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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] New post 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] New post 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] New post 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 :shock:

Thanks
Aj
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Re: An empty pool being filled with water at a constant rate [#permalink] New post 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] New post 20 Dec 2012, 04:56
Expert's post
Walkabout wrote:
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.

Answer: B.
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Re: An empty pool being filled with water at a constant rate [#permalink] New post 26 Apr 2014, 18:12
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Re: An empty pool being filled with water at a constant rate [#permalink] New post 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] New post 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] New post 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

Final Answer:
[Reveal] Spoiler:
B


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An empty pool being filled with water at a constant rate takes 8 hours [#permalink] New post 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] New post 29 May 2015, 20:53
Expert's post
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] New post 29 May 2015, 21:05
Awesome! Thank you so much all! :-D
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Re: An empty pool being filled with water at a constant rate [#permalink] New post 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] New post 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] 02 Jun 2015, 08:13
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