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?
Answers
SOLUTION :
Let the faster pipe fill the reservoir in x h.
Then, the slower pipe the reservoir in (x + 10) h
In 1 hour the faster pipe fills the portion of the reservoir : 1/x
In 12 hour the faster pipe fills the portion of the reservoir : 12 × 1/x = 12/x
In 1 hour the slower pipe fills the portion of the reservoir : 1/(x + 10)
In 12 hour the faster pipe fills the portion of the reservoir : 12 × 1/(x + 10) = 12/(x +10)
A.T.Q
12/x + 12/(x +10) = 1
12(1/x + 1/(x + 10) ) = 1
1/x + 1/(x + 10) = 1/12
(x + 10 + x ) / [x(x + 10)] = 1/12
[By taking LCM]
2x +10 /(x² + 10x) = 1/12
x² + 10x = 12(2x +10)
x² + 10x = 24x + 120
x² + 10x - 24x - 120 = 0
x² - 14x - 120 = 0
x² + 6x - 20x - 120 = 0
[By splitting middle term]
x(x + 6) - 20(x + 6) = 0
(x - 20) (x + 6) = 0
(x - 20) or (x + 6) = 0
x = 20 or x = - 6
Since, time cannot be negative, so x ≠ - 6
Therefore, x = 20
The faster pipe takes 20 hours to fill the reservoir
Hence, the slower pipe (second) takes (x + 20) = 30 hours to fill the reservoir.
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