There are two pipes used to fill a tank and when operated together, they can fill the tank in 20 minutes. If one pipe can fill the tank two and a half times as quickly as the other, then the faster pipe alone can fill the tank in how many minutes?

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 2.5 times as quickly, so its time is x / 2.5 = x / (5/2) = (2x)/5 minutes.

The rates of filling the tank are:
- Slower pipe: 1/x tank per minute
- Faster pipe: 1/((2x)/5) = 5/(2x) tank per minute

Together, they fill the tank in 20 minutes, so their combined rate is 1/20 tank per minute.

Therefore:
(1/x) + (5/(2x)) = 1/20

Multiply both sides by 20x:
20 + 50 = x

So, x = 70 minutes (time taken by slower pipe).

Time taken by faster pipe = (2x)/5 = (2*70)/5 = 140/5 = 28 minutes.

But this contradicts the options, so let's re-check calculations carefully.

Re-calculation:
(1/x) + (5/(2x)) = 1/20

Find common denominator on left:
(2 + 5) / (2x) = 1/20

So:
7 / (2x) = 1/20

Cross multiply:
7 * 20 = 2x

140 = 2x

x = 70 minutes (slower pipe)

Faster pipe time = (2x)/5 = (2*70)/5 = 28 minutes

Therefore, the faster pipe alone can fill the tank in 28 minutes.

Hence, the correct answer is 28 minutes.

Answer

Answer

Answer