## exercise 4.2 class 10 maths

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

(i) xÂ² – 3x – 10 = 0

(ii) 2xÂ² + x – 6 = 0

(iii) âˆš2 xÂ² + 7x + 5âˆš2 = 0

(iv) 2xÂ² – x + 1/8 = 0

(v) 100xÂ² – 20x + 1 = 0

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

(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 the day. On a particular day the total cost of production was rupees 750. We would like to find out the number of toys produced on that day.

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

Solution: let one number be x then other number is 27 – x.

A.T.Q. x(27-x) = 182

27x – xÂ² – 182 = 0

xÂ² – 27x + 182 = 0

xÂ² – 14x – 13x + 182 = 0

x(x – 14) -13(x -14) = 0

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

Either x-13 =0 and x = 13

Or x – 14 = 0 and x = 14

Hans the numbers are 13 and 14.

**4. Find to consecutive positive integers sum of whose squares is 365.**

Solution: let the two consecutive positive integers be x and x + 1.

ATQ. xÂ² + (x+1)Â² = 365

xÂ² + xÂ² + 1 + 2x = 365

2xÂ² + 2x – 364 = 0

xÂ² + x -182 = 0

xÂ² + 14x – 13x – 182 = 0

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

( x – 13)(x + 14) = 0

Either x-13= 0 so. x = 13

Or. x+ 14 = 0 so x = -14 (rejected because it is negative integer)

So x = 13 hence the two consecutive positive integers 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.**

Solutions: Let base of right angled triangle be x cm and altitude be (x-7)m .

Hypotenuse= 13cm

Using Pythagoras theorem

BaseÂ² + altitudeÂ² = hypotenuseÂ²

xÂ² + (x -7)Â² = 13Â²

xÂ² + xÂ² – 14x + 49 = 169

2xÂ² – 14x – 120 = 0

xÂ² – 7x – 60 = 0

xÂ² – 12x + 5x – 60 = 0

x(x – 12) + 5(x – 12) = 0

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

Either x-12 = 0 or x+5 = 0

So x= 12 and x = -5 (rejected because sude can not be negative.

Therefore base is 12cm and altitude is 12 – 7 = 5cm

**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 rupees 90. Find the number of articles produced and the cost of each article.**

Solution: Let the number of articles produced be x and cost of each article is Rs.(2x+3).

Total amount = Rs 90

Therefore x(2x +3) = 90

2xÂ² + 3x – 90 =0

2xÂ² + 15x – 12x – 90 = 0

x(2x + 15) – 6(2 + 15)

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

Either x-6 =0 so x = 6

Or 2x+15 = 0 so x = – 15/2 (rejected)

So number of articles produced are 6 and cost of each is 2 x 6 + 3 = Rs. 15

Ncert Solutions Class 10 Maths Exercise 1.1 https://10thmathsguide.com/exercise-11-class-10-maths/

Ncert Solutions Class 10 Maths Exercise 1.2

Ncert Solutions Class 10 Maths Exercise 2.1 https://10thmathsguide.com/exercise-21-class-10-maths/

Ncert Solutions Class 10 Maths Exercise 2.2

Ncert Solutions Class 10 Maths Exercise 3.1 https://10thmathsguide.com/exercise-31-class-10-maths/

Ncert Solutions Class 10 Maths Exercise 3.2 https://10thmathsguide.com/exercise-32-class-10-maths/

Ncert Solutions Class 10 Maths Exercise 3.3 https://10thmathsguide.com/exercise-33-class-10-maths/

Ncert Solutions Class 10 Maths Exercise 4.1 https://10thmathsguide.com/exercise-41-class-10-maths/

Ncert Solutions Class 10 Maths Exercise 4.2 https://10thmathsguide.com/exercise-42-class-10-maths/

Ncert Solutions Class 10 Maths Exercise 4.3 https://10thmathsguide.com/exercise-43-class-10-maths/

Click below for case study based questions class 10 Maths ðŸ‘‡ðŸ‘‡ðŸ‘‡ðŸ‘‡

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