In this post, you will find the NCERT solutions for class 10 maths ex 3.2. These solutions are based on the latest curriculum of NCERT Maths class 10.
NCERT Solutions for Class 10 Maths Ex 3.2
1. Solve the following pair of linear equations by the substitution method.
(i) x + y = 14
x – y = 4
(ii) s – t = 3
s/3 + t/2 = 6
(iii) 3x – y = 3
9x − 3y = 9
(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
(v) √2x + √3y = 0
√3x + √8y = 0
(vi) (3/2)x – (5/3)y = −2
x/3 + y/2 = 13/6
Solution: (i) x + y = 14 ……(1)
x – y = 4 ….… (2)
From equation (2), we get x = 4 + y
Putting this value of x in equation (1), we get
4 + y + y = 14
⇒ 2y = 10
⇒ y = 5
Putting the value of y in equation (1), we get
x + 5 = 14
⇒ x = 14 – 5
⇒ x = 9
Therefore, x = 9 and y = 5.
(ii) s – t = 3 ...... (1)
s/3 + t/2 = 6
From equation (1), we get s = 3 + t
Putting this value of s in equation (2), we get
(3 + t)/3 + t/2 = 6
⇒ (6 + 2t + 3t)/6 = 6
⇒ 5t + 6 = 36
⇒ 5t = 30
⇒ t = 6
Putting the value of t in equation (1), we get
s – 6 = 3
⇒ s = 3 + 6
⇒ s = 9
Therefore, t = 6 and s = 9.
(iii) 3x – y = 3 … (1)
9x − 3y = 9 … (2)
Comparing equation 3x – y = 3 with a1x + b1y + c1 = 0 and equation 9x − 3y = 9 with a2x + b2y + c2 = 0, we get
a1 = 3, b1 = –1, c1 = –3, a2 = 9, b2 = –3 and c2 = –9
Here, a1/a2 = b1/b2 = c1/c2
Therefore, we have infinitely many solutions for x and y.
(iv) 0.2x + 0.3y = 1.3 … (1)
0.4x + 0.5y = 2.3 … (2)
From equation (1), we have
0.2x = 1.3 − 0.3y
⇒ x = (1.3 − 0.3y)/0.2
Putting this value of x in equation (2), we get
0.4(1.3 − 0.3y)/0.2 + 0.5y = 2.3
⇒ 2.6 − 0.6y + 0.5y = 2.3
⇒ −0.1y = 2.3 – 2.6
⇒ −0.1y =−0.3
⇒ y = 3
Putting the value of y in equation (1), we get
0.2x + 0.3(3) = 1.3
⇒ 0.2x + 0.9 = 1.3
⇒ 0.2x = 1.3 − 0.9
⇒ 0.2x = 0.4
⇒ x = 2
Therefore, x = 2 and y = 3.
(v) √2x + √3y = 0 ……….(1)
√3x − √8y = 0 ……….(2)
From equation (1), we get
x = −√3y/√2
Putting this value of x in equation (2), we get
√3(−√3y/√2) − √8y = 0
⇒ −3y/√2 −√8y = 0
⇒ y = 0
Putting the value of y in equation (1), we get x = 0.
Therefore, x = 0 and y = 0.
(vi) (3/2)x – (5/3)y = −2
x/3 + y/2 = 13/6
From equation (2), we have
x = (13/6 – y/2) × 3
⇒ x = 13/2 – 3y/2
Putting this value of x in equation (1), we get
⇒ y = 12/4
⇒ y = 3
Putting the value of y in equation (2), we get
⇒ x/3 = 2/3
⇒ x = 2
Therefore, x = 2 and y = 3.
2. Solve 2x + 3y = 11 and 2x − 4y = −24 and hence find the value of ‘m’ for which
y = mx + 3.
Solution: 2x + 3y = 11 … (1)
2x − 4y = −24 … (2)
From equation (2), we get
2x = −24 + 4y
⇒ x = −12 + 2y
Putting this value of x in equation (1), we get
2(−12 + 2y) + 3y = 11
⇒ −24 + 4y + 3y = 11
⇒ 4y + 3y = 11 + 24
⇒ 7y = 35
⇒ y = 5
Putting this value of y in equation (1), we get
2x + 3(5) = 11
⇒ 2x + 15 = 11
⇒ 2x = 11 – 15 = −4
⇒ x = −2
Therefore, x = −2 and y = 5.
