If two pipes function simultaneously, the reservoir will be filled in 12 hours. first pipe fills the reservoir 10 hours faster than the second pipe. how many hours does it take the second pipe to fill the reservoir?
Answers
Let the 1st papi takes x to fill alone
2nd takes x-10 ------------—------------
1st. in x hrs the water filled is 1
in 1hr 1st tap fills=(1/x)
similarly in 1hr 2nd tap fills 1/x-10
(1/x +1/x-10)12=1
(x-10+x)/x²-10x=1/12
2x-10=(x²-10x)/12
(2x-10)12=x²-10x
24x-120=x²-10x
x²-34x+120=0
x²-30x-4x+120=0
x(x-30)-4(x-30)=0
x=30
x-10=20
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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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