NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4.

Board CBSE
Textbook NCERT
Class Class 10
Subject Maths
Chapter Chapter 5
Chapter Name Arithmetic Progressions
Exercise Ex 5.4
Number of Questions Solved 5
Category NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4

Ex 5.4 Class 10 Question 1.
Which term of the AP: 121, 117. 113, ….., is its first negative term?
Solution:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4 1

Ex 5.4 Class 10 NCERT Solutions Question 2.
The sum of the third term and the seventh term of an AP is 6 and their product is 8. Fibd the sum of first sixteen terms of the AP?
Solution:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4 2
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4 3

Class 10 Ex 5.4 Question 3.
A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 at the top. If the top and the bottom rungs are \({ 2 }\frac { 1 }{ 2 } m\) apart, what is the length of the wood required for the rungs?
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4 4
Solution:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4 4
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4 6

Ex 5.4 Class 10 Question 4.
The houses of a row are numbered consecutively from 1 to 49. Show that there is value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.
Solution:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4 7
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4 7a

Ex 5.4 Class 10 Question 5.
A small terrace at a football ground comprises of 15 steps each of which is 50  m long and built of solid concrete. Each step has a rise of \(\frac { 1 }{ 2 } m\) and a tread of \(\frac { 1 }{ 2 } m\). Calculate the total volume of concrete required to build the terrace.
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4 8
Solution:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4 9
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4 10

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NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1.

Board CBSE
Textbook NCERT
Class Class 10
Subject Maths
Chapter Chapter 8
Chapter Name Introduction to Trigonometry
Exercise Ex 8.1
Number of Questions Solved 11
Category NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1

Exercise 8.1 Class 10 Question 1.
In ∆ABC right angled at B, AB = 24 cm, BC = 7 cm. Determine:
(i) sin A, cos A
(ii) sin C, cos C
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 1

Class 10 Ex 8.1 Question 2.
In given figure, find tan P – cot R.
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 2

Trigonometry Class 10 Question 3.
If sin A = \(\frac { 3 }{ 4 }\) , calculate cos A and tan A.
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 3
sin A = \(\frac { 3 }{ 4 }\)
sin A = \(\frac { BC }{ AC }\) = sin A = \(\frac { 3 }{ 4 }\)
Let BC = 3k and AC = 4k
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 4

Ex 8.1 Class 10 Question 4.
Given 15 cot A = 8, find sin A and sec A.
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 5

Class 10 Trigonometry Ex 8.1 Question 5.
Given sec θ = \(\frac { 13 }{ 12 }\) , calculate all other trigonometric ratios.
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 6

Ex 8.1 Class 10 Question 6.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 7

Ex 8.1 Class 10 Question 7.
If cot θ = \(\frac { 7 }{ 8 }\), evaluate:
(i) \(\frac { \left( 1+sin\theta \right) \left( 1-sin\theta \right) }{ \left( 1+cos\theta \right) \left( 1-cos\theta \right)}\)
(ii) cot²θ
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 8

Trigonometry NCERT Class 10 Ex 8.1 Question 8.
If 3 cot A = 4, check whether \(\frac { 1-tan^{ 2 }A }{ 1+tan^{ 2 }A }\) = cos² A – sin² A or not.
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 9

Ex 8.1 Class 10 Maths Solutions Question 9.
In triangle ABC, right angled at B, if tan A = \(\frac { 1 }{ \surd 3 }\), find the value of:
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 10

Class 10 Ex 8.1 Question 10.
In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 11
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 12

NCERT Solutions For Class 10 Maths Chapter 8 Question 11.
State whether the following statements are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A = \(\frac { 12 }{ 5 }\) for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin θ = \(\frac { 4 }{ 3 }\) for some angle.
Solution:
(i) tan 60° = √3 , Since √3 > 1. (False)
(ii) sec A is always ≥ 1. (True)
(iii) cos A is the abbreviation for cosine A. (False)
(iv) cot without ∠A is meaningless. (False)
(v) sin θ can never be greater than 1.
∴ sin θ = \(\frac { P }{ H }\) , hypotenuse is always greater than other two sides. (False)

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NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3.

Board CBSE
Textbook NCERT
Class Class 10
Subject Maths
Chapter Chapter 2
Chapter Name Polynomials
Exercise Ex 2.3
Number of Questions Solved 4
Category NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3

Ex 2.3 Class 10 Question 1.
Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder, in each of the following:
(i) p(x) = x3 – 3x2 + 5x -3, g(x) = x2-2
(ii) p(x) =x4 – 3x2 + 4x + 5, g(x) = x2 + 1 -x
(iii) p(x) = x4 – 5x + 6, g(x) = 2 -x2
Solution:
(i) Here p(x) = x3 -3x2 + 5x – 3 and g(x) = x2 -2
dividing p(x) by g(x) ⇒
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3 1
Quotient = x – 3, Remainder = 7x – 9

(ii) Here p(x) = x4– 3x2 + 4x + 5 and g(x) = x2 + 1 -x
dividing p(x) by g(x) ⇒
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3 2
Quotient = x2 + x – 3, Remainder = 8

(iii) Herep(x) = x4– 5x + 6 and g(x) = 2-x2
Rearranging g(x) = -x2 + 2
dividing p(x) by g(x) ⇒
Quotient = -x2 – 2
Remainder = -5x + 10.
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3 3

Ex 2.3 Class 10 Question 2.
Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
(i) t2-3, 2t4 + t3 – 2t2 – 9t – 12
(ii) x2 + 3x + 1,3x4+5x3-7x2+2x + 2
(iii) x3 -3x + 1, x5 – 4x3 + x2 + 3x + l
Solution:
(i) First polynomial = t2 – 3,
Second polynomial = 2t4 + 3t3 – 2t2 – 9t – 12
dividing second polynomial
by first polynomial ⇒
∵ Remainder is zero.
∴First polynomial is a factor of second polynomial.
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3 4

(ii) First polynomial = x2 + 3x + 1
Second polynomial = 3x4 + 5x3 – 7x2 + 2x + 2
dividing second polynomial
by first polynomial ⇒
∵ Remainder is zero.
∴ First polynomial is a factor of second polynomial.
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3 5

(iii) First polynomial = x3 – 3x + 1
Second polynomial = x5 – 4x3 + x2 + 3x + 1.
∵ Remainder ≠ 0.
∴ First polynomial is not a factor of second polynomial.
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3 6

Ex 2.3 Solutions Class 10 Question 3.
Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are \(\sqrt { \frac { 5 }{ 3 } }\) and –\(\sqrt { \frac { 5 }{ 3 } }\)
Solution:
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3 7

