To fill a swimming pool two pipes are used if the pipe of larger diameter is used for 4 hours and pipe of smaller diameter is used for 9 hours only half of the pool can be filled find the how long it will take for each pipe to fill the pool separately the pipe of the smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool ?
Answers
Answer:
Step-by-step explanation:
Let the bigger pipe fill the tank in x hours.
the smaller pipe fills the tanks in (x+10) hours.
∴(4/x) + (9/(x+10)) = 1/2
(4x+40+9x) / (x2+10x) = 1/2
2(13x+40) = x2+10x
26x+80 = x2+10x
x2-16x-80 =0
x2-20x+4x-80 =0
x(x-20) + 4(x-20) =0
(x+4)(x-20) = 0
∴ x = 20 and x = -4 ( dont consider)
Hence, the pipe with larger diameter fills the tank in 20 hours.
and the pipe with smaller diameter fills the tank in 30 hours.
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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