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Question
Two pipes running together can fill a tank in `11 1/9` minutes. If one pipe takes 5 minutes more than the other to fill the tank separately, find the time in which each pipe would fill the tank separately.
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Solution
Let the first pipe takes x minutes to fill the tank. Then the second pipe will takes (x + 5)minutes to fill the tank.
Since, the first pipe takes x minutes to fill the tank.
Therefore, portion of the tank filled by the first pipe in one minutes = 1/x
So, portion of the tank filled by the first pipe in `11 1/9` minutes `=100/(9x)`
Similarly,
Portion of the tank filled by the second pipe in `11 1/9` minutes `=100/(9(x+5))`
It is given that the tank is filled in `11 1/9`minutes.
So,
`100/(9x)+100(9(x+5))=1`
`(100(x+5)+100x)/(9x(x+5))=1`
100x + 500 + 100x = 9x2 - 45x
9x2 + 45x - 200x - 500 = 0
9x2 - 155x - 500 = 0
9x2 - 180x + 25x - 500 = 0
9x(x - 20) + 25(x - 20) = 0
(x - 20)(9x + 25) = 0
x - 20 = 0
x = 20
Or
9x + 25 = 0
9x = -25
x = -25/9
But, x cannot be negative.
Therefore, when x = 20 then
x + 5 = 20 + 5 = 25
Hence, the first water tape will takes 20 min to fill the tank, and the second water tape will take 25 min to fill the tank.
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