Question

Two water taps together can fill a tank in 9 3/8 hours the tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately find the time in which each type can separately fill the tank

Answer 1

Answer:bigger tap=15 hrs

small tap=25 hrs

Step-by-step explanation:

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Answer 2

Let the smaller tank fill the tank in x hours

then the larger fills the tank in (x-10) hours

Time taken by both taps together to fill the tank =$\sf{9 \frac{3}{8} h = \frac{75}{8} h}$

Part of tank filled by the smaller tap in 1h = $\sf{\frac{1}{x}}$

Part of tank filled by the larger tap in 1h = $\sf{\frac{1}{x-10}}$

Part of tank filled by both taps together in 1h = $\sf{\frac{8}{75}}$

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

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

$\sf{\frac{2x - 10}{ {x}^{2} - 10x } = \frac{8}{75}}$

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

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

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

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

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

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

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

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

$\sf{x = \frac{15}{4}}$

$\sf{x - 10 < 0}$

i.e, the time taken by the larger pipe is negative which is not possible, So x= 25.

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

And time taken by the larger tap to fill the tank = 25 - 10 = 15h