Question

A tank fills in 2 hrs if both the taps are open.if only one of the taps is open at the given time, the smaller tap takes 3 hours more than the larger one to fill the tank. How muchtime does each tap take to fill the tank completely?
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Answer 1

Answer:

The time taken by the larger tap to fill the tank completely alone is 3 hours.

The time taken by the smaller tap to fill the tank completely alone is 6 hours.

Step-by-step-explanation:

NOTE: Kindly refer to the attachment first.

Let the larger tap takes x hours to fill the tank completely alone.

And the smaller tap takes ( x + 3 ) hours to fill the tank completely alone.

Both taps fill the tank in 2 hours if they both are open.

$\therefore$ Amount of water filled by the larger tap in 1 hour = $\sf\frac{1}{x}$.

Also, amount of water filled by smaller tap in 1 hour = $\sf\frac{1}{x+3}$

If both taps are open they fill the tank completely in 2 hours.

$\therefore$ Amount of water filled in 1 hour is $\sf\frac{1}{2}$

$\therefore\sf\frac{1}{x}\:+\:\frac{1}{x\:+\:3}\:=\:\frac{1}{2}\\\\\implies\sf\frac{x\:+\:3\:+\:x}{x^{2}\:+\:3x}\:=\:\frac{1}{2}$

$\implies\sf\frac{2x\:+\:3}{x^{2}\:+\:3x}\:=\:\frac{1}{2}\\\\\implies\sf\:2\:(2x\:+\:3)\:=\:x^{2}\:+\:3x\\\\\implies\sf\:4x\:+\:6\:=\:x^{2}\:+\:3x\\\\\implies\sf\:x^{2}\:+\:3x\:-4x\:-6\:=\:0\\\\\implies\sf\:x^{2}\:-\:x\:-\:6\:=\:0\\\\\implies\sf\:x^{2}\:-\:3x\:+\:2x\:-\:6\:=\:0\\\\\implies\sf\:x(x\:-\:3)\:+\:2\:(x\:-\:3)\:=\:0\\\\\implies\sf\:(x\:-\:3)\:(x\:+\:2)\:=\:0\\\\\implies\sf\:x\:-\:3\:=\:0\:\:\:or\:\:\:x\:+\:2\:=\:0\\\\\implies\red{\boxed{\sf\:x\:=\:3}}\:\:\:\sf\:or\:\:\pink{\boxed{\sf\:x\:=\:-2}}$

But, time can't be negative.

$\therefore\sf\:x\:=\:-2\:is\:unacceptable.\\\\\therefore\red{\boxed{\sf\:x\:=\:3}}$

Ans.: 1. The time taken by the larger tap to fill the tank completely alone ( x ) = 3 hours.

2. The time taken by the smaller tap to fill the tank completely alone ( x + 3 ) = ( 3 + 3 ) = 6 hours.

Answer 2

Answer:

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