# Class 10 RD Sharma Solutions – Chapter 8 Quadratic Equations – Exercise 8.12

**Question 1: A takes 10 days less than the time taken by B to finish a piece of work. If both A and B together can finish the work in 12 days, find the time taken by B to finish the work.**

**Solution:**

Let us assume that B takes ‘x’ days to complete a piece of work.

So, B’s 1 day work = 1/x

Now, A takes 10 days less than that of B to finish the same piece of work so,

i.e., (x –10) days

⇒ A’s 1 day work = 1/(x – 10)

If both A and B work together for 12 days, then

(A and B)’s 1 day’s work = 1/12

Given:

A’s 1 day of work + B’s 1 day of work = 1/(x – 10) + 1/x

⇒

1+1=1

x x – 10 12

⇒x – 10 + x=1

x(x – 10) 12

⇒ 12(2x – 10) = x(x – 10)⇒ 24x – 120 = x

^{2}– 10x⇒ x

^{2}– 10x – 24x + 120 = 0⇒ x

^{2}– 34x + 120 = 0⇒ x

^{2}– 30x – 4x + 120 = 0⇒ x(x – 30) – 4(x – 30) = 0

⇒ (x – 30)(x – 4) = 0

Now, either x – 30 = 0

⇒ x = 30

Or, x – 4 = 0 ⇒ x = 4

It’s clear that the value of x cannot be less than 10, so the value of x = 30 is chosen.

Hence, the time taken by B to finish the piece of work is 30 days.

**Question 2: If two pipes function simultaneously, a reservoir will be filled in 12 hours. One pipe fills the reservoir 10 hours faster than the other. How many hours will the second pipe take to fill the reservoir?**

**Solution:**

Let us assume that the faster pipe takes x hours to fill the reservoir.

So, the portion of reservoir filled by faster pipe in one hour = 1/x

Given:

The slower pipe takes 10 hours more than that of faster pipe to fill the reservoir that is (x + 10) hours

So, Portion of reservoir filled by slower pipe = 1/(x + 10)

Now, it’s given that if both the pipes function simultaneously, the same reservoir can be filled in 12 hours

Therefore, as we know the portion of the reservoir filled by both pipes in one hour = 1/12

Now,

Portion of reservoir filled by slower pipe in one hour + Portion of reservoir filled by faster pipe in one hour = 1/x + 1/(x + 10)

Thus, the portion of reservoir filled by both pipes = 1/12

Hence,

⇒

1+1=1

x x + 10 12⇒ 12(2x + 10) = x(x + 10)

⇒ x

^{2}– 14x – 120 = 0⇒ x

^{2}– 20x + 6x – 120 = 0⇒ x(x – 20) + 6(x – 20) = 0

⇒ (x – 20)(x + 6) = 0

Now, either x – 20 = 0 ⇒ x = 20

Or, x + 6 = 0 ⇒ x = – 6 (can be neglected)

Since the value of time cannot be negative.

Thus, the value of x is taken as 20 hours.

Therefore, the time taken by the slower pipe to fill the reservoir = x + 10 = 30 hours.