Question

Working alone at its constant rate, pump X pumped out ¼ of the water in a tank in 2 hours. Then pumps Y and Z started working and the three pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 3 hours.
If pump Y, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water, how many hours would it have taken pump Z, working alone at its constant rate, to pump out all of the water that was pumped out of the tank?

A) 6 B) 12 C) 15 D) 18 E) 24

Answer 1

Answer:

The correct option is B) 12.

Step-by-step explanation:

Consider the provided information.

Working alone at its constant rate, pump X pumped out ¼ of the water in a tank in 2 hours.

As we know: $rate =\dfrac{ work}{time}$,

The rate of pump X is $\frac{\frac{1}{4}}{2} = \frac{1}{8}$

$\frac{1}{4}$ of the water is pumped out of the tank, that means only $\frac{3}{4}$ is left to be pumped out.  

All 3 pumps pumped out the remaining $\frac{3}{4}$ of the water out in 3 hours.

The combined rate of all three pumps is: $\frac{\frac{3}{4}}{3} = \frac{1}{4}$

Pump Y, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water.

The rate of pump Y = $\frac{\frac{3}{4}}{18} = \frac{1}{24}$

Let z is the time taken by pump Z, then the rate of pump Z is $\frac{1}{z}$.

Therefore,

$\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{z}=\dfrac{1}{4}$

Multiplying both sides by 24 z.

$3z + z + 24 = 6z\\24 = 2z\\12 = z$

Hence, the correct option is B) 12.