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# An empty pool being filled with water at a constant rate

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An empty pool being filled with water at a constant rate [#permalink]

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20 Dec 2012, 04:53
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An empty pool being filled with water at a constant rate takes 8 hours to fill to 3/5 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 hr 12 min
(E) 2 hr 40 min
[Reveal] Spoiler: OA

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Re: An empty pool being filled with water at a constant rate [#permalink]

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20 Dec 2012, 04:56
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An empty pool being filled with water at a constant rate takes 8 hours to fill to 3/5 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 hr 12 min
(E) 2 hr 40 min

As pool is filled to 3/5 of its capacity then 2/5 of its capacity is left to fill.

Since it takes 8 hours to fill 3/5 of the pool, then to fill 2/5 of the pool it will take 8/(3/5)*2/5 = 16/3 hours = 5 hours 20 minutes (because if t is the time needed to fill the pool then t*3/5=8 --> t=8*5/3 hours --> to fill 2/5 of the pool 8*5/3*2/5=16/3 hours will be needed).

Or plug values: take the capacity of the pool to be 5 liters --> 3/5 of the pool or 3 liters is filled in 8 hours, which gives the rate of 3/8 liters per hour --> remaining 2 liters will require: time = job/rate = 2/(3/8) = 16/3 hours = 5 hours 20 minutes.

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Re: An empty pool being filled with water at a constant rate [#permalink]

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26 Jan 2013, 04:33
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I tried using just ratios to solve this (because rate is constant):

8Hrs to fill 3/5 tank i.e 0.6 of the tank
X hrs to fill 2/5 tank i.e 0.4 of the tank

=> 8/0.6 = X/0.4
=> X=3.2/0.6 = 32/6 = 16/3 = 5.33hrs i.e 5hr20min

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Re: An empty pool being filled with water at a constant rate [#permalink]

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28 Jan 2013, 13:28
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Hey,
Is'nt 16/3 = 5.3 ~ . This might be silly, but i chose 5 hours 30 mins . I don't want to make this mistake again, kindly help me

Thanks
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Re: An empty pool being filled with water at a constant rate [#permalink]

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28 Jan 2013, 14:11
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30 minutes/60 minutes=.5, so 5 and a half hours would be 5.5, not 5.33.

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Re: An empty pool being filled with water at a constant rate [#permalink]

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21 May 2015, 03:55
Let the total amount of water eq l

then rate : ( 3/5 * l ) / 8 = 3/40 * l

So time required to fill the full tank = l / ( (3/40) * l ) = 40/3

So time required to fill the remaining section = ( 40/3 - 8 ) hrs = 16/3 hrs = 5 1/3 hrs = 5 hr 20 min

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Re: An empty pool being filled with water at a constant rate takes 8 hours [#permalink]

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29 May 2015, 20:03
As per what I did,

8 Hours - 3/5 of capacity
? Time - to fill remaining - 2/5 of capacity.

But the answer I am getting is incorrect. Could someone share the appropriate process to work on this?
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Re: An empty pool being filled with water at a constant rate takes 8 hours [#permalink]

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29 May 2015, 20:08
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Hi Pretz,

There are a couple of different ways to approach the math in this question. It looks like you started to set up a ratio, but didn't complete the work. Here's one way to go about it:

Since it takes 8 hours to fill 3/5 of the pool and X hours to fill 2/5 of the pool.....

8/X = (3/5)/(2/5)

Since both fractions are "over 5", we can multiply those 5s out (by multiplying the numerator and denominator by 5)....

8/X = 3/2

Now we can cross-multiply and solve for X....

16 = 3X
16/3 = X
5 1/3 hours = X

5 1/3 hours = 5 hours 20 minutes

[Reveal] Spoiler:
B

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An empty pool being filled with water at a constant rate takes 8 hours [#permalink]

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29 May 2015, 20:46
Pretz wrote:
As per what I did,

8 Hours - 3/5 of capacity
? Time - to fill remaining - 2/5 of capacity.

But the answer I am getting is incorrect. Could someone share the appropriate process to work on this?