NCERTSolutions Ex 2.3 Class 10 Question 4.
On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and -2x + 4, respectively. Find g(x).
Solution:
p(x) = x3 – 3 x 2 + x + 2 g(x) = ?
Quotient = x – 2; Remainder = -2x + 4
On dividing p(x) by g(x), we have
p(x) = g(x) x quotient + remainder
⇒ x3– 3x2 + x + 2 = g(x) (x – 2) + (-2x + 4)
⇒ x3 – 3x2 + x + 2 + 2 x- 4 = g(x) x (x-2)
⇒ x3 – 3x2 + 3x – 2 = g(x) (x – 2)
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3 8

Exercise 2.3 Class 10 Maths Question 5.
Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0
Solution:
(i) p(x),g(x),q(x),r(x)
deg p(x) = deg q(x)
∴ both g(x) and r(x) are constant terms.
p(x) = 2x2– 2x + 14; g(x) = 2
q(x) = x2 – x + 7; r(x) = 0

(ii) deg q(x) = deg r(x)
∴ this is possible when
deg of both q(x) and r(x) should be less than p(x) and g(x).
p(x) = x3+ x2 + x + 1; g(x) = x2 – 1
q(x) = x + 1, r(x) = 2c + 2

(iii) deg r(x) is 0.
This is possible when product of q(x) and g(c) form a polynomial whose degree is equal to degree of p(x) and constant term.

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NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2.

Board CBSE
Textbook NCERT
Class Class 10
Subject Maths
Chapter Chapter 8
Chapter Name Introduction to Trigonometry
Exercise Ex 8.2
Number of Questions Solved 4
Category NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2

Class 10 Ex 8.2 Question 1.
Evaluate the following:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2 1
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2 2
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2 3

Ex 8.2 Class 10 Question 2.
Choose the correct option and justify your choice:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2 4
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2 5
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2 6

Exercise 8.2 Class 10 Question 3.
If tan (A + B) = √3 and tan (A – B) = \(\frac { 1 }{ \surd 3 }\); 0° < A + B ≤ 90°; A > B, find A and B.
Solution:
tan (A + B) = √3
⇒ tan (A + B) = tan 60°
⇒ A + B = 60° ……(i)
tan (A – B) = \(\frac { 1 }{ \surd 3 }\)
⇒ tan (A – B) = tan 30°
⇒ A – B = 30° ……..(ii)
Adding equation (i) and (ii), we get
2A = 90° ⇒ A = 45°
From (i), 45° + B = 60° ⇒ B = 60° – 45° = 15°
Hence, ∠A = 45°, ∠B = 15°

Exercise 8.2 Class 10 NCERT Solution Question 4.
State whether the following statements are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin θ increases as θ increases.
(iii) The value of cos θ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2 7
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2 8

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NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1

NCERT Solutions for Class 10 Mathematics Chapter 1 Real Numbers Ex 1.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Mathematics Chapter 1 Real Numbers Ex 1.1.

Board CBSE
Textbook NCERT
Class Class 10
Subject Maths
Chapter Chapter 1
Chapter Name Real Numbers
Exercise Ex 1.1
Number of Questions Solved 5
Category NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1

Ex 1.1 Class 10 Question 1.
Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255.
Solutions:
(i) Given numbers are 135 and 225.
On applying Euclid’s division algorithm, we have
225 = 135 x 1 + 90
Since the remainder 90 ≠ 0, so again we apply Euclid’s division algorithm to 135 and 90, to get
135 = 90 x 1 + 45
Since the remainder 45 ≠ 0, so again we apply Euclid’s division algorithm to 90 and 45, to get
90 = 45 x 2 + 0
The remainder has now become zero, so we stop.
∵ At the last stage, the divisor is 45
∴ The HCF of 135 and 225 is 45.
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1 1
(ii) Given numbers are 196 and 38220
On applying Euclid’s division algorithm, we have
38220 = 196 x 195 + 0
Since we get the remainder zero in the first step, so we stop.
∵ At the above stage, the divisor is 196
∴ The HCF of 196 and 38220 is 196.
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1 2
(iii) Given numbers are 867 and 255
On applying Euclid’s division algorithm, we have
867 = 255 x 3 + 102
Since the remainder 102 ≠ 0, so again we apply Euclid’s division algorithm to 255 and 102. to get
255 = 102 x 2 + 51
Since the remainder 51 ≠ 0, so again we apply Euclid’s division algorithm to 102 and 51, to get
102 = 51 x 2 + 0
We find the remainder is 0 and the divisor is 51
∴ The HCF of 867 and 255 is 51.

Exercise 1.1 Class 10 Maths NCERT Solutions Question 2.
Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Solutions:
Let ‘a’ be any positive integer and b = 6.
∴ By Euclid’s division algorithm, we have
a = bq + r, 0 ≤ r ≤ b
a = 6q + r, 0 ≤ r ≤ b [ ∵ b = 6] where q ≥ 0 and r = 0,1, 2, 3, 4,5
Now, ‘a’ may be of the form of 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4, 6q + 5
If ‘a’ is of the form 6q, 6q + 2, 6q + 4 then ‘a’ is an even.
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1 3
In above we can see clearly that the numbers of the form 6q, 6q + 2, 6q + 4 are having the factor 2.
∴ The numbers of the form 6q, 6q + 2, 6q + 4 are even.
If ‘a’ is of the form 6q +1,6q +3, 6q + 5 then ‘a’ is an odd.
As if
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1 4
∵ We know that the number of the form 2k + 1 is odd.
∴ The numbers of the form 6q + 1, 6q + 3, 6q + 5 are odd.

Class 10 Maths Chapter 1 Question 3.
An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Solutions:
Maximum number of columns = HCF of (616, 32)
For finding the HCF we should apply Euclid’s division algorithm
Given numbers are 616 and 32
On applying Euclid’s division algorithm, we have
616 = 32 x 19 + 8
Since the remainder 8 ≠ 0, so again we apply Euclid’s division algorithm to 32 and 8, to get
32 = 8 x 4 + 0
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1 5
The remainder has now become zero, so we stop,
∵ At the last stage, the divisor is 8
∴ The HCF of 616 and 32 is 8.
Therefore, the maximum number of columns in which an army contingent of 616 members can march behind an army band of 32 members in a parade is 8.

