PrasannaStudents can Download Maths Chapter 9 Circle Measures Questions and Answers, Summary, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 9th Standard Maths Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.
Public administration is concerned with the administration of the government. In all countries, there exists administrative institutions to implement the laws and developmental projects formed by govern-ment. The history of public administration begins with the formation of state. Democratic administration becomes more effective and efficient through public administration. E-governance, Information Commission and Right to Service are the steps taken for administrative reforms. In order to make public administration more efficient and to prevent corruption, institutions like Lokpal, Lokayukta, Central Vigilance Commission and Ombudsman are formed.
→ Public administration: Public administration is the effective utilization of man and materials for the implementation of existing laws, governmental policies, programmes and developmental projects.
→ Bureaucracy : The employees who work under public administrative system and administer the country are together known as bureaucracy.
→ Union Public Service Commission : The institution that selects and recruits candidates to all India services and Central services. It functions at all India level.
→ Public Service Commissions : The institutions that function at state levels and select and recruit candidates for state services.
→ Administrative reforms : The number of steps taken by the government for increasing the efficiency of the services and to provide service to people in a time bound manner are known as administrative reforms.
→ E-governance : E-governance is the use of electronic technology in administration.
→ Right to Information: The Right to Information is the right to examine and demand copies of files, documents, circulars, orders, statistics, etc under the disposal of government institutions and institutions which receive government funds.
Social Science Short Notes For Class 10 Kerala Syllabus
→ Right to Service Act : A law which ensures service to people.
→ Lokpal: The institution constituted at the national level to prevent corruption at administrative, bureaucratic and political levels.
→ Lokayukta : The institution constituted at the state level to prevent corruption at administrative, bureaucratic and political levels.
→ Central Vigilance Commission : The institution constituted at the national level in 1964 to prevent corruption.
→ Ombudsman : Ombudsman is a system constituted for the purpose of filing complaints against the corruption, nepotism, financial misappropriation and negligence of duty of elected representatives and bureaucrats.
Kerala Syllabus 10th Standard Social Science Notes
Students can Download Basic Science Chapter 4 Properties of Matter Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 8th Standard Basic Science Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.
Kerala State Board New Syllabus Plus Two Malayalam Textbook Answers Unit 3 Chapter 3 Thenga Text Book Questions and Answers, Summary, Notes.
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You can Download Polynomialss Questions and Answers, Activity, Notes, Kerala Syllabus 9th Standard Maths Solutions Chapter 8 help you to revise complete Syllabus and score more marks in your examinations.
Textbook Page No. 123
Polynomial Root Calculator solves the roots of polynomials easily.
Polynomials Class 9 Kerala Syllabus Chapter 8 Question 1.
In rectangles with one side 1 centimetre shorter than the other, take the length of the shorter side as x centimetres.
i. Taking their perimeters as p(x) centimetres, write the relation between p(x) and x as an equation,
ii. Taking their areas as a(x) square centimetres, write the relation between a(x) and x as an equation.
iii Calculate p(l), p(2), p(3), p(4), p(5). Do you see any pattern?
iv. Calculate a(l), a(2), a(3), a(4), a(5). Do you see any pattern?
Answer:
Let x be the shorter side, then the other side will be (x+ 1).
i. Perimeter = 2[x + (x + 1)] = 2(2x + 1) = 4x + 2
That is, p(x) = 4x + 2
ii. Area = x(x + 1)
a(x) = x2 + x
Area, a(x) = x2 + x
iii. p(x) = 4x + 2
p(1) = 4 × 1 + 2 = 6
p(2) = 4 × 2 + 2 = 10
p(3) = 4 × 3 + 2= 14
p(4) = 4 × 4 + 2= 18
p(5) = 4 × 5 + 2 = 22
Perimeter is a sequence increasing by 4.
iv. a(x) = x2 + x
a(1) = 12 + 1 = 2 = 1 × 2 = 2
a(2) = 22 + 2 = 6 = 2 × 3 = 6
a(3) = 32 + 3 = 12 = 3 × 4 = 12
a(4) = 42 + 4 = 20 = 4 × 5 = 20
a(5) = 52 + 5 = 30 = 5 × 6 = 30
Area is the product of x and the number one more than x.