Hello there Pretz

I think your approach is fine.

$$8$$ $$hours$$ ----- $$\frac{3}{5}$$

multiply by $$\frac{5}{3}$$ on both sides

$$8*\frac{5}{3}$$ $$hours$$ ----- $$\frac{3}{5} *\frac{5}{3}$$

$$\frac{40}{3}$$ $$hours$$ ----- $$1$$

So it takes $$\frac{40}{3}$$ $$hours$$ to fill the tank

The additional time required to fill $$\frac{2}{3}$$rd of the tank will be

$$\frac{40}{3} - 8$$ $$hours$$

$$\frac{16}{3}$$ $$hours$$

$$5 \frac{1}{3}$$ $$hours$$

$$5$$ $$hours$$ $$20$$ $$minutes$$

I hope that explains it
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Re: An empty pool being filled with water at a constant rate takes 8 hours [#permalink]

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29 May 2015, 20:53
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If it takes 8 hours to fill 3/5 of the pool, then
it takes 8/3 hours to fill 1/5 of the pool, and
it thus takes 16/3 hours to fill 2/5 of the pool, which is what we need to do.
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Re: An empty pool being filled with water at a constant rate takes 8 hours [#permalink]

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29 May 2015, 21:05
Awesome! Thank you so much all!
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Re: An empty pool being filled with water at a constant rate [#permalink]

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01 Jun 2015, 01:53
B
if the rates are constant the ratio of capacity is proportional to that of time
x/8=2/3 ------- x = 5 hr 20 min
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Re: An empty pool being filled with water at a constant rate [#permalink]

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02 Jun 2015, 08:13
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An empty pool being filled with water at a constant rate takes 8 hours to fill to 3/5 of its capacity. How much more time will it take to finish filling the pool?

(3/5) An empty pool being filled with water at a constant rate takes 8 hours to fill to 3/5 of its capacity. How much more time will it take to finish filling the pool?
(3/5) of a pool/ 8 hours = 3/40 (the rate)

(3 pools/40 hours) = (2/5* pool)/ x hours
Cross multiply 3x = (2/5) 40
3x = (2/5) (8) (5)
3x = 16
x = 16/3 or 5 1/3
1/3 of an hour = 20 minutes

* The pool is 3/5 full so 2/5 remains.

(A) 5 hr 30 min
(B) 5 hr 20 min
(C) 4 hr 48 min
(D) 3 hr 12 min
(E) 2 hr 40 min

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Re: An empty pool being filled with water at a constant rate [#permalink]

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17 May 2016, 04:59
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We know,
R * T = Q , suppose the Q=100ltr. then,
R * 8 = 3/5 of 100 ==> 60/8 ==> 15/2 ltr per hour.

now, remaning 40ltr(which is 100 - 60 or 2/5 of 100)
R * t = Q
or, 15/2 * t = 40
or, t = 40 * 2/15
t = 16/3 = 5.33 = 5 hour + 1/3 and 1/3 of 60 = 20. so, 5 hour 20 mins.

If only such questions were asked in GMAT.

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Re: An empty pool being filled with water at a constant rate [#permalink]

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17 May 2016, 16:01
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An empty pool being filled with water at a constant rate takes 8 hours to fill to 3/5 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 hr 12 min
(E) 2 hr 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 same as a time of x number of hours is to filling up (the remaining) 2/5 of the pool. Setting this up mathematically we have:

8/(3/5) = x/(2/5)

8/(3/5) = x/(2/5)

40/3= 5x/2

Cross multiplying, we get:

80 = 15x

16 = 3x

16/3 = x

5 1/3 hours = x

5 hours and 20 minutes = x

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Re: An empty pool being filled with water at a constant rate [#permalink]

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28 Oct 2017, 18:27
I don't see anyone converting 1/3 to minutes, this is why a lot of people are being confused with the answer. if you have 1/3 * 60 (minutes) then you get 20 minutes. Just wanted to clarify it for me. 5 + 1/3 = 5 + 20 minutes = 5 hrs and 20 minutes

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Re: An empty pool being filled with water at a constant rate [#permalink]

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08 Nov 2017, 21:03
An easier approach is to break up the math.

$$\frac{16}{3}$$ hrs = $$\frac{16}{3}*60$$ mins

= 16 * 20 mins

= (10 + 6) * 20

= 200 + 120

= 300 + 20

= 5 hrs and 20 mins.
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Re: An empty pool being filled with water at a constant rate [#permalink]

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08 Nov 2017, 22:01
8 hours for 3/5
then 8/3 to fill 1/5
8/3*2 for to fill remaining 2/5
it equals to 5 hours 20 minutes
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Re: An empty pool being filled with water at a constant rate   [#permalink] 08 Nov 2017, 22:01
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