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Question
A tank fills completely in 2 hours if both the taps are open. If only one of the taps is open at the given time, the smaller tap takes 3 hours more than the larger one to fill the tank. How much time does each tap take to fill the tank completely?
Solution
Let the larger tap take x hours.
Smaller tap takes x + 3 hours.
Both fill the tank in 2 hours if they are open together.
Amount of water filled by larger tap in 1 hour = `1/x`
Amount of water filled by smaller tap in 1 hour = `1/(x + 3)`
Amount of water filled in 1 hour = `1/2`
`1/x + 1/(x + 3) = 1/2`
`(x + 3 + x)/(x(x + 3)) = 1/2`
`(2x + 3)/(x^2 + 3x) = 1/2`
2(2x+ 3) = x2 + 3x
4x + 6 = x2 + 3x
0 = x2 + 3x − 4x − 6
x2 − x − 6 = 0
x2 − 3x + 2x − 6 = 0 ...`[(-3 xx 2 = -6),(-3 + 2 = -1)]`
x(x − 3) + 2(x − 3) = 0
(x − 3) (x + 2) = 0
x − 3 = 0 or x + 2 = 0
∴ x = 3 = 0 or x = −2
Thus, the larger tap takes 3 hours while the smaller tap takes 3 + 3 = 6 hours.
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