Two pipes function simultaneously the Reservoir will be filled in 12 hours . One pipe fills the reservior in 10 hours faster than other pipe.in how many hours will the second pipe take to fill the reservior alone.
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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