Two taps 'A' and 'B' fill a swimming pool in 15 days. Taps 'A' and 'B' are kept open for 12 days and then tap 'B' is closed. It takes another 8 days for the swimming pool to be filled. How many days each tap required to fill the swimming pool?
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let the fully filled swimming pool is S.
Two taps A and B fill the swimming pool in 15 days
⇒ A and B fill the swimming pool of S/15 in one day
⇒ A + B = S/15
Given that A and B are kept open for 12 days
if A and B continuously fill the pool for 12 days then the fill 12 × S/15 of the pool
⇒ A and B fill 4S/5 of the pool
⇒ remaining pool to filled is S - 4S/5 = S/5
S/5 of the pool is filled by A in 8 days
⇒ A fills S/40 of the pool in one day
⇒ A fills the entire pool S in 40 days
A fills S/40 of the pool in one day
⇒ B fills S/15 - S/40 of the pool in one day
⇒ B fills 5S/120 of the pool in one day
⇒ B fills S/24 of the pool in one day
⇒ B fills the entire pool S in 24 days
so A tap takes 40 days and B tap takes 24 days to fill the entire pool.
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