A Swimming Pool is Being Filled With Water.
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An empty pool existence filled with h2o at a constant charge per unit takes 8 hours
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07 Feb 2011, 00:20
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An empty pool existence filled with water at a abiding charge per unit takes 8 hours to fill to 3/5 of its capacity. How much more time will it take to finish filling the puddle?
(A) 5 hour 30 min
(B) 5 60 minutes 20 min
(C) 4 60 minutes 48 min
(D) 3 60 minutes 12 min
(E) 2 hour 40 min
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An empty pool being filled with h2o at a constant rate takes 8 hours
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20 Dec 2012, 04:53
An empty puddle being filled with water at a constant rate takes eight hours to fill to 3/5 of its capacity. How much more fourth dimension volition it take to cease filling the pool?
(A) v hr xxx min
(B) five 60 minutes 20 min
(C) 4 hr 48 min
(D) iii 60 minutes 12 min
(E) 2 hr 40 min
Every bit puddle is filled to iii/5 of its chapters then 2/5 of its capacity is left to fill.
Since it takes eight hours to fill 3/v of the pool, then to fill 2/5 of the pool it will take \(\frac{8}{(\frac{3}{five})}*\frac{2}{5} = \frac{sixteen}{3} \ hours = 5 \ hours \ twenty \ minutes\) (because if t is the fourth dimension needed to fill the pool then \(t*\frac{three}{5}=8\) –> \(t=8*\frac{5}{3} \ hours\) –> to make full 2/5 of the pool \(8*\frac{5}{three}*\frac{2}{5}=\frac{16}{3}\) hours will be needed).
Or plug values: take the capacity of the pool to be 5 liters –> iii/v of the pool or iii liters is filled in eight hours, which gives the rate of three/8 liters per hour –> remaining ii liters volition require: \(time = \frac{task}{rate} = \frac{two}{(\frac{3}{8})} = \frac{16}{3} \ hours = v \ hours \ 20 \ minutes\).
Respond: B.
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Re: An empty puddle being filled with water at a abiding charge per unit takes 8 hours
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05 Mar 2018, 08:31
Walkabout wrote:
An empty pool beingness filled with water at a constant rate takes eight hours to fill to 3/5 of its capacity. How much more time will it have to finish filling the pool?
(A) 5 hr 30 min
(B) five 60 minutes 20 min
(C) 4 hr 48 min
(D) 3 hour 12 min
(E) 2 hr 40 min
Here’s another approach:
Important CONCEPT: Subsequently 8 hours, 3/5 of the job is finished and ii/v of the job is remaining.
This means
the remaining part of the job is ii/three the size of the first function of the task.
Think of it this fashion:
If the pool had a capacity of 5 gallons, and so the commencement function of the job would be filling 3 gallons, and the remaining part of the job is filling ii gallons.
So, the remaining part of the job (filling ii gallons) is 2/3 the size of the first part of the job (filling 3 gallons)
And so, if information technology takes eight hours to do the first function of the job, and so the time to practice the remaining part = 2/iii of 8 hours
two/3 of 8 hours = 2/iii x 8 hours
= 16/3 hours
= 5 1/three hours
= 5 hours twenty minutes
Reply: B
Thank you,
Brent
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Re: An empty pool being filled with h2o at a abiding charge per unit takes eight hours
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Updated on: 10 Jun 2011, 23:08
Its B
I would use percentages to do this … piece of cake for me
3/5 work = 60% = is done in 8 hrs
remaining will be 40%( percentages are always 100) exist washed in viii*40/sixty = 16/3 = 5 1/3 hrs = 5+i/3*60 = 5.twenty hrs
PS:Edited for a typo
Originally posted by sudhir18n on 10 Jun 2011, 23:05.
Last edited past sudhir18n on 10 Jun 2011, 23:08, edited two times in total.
