Question

In a tank four taps of equal efficiency are fitted on equal intervals. the first pipe is at the base of the tank. and the 4th pipe is at 3/4 th of height of the tank. then calculate in how much time the whole tank will empty. if the first pipe can empty the tank in 12 hours.

Answer 1

12 hr 1 unit= 12 capacity

so 4 part mean 3 each

now 3/4th the hight there is 4 tap with 1 unit each .

3/4+3/3+3/2+3/1= 75/12= 6 hr 15 min

Answer 2

Given that,

The first pipe is at the base of the tank. and the 4th pipe is at 3/4 th of height of the tank.

If the first pipe can empty the tank in 12 hours.

It means  12 hours one unit = 12 capacity

Four part means 3 each

We know that,

The fourth pipe is at $\dfrac{3}{4}$ th of height of the tank.

Now, $\dfrac{3}{4}$ the height there is fourth tap with 1 unit each.

So, time for third pipe $t_{3}=\dfrac{3}{3}$

time for second pipe $t_{2}=\dfrac{3}{2}$

time for first pipe $t_{2}=\dfrac{3}{1}$

We need to calculate the time

Using formula for time

$t=t_{1}+t_{2}+t_{3}+t_{4}$

Put the value into the formula

$t=\dfrac{3}{1}+\dfrac{3}{2}+\dfrac{3}{3}+\dfrac{3}{4}$

$t=6.25$

Hence, The tank will empty in 6 hours 25 min.