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A pump started filling an empty pool with water and continue

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

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New post Updated on: 11 Apr 2014, 02:29
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Question Stats:

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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. \(3\frac{2}{3}\)

Originally posted by Jcpenny on 20 Nov 2008, 10:30.
Last edited by Bunuel on 11 Apr 2014, 02:29, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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Re: A pump started filling an empty pool with water and  [#permalink]

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New post 10 Apr 2014, 19:13
2
3
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+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: pump--25  [#permalink]

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New post 20 Nov 2008, 22:18
1
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+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.
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Re: pump--25  [#permalink]

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New post 10 Apr 2014, 16:22
alpha_plus_gamma wrote:
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+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?
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A pump started filling an empty pool with water and continue  [#permalink]

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New post 02 Jun 2015, 06:04
2
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

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

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

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New post 02 Jun 2015, 07:54
2
1
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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New post 24 Sep 2016, 04:14
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
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Re: A pump started filling an empty pool with water and continue  [#permalink]

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New post 16 Mar 2017, 09:46
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?
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Re: A pump started filling an empty pool with water and continue  [#permalink]

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New post 17 Mar 2017, 02:34
let the total volume be x and rate of flow be r

3x/4 - x/3 = r *5/4
or 5x/12 = r*5/4
or x = r*3

therefre in three hours, the pool will be filled
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Re: A pump started filling an empty pool with water and continue  [#permalink]

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New post 21 Mar 2017, 05:22
2
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. \(3\frac{2}{3}\)


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

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New post 20 Jan 2018, 15:49
VeritasPrepKarishma wrote:
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+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


Hi Karishma,

I was able to follow till the last portion. Can you clarify why we are able to divide the time by rate to figure out how long it will take to fill 1 pool? I was able to solve this correctly however I am looking for a faster way to solve these type of problems.

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

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New post 02 Jan 2019, 11:04
We have only one variable (volume of the pool) so let us assume the volume of the pool is 12 liters

At noon, the pool is 4 L (1/3 x 12 L) full, we do not know the time it took to fill 4 L of the pool at this point

At 1:15 pm (75 mins later), the pool is 9 L (3/4 x 12 L) full =====> Rate = (9 - 4) L/75 mins = 1/15 L/min

Therefore, it took 60 mins (4 L x 15 L/min) to fill the 4 L at noon

Time it took to fill the remaining 3 L (remember we assumed the pool is 12 L?) is 45 mins (3 L * 15 L/min)

Therefore, the total time it took to fill the pool = 60 + 75 + 45 = 180 minutes = 3 hours (Answer choice C)


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. \(3\frac{2}{3}\)
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Re: A pump started filling an empty pool with water and continue &nbs [#permalink] 02 Jan 2019, 11:04
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