Polynomial Root Calculator solves the roots of polynomials easily.
Polynomials Class 9 State Syllabus Chapter 8 Question 2.
From the four corners of a rect-angle, small squares are cut off and the sides are folded up to make a box, as shown below:
i. Taking a side of the square as x centimetres, write the dimensions of the box in terms of x.
ii. Taking the volume of the box as vfojcubic centimetres, write the relation between v(x) and x as an equation.
iii. Calculate \(\mathrm{V}\left(\frac{1}{2}\right) \quad, \mathrm{V}(1), \quad \mathrm{V}\left(1 \frac{1}{2}\right)\)
Answer:
If 1cm is the length of the small squares they are cut off, then the length of the maked box by folding it up = 7 – 1 – 1 = 5 cm Width = 5 – 1 – 1 = 3 cm, Height = 1 cm
If 2 cm is the length of the small squares that are cut off, then the length of the maked box by folding it up = 7 – 2 – 2 = 3 cm Width = 5 – 2 – 2 = 1 cm, Height = 2cm
i. If x cm is the length of the small squares that are cut off, then the length of the maked box by folding it up = 7 – x – x = (7 – 2x) cm
Width = 5 – x – x = (5 – 2x) cm
Height = x cm
ii. If volume of the box be v(x), then
v(x) = Length × Width × Height v(x) = (7 – 2x) (5 – 2x)x cm3
v (1) = (7 – 2 × 1) (5 – 2 × 1) 1 = (7 – 2) (5 – 2) × 1 = 5 × 3 × 1 = 15
Hss Live Guru 9th Maths Kerala Syllabus Chapter 8 Question 3.
Consider all rectangles that can be made with a 1-metre long rope. Take one of its sides as x centimetres and the area enclosed as a(x) square centimetres.
i. Write the relation between a(x) and x as an equation.
ii. Why are the numbers a(10) and a(40) equal?
iii. To get the same number as a(x), for two different numbers as x, what must be the relation between the numbers?
Answer:
i. 1 m = 100 cm
If one side is x cm, then the other side is 50 – x.
Area of rectangle = a(x) = x(50 – x)
= 50x – x2 cm2
a(x) = 50x – x2
ii. a(10) = 50 × 10 – 102 = 500 – 100 = 400
a(40) = 50 × 40 – 402 = 2000 – 1600 = 400
10 and 40 are the sides of rectangle, so a(10) and a(40) are same .
iii. Numbers must be sides of rectangle. If we add the two numbers together we get the sum as half the length of the wire (50 cm) used to make it.
Textbook Page No. 126
Free Is Polynomial Calculator – Check whether a function is a polynomial step-by-step.
Hsslive Guru 9th Maths Kerala Syllabus Chapter 8 Question 1.
Write each of the relations below in algebra and see if it gives a polynomial. Also, give reasons for your conclusion.
i. A 1-metre wide path goes around a square ground. The relation between the length of a side of the ground and the area of the path.
ii A liquid contains 7 litres of water and 3 litres of acid. More acid is added to it. The relation between the amount of acid added and the change in the percentage of acid in the liquid.
iii. Two poles of heights 3 metres and 4metres are erected upright on the ground, 5 metres apart. A rope is to be stretched from the top of one pole to some point on the ground and from there to the top of the other pole:
The relation between the distance of the point on the ground from the foot of a pole and the total length of the rope.
Answer:
i. Let the side of the ground be x then the side of the square including the path is x + 2 metres.
