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M04-08

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A 10 liter mixture of cranberry juice and water contains juice and water in the ratio of \(3:2\). 5 liters of the mixture are removed and replaced with pure juice and the operation is repeated once more. At the end of the two removals and replacements, what is the ratio of juice to water in the resulting mixture?

A. 5 : 3
B. 6 : 4
C. 8 : 2
D. 17 : 3
E. 9 : 1
[Reveal] Spoiler: OA

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Official Solution:

A 10 liter mixture of cranberry juice and water contains juice and water in the ratio of \(3:2\). 5 liters of the mixture are removed and replaced with pure juice and the operation is repeated once more. At the end of the two removals and replacements, what is the ratio of juice to water in the resulting mixture?

A. 5 : 3
B. 6 : 4
C. 8 : 2
D. 17 : 3
E. 9 : 1


The initial mixture contains 6 liters of juice and 4 of water. 50% of the mixture is removed each time (5 liters is 50% of 10 liters of initial mixture). This means we have to remove 50% of juice as well as 50% of water. After the first removal of 5 liters it has 3 liters of juice and 2 of water. Adding 5 liters of juice gives us a mixture of 8 liters of juice and 2 of water. Removing 5 liters of the mixture, we have 4 liters of juice and 1 of water. Adding 5 liters of juice again gives us a 9 : 1 ratio of juice to water.


Answer: E
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Re: M04-08 [#permalink]

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New post 02 Nov 2014, 02:11
Bunuel wrote:
Official Solution:

A 10 liter mixture of cranberry juice and water contains juice and water in the ratio of 3 : 2. 5 liters of the mixture are removed and replaced with pure juice and the operation is repeated once more. At the end of the two removals and replacements, what is the ratio of juice to water in the resulting mixture?

A. 5 : 3
B. 6 : 4
C. 8 : 2
D. 17 : 3
E. 9 : 1


The initial mixture contains 6 liters of juice and 4 of water. 50% of the mixture is removed each time (5 liters is 50% of 10 liters of initial mixture). This means we have to remove 50% of juice as well as 50% of water. After the first removal of 5 liters it has 3 liters of juice and 2 of water. Adding 5 liters of juice gives us a mixture of 8 liters of juice and 2 of water. Removing 5 liters of the mixture, we have 4 liters of juice and 1 of water. Adding 5 liters of juice again gives us a 9 : 1 ratio of juice to water.


Answer: E


Hi Bunuel,
great solution..Could u please tell me why doesnt this mixture replacement formula work here ?Its working fine with other problem..m i missing something ?

Suppose a container contains x of liquid from which y units are taken out and replaced by water.
After n operations, the quantity of pure liquid = \(A = Q( 1-\frac{q}{Q} )^n\)

Heres how m trying to solve =
For juice=

\(A= 6(1 - \frac{5}{10}) ^2 + 10\)
\(=\frac{23}{2}\)

For water=
\(A2= 4(1-\frac{5}{10})^2\)
\(=1\)

\(\frac{A1}{A2}=\frac{11.5}{1}\)
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Re: M04-08 [#permalink]

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New post 02 Nov 2014, 07:27
Hii Bunuel,
Please help me out with this..m totally confused..
thanks in advance.. :?
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Re: M04-08 [#permalink]

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New post 02 Nov 2014, 08:04
vards wrote:
Bunuel wrote:
Official Solution:

A 10 liter mixture of cranberry juice and water contains juice and water in the ratio of 3 : 2. 5 liters of the mixture are removed and replaced with pure juice and the operation is repeated once more. At the end of the two removals and replacements, what is the ratio of juice to water in the resulting mixture?

A. 5 : 3
B. 6 : 4
C. 8 : 2
D. 17 : 3
E. 9 : 1


The initial mixture contains 6 liters of juice and 4 of water. 50% of the mixture is removed each time (5 liters is 50% of 10 liters of initial mixture). This means we have to remove 50% of juice as well as 50% of water. After the first removal of 5 liters it has 3 liters of juice and 2 of water. Adding 5 liters of juice gives us a mixture of 8 liters of juice and 2 of water. Removing 5 liters of the mixture, we have 4 liters of juice and 1 of water. Adding 5 liters of juice again gives us a 9 : 1 ratio of juice to water.


Answer: E


Hi Bunuel,
great solution..Could u please tell me why doesnt this mixture replacement formula work here ?Its working fine with other problem..m i missing something ?

