Question

2 taps A and B can fill a swimming pool in 3 hours. If turned on alone, it takes tap A 5 hours less than tap B to fill the same pool. How many hours does it take tap A to fill the pool?

Answer 1

Answer:

Tap A will take 2 hours and tap B will take 6 hours to fill the tank when turned on alone.

Step-by-step explanation:

Let tap B fills the pool alone in the time = x hours

So in one hour part of pool will be filled = $\frac{1}{x}$

Another tap A when turned on, it takes time to fill the pool = x-5 hours

So in one hour part of the same pool filled = $\frac{1}{x-5}$

Now both the taps A and B are turned on then time taken to fill the pool = 3 hours.

Part of the pool filled in one hour by both the taps = $\frac{1}{3}$

Now we form an equation

Part of pool filled in one hour by tap A + Part of pool filled in one hour by tap B = Part of pool filled in one hour by both the taps when turned on

$\frac{1}{(x-5)}+\frac{1}{x}=\frac{1}{3}$

$\frac{x-5+1}{x(x-5)}=\frac{1}{3}$

$\frac{x-4}{x(x-5)}=\frac{1}{3}$

3(x - 4) = x(x - 5)

x² -5x = 3x - 12

x² - 8x + 12 = 0

x² - 6x - 2x + 12 = 0

x(x - 6) - 2(x - 6) = 0

(x -2)(x - 6) = 0

x = 2, 6 hours

We will take higher value of x as x = 6 hours for tap B.

Time taken by tap A = 6 - 4 = 2 hours.

Therefore, Tap A will take 2 hours and tap B will take 6 hours to fill the tank when turned on alone.