22
Two pipes together can fill a tank in $\frac{15}{8}$ hours. The pipe with larger diameter takes $2$ hours less than the pipe with smaller diameter to fill the tank separately. Find the time in which each pipe can fill the tank separately.
Show SolutionHide Solution↓
Let the time taken by smaller diameter tap be $x$ hrs.
Time taken by larger diameter tap is $(x - 2)$ hrs.
Therefore $\frac{1}{x-2} + \frac{1}{x} = \frac{8}{15}$
$\Rightarrow 15(2x - 2) = 8x(x - 2)$
$\Rightarrow 8x^2 - 46x + 30 = 0$
$\Rightarrow 4x^2 - 23x + 15 = 0$
$\Rightarrow (4x - 3)(x - 5) = 0$
$\Rightarrow x = \frac{3}{4}$ as $x-2 < 0$ (rejected) or $x = 5$
Smaller diameter tap fills in $5$ hrs.
Larger diameter tap fills in $3$ hrs.
Time taken by larger diameter tap is $(x - 2)$ hrs.
Therefore $\frac{1}{x-2} + \frac{1}{x} = \frac{8}{15}$
$\Rightarrow 15(2x - 2) = 8x(x - 2)$
$\Rightarrow 8x^2 - 46x + 30 = 0$
$\Rightarrow 4x^2 - 23x + 15 = 0$
$\Rightarrow (4x - 3)(x - 5) = 0$
$\Rightarrow x = \frac{3}{4}$ as $x-2 < 0$ (rejected) or $x = 5$
Smaller diameter tap fills in $5$ hrs.
Larger diameter tap fills in $3$ hrs.