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The ratio of the volumes of two empty pools is 5 : 3. If the smaller

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The ratio of the volumes of two empty pools is 5 : 3. If the smaller [#permalink]

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New post 02 Jan 2018, 01:39
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The ratio of the volumes of two empty pools is 5 : 3. If the smaller pool is filled and emptied into the larger pool, what is the ratio of filled volume to empty volume in the pools?

A. 1 : 4
B. 2 : 5
C. 3 : 5
D. 3 : 8
E. 5 : 8
[Reveal] Spoiler: OA

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The ratio of the volumes of two empty pools is 5 : 3. If the smaller [#permalink]

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New post 02 Jan 2018, 09:09
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Bunuel wrote:
The ratio of the volumes of two empty pools is 5 : 3. If the smaller pool is filled and emptied into the larger pool, what is the ratio of filled volume to empty volume in the pools?

A. 1 : 4
B. 2 : 5
C. 3 : 5
D. 3 : 8
E. 5 : 8


The ratio of empty pools is \(5:3\) and as the smaller pool is filled and then emptied into the larger pool

so ratio of filled to empty pool will be simply reverse of the original pool i.e. \(3:5\) (essentially larger pool will contain all the contents of smaller pool, hence the filled part will be the volume of smaller pool and the empty part will be volume of smaller pool+difference in volume between the pools)

Option C
---------------------------------------------------------------------------
Alternatively, Let the volume of larger pool be \(5x\) & that of smaller pool be \(3x\)

so \(3x\) is filled and then emptied into larger pool. Hence filled volume will be \(3x\)

and empty volume = volume of smaller pool + remaining volume of larger pool \(= 3x+(5x-3x)=5x\)

So ratio of filled to empty \(= 3x:5x=3:5\)
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Re: The ratio of the volumes of two empty pools is 5 : 3. If the smaller [#permalink]

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New post 03 Jan 2018, 02:32
Not very sure, but feel the ans is option A.

if you set E1:E2 to be equal to 5:3 (represent the ratio of empty pools).

Then 5x is the maximum possible volume E1 can take while 3x is the maximum volume E2 can take, then the total combined volume when empty becomes

5x+3x=8x

When filled, let F1:F2 be the ratio when both pools are filled. maximum filling capacity in pool 1 will be limited by capacity of pool 2 which is 3X

meaning F1:F2 can be expressed as 3x:3x which is the same as 1:1

In terms of totals it will become 1X+1X=2X

Since the question is asking for F:E

that means 2x:8x

which is 1:4
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Re: The ratio of the volumes of two empty pools is 5 : 3. If the smaller [#permalink]

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New post 03 Jan 2018, 02:46
Aussy2000 wrote:
Not very sure, but feel the ans is option A.

if you set E1:E2 to be equal to 5:3 (represent the ratio of empty pools).

Then 5x is the maximum possible volume E1 can take while 3x is the maximum volume E2 can take, then the total combined volume when empty becomes

5x+3x=8x

When filled, let F1:F2 be the ratio when both pools are filled. maximum filling capacity in pool 1 will be limited by capacity of pool 2 which is 3X

meaning F1:F2 can be expressed as 3x:3x which is the same as 1:1


In terms of totals it will become 1X+1X=2X

Since the question is asking for F:E

that means 2x:8x

which is 1:4


Hi Aussy2000

The highlighted portion is not in sync with what the question says. The question does not mention that both were filled simultaneously. First smaller was filled to its capacity and then it's content were emptied into larger pool. So essentially at any point of time the filled capacity is only 3x.

And as you have mentioned total capacity is 8x so empty capacity has to be 8x-3x=5x

So shouldn't the ratio of filled:empty be 3x:5x=3:5
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The ratio of the volumes of two empty pools is 5 : 3. If the smaller [#permalink]

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New post 04 Jan 2018, 10:35
Bunuel wrote:
The ratio of the volumes of two empty pools is 5 : 3. If the smaller pool is filled and emptied into the larger pool, what is the ratio of filled volume to empty volume in the pools?

A. 1 : 4
B. 2 : 5
C. 3 : 5
D. 3 : 8
E. 5 : 8

If a ratio approach is used, one way to avoid confusion: the ratio of the two empty pools' volumes needs to have its parts totaled.

If \(\frac{Big}{Small} = \frac{5x}{3x}\), then total empty capacity is \(8x\).

If \(3x\) (the capacity of the small pool) is put into the big pool, the filled ratio/fraction is not \(\frac{3x}{5x}\)

The filled ratio/fraction is \(\frac{3x}{8x}\) (i.e., \(\frac{3}{8}\) of total capacity is filled, 3 of 8 parts are filled)

Then
(Total Empty Capacity) - (Fraction Filled) =
(Fraction NOT filled, i.e., Fraction that is Empty)

\((\frac{8x}{8x} - \frac{3x}{8x}) = \frac{5x}{8x}=\frac{5}{8}\) of total capacity is NOT filled. Fraction that is empty = \(\frac{5}{8}\)

Ratio of filled to empty:

\(\frac{\frac{3}{8}}{\frac{5}{8}} = \frac{3}{8}*\frac{8}{5}=\frac{3}{5}\)

Answer C
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Re: The ratio of the volumes of two empty pools is 5 : 3. If the smaller [#permalink]

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New post 04 Jan 2018, 12:24
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Hi niks18 .. really loved the simple explanation u gave for the 1st explanation of the solution


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Re: The ratio of the volumes of two empty pools is 5 : 3. If the smaller [#permalink]

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New post 16 Feb 2018, 09:54
Volume of larger pool = 50lt
Volume of smaller pool = 30lt

The larger pool is filled with 30lt and empty for 20lt
The smaller pool is 30lt empty

So: 30/20+30 = 3/5
Re: The ratio of the volumes of two empty pools is 5 : 3. If the smaller   [#permalink] 16 Feb 2018, 09:54
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