Question

Two taps A and B can together fill a swimming pool in a 15 days. A and B are kept open for 12 days and then B is closed. It takes another 8 days for the pool to be filled. How many days each tap requires to fill the pool ?

Answer 1

Answer:

Tap A can fill the pool in 40 days

Tap B fill the pool in 24 days

Given:

A and B together fill pool in 15 days

They kept open for 12 days and then B is closed now It takes 8 days to fill pool

To find:

In how many days each tap requires to fill

Solution:-

Let A work be x

And B work be Y

A and B can full together in 1/15 days

They worked for 12 days

$= 12 \times \frac{1}{15}$

$= \frac{4}{5}$

Work left=

$1 - \frac{1}{4} = \frac{1}{5}$

Then Tap B is closed

Time required for doing 1/5 work for A=8 days

Time required for doing 1 work=8 ÷1/5

=40

Now

$a + b = \frac{1}{15}$

$b = \frac{1}{15} - \frac{1}{40}$

$b = \frac{1}{24}$

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A take 40 days to fill the pool

B take 24 days to fill the pool

Answer 2

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Since taps a and b fill the pool together in 15 days.

The amount of water filled by both the tap in 1 day = 1/15 of pool.

The water filled by both the taps in 12 days = 12/15 = 4/5

The remaining = 1 - 4/5 = 1/5 of the pool

Now tap a fill 1/5 of the pool in 8 days.

Therefore number of days in which tap "a" fill the pool = 81/5=40 days

The water filled by tap "a" in 1 day = 1/40

The water filled by tap "b" in 1 day = 115−140=8−3120

=5120=124

Thus the number of days in which tap "b" fill the pool = 24 days