Question

Two water taps together can fill a tank in $9\frac{3}{8}$ hours. The tap of larger diameter takes $10$ hours less than smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Answer 1

Let $x$ be speed of water from pipe with layer diameter and $y$ be speed of water from other pipe.
here speed refer to the volume flowing per hour.$V$ be volume of tank.Time taken when both pipes are used=$\frac{V}{x+y}=9\frac{3}{8}=\frac{75}{8}$
When pipe with larger diameter is used,time=$\frac{V}{x}$
when other pipe is uses,time=$\frac{V}{y}$
Now,$\frac{V}{y}+10=\frac{V}{x}$
$\frac{V}{y}-\frac{V}{x}=10$ ..(1)
and $\frac{V}{x+y}=\frac{75}{8}$
$8V=75(x+y)$
$8\frac{V}{x}=75(1+\frac{y}{x})$
$\frac{V}{x}=\frac{75}{8}(1+\frac{y}{x})$..(2)
similarly $\frac{V}{y}=\frac{75}{8}(1+\frac{x}{y})$ ..(3)
from (1),(2),(3)
$\frac{75}{8}(\frac{y}{x}-\frac{x}{y})=10$
$\frac{y}{x}-\frac{x}{y}=\frac{80}{75}=\frac{16}{15}$
let $\frac{y}{x}=t$
$t-\frac{1}{t}=\frac{16}{15}$
$15t^2-15=16t$
$15t^2-16t-15=0$
$t=-\frac{5}{3}$ or $t=+0.6$
$t\ne -\frac{5}{3}\ne \frac{y}{x}$
$\frac{V}{y}=\frac{75}{8}(1+\frac{x}{y})=\frac{75}{8}(1+\frac{5}{3})=25$ hours
$\frac{V}{x}=\frac{75}{8}(1+\frac{y}{x})=\frac{75}{8}(1+\frac{3}{5})=15$hours
Time by pipe B=25 hours
Time by pipe A=15 hours