Drains

[BigGlenntheHeavy]

\(\displaystyle A \ drain \ takes \ four \ times \ as \ long \ to \ empty \ a \ tank \ as \ a \ pipe \ can \ fill \ it.\)

\(\displaystyle When \ both \ the \ pipe \ and \ the \ drain \ are \ open, \ it \ takes \ 48 \ minutes \ to \ fill \ the \ tank.\)

\(\displaystyle How \ long \ would \ it \ take \ the \ drain \ to \ empty \ half \ the \ tank?\)

 

[soroban]

Hello, BigGlenn!

A drain takes four times as long to empty a tank as a pipe can fill it.

When both the pipe and the drain are open, it takes 48 minutes to fill the tank.

How long would it take the drain to empty half the tank?

\(\displaystyle \text{Let }P = \text{ minutes for the pipe to fill the tank (alone).}\)

\(\displaystyle \text{Let }4P = \text{minutes for the drain to empty the tank (alone).}\)


\(\displaystyle \text{The pipe can fill the tank in }P\text{ minutes.}\)
. . \(\displaystyle \text{In one minute, the pipe can fill }\tfrac{1}{P}\text{ of the tank.}\)
. . . . \(\displaystyle \text{In 48 minutes, the pipe can fill }\tfrac{48}{P}\text{ of the tank.}\)

\(\displaystyle \text{The drain can empty the tank in }4P\text{ minutes.}\)
. . \(\displaystyle \text{In one minute, the drain can empty }\tfrac{1}{4P}\text{ of the tank.}\)
. . . . \(\displaystyle \text{In 48 minutes, the drain can empty }\tfrac{48}{4P} = \tfrac{P}{12}\text{ of the tank.}\)


\(\displaystyle \text{Working together, in 48 minutes, they will fill the tank (1 tank).}\)

\(\displaystyle \text{There is our equation: }\;\;\frac{48}{P} - \frac{12}{P} \;=\;1\)


\(\displaystyle \text{Multiply by }P\!:\;\;48 - 12 \:=\\)

\(\displaystyle \text{And we have: }\;P \,=\,36,\;\;D \,=\,144\)


\(\displaystyle \text{Working alone, the drain takes 144 minutes to empty the tank.}\)

\(\displaystyle \text{Therefore, to empty half the tank, the drain takes 72 minutes.}\)

 

[BigGlenntheHeavy]

\(\displaystyle Right \ on \ soroban, \ good \ show.\)

\(\displaystyle My \ way, \ note: \ Rate \ X \ Time \ = \ Fraction \ of \ Work \ Done\)

\(\displaystyle Let \ rate \ of \ pipe \ = \ \frac{1}{x}, \ then \ rate \ of \ drain \ = \ \frac{1}{4x}\)

\(\displaystyle Now, \ since \ the \ drain \ is \ nullifying \ part \ of \ the \ intake,\)

\(\displaystyle \frac{48}{x}-\frac{48}{4x} \ = \ 1, \ \implies \ x \ = \ 36, \ \implies \ 2x \ = \ 72 \ = \ 1 \ hr. \ 12 \ min.,\)

\(\displaystyle the \ time \ it \ takes \ for \ the \ drain \ to \ empty \ half \ the \ tank.\)