Question

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

Answer 1

LET THE NO. OF DAYS TOOK BY TAP A BE X & that of by B be y.

IN ONE DAY TAP A MAY FILL 1/x th part of the pool.
AND TAP B MAY FILL 1/y part of the pool.
BY THE FIRST GIVEN CONDITION,

THEY TOGETHER CAN FILL THE POOL IN 15 DAYS.
$\frac{1}{x} + \frac{1}{y} = \frac{1}{15} ..................(1)$
TAP A AND B ARE KEPT OPEN FOR 12 days and TAP B IS CLOSED AND TAP A TOOK ANOTHER 8 days .
$\frac {12 + 8}{x} + \frac{12}{y} = 1$
$\frac{20}{x} + \frac{12}{y} = 1.....................(2)$
LET 1/x be a and 1/y be b
a+b=1/15
15a+15b=1..............(multiplying by 15)...(3)
20a+12b=1...............................................(4)
(multiplying eq (3) by 4 & eq (4) by 3)
60a+60b=4...........(5)
60a+36b=3.............(6)
subtracting eq no. (6) from (5)
24b=1
b=1/24
substitute b=1/24 in eq (3)
15a+15b=1
15a+15(1/24)=1
15a+5/8=1
120a+5=8
120a=3
a=3/120
a=1/40
RESUBSTITUTING,

1/x=a 1/y=b
1/x=1/40 ,1/y=1/24
we get
x=40, y=24
THEREFORE,
TAP A REQUIRE 40 days and TAP B REQUIRED 24 DAYS TO FILL THE SWIMMING POOL

HOPE THIS HELPS YOU!!!