Question

One tap can fill a tank in 60 minutes while the other can empty it in 50 minutes. If the tank is full and both the taps are opened, in what time will the tank be emptied?

Answer 1

Given a tap can fill in 60 mins
In 1 min it fills : 1/60 part
Another tap empties it in 50 mins
In 1 min it empties: 1/50 part
Both pipes opened together(when the tank is full):
In 1 min, part emptied = 1/50 - 1/60 = 6-5/300 = 1/300 part emptied
1 min == 1/300 part
X mins == 1 part emptied
X = 300 minutes or 5 hours.
Hence the tank will be emptied in 300 mins or 5 hours.
Hope it helps.

Answer 2

The time in which the tank will be emptied is 300 minutes or, 5 hours.

Given,

A tap can fill a tank in 60 minutes,

another can empty it in 50 minutes.

To find,

Time in which tank will be emptied, if it is full and both taps are opened.

Solution,

Here, it is given that one tap can fill a tank in 60 minutes.

It means the part of the tank that is filled in 1 minute

$=\frac{1}{60} .$

Another tap can empty the tank in 50 minutes.

So, the part of the tank filled in 1 minute

$=\frac{1}{50} .$

Here, it can be observed that the rate of filling the tank is lower than that of emptying it.

Now, if we combine the above information, we can see that part of the tank that will be emptied every minute is

$=\frac{1}{50} -\frac{1}{60}$

On simplifying, we get,

part of the tank emptied in 1 minute

$=\frac{6-5}{300}=\frac{1}{300}$

Since $\frac{1}{300}th$ part of the tank is emptied in 1 minute,

the time taken to empty the tank will be = 300 minutes or 5 hours.

⇒ the time in which the tank will be emptied = 5 hours.

Therefore, the time in which the tank will be emptied is 300 minutes or, 5 hours.

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