To fill a swimming pool two pipes are used. If the pipe of larger diameter used for 4 hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. Find, how many long it would take for each pipe to fill the pool separately, if the pipe smaller diameter takes 10 hours more than the pipe of larger diameter 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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