Area of path = Area of large square – Area of small square.
a(x) = (x + 2)2 – x2 = x2 + 4x +4 – x2
= 4x + 4
Here a(x) is a polynomial. Here x is multiplied by 4 and 4 is added to it.
ii. Ratio of acid in the first fluid = 3/10 litre that is 30%
If x litres of acid is added to it, then change in the amount of acid is
\(=\frac{3+x}{10+x}\)
Change in the percentage of acid is = \(\frac{3+x}{10+x} \times 100 \%\)
\(\mathrm{b}(\mathrm{x})=\frac{300+100 x}{10+x} \%\)
iii.
In figure, AB = x m, BD = 5 – x
Total length of the wire = BC + BE
It is not a polynomial since it involves square root of variable x.
Sum and difference of cubes calculator, multiplying and dividing rational exponents not possible.
Kerala Syllabus 9th Standard Maths Notes Chapter 8 Question 2.
Write each of the operations below as an algebraic expression, find out which are polynomials and explain why.
i. Sum of number and it’s reciprocal
ii. Sum of a number and its square root.
iii. Product of the sum and difference of a number and its square root.
Answer:
i. Let the number be x, then the reciprocal is 1/x
sum = \(x+\frac{1}{x}\)
This is not a polynomial, because here the operation of reciprocal is involved.
ii. Let the number be x, then the square root is √x
sum = x + √x
This not a polynomial because here the square root is taken.
iii. Let x be the number
(x + √x) (x – √x) = x2– x
This is a polynomial.
Std 9 Maths Notes Kerala Syllabus Chapter 8 Question 3.
Find p(1) and p(10) in the following polynomials,
i. p(x) = 2x + 5
ii. p(x) = 3x2 + 6x + 1
iii. p(x) = 4x3 + 2x2 + 3x + 7
Answer:
i. p(x) = 2x + 5
p(1) = 2 × 1 + 5 = 2 + 5 = 7
p(10) = 2 × 10 + 5 = 20 + 5 = 25
ii. p(x) = 3x2 + 6x + 1
p(1) = 3 × 12 + 6 x 1 + 1
= 3 + 6 + 1 = 10
p(10) = 3 × 102 + 6 × 10 + 1 = 300 + 60 + 1 = 361
iii. p(x) = 4x3 + 2x2 + 3x + 7
p(1) = 4 × 13 + 2 × 12 + 3 × 1 +7
=4 + 2 + 3 + 7 = 16
p(10) = 4 × 103 + 2 × 102 + 3 × 10 + 7
= 4000 + 200 + 30 + 7 = 4237
Hss Live Maths 9th Kerala Syllabus Chapter 8 Question 4.
Find p(0), p(1) and p(-1) in the fol-lowing polynomials,
i. p(x) = 3x + 5
ii. p(x) = 3x2 + 6x + 1
iii. p(x) = 2x2 – 3x + 4
iv. p(x) = 4x3 + 2×2 + 3x + 7
v. p(x) = 5x3 – x2 + 2x – 3
Answer:
i. p(x) = 3x + 5
p(0) = 3 × 0 + 5 = 5
p(1) = 3 × 1 + 5 = 8
p(-1) = 3 x-1 + 5 = 2
ii. p(x) = 3x2 + 6x + 1
p(0) = 3 × 02 + 6 × 0 + 1 = 1
p(1) = 3 × 12 + 6 × 1 + 1 = 10
p(-1) = 3 × (-1)2 + 6 × (-1) + 1 = -2
iii. p(x) = 2x2 – 3x + 4
p(0) = 2 × 02 – 3 × 0 + 4 = 4
p(1) = 2 × 12 – 3 × 1 + 4 = 3
p(-1) =2 × (-1)2 – 3 × (-1) + 4 = 9
iv. p(x) = 4x3 + 2x2 + 3x + 7
p(0) = 4 × 03 + 2 × 02 + 3 × 0 +7 = 7
p(1) = 4 × 13 + 2 × 12 + 3 × 1 + 7 = 16
p(-1) = 4 × (-1)3 + 2 × (-1)2 + 3 × (-1) +7 =2
v. p(x) = 5x3 – x2 + 2x – 3
p(0) = 5 × 03 – 02 + 2 × 0 – 3 =-3
p(1) = 5 × 13 – 12 + 2 × 1 – 3 = 3
p(-1) = 5 × (-1)3 – (-1)2 + 2 × (-1) – 3 = -11
Scert Class 9 Maths Solutions Kerala Syllabus Chapter 8 Question 5.
