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

Answer:

it may take 5 hrs to fill tank by tap b alone

Answer 2

Answer:

6 hours

Step-by-step explanation:

Let x be the time ( in hours ) taken by tap b,

Thus, one hour work of tab b = $\frac{1}{x}$

Also, the time taken by tap a = ( x + 2 ) hours

So, one hour work of tab a = $\frac{1}{x+2}$

Thus, the total work in one hour when they work simultaneously = $\frac{1}{x}+\frac{1}{x+2}$

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

∵ 1 hour = 60 minutes ⇒ 1 minute = 1/60 hour ⇒ 20 minutes = 1/3 hours,

So, the time taken by them when they work together = 3 + 1/3 = 10/3 hours,

∴ One hour work when they work simultaneously = $\frac{3}{10}$

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

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

$10(2x+2) = 3x^2 + 6x$

$3x^2 + 6x - 20x - 20=0$

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

By quadratic formula,

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

$\implies x=5.813 \approx 6 \text{ or }x=-1.147\approx -1$

Since, the time can not be negative,

Hence, the time taken by tap b is 6 hours.