Question

Three taps P, Q and R are fitted in a tank manut catkey. The tap R is the tap emptying tank. Nal P and tank have the capacity to fill the tank separately in 12 minutes and 16 minutes. Shortly after opening the three taps simultaneously, tap R is closed, due to which the tank is filled in 20 minutes. In what time will Null R empty the entire tank?​

Answer 1

Given:

P and Q taps can fill the tank separately in 12 minutes and 16 minutes

If the 3 taps are opened simultaneously and tap R is closed after some time, the tank is filled in 20 minutes

To find:

In what time will tap R empty the entire tank?​

Solution:

Let's assume,

"P" → represents the time taken by tap P to fill the entire tank

"Q" → represents the time taken by tap Q to fill the entire tank

"R" → represents the time taken by tap R to empty the entire tank

So, we have,

In 1 hr, the part of the tank filled by tap P = $\frac{1}{P} = \frac{1}{12}$

In 1 hr, the part of the tank filled by tap Q = $\frac{1}{Q} = \frac{1}{16}$

In 1 hr, the part of the tank emptied by tap R = $\frac{1}{R}$

In 1 hr, the part of the tank filled by tap P, Q & R when opened simultaneously is = $\frac{1}{P}+\frac{1}{Q} - \frac{1}{R} = \frac{1}{20}$

∴ $\frac{1}{P}+\frac{1}{Q} - \frac{1}{R} = \frac{1}{20}$

substituting the values

$\implies \frac{1}{12}+\frac{1}{16} - \frac{1}{R} = \frac{1}{20}$

$\implies \frac{1}{12}+\frac{1}{16} - \frac{1}{20} = \frac{1}{R}$

$\implies \frac{1}{R} = \frac{20\:+\:15\:-\:12}{240}$

$\implies \frac{1}{R} = \frac{23}{240}$

taking reciprocal on both sides

$\implies R = \frac{240}{23}$

$\implies \bold{R = 10.43\:minutes}$

Thus, tap R will empty the entire tank in 10.43 minutes.

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