Find polynomials p(x) satisfying
each set of conditions below.
i. First degree polynomials with p(1) = 1 and p(2) = 3
ii. First degree polynomials with p(1) = -1 and p(-2) = 3
iii. Second degree polynomials with p(0) = 0, p(1) = 2 and p(2) = 6.
iv. Three different second degree polynomials with p(0) = 0 and p(1) = 2.
Answer:
i. General form of a first degree poly-nomial is
p(x) = ax + b
Let p(1) = 1, then a × 1 + b = l
a + b = 1 ………. (1)
Let p(2) = 3, then a × 2 + b = 3
2a + b = 3 …….. (2)
(1) × 2, 2a + 2b = 2 …… (3)
(3) – (2), b = -1
From (1), a + -1 = 1,
a = 1 + 1 = 2 Polynomial p(x) = 2x – 1
ii. General form of a first degree poly-nomial is p(x) = ax + b
Let p(1) = -1, then a × 1 + b = -l
a + b = -1 ……….. (1)
Let p(-2) =3 , then a × (-2) + b = 3
-2a + b = 3 ………. (2)
(1) × 2, 2a + 2b = -2 ………. (3)
(2) + (3), 3b = 1, b = 1/3
From (1), a = 1/3 = -1
iii. General form of a second degree polynomial is
p(x) = ax2 + bx + c
Letp(0) = 0, then a × 02 + b × 0 + c = 0
0a + 0b + c = 0.
c = 0 (1)
Letp(1) = 2 ,then a × 12 + b × 1 + c = 2
a + b + 0 = 2
a + b = 2 ………. (2)
Letp(2)= 6, then a × 22 + b × 2 + c = 6
4a + 2b = 6
2a + b = 3 (3)
(3) – (2), a = 1
From (2), 1 + b = 2, b = 2 – 1 = 1
Polynomial p(x) = x2 + x
iv. General form of a second degree polynomial is p(x) = ax2 + bx + c
Letp(0) = 0, then a x 0 + b x 0 + c = 0
0 + 0 + c = 0
c = 0
Letp(1) = 2, then a × 12 + b × 1 + c = 2
a + b + c = 2
a + b = 2
Selecting a and b such that a + b = 2 will give different polynomials.
a = 1, b = 1
a= 3, b =-l
a=4, b =-2
Three different second degree poly¬nomials are
p(x) = x2 + x
p(x) = 3x2 – x
p(x) = 4x2 – 2x
Hss Live 9th Maths Kerala Syllabus Chapter 8 Question 1.
In a polynomial p(x) = 2x3 + ax2 – 7x + b.
p(1) = 3, p(2) = 19. Then find the value of ‘a’ and ‘b’?
Answer:
p(x) = 2x3 + ax2 – 7x + b
We have p(1) = 3.
p(1) = 2 x 13 + a x 12 – 7 x 1 + b = 3
2 + a – 7 + b = 3
a + b – 5 = 3
a + b =8 ………. (1)
We have p(2) = 19.
p(2 ) = 2 x 23 + a x 22– 7 x 2 + b = 19
16 + 4a – 14 + b = 19
4a + b + 2 = 19
4a + b = 17…….. (2)
From equation (1), (2)
(2) – (1) 4a + b = 17
\(\frac{a+b=8}{3 a=9}\)
a = 9/3 = 3
From equation (1)
a + b = 8
3+b =8
b =5
Hsslive Class 9 Maths Kerala Syllabus Chapter 8 Question 2.
