NCERT Solutions Maths Class 10 Chapter 4

# NCERT Solutions Maths Class 10 Chapter 4

## NCERT Solutions Maths Class 10 Exercise 4.2

1. Find the roots of the following quadratic equations by factorization:

(i) x2 − 3x − 10 = 0

(ii) 2x2 + x − 6 = 0

(iii) √2x2 + 7x + 5√2 = 0

(iv) 2x2 x + 1/8 = 0

(v) 100x2 − 20x + 1 = 0

Solution: (i) x2 − 3x − 10 = 0

x2 − 5x + 2x − 10 = 0

x(− 5) + 2(− 5) = 0

(− 5) (+ 2) = 0

(− 5) = 0 or (+ 2) = 0

= 5 or = −2

Thus, = 5, −2

(ii) 2x2 + x − 6 = 0

2x2 + 4x − 3x − 6 = 0

2(+ 2) – 3(+ 2) = 0

(2− 3) (+ 2) = 0

(2− 3) = 0 or (+ 2) = 0

= 3/2 or = −2

Thus, = 3/2, −2

(iii) √2x2 + 7x + 5√2 = 0

√2x2 + 2x + 5x + 5√2 = 0

√2x(x + √2) + 5(x + √2) = 0

(√2x + 5)(x + √2) = 0

x = −5/√2, √2

or x = −5√2/2, √2

(iv) 2x2 x + 1/8 = 0

(16x2 – 8x + 1)/8 = 0

16x2 − 8x + 1 = 0

16x2 − 4x − 4x + 1 = 0

4x(4− 1) – 1(4− 1) = 0

(4− 1) (4− 1) = 0

x = 1/4, 1/4

(v) 100x2 − 20x + 1 = 0

100x2 − 10x – 10x + 1 = 0

10x(10− 1) – 1(10− 1) = 0

(10− 1) (10− 1) = 0

= 1/10, 1/10

2. Solve the following problems given:

(i) x2 – 45x + 324 = 0

(ii) x2 − 55x + 750 = 0

Solution: (i) x2 – 45x + 324 = 0

x2 – 36x – 9x + 324 = 0

x(− 36) – 9(− 36) = 0

(− 9) (− 36) = 0

= 9, 36

(ii) x2 − 55x + 750 = 0

x2 – 25x – 30x + 750 = 0

x(− 25) – 30(− 25) = 0

(− 30) (− 25) = 0

= 30, 25

3. Find two numbers whose sum is 27 and product is 182.

Solution: Let the first number be x.

Then, the second number is (27 − x).

According to the question, the product of the two numbers is 182.

Therefore, x(27 − x) = 182

27x x2 = 182

x2 – 27x + 182 = 0

x2 – 14x – 13x + 182 = 0

x(− 14) – 13(− 14) = 0

(− 14) (− 13) = 0

= 14, 13

Therefore, the first number is 14 or 13.

If the first number = 14, then, the second number = 27 – = 27 – 14 = 13

If the first number = 13, then, the second number = 27 – = 27 – 13 = 14

Therefore, the two numbers are 13 and 14.

4. Find two consecutive positive integers, sum of whose squares is 365.

Solution:  Let the two consecutive positive integers be and (+ 1).

According to the question, we have

x2 + (x + 1)2 = 365

x2 + x2 + 1 + 2x = 365            [Since (a + b)2 = a2 + 2ab + b2]

2x2 + 2x – 364 = 0

Dividing the above equation by 2, we get

x2 + x – 182 = 0

x2 + 14x – 13x – 182 = 0

x(+ 14) – 13(+ 14) = 0

(+ 14) (− 13) = 0

= 13, −14

Therefore, the first number is 13.         [We discard −14 because it is a negative number.]

Thus, the second number = + 1 = 13 + 1 = 14

Hence, the two consecutive positive integers are 13 and 14, sum of whose squares is 365.

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 the base of the triangle be x cm.

Then, the altitude of the triangle is (− 7) cm.

It is given that the hypotenuse of the triangle is 13 cm.

According to Pythagoras Theorem, we have

(Hypotenuse)2 = (Base)2 + (Altitude)2

132 = x2 + (x – 7)2

169 = x2 + x2 + 49 – 14x                     [Since (a + b)2 = a2 + b2 + 2ab]

169 = 2x2 14x + 49

2x2 – 14x – 120 = 0

Dividing the above equation by 2, we get

x2 – 7x – 60 = 0

x2 – 12x + 5x – 60 = 0

x(− 12) + 5(− 12) = 0

(− 12) (+ 5) = 0

= −5, 12

Therefore, the base of the triangle = 12 cm

[We discard = −5, because the length of the side of a triangle cannot be negative.]

Thus, altitude of the triangle = (− 7) = 12 – 7 = 5 cm

Hence, the base is 12 cm and the altitude is 5 cm.

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 Rs 90, find the number of articles produced and the cost of each article.

Solution: Let the cost of production of one article be Rs x.

It is given that the total cost of production on that particular day = Rs 90

Therefore, the total number of articles produced on that day = 90/x

According to the question, we have

x = 2(90/x) + 3

x = 180/x + 3

x = (180 + 3x)/x

x2 = 180 + 3x

x2 − 3x180 = 0

x2 − 15x + 12x − 180 = 0

x(− 15) + 12(− 15) = 0

(− 15) (+ 12) = 0

= 15, −12

Therefore, Rs 15, which is the cost of production of one article.

[We discard = −12, because the cost cannot be negative.]

The number of articles produced on that particular day = 90/15 = 6

Hence, the number of articles produced is 6 and the cost of each article is Rs 12.