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Re: An empty pool being filled with water at a constant rate takes eight hours
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11 Jun 2011, 09:47
1st convert to mins to avoid the confusion
threescore% = 480 mins
therefore 40% (remaining) = 320 mins (using unitary method)
320 mins = 5 hrs twenty mins
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Re: An empty pool being filled with h2o at a constant rate takes eight hours
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26 Jan 2013, 04:33
I tried using only ratios to solve this (considering rate is constant):
8Hrs to fill three/5 tank i.e 0.half dozen of the tank
X hrs to fill 2/5 tank i.eastward 0.4 of the tank
=> eight/0.6 = 10/0.four
=> 10=3.2/0.half dozen = 32/halfdozen = 16/3 = 5.33hrs i.e 5hr20min
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Re: An empty pool being filled with water at a abiding rate takes viii hours
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29 May 2015, 20:03
As per what I did,
8 Hours – 3/v of capacity
? Fourth dimension – to fill remaining – 2/5 of chapters.
Merely the answer I am getting is incorrect. Could someone share the appropriate process to work on this?
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Re: An empty puddle existence filled with h2o at a constant charge per unit takes 8 hours
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29 May 2015, xx:08
Hi Pretz,
In that location are a couple of different ways to approach the math in this question. It looks similar you started to set up a ratio, but didn’t consummate the work. Here’s ane fashion to become about it:
Since it takes 8 hours to fill 3/5 of the pool and X hours to make full two/5 of the pool…..
8/X = (3/v)/(2/5)
Since both fractions are “over 5”, we tin multiply those 5s out (by multiplying the numerator and denominator by 5)….
8/X = 3/two
Now we can cantankerousmultiply and solve for X….
16 = 3X
16/3 = Ten
five 1/iii hours = X
five 1/3 hours = 5 hours 20 minutes
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Re: An empty pool beingness filled with water at a constant rate takes 8 hours
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29 May 2015, 20:53
If it takes 8 hours to fill up 3/5 of the pool, and so
information technology takes 8/3 hours to make full i/5 of the pool, and
it thus takes 16/3 hours to fill 2/5 of the pool, which is what we need to practice.
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Re: An empty puddle being filled with water at a abiding rate takes viii hours
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02 Jun 2015, 08:xiii
An empty pool being filled with water at a constant rate takes eight hours to fill to 3/5 of its capacity. How much more time will it take to finish filling the pool?
(iii/5) An empty pool being filled with h2o at a constant rate takes 8 hours to fill to 3/5 of its capacity. How much more fourth dimension will it take to cease filling the pool?
(three/5) of a pool/ 8 hours = 3/40 (the rate)
(three pools/40 hours) = (two/5* puddle)/ x hours
Cross multiply 3x = (2/5) forty
3x = (2/v) (eight) (five)
3x = 16
x = 16/3 or five 1/three
1/three of an hour = twenty minutes
* The puddle is 3/5 full so 2/5 remains.
(A) 5 hr 30 min
(B) five hr 20 min
(C) 4 hr 48 min
(D) three hour 12 min
(Due east) 2 hour twoscore min
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Re: An empty pool being filled with water at a abiding rate takes eight hours
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17 May 2016, xvi:01
Walkabout wrote:
An empty pool being filled with h2o at a constant rate takes 8 hours to fill to 3/v of its capacity. How much more time will it take to finish filling the pool?
(A) 5 hr 30 min
(B) 5 hr 20 min
(C) 4 hr 48 min
(D) 3 hour 12 min
(Eastward) 2 hour 40 min
Solution:
To solve we can setup a proportion. The proportion will read:
A time of 8 hours is to filling up 3/5 of the pool is the aforementioned as a fourth dimension of x number of hours is to filling up (the remaining) 2/5 of the pool. Setting this up mathematically we have:
eight/(3/5) = 10/(2/5)
8/(3/5) = x/(ii/5)
40/iii= 5x/two
Cross multiplying, we get:
80 = 15x
sixteen = 3x
sixteen/3 = x
5 1/3 hours = x
5 hours and 20 minutes = x
Respond: B
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Re: An empty pool being filled with water at a constant rate takes eight hours
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22 Oct 2017, 09:34
GMATD11 wrote:
An empty pool being filled with water at a constant rate takes 8hours to fill to 3/v of its capacity.how much more fourth dimension volition it take to finish filling the puddle?
A. 5hr 30min
B. 5hr 20min
C. 4hr 48min
D. iii hour 12min
E. 2hr 40 min
45second approach: Detect rate of filling. Detect work remaining. Observe fourth dimension needed to finish: Work divided past charge per unit equals time (to fill book that needs filling).
What is charge per unit of filling?
