi Solutions Unit 1 Chapter 3 प्रश्ना 3.
लगने दो बहिन! पोस्टमैन का काम जिस-तिस की डाक जिस-तिस के हाथ में सौंपने का है- अहाते में फेंकने का नहीं।” यहाँ पोस्टमैन का कौन-सा मनोभाव प्रकट होता है?

उत्तर:
इससे स्पष्ट होता है कि पोस्टमैन अपने काम के प्रति ईमानदार है। वह अपना दायित्व समझता है। उसे ईमानदारी के साथ निभाता है। अपने काम के प्रति पोस्टमैन का सकारात्मक मनोभाव यहाँ हम देख सकते हैं।

Kerala Syllabus 9th Standard Hindi Solutions Unit 3 Chapter 1 प्रश्ना 4.
बेबी ने मन ही मन निश्चय किया कि वह उस पोस्टमैन को नंगे पैर चलने नहीं देगी । यह तय करते ही उसे सुखद अनुभूति हुई’ कारण क्या होगा?’

उत्तर:
दूसरों की खुशी में खुशी ढूँढ़ना बड़ी बात है। यही सबसे बड़ी खुशी है। बेबी ने पोस्टमैन को जूते देकर खुश करने का निश्चय किया। इससे ही वह खुश हो जाती है। यहाँ बेबी के खुले दिल का परिचय मिलता है। इससे ही उसे सुखद अनुभूति होती है।

Hss Live Guru 9th Hindi Kerala Syllabus प्रश्ना 5.
‘तब होली का त्योहार मनाया जा चुका था।’ इस प्रस्ताव का तात्पर्य क्या है?

उत्तर:
बेबी पोस्टमैन को चप्पल सीधे देना नहीं चाहती है। वह उपहार के रूप में देने का बहाना ढूँढ़ती है। लेकिन होली का त्योहार बीत गया था।

9th Standard Hindi Notes Pdf Kerala Syllabus प्रश्ना 6.
‘साहब मेरी लाइन बदल दीजिए, सिटी में कहीं भी बदली दीजिए’ ऐसा क्यों कहा होगा?

उत्तर:
पोस्टमैन बिना चप्पल चलता था। यह देखकर बेबी ने उसे जूते दिए। बेबी के एक पैर ही नहीं है। लेकिन पोस्टमैन उस अभाव की पूर्ति करने में असमर्थ है। उसे यह हाल सह नहीं पाता होगा। इससे बदली माँगी होगी।

नंगे पैर Text Book Activities

नंगे पैर अभ्यास के प्रश्न

Hsslive Hindi Class 9 Kerala Syllabus प्रश्ना 1.
‘पोस्टमैन की आवाज़ उसके कानों में आई, पर फर्श पर डाक गिरने की आवाज़ सुनाई नहीं दी। बेबी पीछे मुड़ी। सँभालकर पलंग से उतरी। फिर बैसाखियों के सहारे एक-एक कदम चलकर दरवाजे तक आई।’

i. बेबी हर बात का सूक्ष्म निरीक्षण करती है। उसके मुताबिक व्यवहार भी करती है।

जैसे,

9th Standard Hindi Notes Kerala Syllabus प्रश्ना 2.
लिखें, इसपर आपकी प्रतिक्रिया।

उत्तर:

नंगे पैर विधात्मक प्रश्न

Kerala Syllabus 9th Standard Notes Hindi प्रश्ना 1.
डायरी लिखें :
‘बेबी ने मन ही मन निश्चय किया कि वह उस पोस्टमैन को नंगे पैर चलने नहीं देगी। यह तय करते ही उसे सुख अनुभूति हुई।’ सोचें, उस दिन बेबी के मन में कौन-कौन से विचार आए होंगे? बेबी के विचारों को डायरी के रूप में लिखें।