Class 10 Maths Ex 1.1 Question 4.
Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Solutions:
Let ‘a’ be any positive integer and b = 3.
∴ By Euclid’s division algorithm, we have a = 3q + r, 0 ≤ r < b
a = 3q + r, 0 ≤ r < 3 [ ∵ b = 3] where q ≥ 0 and r = 0,1, 2
∴ a = 3q or 3q + 1 or 3q + 2
Now
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1 6
Thus, the square of any positive integer is either of the form 3m or 3m + 1.

Maths 1.1 Class 10 Question 5.
Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Solutions:
Let ‘a’ be any positive integer and b = 3.
∴ By Euclid’s division algorithm, we have a = bq + r,0 ≤ r ≤ b
a = 3q + r,0 ≤ r < 3 [ ∵ b = 3] where q ≥ 0 and r = 0. 1, 2
∴ a = 3q or 3q + 1 or 3q + 2
Now
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1 7
Thus, the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

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NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2.

Board CBSE
Textbook NCERT
Class Class 10
Subject Maths
Chapter Chapter 4
Chapter Name Quadratic Equations
Exercise Ex 4.2
Number of Questions Solved 6
Category NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2

Ex 4.2 Class 10 NCERT Solutions Question 1.
Find the roots of the following quadratic equations by factorisation:
(i) x2 -3x – 10 = 0
(ii) 2x2 + x – 6 = 0
(iii) √2x2 + 7x + 5√2 = 0
(iv) 2x2 – x + \(\frac { 1 }{ 8 }\) = 0 8
(v) 100 x2 – 20 X + 1 = 0
Solution:
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2 1
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2 2

Class 10 Maths 4.2 NCERT Solutions Question 2.
Solve the following situations mathematically:

(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹750. We would like to find out the number of toys produced on that day.
Solution:
(i) Let the number of marbles John had be x
Then, the number of marbles Jivanti had = 45 -x
The number of marbles left with John, when he lost 5 marbles = x – 5
The number of marbles left with Jivanti, when she lost 5 marbles = 45 -x – 5 = 40 -x
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2 3

Ex 4.2 Class 10 Question 3.
Find two numbers whose sum is 27 and product is 182.
Solution:
Let one number = x ,
∴ other number = 21 -x
According to question,
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2 4

Question 4.
Find two consecutive positive integers, sum of whose squares is 365.
Solution:
Let the two consecutive integers be x and x + 1.
According to question,
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2 5

Ex 4.2 Class 10 Question 5.
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Solution:
Let base = x cm,
∴ height = (x – 7) cm
By Pythagoras Theorem,
(base)² + (height)² = (hypotenuse)²
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2 6

Ex 4.2 Class 10 Question 6.
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹90, find the number of articles produced and the cost of each article.
Solution:
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2 7

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NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.1

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.1.

Board CBSE
Textbook NCERT
Class Class 10
Subject Maths
Chapter Chapter 4
Chapter Name Quadratic Equations
Exercise Ex 4.1
Number of Questions Solved 2
Category NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.1

NCERT Solutions For Class 10 Maths Chapter 4 Question 1.
Check whether the following are quadratic equations:
(i) (x+ 1)2=2(x-3)
(ii) x – 2x = (- 2) (3-x)
(iii) (x – 2) (x + 1) = (x – 1) (x + 3)
(iv) (x – 3) (2x + 1) = x (x + 5)
(v) (2x – 1) (x – 3) = (x + 5) (x – 1)
(vi) x2 + 3x + 1 = (x – 2)2
(vii) (x + 2)3 = 2x(x2 – 1)
(viii) x3 -4x2 -x + 1 = (x-2)3
Solution:
(i) (x+ 1)2=2(x-3)
⇒ x2 + 2x +1 = 2x – 6
⇒ x2 + 2x – 2x+1 + 6 = 0
⇒ x2 + 7 = 0
⇒ x2 + 0x + 1 = 0
Which is of the form
ax2 + bx + c = 0
Hence, the given equation is a quadratic equation.

(ii) x2 – 2x = (- 2) (3 -x)
⇒ x2 -2x = -6 + 2x
⇒ x2 -4x + 6 = 0
Which is of the form
ax2 + bx + c = 0
Hence, the given equation is a quadratic equation.

(iii) (x – 2) (x + 1) = (x – 1) (x + 3)
⇒ x2 + x-2x-2 = x2 + 3x-x -3
⇒ x2 + x – 2x – 2 = x2 – 3x + x + 3 = 0
⇒ – 3x + 1 = 0 ⇒ 3x – 1 = 0
Since degree of equation is 1, hence, given equation is not a quadratic equation.

(iv) (x-3) (2x+ 1) = x (x + 5)
⇒ 2x2 + x – 6x – 3 = x2 + 5x
2x2 + x – 6x – 3-x2 – 5x = 0
⇒ x2 – 10x -3 = 0
Which is of the form
ax2 + bx + c 0
Hence, the given equation is a quadratic equation.

(v) (2x-1)(x-3) = (x + 5)(x-1)
⇒ 2x2 – 6x-x + 3 = x2 -x + 5x – 5
2x2 – 6x-x + 3 = x2 + x – 5x + 5 = 0
⇒ x2 – 11x + 8 = 0
Which is of the form
ax2 + bx + c = 0
Hence, the given equation is a quadratic equation.

(vi) x2 + 3x + 1 = (x-2)2
⇒ x2 + 3x + 1 = x2 + 4 – 4x
⇒ x2 + 3x + 1 = x2– 4 + 4c = 0
⇒ 7x – 3 = 0
Since degree of equation is 1, hence, the given equation is not a quadratic equation,

(vii) (x + 2)3 = 2x (x2 – 1)
⇒ x3 + 8 + 3.x.2 (x + 2) = 2x3 – 2x
⇒ x3 + 8 + 6x2 + 12x = 2x3 – 2x
⇒ x3 – 6x2 – 14x – 8 = 0
Which is not of the form
ax2 + bx + c = 0
Hence, the given equation is not a quadratic equation.

(viii) x3 – 4x2 – x+1 = (x-2)3
⇒ x3 – 4x2 – x + 1 = x3-8 + 3x(-2)(x – 2)
⇒ x3 – 4x2 -x + 1 = x3 – 6x2 + 12x – 8
⇒ 2x2 – 13x + 9 = 0
Which is of the form
ax2 + bx + c = 0
Hence, the given equation is a quadratic equation.

Exercise 4.1 Maths Class 10 Solutions Question 2.
Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Solution:
(i) Let breadth of the rectangular plot = x m
Then, length of the plot = (2x + 1)m
Area of a rectangular plot = l x b ,
⇒ 528 (2x + 1)x
⇒ 528 = 2x2 +x
⇒ 2x2 + x – 528 = 0
Which is the required quadratic equation.

