A pool is fitted with two taps. When operating independently, tap b needs 6
hours more than tap a to fill the pool. Tap a needs 2 hours more than the time
taken when both taps are turned on at the same time. How long does it take
to fill the pool if two taps were turned on together at the same time?
+128408 Let the time for Tap A to fill the pool = A hours
Let the time for Tap B to fill the pool = A + 6 hours
So Tap A fills 1/A of the pool in every hour
And Tap B fills 1/ {A + 6) of the pool every hour
Add these and we get that
1 / A + 1 / ( A + 6) = [ A + 6 + A ] / [ A (A + 6)] = [ 2A + 6 ] / [ A^2 + 6A ]
The reciprocal of this is the time it takes both taps working together to fill the pool =
[ A^2 + 6A ] / [2A + 6 ] [ in hours ]
And we know that
Time for both taps working together + 2 hours = = time for Tap A to fill the pool
So we have that
[ A^2 + 6A ] / [2A + 6 } + 2 = A
Simplify
A^2 + 6A / [ 2A + 6 ] = A - 2
A^2 + 6A = [ A - 2 ] [ 2A + 6]
A^2 + 6A = 2A^2 - 4A + 6A - 12
A^2 + - 4A - 12 = 0 factor
(A - 6) ( A + 2) = 0
Setting the first factor to 0 and solving for A gives us the positive answer for A = 6 hours to fill pool
And B takes 6 + 6 = 12 hours to fill the pool
So....the time for both to fill the pool is
1/6 + 1/12
[12 + 6 ] / [ 6 * 12 ]
18/ 72 take the reciprocal
72 / 18 = 4 hours for both to fill the pool
