one pipe 2 feet in diameter takes 12 hours to fill a reservoir. The second pipe 3 feet in diameter takes 6 hours to fill the reservoir. How long would it take for both pipes working at the same time to fill the reservoir?
Two ways to solve it:
1. In your head
2. By algebra.
First way: In your head
Ignore the diameters. Pretend there were several
identical reservoirs that could be filled.
In 12 hours the smaller pipe would fill 1 reservoir and the
larger pipe could fill 2 reservoirs (since it fills 1 in 6 hours)
So in 12 hours they both could fill 2+1 or 3 reservoirs.
So it would only take 1/3 as long for them both to fill just 1
reservoir,
so the answer is 1/3 of 12 hours, or 4 hours.
Second way: by algebra:
Make this chart
reservoirs filled time rate in reservoirs/hour
small pipe
large pipe
both pipes
Let the answer be t. So put t for the time it takes both
pipes to fill 1 reservoir when both are open. So put t
in the bottom middle slot and 1 for the number of reservoirs
filled by both pipes.
reservoirs filled time rate in reservoirs/hour
small pipe
large pipe
both pipes 1 t
Fill in the two times, 12 hours and 6 hours, for the two pipes'
times and 1 for the number of reservoirs they can fill in those
times
reservoirs filled time rate in reservoirs/hour
small pipe 1 12
large pipe 1 6
both pipes 1 t
Next put in the rates in reservoirs/hour by dividing the reservoirs
filled by the number of hours:
reservoirs filled time rate in reservoirs/hour
small pipe 1 12 1/12
large pipe 1 6 1/6
both pipes 1 t 1/t
The equation comes from:
Small pipe's rate + Large pipe's rate = Their combined rate
1/12 + 1/6 = 1/t
Multiply through by LCD of 12t
12t(1/12) + 12t(1/6) = 12t(1/t)
t + 2t = 12
3t = 12
t = 4
answer: 4 hours.
Edwin