(ii) Let the two consecutive integers be x and x + 1
Then, x(x+l) = 306
⇒ x2 +x-306 = 0
Which is the required quadratic equation.

(iii) Let the present age of Rohan = x years
Rohan’s mother’s present age = (x + 26) years
After 3 years, Rohan’s age = (x + 3) years
After 3 years, Rohan’s mother’s age = (x + 26 + 3) years
According to question,
(x + 3) (x + 29) = 360
⇒ x2 + 29x + 3x + 87 – 360 = 0
⇒ x2 + 32x – 273 = 0
Which is the required quadratic equation.

(iv)
Let speed of the train = x km/h
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.1 1

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NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2

NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2.

Board CBSE
Textbook NCERT
Class Class 10
Subject Maths
Chapter Chapter 3
Chapter Name Pair of Linear Equations in Two Variables
Exercise Ex 3.2
Number of Questions Solved 7
Category NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2

Class 10th Exercise 3.2 Question 1.
Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.
Solution:
(i) Let the number of girls be x and number of boys be y.
A.T.Q.
1st Condition :
x +y =10
Table :
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 1
2nd Condition :
x = y + 4 ⇒ x – y = 4
Table :
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 2
Solving
(i) and
(ii) graphically
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 3
Both the lines cut at (7, 3)
Hence, solutions is (7, 3), i.e.x = 7,y = 3
Number of girls – 7 and Number of boys = 3

(ii) Let cost of 1 pencil = ₹ x and cost of 1 pen = ₹ y.
A.T.Q.
1st Condition :
5x + 7y = 50
Table :
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 4
2nd Condition :
7x + 5y = 46
Table :
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 5
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 6

Class 10 Maths Ex 3.2 Question 2.
On comparing the ratios and find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) 5x – 4y + 8 = 7x + 6y – 9 = 0
(ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0
(iii) 6 – 3y + 10 = 0, Zx – v + 9 = 0
Solution:
(i) Equations are 5x – 4y + 8 = 7x + 6y – 9 = 0

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 7
∴ Pair of lines represented by given equations intersect at one point. So, the system has exactly one solution.

(ii)
9x + 3y + 12 = 0, 18 + 6y + 24 = 0
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 8
∴ Pair of equations represents coincident lines and having infinitely many solutions.


(iii)
6x – 3y + 10 = 0, 2x -y + 9 = 0
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 9
∴ It represents parallel lines and having no solution.

Ex 3.2 NCERT Class 10 Question 3.
On comparing the ratios and find out whether the following pair of linear equations are consistent, or inconsistent,

(i) 3x + 2y = 5; 2x – 3y = 7
(ii) 2x – 3y = 8; 4x – 6y = 9
(iii) 3/2x + 5/3y = 7; 9x – 10y = 14
(iv) 5x-3y = 11; -10c + 6y = -22
(v) 4/3x + 2y = 8; 2x + 3y = 12

Solution:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 10
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 11
∴ Pair of equations is consistent with infinitely many solutions.

Ex 3.2 Class 10 NCERT Solutions Question 4.
Which of the following pairs of linear equations are consistent/inconsistent If consistent, obtain the solution graphically:
(i) x + y = 5, 2x + 2y = 10
(ii) x – y = 8, 3x – 3y = 16
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

Solution:
(i) x + y – 5, 2x + 2y = 10
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 12
Here,
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 14

∴ System of equations is consistent and the graph gr represents coincident lines.
Table for equation (i),
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 15

Table for equation (ii),
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 15a

(ii) x – y = 8, 3x – 3y = 16
Here \(\frac { a_{ 1 } }{ a_{ 2 } } =\frac { 1 }{ 3 } ,\)
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 16

Pair of equations is inconsistent. Hence, lines are parallel and
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 17


(iii)
2x + y – 6 = 0, 4x – 2y – 4 = 0
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 18
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 19
Pair of equations is consistent.
Table for equation 2x + y – 6 = ()
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 20
Table for equation 4x – 2y – 4 = 0
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 21

(iv)
2x – 2y – 2 = 0, 4x – 4y – 5 = 0
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 22
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 23

∴ Pair of equations is inconsistent. Hence, lines are parallel and system has no solution.

Class 10 Maths Chapter 3 Exercise 3.2 Question 5.
Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Solution:
Let length of garden = x m and width of garden = y m
Perimeter of rectangular garden = 2(x + y)
A.T.Q.
1st Condition :
\(\frac { 2(x+y) }{ 2 }\) = 36 ⇒ x + y = 36
2nd Condition :
x = y + 4 ⇒ x -y = 4 … (ii)
Adding equation (i) and (ii), we get
2x = 40 ⇒ x = 20
Putting x = 20 in equation (i), we get
20 + y = 36
y = 16
Hence, dimensions of the garden are 20 m and 16 m.

Exercise 3.2 Class 10 Question 6.
Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables, such that the geometrical representation of the pair so formed is:
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines
Solution:
Given equation is 2x + 3y – 8 – 0
We have 2x + 3y = 8
Let required equation be ax + by = c
Condition :
(i) For intersecting lines
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 24
\(\frac { 2 }{ a } \neq \frac { 3 }{ b } \neq \frac { 8 }{ c }\) where a, b, c can have any value which satisfy the above condition.
Let a = 3, b = 2, c = 4
so, \(\frac { 2 }{ 3 } \neq \frac { 3 }{ 2 } \neq \frac { 8 }{ 4 }\)
∴ Equations are 2v + 3y = 8 and 3A + 2y = 4 have unique solution and their geometrical representation shows intersecting lines.

(ii) Given equation is 2x + 3y = 8 Required equation be ax + by = c
Condition :
For parallel lines
\(\frac { 2 }{ a } \neq \frac { 3 }{ b } \neq \frac { 8 }{ c }\)
where a, b, c can have any value which satisfy the above condition.
Let, a = 2, b = 3, c = 4
Required equation will be 2x + 3y = 4
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 25
Equations 2x + 3y = 8 and 2x + 3y = 4 have no solution and their geometrical representation shows parallel lines.

(iii)
For coincident lines:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 26
Given equation is 2x + 3y = 8 Let required equation be ax + by = c For coincident lines.
\(\frac { 2 }{ 3 } =\frac { 3 }{ 2 } =\frac { 8 }{ 4 } \) where a, b, c can have any a b c possible value which satisfy the above condition.
Let a = 4, b = 6, c = 16 Required equation will be 4x + 6y = 16.
Equations 2x + 3y = 8 and 4x + 6y = 16 have infinitely many solutions and their geometrical representation shows coincident lines.

