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 tap can separately fill the tank.​

Answer 1

$\huge\boxed{\fcolorbox{white}{pink}{Answer}}$

Let the time taken by the smaller pipe to fill the tank = x hr.

Time taken by the larger pipe = (x – 10) hr

Part of tank filled by smaller pipe in 1 hour = 1/x

Part of tank filled by larger pipe in 1 hour = 1/(x – 10)

As given, the tank can be filled in 9 [tex]\frac{3}{8}[/tex]=

75/8 hours by both the pipes together.

75/8 hours by both the pipes together.Therefore,

1/x + 1/x-10 = 8/75

x-10+x/x(x-10) = 8/75

⇒ 2x-10/x(x-10) = 8/75

⇒ 75(2x – 10) = 8x2 – 80x

⇒ 150x – 750 = 8x2 – 80x

⇒ 8x^2 – 230x +750 = 0

⇒ 8x^2 – 200x – 30x + 750 = 0

⇒ 8x(x – 25) -30(x – 25) = 0

⇒ (x – 25)(8x -30) = 0

⇒ x = 25, 30/8

Time taken by the smaller pipe cannot be 30/8 = 3.75 hours,

as the time taken by the larger pipe will become negative,

which is logically not possible.

Therefore,

time taken individually by the smaller pipe and the larger pipe will be 25 and 25 – 10 =15 hours respectively.

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