In a polynomial p(x) = 2x3 + 9x2 + kx + 3, p(-2) = p(-3). Find the value of k.
Answer:
p(x) = 2x3 + 9x2 + kx + 3
p(-2) = 2(-2)3 + 9(-2)2 + k(-2) + 3
= -16 + 36 – 2k + 3 = 23 – 2k
p(-3) = 2(-3)3 + 9(-3)2 + k(-3) + 3
= -54 + 81 – 3k + 3 = 30 – 3k
We have p(-2) = p(-3),
23 -2k = 30 -3k
-2k + 3k = 30 – 23
k = 7
Maths Kerala Syllabus Std 9 Notes Chapter 8 Question 3.
From the polynomial p(x) = 2x – 3x +1, find p(0), p(1) and p(-1).
Answer:
p(x) = 2x2 – 3x +1
p(0) = 2(0)2 – 3(0) + 1 = 1
p(1) = 2(1)2 – 3(1) +1 = 2 – 3 + 1 = 0
p(-1) = 2(-1)2 – 3(-1) + 1 = 2 + 3 + 1 = 6
Kerala Syllabus 9th Standard Maths Notes Pdf Chapter 8 Question 4.
In a polynomial p(x) = 2x3 – 7x2 + kx + 20,
p(2) = p(3).
a. Find the value of k.
b. Using the value of k, write the polynomial.
c. Find p(1).
Answer:
a. p(2) = 2 × 22 – 7 × 22 + k × 2 + 20
= 16 – 28 + 2k + 20 = 8+ 2k
p(3) = 2 × 33 – 7 × 32+ k × 3 + 20
= 54 – 63 + 3k + 20 = 11 +3k
P(2) = P(3)
8 + 2k = 11 + 3k
k = -3
b. p(x) = 2x3 – 7x2 -3x + 20
c. p(1) = 2 – 7 – 3 + 20 = 12
Chapter 8 Polynomials Answers Kerala Syllabus Question 5.
Simplify the followi ng
i. (2x + 1) (3x + 4) (4x + 3) (3x + 4)
ii. (3x + 4)2 – (2x – 1) (3x + 4)
Answer:
i. (2x + 1) (3x + 4) (4x + 3) (3x + 4)
= (3x + 4) [(2x + 1 + 4x + 3)]
= (3x + 4)[6x + 4]
= (3x x (6x) + (3x) x (4)+ 4x(6x) + 4 x 4
= 18x2 + 12x + 24x + 16
= 18x2 + 36x + 16
ii. (3x + 4)2 – (2x – 1) (3x + 4)
= (3x + 4) [3x + 4 – (2x – 1)]
= (3x + 4)[3x + 4 – 2x + 1] = (3x + 4) (x + 5)
= 3x2 + 15x + 4x + 20
= 3x2 + 19x + 20
Hsslive Guru Maths 9th Kerala Syllabus Chapter 8 Question 6.
7x3 – 4x2 – x + 4 is a polynomial.
a. Write the terms of the polynomial.
b. Write the coefficient of x2.
c. Write the constant terms of the polynomial.
d. What is the degree of the polynomial ?
Answer:
a. Terms = 7x2, -4x2, -x, 4
b. Coefficient of x2 = -4
c. Constant term = 4
d. Degree of the polynomial = 3
Hss Live Guru Class 9 Maths Kerala Syllabus Chapter 8 Question 7.
In the polynomial p(x)=3x2 – ax + 1,
Find ‘a’ satisfying p(1) = 2.
Answer:
p(x)=3x2 – ax + 1
p(1) = 3(1)2 – (a × 1) + 1 = 3 – a + 1 =4 – a
Given p(1) = 2
That is, 4 – a = 2 , a = 4 – 2 = 2
Hsslive Maths Class 9 Kerala Syllabus Chapter 8 Question 8.
