There are two pipes to fill a tank. Together, they can fill the tank in 15 minutes. If one pipe can fill the tank one and a half times as fast as the other, find the time taken by the faster pipe alone to fill the tank.

Question

Accepted Answer

Let the time taken by the slower pipe to fill the tank be x minutes.

Then, the faster pipe fills the tank 1.5 times as fast, so its time is x / 1.5 = (2x)/3 minutes.

Rates of filling:
- Slower pipe rate = 1/x tank per minute
- Faster pipe rate = 1/((2x)/3) = 3/(2x) tank per minute

Together, they fill the tank in 15 minutes, so combined rate = 1/15 tank per minute.

Sum of rates:
1/x + 3/(2x) = 1/15

Find common denominator:
(2 + 3) / (2x) = 1/15

5 / (2x) = 1/15

Cross multiply:
5 * 15 = 2x

75 = 2x

x = 75 / 2 = 37.5 minutes (time taken by slower pipe)

Time taken by faster pipe:
(2x)/3 = (2 * 37.5) / 3 = 75 / 3 = 25 minutes

But this contradicts the options and initial assumption about faster pipe time. Let's re-examine the assumption.

The problem states: "one pipe can fill the tank in one and a half times as fast as the other."

This means the faster pipe's rate is 1.5 times the slower pipe's rate.

Let the slower pipe's rate be r.
Then, faster pipe's rate = 1.5r.

Combined rate = r + 1.5r = 2.5r

Given combined rate = 1/15

So, 2.5r = 1/15

r = 1/(15 * 2.5) = 1/37.5

Slower pipe's time = 1 / r = 37.5 minutes

Faster pipe's rate = 1.5r = 1.5 * (1/37.5) = 1/25

Faster pipe's time = 25 minutes

Therefore, the faster pipe alone can fill the tank in 25 minutes.

Check options: 25 minutes is option 2.

Hence, the correct answer is 25 minutes.

Answer

Answer

Answer