Two taps running together can fill a tank in . . .

Question : Two taps running together can fill a tank in 3.(1/13) hours. If one tap takes 3 hours more than the other to fill the tank, then how much time will each tap take to fill the tank?

Doubt by Jaskirat

Solution : 

Let the time taken by Tap 1 to fill the tank = x hours

and the time taken by the Tap 2 to fill the tank = y hours

ATQ

x=3+y
x-y=3 — (1)

Fraction of tank filled by Tap 1 in 1 hour = 1/x
Fraction of tank filled by Tap 1 in 40/13 hours = 40/13x 

Fraction of tank filled by Tap 1 in 1 hour = 1/y
Fraction of tank filled by Tap 1 in 40/13 hours = 40/13y 

ATQ

40/13x + 40/13y = 1
40/13[1/x+1/y] = 1
1/x+1/y=13/40 — (2)

Using equation (1)

x-y=3
x-3=y
y=x-3 — (3)

putting in equation (2)

1/x+1/(x-3) = 13/40

x=8 but
x≠ 15/13

Because if x=15/13 then the value of y will be negative which is not possible.

Hence, x=8 

putting in equation (3) 

y=x-3

y=8-3
y=5 

So the time taken by Tap 1 to fill the tank = 8 Hours and the time taken by the Tap 2 to fill the tank = 5 Hours