Suppose a container contains x of liquid from which y units are taken out and replaced by water.
After n operations, the quantity of pure liquid = \(A = Q( 1-\frac{q}{Q} )^n\)

Heres how m trying to solve =
For juice=

\(A= 6(1 - \frac{5}{10}) ^2 + 10\)
\(=\frac{23}{2}\)

For water=
\(A2= 4(1-\frac{5}{10})^2\)
\(=1\)

\(\frac{A1}{A2}=\frac{11.5}{1}\)


Not familiar with this formula but for juice it should be \((6(1 - \frac{5}{10})+5)(1 - \frac{5}{10})+5\)
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M04-08 [#permalink]

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New post 01 Jul 2015, 21:18
May be you wrongly applied the formula. The original statement said: "Remove the liquid and replace by water", or in general "remove one, and replace by another"

Therefore, the formula in this case can be applied for the water only (remove water and replace by juice), and cannot be applied for the juice (remove juice and replace by juice!!!!).

Once you get the amount of water remained 1l, you know that juice remain 9l (10-1).


vards wrote:
Bunuel wrote:
Official Solution:

A 10 liter mixture of cranberry juice and water contains juice and water in the ratio of 3 : 2. 5 liters of the mixture are removed and replaced with pure juice and the operation is repeated once more. At the end of the two removals and replacements, what is the ratio of juice to water in the resulting mixture?

A. 5 : 3
B. 6 : 4
C. 8 : 2
D. 17 : 3
E. 9 : 1


The initial mixture contains 6 liters of juice and 4 of water. 50% of the mixture is removed each time (5 liters is 50% of 10 liters of initial mixture). This means we have to remove 50% of juice as well as 50% of water. After the first removal of 5 liters it has 3 liters of juice and 2 of water. Adding 5 liters of juice gives us a mixture of 8 liters of juice and 2 of water. Removing 5 liters of the mixture, we have 4 liters of juice and 1 of water. Adding 5 liters of juice again gives us a 9 : 1 ratio of juice to water.


Answer: E


Hi Bunuel,
great solution..Could u please tell me why doesnt this mixture replacement formula work here ?Its working fine with other problem..m i missing something ?

Suppose a container contains x of liquid from which y units are taken out and replaced by water.
After n operations, the quantity of pure liquid = \(A = Q( 1-\frac{q}{Q} )^n\)

Heres how m trying to solve =
For juice=

\(A= 6(1 - \frac{5}{10}) ^2 + 10\)
\(=\frac{23}{2}\)

For water=
\(A2= 4(1-\frac{5}{10})^2\)
\(=1\)

\(\frac{A1}{A2}=\frac{11.5}{1}\)

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Re: M04-08 [#permalink]

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New post 27 Apr 2016, 00:08
kimnguyen wrote:
May be you wrongly applied the formula. The original statement said: "Remove the liquid and replace by water", or in general "remove one, and replace by another"

Therefore, the formula in this case can be applied for the water only (remove water and replace by juice), and cannot be applied for the juice (remove juice and replace by juice!!!!).

Once you get the amount of water remained 1l, you know that juice remain 9l (10-1).


vards wrote:
Bunuel wrote:
Official Solution:

A 10 liter mixture of cranberry juice and water contains juice and water in the ratio of 3 : 2. 5 liters of the mixture are removed and replaced with pure juice and the operation is repeated once more. At the end of the two removals and replacements, what is the ratio of juice to water in the resulting mixture?

A. 5 : 3
B. 6 : 4
C. 8 : 2
D. 17 : 3
E. 9 : 1


The initial mixture contains 6 liters of juice and 4 of water. 50% of the mixture is removed each time (5 liters is 50% of 10 liters of initial mixture). This means we have to remove 50% of juice as well as 50% of water. After the first removal of 5 liters it has 3 liters of juice and 2 of water. Adding 5 liters of juice gives us a mixture of 8 liters of juice and 2 of water. Removing 5 liters of the mixture, we have 4 liters of juice and 1 of water. Adding 5 liters of juice again gives us a 9 : 1 ratio of juice to water.


Answer: E


Hi Bunuel,
great solution..Could u please tell me why doesnt this mixture replacement formula work here ?Its working fine with other problem..m i missing something ?

Suppose a container contains x of liquid from which y units are taken out and replaced by water.
After n operations, the quantity of pure liquid = \(A = Q( 1-\frac{q}{Q} )^n\)

Heres how m trying to solve =
For juice=

\(A= 6(1 - \frac{5}{10}) ^2 + 10\)
\(=\frac{23}{2}\)

For water=
\(A2= 4(1-\frac{5}{10})^2\)
\(=1\)

\(\frac{A1}{A2}=\frac{11.5}{1}\)



Trying to remember formulas on the day might be an issue, so I would recommend understanding the solution. Then you can solve the question every time.

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Re: M04-08 [#permalink]

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New post 01 Apr 2017, 08:28
My 2 cents:
>> !!!

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Re: M04-08 [#permalink]

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New post 17 Aug 2017, 06:12
Please change how the question is written, 2 . 5 is totally misleading, & providing multiple meanings whether 5 litre of solution was removed or .5 . It is also not giving clarity on ratio.

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Re: M04-08 [#permalink]

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Re: M04-08   [#permalink] 17 Aug 2017, 06:20
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