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.

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 xx 11 1/9`

= `V/x xx 100/9` 

Similarly, Volume of the tank filled by the other pipe in `11 1/9` minutes = `V/(x + 5) xx 11 1/9`

= `V/(x + 5) xx 100/9`  

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)) xx 100 = 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 = 9x² + 45x 

⇒ 9x² – 155x – 500 = 0 

⇒ 9x² – 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