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A pump started filling an empty pool with water and continue [#permalink]

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20 Nov 2008, 11:30

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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 1/3 full, and \(1\frac{1}{4}\) hours later it was 3/4 full. What was the total number of hours that it took the pump to fill the pool?

A. \(2\frac{1}{3}\) B. \(2\frac{2}{3}\) C. 3 D. \(3\frac{1}{2}\) E. \(3\frac{2}{3}\)

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 1/3 full, and 1+1/4 hours later it was 3/4 full. What was the total number of hours that it took the pump to fill the pool? A. 2+1/3 B. 2+2/3 C. 3 D. 3+1/2 E. 3+2/3

in 1+1/4 =5/4 hrs the pump filled 3/4-1/3 = 5/12th of the pool in 1 hr it fills 1/3 of the pool so in 3 hrs it will fill the entire pool.

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 1/3 full, and 1+1/4 hours later it was 3/4 full. What was the total number of hours that it took the pump to fill the pool? A. 2+1/3 B. 2+2/3 C. 3 D. 3+1/2 E. 3+2/3

in 1+1/4 =5/4 hrs the pump filled 3/4-1/3 = 5/12th of the pool in 1 hr it fills 1/3 of the pool so in 3 hrs it will fill the entire pool.

I was with you up until 5/12th of the pool. How did you get the in 1 hour it fills 1/3 of the poor?

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 1/3 full, and 1+1/4 hours later it was 3/4 full. What was the total number of hours that it took the pump to fill the pool? A. 2+1/3 B. 2+2/3 C. 3 D. 3+1/2 E. 3+2/3

The time elapsed between 1/3 full pool and 3/4 full pool is 5/4 hrs. This means 3/4 - 1/3 = 5/12 of the pool was filled in 5/4 hrs.

Again, (5/12)th of the pool will be filled in 5/4 hrs

1 pool will be filled in \(\frac{5/4}{5/12} * 1 = 3\) hrs
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Re: A pump started filling an empty pool with water and continue [#permalink]

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12 Apr 2015, 22:07

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A pump started filling an empty pool with water and continue [#permalink]

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02 Jun 2015, 07:04

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Jcpenny wrote:

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 1/3 full, and \(1\frac{1}{4}\) hours later it was 3/4 full. What was the total number of hours that it took the pump to fill the pool?

A. \(2\frac{1}{3}\) B. \(2\frac{2}{3}\) C. 3 D. \(3\frac{1}{2}\) E. \(3\frac{2}{3}\)

Taking a smart number. Let the capacity of the pool be 12 Liters

At 12 PM = \(\frac{1}{3}*12\) = 4 Liters of water was in the pool. After \(1\frac{1}{4}\) Hours i.e. at 13:15 Hrs = \(\frac{3}{4}*12\) = 9 liters of water was in the pool.

So in 75 mins the pump fills 9-4 = 5 liters of water in the pool.

setting up ratio and proportion

\(\frac{X}{75}=\frac{12}{5}\) \(X = 180 mins = 3 Hours\)

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 1/3 full, and 1 1/4 hours later it was 3/4 full. What was the total number of hours that it took the pump to fill the pool?

(3/4) - (1/3) = 5/12. 5/12 of the pool was full in 1 1/4 hours time (5/4 hours). To determine a rate of filling we divide how much is full by the time. That is (5/12) ÷ (5/4) = 1/3. This means it fills a pool in 3 hours.

A. 2 1/3 B. 2 2/3 C. 3 D. 3 1/2 E. 3 2/3
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Re: A pump started filling an empty pool with water and continue [#permalink]

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24 Sep 2016, 01:59

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: A pump started filling an empty pool with water and continue [#permalink]

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24 Sep 2016, 05:14

at noon or 12:00PM it is filled 1/3 of total, say x = 1/3x 1hour 15 mins later it is 3/4x filled so amount it filled in 75mins = (3/4)x - (1/3)x => 15*12 mins => 3 hours C is the answer

Re: A pump started filling an empty pool with water and continue [#permalink]

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16 Mar 2017, 10:46

OptimusPrepJanielle wrote:

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 1/3 full, and 1 1/4 hours later it was 3/4 full. What was the total number of hours that it took the pump to fill the pool?

(3/4) - (1/3) = 5/12. 5/12 of the pool was full in 1 1/4 hours time (5/4 hours). To determine a rate of filling we divide how much is full by the time. That is (5/12) ÷ (5/4) = 1/3. This means it fills a pool in 3 hours.

A. 2 1/3 B. 2 2/3 C. 3 D. 3 1/2 E. 3 2/3

I got as far as finding 5/12- though where are you getting 3 hours? Is it the reciprocal of 1/3?

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 1/3 full, and \(1\frac{1}{4}\) hours later it was 3/4 full. What was the total number of hours that it took the pump to fill the pool?

A. \(2\frac{1}{3}\) B. \(2\frac{2}{3}\) C. 3 D. \(3\frac{1}{2}\) E. \(3\frac{2}{3}\)

Since it took 1¼ = 5/4 hours to fill 3/4 - 1/3 = 9/12 - 4/12 = 5/12 of the pool, we can let x = the number of hours to fill the pool and create a proportion to determine how long it will take to fill the entire pool.

(5/4)/(5/12) = x/1

5/4 = 5x/12

60 = 20x

x = 3

Answer: C
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