It takes 12 hours to fill a swimming pool using two pipes. If pipe of larger diameter is used for 4 hours and pipe of smaller diameter is used for 9 hours, only half of pool is filled. How long would each take to fill the pool?
Answers
SOLUTION :
Let the pipe of larger diameter fill the tank in x hours.
the pipe of Smaller diameter fills the tanks in (x +10) hours.
In 1 hr the part of the pool filled by the pipe of larger diameter = 1/x
In 4 hr the part of the pool filled by the pipe of larger diameter = 4 × 1/x = 4/x
In 1 hr the part of the pool filled by the pipe of Smaller diameter = 1/(x + 10)
In 9 hr the part of the pool filled by the pipe of Smaller diameter = 9 × 1/(x + 10) = 9/(x + 10)
A.T.Q
Given : Half of the pool can be filled
(4/x) + (9/(x+10)) = ½
[4(x + 10) + 9x] / [(x) (x + 10)] = ½
[By taking LCM]
(4x + 40 + 9x) / (x² + 10x) = ½
(13x + 40 ) / (x² + 10x) = ½
2(13x + 40 ) = (x² + 10x)
26x + 80 = x² + 10x
x² + 10x - 26x - 80 = 0
x² -16x - 80 = 0
x² - 20x + 4x - 80 = 0
[By middle term splitting]
x(x-20) + 4(x-20) =0
(x + 4)(x - 20) = 0
(x + 4) = 0 or (x - 20) = 0
x = - 4 or x = 20
Since, Time can't be negative , so x ≠ - 4
Therefore, x = 20
Time taken by pipe of larger diameter to fill the tank = 20 minutes.
Time taken by pipe of Smaller diameter to fill the tank = (x + 10 ) = 20 + 10 = 30 minutes.
Hence, pipe of larger diameter takes 20 minutes to fill the tank while pipe of Smaller diameter takes 30 minutes to fill the tank separately.
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