hours. The tap of a larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time which in each tap
can separately fill the tank.
Answer:
Let the time taken to fill the tank by smaller tap = x (hours)
So the part filled by smaller tap in
1 hour=
1
x
×
75
8
=
75
8x
----------(1)
Again than the time taken to fill the tank by larger tap = (x - 10) hours
∴ the part of the tank that can be filled by the larger tap alone in one hour of time =
1
x-10
∴ In
75
8
hours the part filled by larger tap =
75
8
(
1
x-10
)
∴By both taps togather
=
75
8
[
1
x
+
1
x - 10
]=1
x - 10 + x
x(x - 10)
=
8
75
2x - 10
x2 - 10x
=
8
75
150x - = 8x2 - 80x
8x2 - 80x - 150x + = 0
8x2 - 230x + = 0
4x2 - 115x + = 0
4x2 - x - 15x + = 0
4x(x - ) - 15(x - 25) = 0
∴ (4x - 15) (x -) = 0
4x = 15 x =
15
4
or x =
(x cannot be
15
4
since we have considered 'x' has time taken by smaller tap, which is to be higher one)