Putting the values of x and y in y = mx + 3, we get
5 = m(−2) + 3
⇒ 5 = −2m + 3
⇒ −2m = 2
⇒ m = −1
3. Form a pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii)The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it becomes 5/6. Find the fraction.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Solution: (i) Let the first number be x and the second number be y. Let x > y.
According to the first condition, we have
x – y = 26 … (1)
According to the second condition, we have
x = 3y … (2)
Putting the value of x from equation (2) in equation (1), we get
3y – y = 26
⇒ 2y = 26
⇒ y = 13
Putting the value of y in equation (2), we get
x = 3y = 3 × 13 = 39
Therefore, the two numbers are 13 and 39.
(ii) Let the two supplementary angles be x and y. Let y > x.
According to the given conditions, we have
y = x + 18 … (1)
Also, x + y = 180° (Sum of supplementary angles is 180°) … (2)
Putting the value of y from equation (1) in equation (2), we get
x + x + 18 = 180
⇒ 2x = 180 – 18 = 162
⇒ x = 81°
Putting the value of x in equation (1), we get
y = x + 18 = 81 + 18 = 99°
Therefore, the two supplementary angles are 81° and 99°.
(iii) Let the cost of one bat be Rs x and the cost of one ball be Rs y.
According to the first condition, we have
7x + 6y = 3800 … (1)
According to the second condition, we have
3x + 5y = 1750 … (2)
From equation (1), we have
7x = 3800 − 6y
⇒ x = (3800 − 6y)/7
Putting the value of x in equation (2), we get
3(3800 − 6y)/7 + 5y = 1750
(11400 − 18y)/7 + 5y = 1750
⇒ 17y = 850
⇒ y = 50
Putting the value of y in equation (2), we get
3x + 250 = 1750
⇒ 3x = 1500
⇒ x = 500
Therefore, the cost of one bat is Rs 500 and the cost of one ball is Rs 50.
(iv) Let the fixed charge be Rs x and the charge per km be Rs y.
According to the given conditions, we have
x + 10y = 105… (1)
x + 15y = 155… (2)
From equation (1), we have
x = 105 − 10y
Putting this value of x in equation (2), we get
105 − 10y + 15y = 155
⇒ 5y = 50
⇒ y = 10
Putting the value of y in equation (1), we get
x + 10(10) = 105
⇒ x = 105 – 100 = 5
⇒ x = 5
Therefore, the fixed charge is Rs 5 and the charge per km is Rs 10.
To travel a distance of 25 km, a person will have to pay = Rs (x + 25y)
= Rs (5 + 25 × 10)
= Rs (5 + 250)
= Rs 255
(v) Let the numerator of the fraction be x and the denominator be y.
According to the first condition, we have
(x + 2)/(y + 2) = 9/11
⇒ 11(x + 2) = 9(y + 2)
⇒ 11x + 22 = 9y + 18
⇒ 11x = 9y – 4
⇒ x = (9y – 4)/11 …… (1)
According to the second condition, we have
(x + 3)/(y + 3) = 5/6
6(x + 3) = 5(y + 3)
6x + 18 = 5y + 15 ……. (2)
Putting the value of x in equation (2) from equation (1), we get
6(9y – 4)/11 + 18 = 5y + 15
54y/11 – 24/11 + 18 = 5y + 15
–24/11 + 18 – 15 = 5y – 54y/11
–24/11 + 3 = (55y – 54y)/11
(–24 + 33)/11 = y/11
y/11 = 9/11
⇒ y = 9
Putting the value of y in equation (1), we get
x = (9 × 9 – 4)/11
⇒ x = 77/11
⇒ x = 7
Therefore, the fraction is 7/9.
(vi) Let the present age of Jacob be x years and the present age of his son be y years.
According to the first condition, we have
(x + 5) = 3(y + 5)
x + 5 = 3y + 15
x = 3y + 15 – 5
x = 3y + 10 ……. (1)
According to the second condition, we have
(x − 5) = 7(y − 5)
x − 5 = 7y – 35
x = 7y – 35 + 5
x = 7y – 30 …….. (2)
From equation (1) and equation (2), we get
3y + 10 = 7y – 30
3y – 7y = −30 − 10
⇒ −4y = −40
⇒ y = 10 years
Putting the value of y in equation (1), we get
X = 3 × 10 + 10
⇒ x = 30 + 10
⇒ x = 40 years
Therefore, the present age of Jacob is 40 years and the present age of his son is 10 years.