Question

the taps a and b can fill a tank together in 3 hours and 20 minutes. when tap a alone is open it takes 2 hours more to fill tank than when b alone is open. assuming uniform flow how long does it take for b alone to fill the tank ?

Answer 1

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Answer 2

Answer:

It take 5.813 hours for b alone to fill the tank .

Step-by-step explanation:

The taps a and b can fill a tank together in 3 hours and 20 minutes.

3 hours 20 minutes = $3+\frac{20}{60}=\frac{10}{3} hours$

So, Tap A and B together can fill tank in $\frac{10}{3}$ hours

Tap A and B together work in 1 hour = $\frac{3}{10}$  ---1

Let B completes work alone in x hours

So, we are given that tap a alone is open it takes 2 hours more to fill tank than when b alone is open.

So, A completes work in x+2 hours

Tap B's 1 hour work = $\frac{1}{x}$

Tap A's 1 hour work = $\frac{1}{x+2}$

So, Tap A and B together work in 1 hour = $\frac{1}{x}+\frac{1}{x+2}$ ---2

On comparing 1 and 2

$\frac{1}{x}+\frac{1}{x+2}=\frac{3}{10}$

$\frac{x+2+x}{(x)(x+2)}=\frac{3}{10}$

$\frac{2x+2}{x^2+2x}=\frac{3}{10}$

$20x+20=3x^2+6x$

$3x^2-14x-20=0$

Use quadratic formula : $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

a = 3

b= -14

c = -20

$x=\frac{14\pm \sqrt{(-14)^2-4(3)(-20)}}{2(3)}$

$x=\frac{14+ \sqrt{(-14)^2-4(3)(-20)}}{2(3)},\frac{14+ \sqrt{(-14)^2-4(3)(-20)}}{2(3)}$

$x=5.81343,-1.14676$

Since hours cannot be negative

So, it take 5.813 hours for b alone to fill the tank .