Question

Two water taps together can fill a tank in one hour and twelve minutes. The tap
of smaller diameter takes 1 hour more than the larger one to fill the tank
separately. Find the time in which each tap can separately fill the tank.​

Answer 1

The time taken by tap 1 to fill the tank separately is 2 hours

The time taken by tap 2 to fill the tank separately is 3 hours

Step-by-step explanation:

Given as :

Two water taps together can fill a tank in one hour and twelve minutes.

Or, Two water taps together can fill a tank = 1 + $\dfrac{1}{5}$ = $\dfrac{6}{5}$  hours

The part of tank filled by tap 1 = $\dfrac{1}{x}$  hours

The part of tank filled by tap 2 = $\dfrac{1}{y}$  hours

The part of tank filled by two taps together = $\dfrac{1}{\dfrac{6}{5} }$ = $\dfrac{5}{6}$  hours

The tap  of smaller diameter takes 1 hour more than the larger one to fill the tank  separately

According to question

Two water taps together can fill a tank =  $\dfrac{5}{6}$  

Or, $\dfrac{1}{x}$   +   $\dfrac{1}{y}$  = $\dfrac{5}{6}$

Or, $\dfrac{1}{x}$   +   $\dfrac{1}{x+1}$  = $\dfrac{5}{6}$

Or, $\dfrac{x+x+1}{x(x+1)}$  = $\dfrac{5}{6}$

Or,    $\dfrac{2x+1}{x(x+1)}$ = $\dfrac{5}{6}$

By cross multiplication

6 × ( 2 x + 1 ) = 5 × ( x² + x )

Or, 12 x + 6 = 5 x² + 5 x

Or, 5 x² + 5 x - 12 x - 6  = 0

Or,  5 x² - 10 x + 3 x - 6  = 0

Or,  5 x ( x - 2 ) + 3 ( x - 2 ) = 0

Or,  ( 5 x + 3 ) ( x - 2 ) = 0

i.e ( x - 2 ) = 0    and (5 x + 3 ) = 0

∴      x = 2   and   x = $\dfrac{-3}{5}$

So, The part of tank filled by tap 1 = $\dfrac{1}{x}$ = $\dfrac{1}{2}$ hours

The part of tank filled by tap 2 = $\dfrac{1}{y}$ = $\dfrac{1}{x+1}$  hours

Or, The part of tank filled by tap 2 = $\dfrac{1}{2+1}$ = $\dfrac{1}{3}$  hours

Hence, The time taken by tap 1 to fill the tank separately is 2 hours

And The time taken by tap 2 to fill the tank separately is 3 hours  Answer