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Question
Two pipes running together can fill a tank in `11 1/9` minutes. If on 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.
Solution
Let the time taken by one pipe to fill the tank be x minutes.
∴Time taken by the other pipe to fill the tank = (x +5)min
Suppose the volume of the tank be V.
Volume of the tank filled by one pipe in x minutes = V
∴Volume of the tank filled by one pipe in 1 minute =`V/x`
⇒ Volume of the tank filled by one pipe in `11 1/9` minutes = `V/x xx11 1/9=V/x xx100/9`
Similarly,
Volume of the tank filled by the other pipe in `11 1/9` minutes= `V/(x+5)xx11 1/9=V/(x+5)xx100/9`
Now,
Volume of the tank filled by one pipe in `11 1/9` minutes + Volume of the tank filled by the other pipe in `11 1/9`=V
∴`V (1/x+1/(x+5))xx100=V`
⇒` 1/x+1/(x+5)=9/100`
⇒ `(x+5+x)/(x(x+5))=9/100`
⇒`(2x+5)/(x^2+5x)=9/100`
⇒`200x+500=9x^2+45x`
⇒`9x^2-155x-500=0`
⇒`9x^2-180x+25x-500=0`
⇒ `9x(x-20)+25(x-20)=0`
⇒`(x-20) (9x+25)=0`
⇒`x-20=0 or 9x+25=0`
⇒` x=20 or x=-25/9`
∴ `x=20` (Time cannot be negative)
Time taken by one pipe to fill the tank = 20 min
Time taken by other pipe to fill the tank = (20+5) 25 min
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