Quadratic Equation NCERT Exercise 4.2 part two Class Ten Mathematics

## Exercise 4.2 Part 2

Question 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.

Solution: Given, John and Jivanti together have number of marbles = 45

After losing of 5 marbles by each of them, number of marble = 45 – 5 – 5 = 45 – 10 = 35

Let us assume, John has x marbles

Hence; marbles with Jivanti = 35 – x

As per question; product of marbles after loss = 124

Therefore; x(35 – x) = 124

Or, 35x – x^2 = 124

Or, - x^2 + 35x – 124 = 0

Or, x^2 – 35x + 124 = 0

Or, x^2 – 4x – 31x + 124 = 0

Or, x(x – 4) – 31 (x – 4) = 0

Or, (x – 31)(x – 4) = 0

Hence, x = 31 and x = 4

One person has 31 marbles and another has 4 marbles

At the beginning; one person had 36 marbles and another had 9 marbles.

(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 Rs 750. We would like to find out the number of toys produced on that day.

Solution: Let us assume, number of toys = x

Then, cost of production of each toy = x – 55

Hence, total cost of production = x(55 – x) = 750

Or, 55x – x^2 = 750

Or, x^2 - 55x + 750 = 0

Or, x^2 - 30x - 25x + 750 = 0

Or, x(x - 30) – 25(x – 30) = 0

Or, (x – 25)(x – 30) = 0

Hence, x = 25 and x = 30

Thus, number of toys is 25 or 30

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

Solution: Let us assume, one of the numbers = x

Hence, second number = 27 – x

As per question; x(27 – x) = 182

Or, 27x – x^2 = 182

Or, 27x – x^2 – 182 = 0

Or, x^2 – 27x + 182 = 0

Or, x^2 – 14x – 13x + 182 = 0

Or, x(x – 14) – 13(x – 14) = 0

Or, (x – 13)(x – 14) = 0

Hence, x = 13 and x = 14

Hence, the numbers are 13 and 14