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Two water taps together can fill a tank in $3\frac{1}{3}$ hours. The tap of larger diameter takes $5$ hours less than the smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.
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Let the time taken by the tap of smaller diameter to fill the tank separately be '$x$' hours and the time taken by the tap of larger diameter to fill the tank separately be $(x - 5)$ hours.
A.T.Q.
$\frac{1}{x} + \frac{1}{x-5} = \frac{3}{10}$
$\Rightarrow 3x^2 - 35x + 50 = 0$
$\Rightarrow (x - 10) (3x - 5) = 0$
$x = 10$ or $x = \frac{5}{3}$
But $x = \frac{5}{3}$ is not possible, so $x = 10$
$\therefore$ time taken by the tap of smaller diameter to fill the tank separately is $10$ hours
and time taken by the tap of larger diameter to fill the tank separately is $10-5=5$ hours
A.T.Q.
$\frac{1}{x} + \frac{1}{x-5} = \frac{3}{10}$
$\Rightarrow 3x^2 - 35x + 50 = 0$
$\Rightarrow (x - 10) (3x - 5) = 0$
$x = 10$ or $x = \frac{5}{3}$
But $x = \frac{5}{3}$ is not possible, so $x = 10$
$\therefore$ time taken by the tap of smaller diameter to fill the tank separately is $10$ hours
and time taken by the tap of larger diameter to fill the tank separately is $10-5=5$ hours