If p(x)= x3 + 2x2 – 3x + 1 and q(x) = x3 – 2x2 + 3x + 5
a) Find p(x) + q(x). What is its de-gree?
b) Find p(x) – q(x). What is its de-gree?
Answer:
a) p(x) = x3 + 2x2 – 3x + 1,
q(x) = x3 – 2x2 + 3x + 5
p(x) + q(x)
= x3 + 2x2 – 3x + 1 + x3 – 2x2 + 3x + 5
= 2x3 + 6
Degree = 3
b) p(x) – q(x)
= x3 + 2x2 – 3x + 1 – (x3 – 2x2 + 3x + 5)
= x3 + 2x2 – 3x + 1 -x3 + 2x2 – 3x – 5
= 4x2 – 6x – 4
Degree = 2
9th Std Maths Notes Kerala Syllabus Chapter 8 Question 9.
A right-angled triangle of perpendicular sides 3 cm and 4 cm are ex-tended equally then get another large right-angled triangle. Write the algebraic form of the hypotenuse of the large right-angled triangle.
Answer:
Hypotenuse 2 = base 2 + height 2
(According to pythagorus theorem)
Let the perpendicular side be x cm Length of other sides = 3 + x cm, 4 + x cm
Question 10.
In the polynomial p(x)= 3x2 – 4x + 7, check whether p(1) + p(2) = p(3) and P(2) × p(3) = p(6).
Answer:
p(x) = 3x2 – 4x + 7
p(1) = 3(1)2 – 4(1) + 7 = 3 – 4 + 7 = 6
p(2) = 3(2)2 -4(2) + 7 = 12 – 8 + 7 = 11
p(3) = 3(3)2 – 4(3) +7 = 27 – 12 + 7 = 22
p(l) + p(2) ≠ p(3)
p(6) = 3(6)2 -4(6) + 7 = 108 – 24 + 7 = 91
P(2) × p(3) ≠ p(6)
Question 11.
In the polynomial p(x)= 2x2 + ax2 – 7x + b,
p(1) = 3 and p(2) = 19. Find a and b.
Answer:
p(x)=2x2 + ax2 – 7x + b
p(l)=2 x 13+ a x 12 – 7 x 1 +b
=2 + a – 7 + b =a + b + 2 – 7 = a + b – 5
p(1) = 3, a + b – 5 = 3
a + b = 8……… (1)
p(2) = 2 x 23 + a x 22 – 7 x 2 + b
= 16 + 4a – 14 + b = 4a + b + 2
p(x) = 19
4a + b + 2 = 19
4a + b = 17 ……….. (2)
(2) – (1)
3a = 9, a = 3 a + b – 5 = 3
From equation (1),
a + b = 8
3 + b = 8
b = 5
Question 12.
If p(x) = x2 + 3x + 1 and q(x) = 2x – 4, then
i. Find the degree of the polyno-mial p(x) q(x).
ii. If the degree of p(x) × r(x) is 5, then find the degree of the polynomial r(x) .
iii. If p(x) is a third degree poly-nomial and q(x) is a fourth de-gree polynomial then find the degree of p(x) × q(x).
Answer:
i. p(x) = x2 + 3x + 1 , q(x) = 2x – 4
p(x) × q(x) = (x2 + 3x + 1) (2x – 4)
The degree of p(x) x q(x) is 3.
ii. p(x) = x2 + 3x + 1 is a second degree polynomial.
If the degree of the polynomial p(x) × r(x) is 5, then p(x) × r(x) must have a term of x5.
x2 × x3 = x5
So, r(x) is a third-degree polynomial.
iii. If p(x) is a third-degree polynomial, then p(x) must have a term of x3.
q(x) is a fourth-degree polynomial, then q(x) must have a term of x4
Then in p(x) × q(x), must have a term of x3 × x4= x7
So p(x) × q(x) is a seventh-degree polynomial. In general,
If p(x) is an mth degree polynomial and q(x) is an n,h degree polynomial then p(x) × q(x) is an (m + n),h degree polynomial.
