A pump started filling an empty pool with water and continued at a constant rate until the pool was full. At noon the po

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A pump started filling an empty pool with water and continued at a constant rate until the pool was full. At noon the pool was \(\dfrac13\) full, and \(1\frac14\) hours later it was \(\dfrac34\) full. What was the total number of hours that it took the pump to fill the pool?

A. \(2\frac13\)
B. \(2\frac23\)
C. \(3\)
D. \(3\frac12\)
E. \(3\frac23\)

Answer: C

Source: GMAT Prep

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Gmat_mission wrote:
Sat Oct 31, 2020 6:24 am
A pump started filling an empty pool with water and continued at a constant rate until the pool was full. At noon the pool was \(\dfrac13\) full, and \(1\frac14\) hours later it was \(\dfrac34\) full. What was the total number of hours that it took the pump to fill the pool?

A. \(2\frac13\)
B. \(2\frac23\)
C. \(3\)
D. \(3\frac12\)
E. \(3\frac23\)

Answer: C

Source: GMAT Prep
For this type of question, I like to assign a "nice value" to the job.

In this case we're looking for a number that works well with 1/3, 3/4 and even 1 1/4
So, let's say the pool has a capacity of 60 liters.

At noon the pool was 1/3 full, . . .
1/3 of 60 liters = 20 liters
So, at 12:00pm, the pool contained 20 liters of water

. . . and 1 1/4 hours later it was 3/4 full.
1 1/4 hours = 1.25 hours = 75 minutes.
3/4 of 60 liters = 45 liters
So, at 1:15pm, the pool contained 45 liters of water

What was the total number of hours that it took the pump to fill the pool?
45 liters - 20 liters = 25 liters
So, in 1.25 hours, 25 liters of water was added to the pool

Rate = output/time
So, the rate at which water is added to the pool = (25 liters)/(1.25 hours) = 20 liters per hour

Time = output/rate
So, at a rate of 20 liters per hour, the time it takes to fill the 60 liter pool = 60/20 liters = 3 hours

Answer: C
Brent Hanneson - Creator of GMATPrepNow.com
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