Two pipes running together can fill a tank in $1\frac{7}{8}$ hours. The tap with longer diameter takes 2 hours less than the tap with the smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.

Given:

Two pipes running together can fill a tank in $1\frac{7}{8}$ hours.

The tap with longer diameter takes 2 hours less than the tap with the smaller one to fill the tank separately. 

To do:

We have to find the time in which each tap can separately fill the tank.

Solution:

The time taken by two taps running together to fill the tank$=1\frac{7}{8}=\frac{1\times8+7}{8}=\frac{15}{8}$ hours.

Let the time taken by the tap of the longer diameter to fill the tank be $x$ hours. 

This implies,

The time taken by the tap of the smaller diameter to fill the tank$=x+2$ hours.

The portion of the tank filled by the longer tap in one hour $=\frac{1}{x}$.

The portion of the tank filled by the smaller tap in one hour $=\frac{1}{x+2}$.

Portion of the tank filled by both the taps in one hour$=\frac{1}{\frac{15}{8}}=\frac{8}{15}$

Therefore,

$\frac{1}{x}+\frac{1}{x+2}=\frac{8}{15}$

$\frac{1(x+2)+1(x)}{(x+2)x}=\frac{8}{15}$

$\frac{x+2+x}{x^2+2x}=\frac{8}{15}$

$\frac{2x+2}{x^2+2x}=\frac{8}{15}$

$15(2)(x+1)=8(x^2+2x)$

$15x+15=4x^2+8x$

$4x^2+8x-15x-15=0$

$4x^2-7x-15=0$

Solving for $x$ by factorization method, we get,

$4x^2-12x+5x-15=0$

$4x(x-3)+5(x-3)=0$

$(x-3)(4x+5)=0$

$x-3=0$ or $4x+5=0$

$x=3$ or $4x=-5$

Therefore, the value of $x=3$.    ($x$ cannot be negative)

$x+2=3+2=5$

The time taken by the tap with longer diameter to fill the tank is $3$ hours and the time taken by the tap with smaller diameter is $5$ hours.

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Updated on: 10-Oct-2022

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