\(\frac{Piece of work}{time} = rate\)
Rate = \(\frac{(\frac{3}{5})}{8} = (\frac{3}{v}*\frac{1}{8})= \frac{three}{twoscore}\)
How much work remaining?
\(1 – (\frac{3}{v}) = (\frac{5}{5} – \frac{3}{5})= \frac{2}{v}\)
How much time to finish remaining work?
\(\frac{Piece of work}{rate} = time\)
Work remaining = \(\frac{two}{5}\)
Charge per unit = \(\frac{3}{40}\)
\(\frac{(\frac{2}{5})}{(\frac{3}{twoscore})}\) =
\(\frac{2}{5} *
\frac{forty}{iii} = 5\frac{one}{iii}hrs\)
Multiply
any
fraction of an hour past 60 to get minutes.* In cases where you already accept the hours, and you demand hours plus minutes, utilize
merely
the fraction. (Do not include the 5 here.)
\(\frac{1}{3}hour * lx =\) 20 minutes
\(5\frac{1}{3}hrs\) = 5 hours, twenty minutes
Answer B
*Because:
\(\frac{1hr}{3} * \frac{60min}{1hr} =\) 20 minutes, where hours cancel
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Re: An empty pool beingness filled with water at a constant rate takes 8 hours
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20 Feb 2018, xv:05
Hi All,
We’re told that viii hours fills iii/5 of the pool; the question asks how long it volition take to fill the remaining 2/5 of the pool. Since the pool is filling at a constant charge per unit, we tin can use a ratio to answer the question. The ratio tin can be written in a number of dissimilar means; I used:
eight/10 = (3/v)/(2/v)
Simplifying, we get…
8/X = 3/2
Now, crossmultiply…
3X = 16
X = 5 1/3 hours
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Re: An empty pool being filled with water at a constant rate takes 8 hours
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06 Mar 2018, 07:06
GMATD11 wrote:
An empty pool existence filled with h2o at a constant rate takes 8hours to fill to 3/five of its capacity.how much more time will it have to finish filling the pool?
A. 5hr 30min
B. 5hr 20min
C. 4hr 48min
D. 3 hr 12min
Due east. 2hr 40 min
We can let due north = the time information technology takes to make full the remaining ii/five of the pool and create the proportion:
8/(iii/5) = north/(two/v)
twoscore/3 = 5n/2
80 = 15n
n = 80/15 = 16/3 = five 1/3 hours = 5 hours and twenty minutes
Respond: B
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Re: An empty pool being filled with water at a constant rate takes 8 hours
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24 Oct 2018, 07:13
3/5 full in 8 hours (rest = 1(3/v) = ii/5)
2/five is filled in (eight*five*20)/3*5 = viii 60 minutes 20 minutes
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Re: An empty pool beingness filled with water at a constant charge per unit takes eight hours
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24 Oct 2018, 08:08
Bunuel wrote:
An empty pool being filled with h2o at a constant charge per unit takes eight hours to fill to 3/5 of its capacity. How much more than time will it accept to finish filling the pool?
(A) v hr 30 min
(B) 5 60 minutes 20 min
(C) four hr 48 min
(D) 3 hr 12 min
(E) ii hr 40 min
Let the capacity of the puddle exist \(40\) Units ( LCM of five & 8)
So, in viii Hours the pipage fills \(24\) units; Thus efficiency of the pipe is 3units/Hour
Time required to fill the entire pool is \(\frac{16}{3}\) hours = \(5\frac{1}{3} = 5 Hours twenty Min\); Reply must be (B)
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Re: An empty pool existence filled with h2o at a constant charge per unit takes 8 hours
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06 May 2019, 19:28
3/5 of the capacity takes =eight hours
So whole chapters will take =8*(five/3)=40/iii hours
More than times needed=(40/iii)8=sixteen/3= 5 hours 20 minutes.
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Re: An empty puddle existence filled with water at a constant rate takes 8 hours
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x Mar 2020, 20:12
Bunuel, pls explain me this part => “because if t is the time needed to fill the puddle then t∗35=8t∗35=eight –> t=viii∗53 hourst=8∗53 hours –> to make full two/five of the pool 8∗53∗25=1638∗53∗25=163 hours will be needed).”
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