NCERT Ex 3.2 Class 10 Question 7.
Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y -12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x – axis, and shade the triangular region.
Solution:
First equation is x – y + 1 = 0.
Table for 1st equation x = y – 1
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 27
Second equation is 3x + 2y = 12
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 28
⇒ 3x = 12 – 2y ⇒ x = \(x=\frac { 12-2y }{ 3 }\)
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 29
Required triangle is ABC. Coordinates of its vertices are A(2, 3), B(-1, 0), C(4, 0).

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NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6

NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6.

Board CBSE
Textbook NCERT
Class Class 10
Subject Maths
Chapter Chapter 3
Chapter Name Pair of Linear Equations in Two Variables
Exercise Ex 3.6
Number of Questions Solved 2
Category NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6

NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.6 Question 1.
Solve the following pairs of equations by reducing them to a pair of linear equations:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 1
Solution:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 2
By cross multiplication method
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 3
By cross multiplication method
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 4
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 5
By cross multiplication method
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 6
By cross multiplication method
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 7
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 8
solving for u and v by cross multiplication method:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 9
By cross multiplication method:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 10
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 11
By cross multiplication method:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 12
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 13
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 14

NCERT Solutions For Class 10th Maths Chapter 3 Exercise 3.6 Question 2.
Formulate the following problems as a pair of equations, and hence find their solutions:
(i) Ritu can row’ downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
Solution:
(i) Let Ritu’s speed in still water = x km/h
Speed of current = y km/h
During downstream, speed = (x + y) km/h
During upstream, speed = (x -y) km/h
A.T.Q.
1st condition :
x + y = 20/2 ⇒ x + y = 10
2nd condition :
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 15
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 16
Time taken by 1 woman to finish the work = 18 days.
Time taken by 1 man to finish the work = 36 days.

(iii) Let speed of train = x km/h and Speed of bus = y km/h
Total distance = 300 km
A.T.Q.
1st condition :
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 17
2nd condition :
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 18

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NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2.

Board CBSE
Textbook NCERT
Class Class 10
Subject Maths
Chapter Chapter 5
Chapter Name Arithmetic Progressions
Exercise Ex 5.2
Number of Questions Solved 20
Category NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

Ex 5.2 Class 10 Maths Solutions Question 1.
Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 1
Solution:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 2

Class 10 Maths Chapter 5 Exercise 5.2 Question 2.
Choose the correct choice in the following and justify:
(i) 30th term of the AP: 10, 7, 4, …, is
(A) 97
(B) 77
(C) -77
(D) -87

(ii) 11th term of the AP: -3, \(\frac { -1 }{ 2 }\) , 2, …, is
(A) 28
(B) 22
(C) -38
(D) -48
Solution:
(i) 10, 7, 4, …,
a = 10, d = 7 – 10 = -3, n = 30
an = a + (n – 1)d
⇒ a30 = a + (30 – 1) d = a + 29 d = 10 + 29 (-3) = 10 – 87 = – 77
Hence, correct option is (C).
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 3

Class 10 Maths Chapter 5 Exercise 5.2 Question 3.
In the following APs, find the missing terms in the boxes:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 4
Solution:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 5
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 6

Class 10 Maths Chapter 5 Exercise 5.2 Question 4.
Which term of the AP: 3, 8, 13, 18, …, is 78?
Solution:
Given: 3, 8, 13, 18, ………,
a = 3, d = 8 – 3 = 5
Let nth term is 78
an = 78
a + (n – 1) d = 78
⇒ 3 + (n – 1) 5 = 78
⇒ (n – 1) 5 = 78 – 3
⇒ (n – 1) 5 = 75
⇒ n – 1 = 15
⇒ n = 15 + 1
⇒ n = 16
Hence, a16 = 78

Class 10 Maths Chapter 5 Exercise 5.2 Question 5.
Find the number of terms in each of the following APs:
(i) 7, 13, 19, …, 205
(ii) 18, 15\(\frac { 1 }{ 2 }\), 13, …, -47
Solution:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 7

Class 10 Ex 5.2 Solutions Question 6.
Check, whether -150 is a term of the AP: 11, 8, 5, 2, ….
Solution:
11, 8, 5, 2, …….
Here, a = 11, d = 8 – 11= -3, an = -150
a + (n – 1) d = an
⇒ 11 + (n – 1) (- 3) = -150
⇒ (n – 1) (- 3) = -150 – 11
⇒ -3 (n – 1) = -161
⇒ n – 1 = \(\frac { -161 }{ -3 }\)
⇒ n = \(\frac { 161 }{ 3 }\) + 1 = \(\frac { 164 }{ 3 }\) = 53\(\frac { 4 }{ 3 }\)
Which is not an integral number.
Hence, -150 is not a term of the AP.

Maths NCERT Solutions Class 10 Arithmetic Progression Exercise 5.2 Question 7.
Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
a11 = 38 and a16 = 73
⇒ a11 = a + (11 – 1) d ⇒ a + 10d = 38 ….. (i)
⇒ a16 = a + (16 – 1 )d ⇒ a + 15d = 73 …(ii)
Subtracting eqn. (i) from (ii), we get
a + 15d – a – 10d = 73 – 38
⇒ 5d = 35
⇒ d = 1
From (i), a + 10 x 7 = 38
⇒ a = 38 – 70 = – 32
a31 = a + (31 – 1) d = a + 30d = – 32 + 30 x 7 = – 32 + 210 = 178

Class 10 Maths Chapter 5 Exercise 5.2 Question 8.
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Solution:
Given:
a50 = 106
a50 = a + (50 – 1) d
⇒ a + 49d = 106 …(i)
and a3 = 12 ⇒ a3 = a + (3 – 1 )d ⇒ a + 2d = 12 …(ii)
Subtracting eqn. (ii) from (i), we get
a + 49d – a – 2d = 106 – 12
⇒ 47d = 94
⇒ d = \(\frac { 94 }{ 47 }\) = 2
a + 2d = 12
⇒ a + 2 x 2 = 12
⇒ a + 4 = 12
⇒ a = 12 – 4 = 8
a29 = a + (29 – 1) d = a + 28d = 8 + 28 x 2 = 8 + 56 = 64

Ex 5.2 Class 10 NCERT Solutions Question 9.
If the 3rd and the 9th term of an AP are 4 and -8 respectively, which term of this AP is zero?
Solution:
Given: a3 = 4 and a9 = – 8
⇒ a3 = a + (3 – 1 )d ⇒ a + 2d = 4 …(i)
a9 = a + (9 – 1) d ⇒ a + 8d = -8 ….(ii)
Subtracting eqn. (i) from (ii), we get
a + 8d – a – 2d = -8 – 4
⇒ 6d = -12.
⇒ d = -2
Now,
a + 2d = 4
⇒ a + 2(-2) = 4
⇒ a – 4 = 4
⇒ a = 4 + 4 = 8
Let an = 0
⇒ a + (n – 1) d = 0
⇒ 8 + (n – 1) (- 2) = 0
⇒ 8 = 2 (n – 1)
⇒ n – 1 = 4
⇒ n = 4 + 1 = 5
Hence, 5th term is zero.

