kirankp wrote:

A tank of volume 432 cubic feet has one inlet pipe and 2 outlet pipes. The inlet pipe fills water into the tank at the rate of 4 cubic inches/min and the 2 outlet pipes empty it out at the rates of 14cubic inches/min and 6 cubic inches/min respectively. If all 3 pipes are opened when the tank is full in how much time is it emptied?(1 foot = 12 inches)

A) 777.6 hrs

B) 3 days

C) 15 days

D)77.76 hrs

E)5 days

The options are such that you don't have to calculate a whole lot.

3 days is 72 hrs (think less than 100 if you will)

15 days is 360 hrs (10 days have 240 hrs so 5 days have 120 so overall, 15 days have 360 hrs)

Look at the question now:

The tank has one inlet pipe and 2 outlet pipes. When all 3 pipes are open, think what is happening in one minute.

4 \(in^3\) is flowing in the tank and (14+6) \(in^3\) is flowing out of the tank. So overall, in every one minute, 20-4 = 16 \(in^3\) is flowing out of the tank.

Now we have the volume of the full tank. In every minute, 16 \(in^3\) flows out of it. How long will it take to empty?

\(\frac{432*12*12*12 in^3}{16}\) minutes

We see that 16*27 = 432 (I wondered if 432 will directly go by 16. I saw that it does. It's good to know the multiplication tables)

Let's convert it to hours now: \(\frac{27*12*12*12 in^3}{60}\) hours

= \(\frac{27*12*12 in^3}{5}\) hours

Now this is painful calculation so I will try to ballpark.

I say that 27/5 is more than 5.

I know that 12*12 = 144

So answer will be more than 5 times 144 which has to be option (A).

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