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Question
If two pipes function simultaneously, a reservoir will be filled in 12 hours. One pipe fills the reservoir 10 hours faster than the other. How many hours will the second pipe take to fill the reservoir?
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Solution
Let the first pipe takes x hours to fill the reservoir. Then the second pipe will takes (x + 10) hours to fill the reservoir.
Since, the faster pipe takes x hours to fill the reservoir.
Therefore, portion of the reservoir filled by the faster pipe in one hour = 1/x
So, portion of the reservoir filled by the faster pipe in 12 hours = 12/x
Similarly,
Portion of the reservoir filled by the slower pipe in 12 hours `=12/(x + 10)`
It is given that the reservoir is filled in 12 hours.
So,
`12/x+12/(x+10)=1`
`(12(x+10)+12x)/(x(x+10))=1`
12x + 120 + 12x = x2 + 10x
x2 + 10x - 24x - 120 = 0
x2 - 14x - 120 = 0
x2 - 20x + 6x - 120 = 0
x(x - 20) + 6(x - 20) = 0
(x - 20)(x + 6) = 0
x - 20 = 0
x = 20
Or
x + 6 = 0
x = -6
But, x cannot be negative.
Therefore, when x = 20then
x + 10 = 20 + 10 = 30
Hence, the second pipe will takes 30hours to fill the reservoir.
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