Exercise 5.2 Class 10 NCERT Solutions Question 10.
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution:
Given: a17 – a10 = 7
⇒ [a + (17 – 1 ) d] – [a + (10 – 1 ) d] = 7
⇒ (a + 16d) – (a + 9d) = 7
⇒ 7d = 7
⇒ d = 1

Class 10 Maths Chapter 5 Exercise 5.2 Question 11.
Which term of the AP: 3, 15, 27, 39, … will be 132 more than its 54th term?
Solution:
3, 15, 27, 39, …..
Here, a = 3, d = 15 – 3 = 12
Let an = 132 + a54
⇒ an – a54 = 132
⇒ [a + (n – 1) d] – [a + (54 – 1) d] = 132
⇒ a + nd – d – a – 53d = 132
⇒ 12n – 54d = 132
⇒ 12n – 54 x 12 = 132
⇒ (n – 54)12 = 132
⇒ n – 54 = 11
⇒ n = 11 + 54 = 65

Class 10 Maths Chapter 5 Exercise 5.2 Question 12.
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Solution:
Let a and A be the first term of two APs and d be the common difference.
Given:
a100 – A100 = 100
⇒ a + 99d – A – 99d = 100
⇒ a – A = 100
⇒ a1000 – A1000 = a + 999d – A – 999d
⇒ a – A = 100
⇒ a1000 – A1000 = 100

NCERT Solutions Class 10 Maths Ch 5 Ex 5.2 Question 13.
How many three-digit numbers are divisible by 7?
Solution:
The three-digit numbers which are divisible by 7 are 105, 112, 119, ………., 994
Here, a = 105, d = 112 – 105 = 7 , an = 994
a + (n – 1) d = 994
⇒ 105 + (n – 1) 7 = 994
⇒ (n – 1) 7 = 994 – 105
⇒ 7 (n – 1) = 889
⇒ n – 1 = 127
⇒ n = 127 + 1 = 128

CBSE Class 10 Maths Ex 5.2 Solutions Question 14.
How many multiples of 4 lie between 10 and 250?
Solution:
The multiples of 4 between 10 and 250 be 12, 16, 20, 24,…., 248
Here, a = 12, d = 16 – 12 = 4, an = 248
an = a + (n – 1) d
⇒ 248 = 12 + (n – 1) 4
⇒ 248 – 12 = (n – 1) 4
⇒ 236 = (n – 1) 4
⇒ 59 = n – 1
⇒ n = 59 + 1 = 60

NCERT Solutions For Class 10 Maths Ch 5 Ex 5.2 Question 15.
For what value of n, the nth term of two APs: 63, 65, 61,… and 3, 10, 17,… are equal?
Solution:
First AP
63, 65, 67,…
Here, a = 63, d = 65 – 63 = 2
an = a + (n – 1) d = 63 + (n – 1) 2 = 63 + 2n – 2 = 61 + 2n
Second AP
3, 10, 17, …
Here, a = 3, d = 10 – 3 = 7
an = a + (n – 1) d = 3 + (n – 1)7 = 3 + 7n – 7 = 7n – 4
Now, an = an
⇒ 61 + 2n = 7n – 4
⇒ 61 + 4 = 7n – 2n
⇒ 65 = 5n
⇒ n = 13

Class 10 Maths Chapter 5.2 Question 16.
Determine the AP whose 3rd term is 16 and 7th term exceeds the 5th term by 12.
Solution:
Given: a3 = 16
⇒ a + (3 – 1)d = 16
⇒ a + 2d = 16
and a7 – a5 = 12
⇒ [a + (7 – 1 )d] – [a + (5 – 1 )d] = 12
⇒ a + 6d – a – 4d = 12
⇒ 2d = 12
⇒ d = 6
Since a + 2d = 16
⇒ a + 2(6) = 16
⇒ a + 12 = 16
⇒ a = 16 – 12 = 4
a1 = a = 4
a2 = a1 + d = a + d = 4 + 6 = 10
a3 = a2 + d = 10 + 6 = 16
a4 = a3 + d = 16 + 6 = 22
Thus, the required AP is a1, a2, a3, a4,…, i.e. 4, 10, 16, 22

Class 10 Maths Chapter 5 Exercise 5.2 Question 17.
Find the 20th term from the last term of the AP: 3, 8, 13, …, 253.
Solution:
Given: AP is 3, 8, 13,…….. ,253
On reversing the given A.P., we have
253, 248, 243 ,………, 13, 8, 3.
Here, a = 253, d = 248 – 253 = -5
a20 = a + (20 – 1)d = a + 19d = 253 + 19 (-5) = 253 – 95 = 158

Exercise 5.2 Class 10 Solutions Question 18.
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Solution:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 8

Solution Of Ex 5.2 Class 10 Question 19.
Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000 ?
Solution:
a = ₹ 5000, d = ₹ 200
Let an = ₹ 7000
We have, a + (n – 1) d = 7000
⇒ 5000 + (n – 1) 200 = 7000
⇒ (n – 1) 200 = 7000 – 5000
⇒ (n – 1) 200 = 2000
⇒ (n- 1) = 10
⇒ n = 11
⇒ 1995 + 11 = 2006
Hence, in 2006 Subba Rao’s income will reach ₹ 7000.

NCERT Solutions For Class 10 Maths 5.2 Question 20.
Ramkali saved ₹ 5 in the first week of a year and then increased her weekly saving by ₹ 1.75. If in the nth week, her weekly saving become ₹ 20.75, find n.
Solution:
Given: a = ₹ 5, d = ₹ 1.75
an = ₹ 20.75
a + (n – 1) d – 20.75
⇒ 5 + (n – 1) 1.75 = 20.75
⇒ (n – 1) x 1.75 = 20.75 – 5
⇒ (n – 1) 1.75 = 15.75
⇒ n – 1 = 9
⇒ n = 9 + 1
⇒ n = 10
Hence, in 10th week Ramkali’s saving will be ₹ 20.75.

