Question

Two water taps together can fill a tank in 6 hours. The tap of larger diameter takes 9 hours less than the smaller one to fill the tank separately. Find the time which each tap can separately fill the tank.

Answer 1

Let the tap of smaller diameter fill the tank in x hours.
∴Time taken by the tap of larger diameter to fill the tank =`(x-9)h`

Suppose the volume of the tank be V.
Volume of the tank filled by the tap of smaller diameter in x hours = V 

∴ Volume of the tank filled by the tap of smaller diameter in 1 hour =`V/x` 

⇒ Volume of the tank filled by the tap of smaller diameter in 6 hours =`V/x xx6` 

Similarly,
Volume of the tank filled by the tap of larger diameter in 6 hours =`V/(x-9)xx6` 

Now,
Volume of the tank filled by the tap of smaller diameter in 6 hours + Volume of the tank filled by the tap of larger diameter in 6 hours = V 

∴`V(1/x+1/(x-9))xx6=V` 

⇒`1/x+1/(x-9)=1/6` 

⇒`(x-9+x)/(x(x-9))=1/6` 

⇒`(2x-9)/(x^2-9x)=1/6`  

⇒`12x-54=x^2-9x` 

⇒`x^2-21x+54=0` 

⇒`x^2-81x-3(x-18)=0` 

⇒`(x-18)(x-3)=0` 

⇒`x-18=0  or  x-3  =0` 

⇒` x=18  or  x=3` 

For x = 3, time taken by the tap of larger diameter to fill the tank is negative which is not possible.

∴` x=18` 

Time taken by the tap of smaller diameter to fill the tank = 18 h
Time taken by the tap of larger diameter to fill the tank=` (18-9)=9h` 

Hence, the time taken by the taps of smaller and larger diameter to fill the tank is 18 hours and 9 hours, respectively.