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.

Answer 1

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