Question 13.
If p(x) = 4x2 + 3x + 5 and q(x) = 3x2 – x – 7, then find p(x) + q(x) and p(x) – q(x).
Answer:
p(x) + q(x) = (4x2 + 3x + 5) + (3x2 – x – 7)
= 4x2 + 3x2 + 3x – x + 5 – 7
= 7x2 + 2x – 2
p(x) – q(x) = (4x2 + 3x + 5) – (3x2 – x – 7)
= 4x2 + 3x + 5 – 3x2 + x + 7
= x2 + 4x + 12
Question 14.
In polynomial p(x) = ax3+ bx2 + cx + d, p(l) = p(-l). Prove that a + c = 0.
Answer:
p(x) = ax3 + bx2 + cx + d
p(l) = a(1)3 + b(1)2 + c(1) + d
= a+ b + c + d
p(-1) = a(-1)3 + b(-1)2 +c(-1) + d
= -a + b – c + d
p(1) = p(-1) is given ie, a + b + c + d = -a + b – c + d
2a + 2c = 0; 2(a + c ) = 0;
ie, a + c = 0
Question 15.
If we divide a polynomial by (x-2) we get quotient as x2 + 1 and remainder as 5. Find the polynomial.
Answer:
Polynomial = (x2 + 1) (x – 2) + 5
= x3 + x – 2x2 – 2 + 5 ,
= x3 – 2x2 + x + 3
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You can Download दौड़ Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Hindi Solutions Unit 4 Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.
दौड़ विधात्मक प्रश्न
Hsslive Guru 9th Hindi प्रश्ना 1.
बारिश का अनुभव सुहाना होता है। आपको बारिश कैसे महसूस होता है? इसपर एक छोटा-सा संस्मरण लिखें।
उत्तर:
बारिश मेरे लिए बहुत खुशी का अवसर था। बारिश देखने के लिए मैं अपने घर की खिड़की के पास खड़ा होता था। बारिश में छाता लेकर स्कूल जाना मुझे बहुत पसंद था। पीठ पर बस्ता टाँगकर, एक हाथ में छाता पकड़कर, दूसरे हाथ से छाते के कोने से आता पानी. को छुआ करता था। खेल घंटी के समय अगर बारिश होता तो कागज़ से नाव बनाकर पानी रख देते थे। बहती नाव के साथ हम भी दौडते थे। बारिश की ठंड में चादर ओढ़कर सोना भी मुझे पसंद था।
दौड़ शब्दार्थ
Class 10 Geography Chapter 3 Kerala Syllabus Question 1.
List out the quantitative and qualitative aspects of human resource. –
Answer:
Human resource refers to people who have the manpower which can be utilised in the production sector.
Std 10 Geography Notes Question 2.
Prepare a note by analysing the importance of population studies.
Answer:
Question 3.
Compare the changes in population due to birth rate, death rate and migration.
Answer:
Birth rate, death rate and migration are the factors that affect the population of a country.
Question 4.
Labour force participation rate and dependency rate as per the census of India 2011 is given below. Prepare a graph based on this
| Labour force participation rate (in percentage) | Dependency rate (in percentage) | |
| Female | 62.8 | 37.2 |
| Male | 62.2 | 37.7 |
Question 5.
List the advantages of the increase in labour force participation rate and disadvantages due to increase in depending rate.
Answer:
Question 6.
What are the factors that improve human resource? How does this influence a country’s development?
Answer:
The factors that improve human resource are
Influence of improved human resource
Question 7.
How does education help in country’s development? Prepare a flowchart.
Answer:
Question 8.
List the existing problems in health sector.
Answer:
Human Resource Development In India Class 10 Textbook Pdf Question 9.
Explain how education and healthcare help in human resource development.