We hope the NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Science Chapter 16 Management of Natural Resources

NCERT Solutions for Class 10 Science Chapter 16 Management of Natural Resources

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 16 Management of Natural Resources. LearnInsta.com provides you the Free PDF download of NCERT Solutions for Class 10 Science (Biology) Chapter 16 – Management of Natural Resources solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 16 – Management of Natural Resources Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Questions

In Text Questions

Question 1.
What changes would you make in your habits to become environment friendly ? (CBSE AI 2009)
Answer:
Three Rs — reduce, recycle and reuse.

  1. Reduce: It is to reduce consumption by preventing wastage.
    1. Switching off unnecessary lights, fans and other electrical appliances,
    2. Repair of leaky taps.
    3. Reducing food wastage,
    4. Walking down to nearby market instead of using vehicle.
  2. Recycle: Separation of recyclable wastes from non-recyclable wastes. The former are taken by rag pickers for sending them to industries involved in recycling, e.g., paper, plastic, metal, glass.
  3. Reuse: Carry bags, packing material, plastic containers and other reusable articles should not be thrown away if the same are uncontaminated. For example, plastic bottles and jars containing various food items brought from market can be washed and used for storing things in the kitchen.

More Resources

Question 2.
What would be advantages of exploiting resources with short term aims ?
Answer:

  1. All the commodities for use and comforts for human society will become available in plenty.
  2. There will be rapid industrialisation and development.

Question 3.
How would these advantages differ from the advantages of using a long term prospective in managing our resources ?
Answer:

Short Term Aims

Long Term Aims

1. All the commodities become available in plenty. 1. Availability of various commodities is restricted.
2. There is a tendency to manufacture articles of larger size. 2. There is a tendency to manufacture articles of smaller size.
3. The mentality is of use and throw. 3. The mentality is reuse, recycle and replace.
4. The resources are likely to get exhausted soon. 4. The resources are likely to remain available for a long time.
5. The ecology is disturbed and harmed. 5. It is eco-friendly.

 

Question 4.
Why do you think there should be equitable distribution of resources ? What forces would be working against an equitable distribution of our resources ?
Answer:
Need for Equitable Distribution. Resources are living and non-living components of nature which are drawn upon to provide food, fodder, shelter, water, energy, articles of daily use and comforts. Every human being has the fundamental right to obtain and use the same. This is possible only when there is equitable distribution of resources.
Forces Against Equitable Distribution,

  1. Restricted availability of resources,
  2. Excessive consumption by the rich,
  3. Profit motto by the persons exploiting resources.

Question 5.
Why should we conserve forests and wildlife ?
Answer:
There are economic and ecological reasons to conserve forests and wildlife.

  1. Economic Reasons:
    1. Forests are a source of shelter, food and fodder to people living in and around forests,
    2. They provide firewood and timber,
    3. A number of industrial raw materials are obtained from forests, e.g., tannins, dyes, resins, perfumes, wood for paper,
    4. Several medicines are got from forest plants and animals,
    5. Wildlife is a gene bank for improvement of domesticated plants and animals,
    6. All plants and animals have not yet been explored for their potential uses.
  2. Ecological Reasons:
    1. They regulate climate,
    2. Forests help in retaining rain water and its storage in aquifers from where water becomes available to springs and streams,
    3. Forests control soil erosion and prevent the occurrence of floods,
    4. Wildlife is essential for maintaining the forests.
    5. CO2-O2 balance is maintained.

Question 6.
Suggest some approaches towards the conservation of forests.
Answer:
Conservation is protection, augmentation and scientific management of a resource so as to maintain it at its optimum level while providing sustainable benefits for the present as well as future generations. Some approaches for conservation of forests are as follows :

  1. Afforestation and Reforestation. Afforestation is developing forest over an area where none existed earlier. Reforestation is developing forest cover which has been cleared during exploitation. Degraded forests are also mended to bring them back to health. All this is done by state forest department either by themselves or with the help of joint forest management committees.
  2. Separation of Commercial Forestry. Useful plants required for meeting commercial needs should be planted separately so that there is no undue pressure on natural forests. Social forestry is growing multipurpose plants on village common lands for meeting requirement of fodder, firewood and small timber. Production plantation is growing industry required plants on wastelands.
  3. Controlled grazing.
  4. Prevention of scraping and litter removal.
  5. Practising block cutting or cutting of forest equal to its regeneration capacity.
  6. Building of national parks, sanctuaries and biosphere reserves.

Question 7.
Find out about traditional systems of water harvesting management in your region.
Answer:
Ponds, pools, tanks, lakes, ditches and wetlands. They get filled up during rain. Larger reservoirs are perennial. The smaller ones dry up after a few months. However, both help in recharge of ground water.

Question 8.
Compare the above system with the probable system in hilly/mountainous areas or plains or plateau regions.
Answer:
Hilly/Mountainous Areas. Water is impounded on the slopes, at the base of slope, on the border of ravine or melting snow into tanks called Kulhs, Khatris, Kuls (Himachal Pardesh, J and K), Zabo (Nagaland) and Zings (Ladakh). Water from tanks is used for different purposes—drinking, washing, animals, irrigation.
Plains,

  1. Roof top harvesting of rain water has been in practice since ages in Rajasthan. The collected water is stored below the ground level for later use.
  2. Other methods of water harvesting are specially constructed tanks, Kuis, jhaslars, booris or bers, paar, vivdas, khadins and nadis (Rajasthan), ponds in Kandi belts (Jammu), bhandaras and tals (Maharashtra), bundhis (MP and UP), ahars and pynes (Bihar).
    Plateau. Water is collected in series of tanks through impounding rain water, building check dams and channels from flowing water. They are called eris (Tamil Nadu), surrangams (Kerala) and kattas (Karnataka).

Question 9.
Find out the source of water in your region!locality. Is water from this source available to all people living in that area ?
Answer:
Water is supplied by municipal corporation through pipes. It is obtained from river/canal (if the same happens to be nearby). In many cases piped water is pumped out from ground by tube wells and stored in raised water tanks.
Municipal water supply is not available to all the residents. Many newly built colonies, slums and unauthorised localities are often without the piped water.

NCERT Chapter End Exercises

Question 1.
What changes would you suggest in your home in order to be environment friendly ?
(CBSE AI 2009, CCE 2012, 2014)
Answer:

  1. Judicious use of electricity by switching off lights and electrical appliances not required,
  2. Replacement of incandescent bulbs with fluorescent, compact fluorescent ones and LED bulbs.
  3. Replacement of electricity or gas operated geysers with solar water heaters,
  4. Replacement of electricity generating sets with solar light,
  5. Having more natural light and ventilation with design supporting warming during winters and cooling during summer,
  6. Reducing wastage of water, food and other articles.
  7. Separation of recyclable waste from non-cyclable waste prior to disposal.
  8. Increasing reuse of containers,
  9. Using cloth bags instead of polythene, plastic or paper bags.

