**Download NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.2**

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

**(i) x ^{2}-3x-10 = 0**

**Solution: **

x^{2}-3x-10 = 0

x^{2}-5x+2x-10 = 0

x(x-5)+2(x-5) = 0

(x-5)(x+2) = 0

⇒ x-5 = 0 or, x+2 = 0

x = 5 or x =-2

Hence the required roots are 5, -2.

**(ii) 2x ^{2}+x-6=0 **

**Solution:**

2x^{2}+x-6 = 0

2x^{2}+4x-3x-6 = 0

2x(x+2)-3(x+2) = 0

(2x-3)(x+2) = 0

⇒ 2x-3 = 0 or, x+2 = 0

x = or x = -2

Hence the required roots are , -2.

**(iii) √2x ^{2}+7x+5√2 = 0**

**Solution: **

√2x^{2}+7x+5√2 = 0

√2x^{2}+2x+5x+ 5√2 = 0

√2x(x+√2)+5(x+√2) = 0

(x+√2)(√2x +5) = 0

⇒ x+√2 = 0 or, √2x +5 = 0

x = -√2 or x = –

Hence the required roots are -√2, – .

**(iv) 2x ^{2}-x+**

**=0**

**Solution: **

2x^{2}-x+ =0

16x^{2}-8x+1 = 0

(4x)^{2}-2.4x.1+1^{2} = 0

(4x-1)^{2}= 0

⇒ 4x-1 = 0 or 4x-1 = 0

x = or x =

Hence the required roots are ,

**(v) 100x ^{2}-20x+1 = 0**

**Solution:**

100x^{2}-20x+1 = 0

(10x)^{2}-2×10x.1+1^{2}= 0

(10x-1)^{2} = 0

10x-1 = 0 or 10x-1 = 0

x = or x =

Hence the required roots are or

**2. Solve the problem given in example 1.**

**(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each others, and product of the marbles they now have is 124. We would like to find how many marbles they had to start with? **

**Solution:** Let the number of marbles John had be x.

Then the number of marbles Jivanti had be 45-x

They lost 5 marbles to each others.

Hence now the number of marbles of John is x-5 and that of Jivanti is 45-x-5 = 40-x.

Then by given condition we have

(x-5)(40-x) = 124

40x-200-x^{2}+5x = 124

x^{2}-45x+324 = 0

x^{2}-36x-9x+324 = 0

x(x-36)-9(x-36) = 0

(x-36)(x-9) = 0

x-36 = 0

x = 36

or x-9 = 0

x = 9

Therefore, they had 36 and 9 marvels respectively.

**(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:** Let the number of toys produced on that day is x.

Therefore, the cost of production of each toy on that day is (55-x)

So, the total cost of production on that day is = x(55-x)

Then using the given condition, we have,

x(55-x) = 750

55x-x^{2} =750

x^{2}-55x+750 = 0

x^{2}-25x-30x+750 = 0

x(x-25)-30(x-25) = 0

(x-25)(x-30) = 0

x-25 = 0

x =25

or x-30 =0

x = 30

Therefore, the number of toys produced on that day is 25 or 30.

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

**Solution:** Let one number is x. Then another number is 27-x.

Then by the given condition,

x(27-x) = 182

27x- x^{2}-182 = 0

x^{2}-27x+182 = 0

x^{2}-14x-13x+182 = 0

x(x-14)-13(x-14) = 0

(x-14)(x-13) = 0

x-14 = 0

x =14

or, x-13 = 0

x =13

Hence the required two numbers are 13 and 14.

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

**Solution:** Let the two consecutive positive integers be x and x+1

Then by the given condition,

x^{2}+(x+1)^{2} = 365

x^{2}+ x^{2}+2x+1 = 365

2x^{2}+2x-364 = 0

x^{2}+x-182 = 0

x^{2}+14x-13x-182 = 0

x(x+14)-13(x+14) = 0

(x+14)(x-13) = 0

x = -14 or 13

Since the numbers are positive so x = -14 not possible.

So, the numbers are 13 and 13+1 = 14.

**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 of the right triangle is x cm. Then the altitude is x-7 cm.

Then by the property of right triangle and using the given property we have

x^{2}+(x-7)^{2} = (13)^{2}x^{2}+ x^{2}-14x+49 = 169

2x^{2}-14x-120 = 0

x^{2}-7x-60 = 0

x^{2}-12x+5x-60 = 0

x(x-12)+5(x-12) = 0

(x-12)(x+5) = 0

x-12 = 0

x =12

or, x+5 = 0

x = -5

Since length can’t be negative, so x = -5 is not possible.

Hence the base is 12 cm and the altitude is (12-7)cm = 5 cm

**6. A cottage industry produces a certain number of pottery articles in a day. It was observed that on a particular day that the cost of production of each articles (in rupees) was 3 more than twice the number of articles produced on that day. If the total number of production on that day was ₹90, find the number of articles produced and the cost of each article.**

**Solution:** Let the number of articles produced on that day is x. then the price on that day is 2x+3

Then using the given condition we have,

x(2x+3) = 90

2x^{2}+3x-90 = 0

2x^{2}+15x-12x-90 = 0

x(2x+15)-6(2x+15)=0

(2x+15)(x-6) = 0

2x+15 = 0

x = –

or, x-6 = 0

x = 6

Since Articles can’t be negative then x = – is not possible.

Therefore, the number of articles produced on that day = 6

Cost of each article = 2×6+3 = ₹15

**Download NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.2**

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