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 the first pipe takes *x* hours to fill the reservoir. Then the second pipe will takes (x + 10) hours to fill the reservoir.

Since, the faster pipe takes *x* hours to fill the reservoir.

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

So, portion of the reservoir filled by the faster pipe in 12 hours = 12/x

Similarly,

Portion of the reservoir filled by the slower pipe in 12 hours `=12/(x + 10)`

It is given that the reservoir is filled in 12 hours.

So,

`12/x+12/(x+10)=1`

`(12(x+10)+12x)/(x(x+10))=1`

12x + 120 + 12x = x^{2} + 10x

x^{2} + 10x - 24x - 120 = 0

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

x^{2} - 20x + 6x - 120 = 0

x(x - 20) + 6(x - 20) = 0

(x - 20)(x + 6) = 0

x - 20 = 0

x = 20

Or

x + 6 = 0

x = -6

But, x cannot be negative.

Therefore, when x = 20then

x + 10 = 20 + 10 = 30

Hence, the second pipe will takes 30hours to fill the reservoir.