Question 2.
Can you suggest some changes in your school which would make it environment friendly ?
Answer:
Yes.

  1. Rainwater harvesting (RWH) by draining roof top rain for underground seepage,
  2. Plantation of trees and shrubs along the boundary for protection against noise and air pollution. Plants will not only provide more oxygen during the school hours but also have a moderating effect on climate.
  3. Prevention of wastage of electricity and water,
  4. Having compost pits for conversion of plant refuse and other biodegradable wastes into useful resource,
  5. Increasing use of solar energy,
  6. Activating ecoclub.

Question 3.
There are four main stakeholders when it comes to forests and wildlife. Which among these should have the authority to decide the management offorest produce ? Why do you think so (CBSE Foreign 2010, CCE 2011)
Answer:
The four stakeholders in forests and wildlife are

  1. People living in and around forests who obtain most of their requirements from forests,
  2. Forest department which owns the forests.
  3. Industrialists who obtain raw materials for their industries,
  4. Wildlife and nature enthusiasts. Among them the persons living in and around the forests have traditional knowledge of how best the forests can be managed so that a sustainable yield can be obtained indefinitely. Therefore, authority for management of forest produce should be handed over to them. However, the forest department must monitor the same so that there is no excessive exploitation.

Question 4.
How can you as an individual contribute or make a difference to the management of
(a) Forests and wildlife
(b) Water resources
(c) Coal and petroleum ?
Answer:
(a) Forests and Wildlife:

  1. I will stress on sustainable use of produce from forests and wildlife,
  2. The area under forest cover has to be increased to control global warming.
  3. Form more joint forest management committees,
  4. Separate commercial forestry from natural forestry.

(b) Water Resources. My major stress will be

  1. Recharging of ground water through rain water harvesting, protection to wetlands and digging sponge pits or wells in the beds of rivulets,
  2. Reducing spoilage of water in agriculture by helping formers to switch over to sprinkler system of irrigation, changing conventional method of rice cultivation with system of rice intensification (SRI),
  3. Judicious use of water for domestic and industrial purposes.

(c) Coal and Petroleum:

  1. Higher level of purification to reduce pollution,
  2. Increased use of solar energy,
  3. More investment to commercialise hydrogen powered transport system.

Question 5.
What can you, as an individual, do to reduce your consumption of various natural resources ?
Answer:
Follow the principle of three Rs, i.e, reduce, recycle, reuse.

Question 6.
List five things you have done over the last one week to
(a) Conserve our natural resources
(b) Increase the pressure on our natural resources ?
Answer:
(a) Conservation of Natural Resources

  1. Reduced consumption of electricity by switching off unwanted fans and bulbs,
  2. Reduced consumption of water by closing the tap to prevent overflow of water in the bucket during taking bath,
  3. Used refills instead of purchasing new pens,
  4. I did not waste food,
  5. I have started going to school by school bus instead of using car.

(b) Increase Pressure on Natural Resources

  1. While reading at night on bed, I fell asleep and forgot to switch off the lights,
  2. While brushing my teeth and later taking bath, I did not close the tap, so a lot of water was wasted,
  3. I left the ball pen uncapped, so the ink dried up. I had to throw three ball pens last week,
  4. I am in the habit of using loose sheets for doing rough work. They are then thrown away,
  5. I bring more food in my tiffin than I eat. Daily I throw a lot of food in the dust bin.

Question 7.
On the basis of environmental issues you studied, what changes would you incorporate in your life style in a move towards a sustainable use of our resources ?
Answer:
1. Follow the principle of three Rs — reduce, recycle and reuse.
2. Undertake tree plantation twice a year during days of Van Mahotsava.
3. Where possible, use of public transport, school, bus and car pools.
4. Use of solar energy for as many purposes as possible.

Selection Type Questions

Alternate Response Type Questions
(True/False, Right(√)/Wrong (x), Yes/No)

Question 1.
Presence of coliform bacteria in a water body indicates contamination by industrial effluents.
Question 2.
An environmentally friendly decision is reuse jam and pickle bottles.
Question 3.
Forests are reservoirs of wildlife.
Question 4.
Chipko andolan originated in Haridwar during 1980s.
Question 5.
Check dams are built along seasonal flooded gullies for water harvesting.
Question 6.
Coal and petroleum are non-renewable inexhaustible resource.
Question 7.
An important protective function of forests is reduction of atmospheric pollution.
Question 8.
Forest department has been able to maintain biodiversity by growing pine, Teak and Eucalyptus on large tracts.

Matching Type Questions

Question 9.
Match the columns I and II (single matching) :

Column I

Column II

(a)    Desertification

(b)    Khejri

(c)   Khadin

(d)   Carbon monoxide

(i) Amrita Devi Bishnoi

(ii) Incomplete combustion

(iii)    Deforestation

(iv)   Water harvesting

Question 10.
Match the contents of columns I, II and III (double matching) :

Column I

Column II

Column III

(a)    Flyash

(b)    Kulhs

(c)    Van Mahotsav

(d)    Water pollution

(i) Coliform count

(ii) Coal

(iii)     Water harvesting

(iv)     Afforestation

p.  Himachal Pradesh

q. BOD

r. Toxic chemicals

s. February, July

Question 11.
Mark the resource as I (inexhaustible), R (renewable) and N (non-renewable) (key or check list items):
NCERT Solutions for Class 10 Science Chapter 16 Management of Natural Resources 1

Question 12.
Match stimulus with appropriate response.

Water Harvesting

Karnataka A Tamil Nadu B Kerala C

(i) Surrangams

(ii) Kattas

(iii) Eris

Fill in the Blanks

Question 13. Forests are ………….. hot spots.
Question 14. Amrita Devi Bishnoi sacrificed her life and that of 363 others in for ……………….. protection of …………….. trees.
Question 15. In blood carbon monoxide forms …………… that is unable to transport oxygen.
Question 16. Ganga runs its course of 2500 km from …………….. in Bay of Bengal.
Question 17. To save energy and prevent warming, I use ……………… instead of incandescent bulbs.

Answers:
NCERT Solutions for Class 10 Science Chapter 16 Management of Natural Resources image - 1

NCERT Solutions for Class 10 Science Chapter 16 Management of Natural Resources

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