उत्तर:
14 सितंबर 2016
सोमवार
आज कितना अच्छा दिन है। जीवन में एक अच्छा निर्णय लिया। कितने दिन से उस पोस्टमैन को नंगे पैर चलते देख रही हूँ। लेकिन उसे ऐसे चलने देना सही नहीं है। बेचारे के पैर गर्मी से तपते होंगे। जो भी हो अपना निर्णय सही है। उसे मैं जूते बनवाकर दूंगी। इससे बढ़िया और क्या मैं कर सकती हूँ।

Kerala Syllabus 9th Standard Hindi Guide Pdf प्रश्ना 2.
टिप्पणी लिखें :
‘नंगे पैर’ शीर्षक इस कहानी के लिए कहाँ तक सार्थक है? अपना मत स्पष्ट करते हुए एक टिप्पणी लिखें।

उत्तर:
इस कहानी में दो विशेष पात्र हैं- पोस्टमैन और बेबी। बेबी अपाहिज है, उसके एक पैर नहीं है। पोस्टमैन बिना चप्पल चलता है। बेबी पोस्टमैन की हालत सह नहीं पाती है। पोस्टमैन के लिए वह जूता बनवाकर देती है। बेबी के पैर नहीं है लेकिन वह नंगे पैर चलनेवाले पोस्टमैन की कठिनाई को महसूसती है। यह कहानी नंगे पैर के इर्द-गिर्द घूमती है। इससे हम यह कह सकते हैं कि यह शीर्षक बिल्कुल सार्थक है।

नंगे पैर Additional Questions and Answers

नंगे पैर आशयग्रहण के प्रश्न

9th Standard Hindi Textbook Pdf Kerala Syllabus प्रश्ना 1.
दुपहर के समय बेबी के कमरे का माहौल कैसा था?

उत्तर:
बीच हॉल में रखे रेडियो में साढ़े बारह बजे के रिकॉर्ड बज रहे थे। शांत माहौल में गीतों के सुर गूंज रहे थे।

Class 9 Hindi Notes Kerala Syllabus प्रश्ना 2.
नए पोस्टमैन को धक्का-सा लगने का कारण क्या था?

उत्तर:
नए पोस्टमैन ने एक नज़र बेबी के पैरों की ओर देखा। बेबी के एक पैर नहीं था। वह बैसाखियों के सहारे चलती थी। यह देखकर उसे धक्का-सा लगा।

Hsslive Guru 9th Hindi Kerala Syllabus प्रश्ना 3.
पोस्टमैन को बिना चप्पल चलने के संबंध में बेबी ने क्या सोचा? ।

उत्तर:
बेबी ने सोचा कि शायद उसके जूते फट गए होंगे, वे मरमत होकर आते होंगे। .

Hss Live Guru Hindi 9th Kerala Syllabus प्रश्ना 4.
पोस्टमैन के अभावों को कहानी में कैसे दर्शाया गया है?

उत्तर:
पोस्टमैन के पास छाता नहीं, साइकिल नहीं और जूता भी नहीं था। धूप से उसका चेहरालाल-पीला होता। पसीने से उसकी कमीज़ पीठ से चिपकी रहती।

9th Standard Hindi Textbook Answers Kerala Syllabus प्रश्ना 5.
बेबी ने पोस्टमैन के पैर का नाप लेने के लिए कौन-सा तरकीब अपनाया?

उत्तर:
पोस्टमैन के डाक देकर चले जाने पर बेबी अहाते के किनारे आई, सीढियों से नीचे उतर आई। फ्रॉक की जेब से परकार निकालकर जानेवाले पोस्टमैन के पैरों के निशान का नापजोख किया।

नंगे पैर Grammar

नंगे पैर व्याकरण के प्रश्न

प्रश्ना 1.
इस वाक्य पर ध्यान दें

‘पैरों की. गोलाई आदि का ठीक-ठीक माप।’

प्रश्ना 2.
इसमें ‘गोलाई’ शब्द ‘गोल’ और ‘आई’ के मेल से बना है। ध्यान दें, इससे अर्थ में कौन-सा परिवर्तन आता है। इस तालिका में इसी प्रकार के और कुछ शब्दों को जोड़ें।

उत्तर:

नंगे पैर Summary in Malayalam and Translation

नंगे पैर शब्दार्थ