Question

Two water taps together can fill a tank in 9 hours 36 minutes. The tap of larger diameter takes 8 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Answer 1

Let the time taken by the tap of smaller diameter = x years
Therefore time taken by the tap of larger diameter = ( x - 9 ) hours
Work done by the tap of smaller diameter in one hour = 1÷ x
and the work done by the tap of larger diameter in one hour = 1 / x - 9
Thus , the work done by the 2 taps together in 1 hour
= 1/x + 1/ x- 9
= x - 9 + x / x ( x - 9)
= 2x - 9/ x( x - 9)
The 2 tapes together can fill the tank in x( x- 9)/ 2x - 9
According to question
x( x- 9)÷ 2x - 9 = 6 hours
x² - 21x + 54 = 0
x² - 3x - 18 + 54 = 0
x( x - 3) - 18 ( x - 3) = 0
(x- 3 ) ( x - 18) = 0
Either x - 3 = 0
or x - 18 =0
Either x = 3 or x = 18
When x = 3, then tap of smaller diameter can fill the tank in 3 hours and the tap of larger diameter can fill the tank in 3 - 9 = -6, which is rejected as a time to fill the tank can't be negative

When x = 18 , then the tap of smaller diameter can fill the tank in 18 hours and the tap of larger diameter can fill the tank in 18 - 9 = 9 hours.

Answer 2

let the time taken by larger tap to fill the tank be x/hr.
let the time taken by smaller tap to fill the tank ba (x+8)..
therefore: position of tank filled by larger tap in 1hr=1/x
position of tank filled by smaller tap in 1hr=1/x+8.
If both taps together, then position of tank filled in 1hr=1/48/5 i.e 5/48.
1/x + 1/x+8 =5/48
x-8+x/x(x-8) =5/48
2x-8/x²-8x =5/48
48(2x-8)=5(x²-8x)
96x-384=5x²-40x
0=5x²-40x-96x-384
=5x²-136x+384
compare with ax²+ bx+c=0
a=5,b=-136,c=384
by quaratic formula
x=-b+-√b²-4ac/2a
=-(-136)+-√(136)²-4(5)(382)/2(5)
=136+-√18496-7680/10
=136+-√10816/10
=136+-104/10
=+ve 136+104/10
=240/10
=24
-be 136-104/10
= 32/10
=16/5
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