Question

Two water taps together can fill a tank in 9 (3/8) hours. The tap of larger diameter takes 10 hrs less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Answer 1

Answer:

Let the smaller tap fill the tank in x hours.

Then, the larger tap fills it in (x - 10) hours.

Time taken by both together to fill the tank = 75/8 hours

Part filled by the smaller tap in 1 hr = 1/x

Part filled by the larger tap in 1 hr = 1/(x - 10)

Part filled by both the taps in 1 hr = 8/75

$\sf \therefore \frac{1}{x} + \frac{1}{(x - 10)} = \frac{8}{75}$

$\sf \longrightarrow \frac{(x - 10) + x}{x(x - 10)} = \frac{8}{75}$

$\sf \longrightarrow \frac{(2x - 10)}{x(x - 10)} = \frac{8}{75}$

By cross multiplication

$\sf \longrightarrow 75(2x - 10) = 8x(x - 10)$

$\sf \longrightarrow 150x - 750 = 8 {x}^{2} - 80x$

$\sf \longrightarrow 8 {x}^{2} - 230x + 750 = 0$

$\sf \longrightarrow 4 {x}^{2} - 115x + 375 = 0$

$\sf \longrightarrow 4 {x}^{2} - 100x - 15x + 375 = 0$

$\sf \longrightarrow 4x(x - 25) - 15(x - 25) = 0$

$\sf \longrightarrow (x - 25)(4x - 15) = 0$

$\sf \longrightarrow x - 25 = 0 \: \: or \: \: 4x - 15 = 0$

$\sf \longrightarrow x = 25 \: \: or \: \: x = \frac{15}{4}$

$\sf \longrightarrow x = 25$

$\because \sf \: x = \frac{15}{4} \\\\ \implies \sf(x - 10) < 0$

Hence, the time taken by the smaller tap to fill the tank = 25 hours

And, the time taken by the larger tap to fill the tank = (25 - 10) = 15 hours.