Answer:
Healthcare and human resource development: Each individual can work for the economic development of a country only if proper healthcare facilities are provided. Healthy persons help in the progress of the country in the following ways.
Question 10.
List the different jobs around us, the goods and services provided by them and the skill needed for them.
Answer:
| Job | Goods/Service | Skill |
| Farmer | Producers agricultural crops | Knowledge about different agricultural crops and the method of cultivation |
| Teacher | Teaches | Knowledge and training on the subject and method of teaching. |
| Doctor | Treats diseases | Knowledge about diseases and medicines |
Question 11.
Human resource development is the development of man‘s physical and mental abilities. Find out the factors that help to develop such abilities.
factors that help to develop such abilities.
Answer:
Question 12.
Complete the flowchart on features of human resource.
Answer:
a. Quantitative features
b. Size of population
c. Growth of population
d. Population structure
e. Quai initiative features
f. Education
g. Healthcare
h. Literacy rate
Question 13.
Examine the table indicating population growth rate in India and answer the questions given below.
| Year | Population (in crores) | Decadal growth rate (in percentage) |
| 1951 | 36.11 | 13.31(1941-51) |
| 1961 | 43.90 | 21.64(1951-61) |
| 1971 | 54.82 | 24.80(1961-71) |
| 1981 | 68.33 | 24.66(1971-81) |
| 1991 | 84.64 | 23.87(1981-91) |
| 2001 | 102.87 | 21.54(1991-2001) |
| 2011 | 121.02 | 17.64(2001-11) |
a. Which decade has marked the maximum population growth?
b. From which year onwards is there a decrease in the population growth rate?
c. How much decrease did the population growth rate record in the decade 2001-2011?
Answer:
a. 1961-71.
b. 1971
c. 3.90 (in percentage)
Question 14.
Given below is a pie diagram showing the age structure based on census of India 2011. Based on this, answer the questions.
What percentage of the total population belongs to the age group 0-14 years?
Answer:
29.4%
What percentage of total population belongs to the age group of 60 years and above?
Answer:
8.09%
What percentage of the total population belongs to the age group of 15-59 years?
Answer:
62.5%. High labour force participation rate is beneficial for the development of the country.
Question 15.
How does the population density of an area influence the availability of human resource?
Answer:
Density of population refers to the number of people residing in a square kilometre area. More people means more manpower. This will increase production.
Question 16.
What are the problems in the economy as a result of a decrease in the labour force participation rate and increase in the dependency ratio?
Answer:
The portion of the population in the age group 15-59 is included in labour force participation rate. This age group has the capability to contribute to the progress of the nation. Decrease in labour force participation rate will adversely affect the progress of the nation.
An increase in dependency ratio decreases the per capita income. The availability of manpower decreases with a fall in the age group of 15-59. This results in low production.
Question 17.
How does human resource development influence economic development?
Answer:
As a result of human resource development:
Sslc Geography Notes Question 18.
The given table depicts the features of Indian population. Compare the sex ratio, life expectancy and literacy rate and prepare a note.
Answer:
Answer:
Question 19.
List the advantages of improved human resource.
Answer:
Question 20.
Find out how education helps in the development of a nation and complete the flowchart.
Answer:
a. Improves the skills of individuals
b. Betters the technological know-how.
c. Helps to secure better job and income.
10th Geography Notes Question 21.
Find out how healthcare helps in national progress and prepare a note.
Answer:
Class 10 Social Science Chapter 3 Kerala Syllabus Question 22.
List the facilities to be ensured for healthcare
Answer:
Kerala Syllabus 10th Standard Social Science Solutions
Students can Download Adisthana Padavali Unit 3 Chapter 1 Prataniti Questions and Answers, Summary, Notes Pdf, Activity, Adisthana Padavali Malayalam Standard 10 Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.
Kerala State Board New Syllabus Plus One Malayalam Textbook Answers Unit 4 Chapter 6 Shasthrakriya Text Book Questions and Answers, Summary, Notes.
