Two water taps together can fill a tank in 9\dfrac{3}{8} hours. The tap of the larger diameter tak

Question 9 Two water taps together can fill a tank in $9\dfrac{3}{8}$ hours. The tap of the larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time at which each tap can separately fill the tank.

Solution:

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 the tank filled by smaller pipe in 1 hour = $\dfrac{1}{x}$

Part of the tank filled by larger pipe in 1 hour =$\dfrac{1}{x-10}$

As given, the tank can be filled in

$9\dfrac{3}{8}$ = $\dfrac{75}{8}$ hours by both the pipes together.

Therefore,

$\dfrac{1}{x} + \dfrac{1}{x}-10 = $\dfrac{8}{75}$

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

⇒ $\dfrac{2x-10}{x(x-10)}$ = $\dfrac{8}{75}$

⇒ 75(2x – 10) = $8x^{2} – 80x$

⇒ 150x – 750 =$ 8x^{2} – 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, $\dfrac{30}{8}$

Time taken by the smaller pipe cannot be $\dfrac{30}{8}$= 3.75 hours, as the time taken by the larger pipe will become negative, which is logically not possible.

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

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