**Question:**

Two pipes together can fill a tank in 12 hours. If the first pipe can fill the tank 10 hours faster than the second then how many hours will the second pipe take to fill the tank?

**Solution:**

Let the time required to fill the tank by second pipe be *x* hours.

Then, the time required to fill the tank by first pipe is (*x *− 10) hours.

Given:

Two pipes together can fill a tank in 12 hours.

According to the question,

$\frac{1}{x}+\frac{1}{x-10}=\frac{1}{12}$

$\Rightarrow \frac{(x-10)+(x)}{(x)(x-10)}=\frac{1}{12}$

$\Rightarrow \frac{2 x-10}{x^{2}-10 x}=\frac{1}{12}$

$\Rightarrow 24 x-120=x^{2}-10 x$

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

$\Rightarrow x^{2}-34 x+120=0$

$\Rightarrow x^{2}-30 x-4 x+120=0$

$\Rightarrow x(x-30)-4(x-30)=0$

$\Rightarrow(x-30)(x-4)=0$

$\Rightarrow x=30,4$

But $(x-10)$ is the time required by the first pipe to fill the tank, which is always positive.

Thus, $x=30$ and $x \neq 4$

Hence, the second pipe will take 30 hours to fill the tank.

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