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# M04-08

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Math Expert
Joined: 02 Sep 2009
Posts: 47183

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16 Sep 2014, 00:22
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15% (low)

Question Stats:

78% (01:20) correct 22% (01:15) wrong based on 186 sessions

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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

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Joined: 02 Sep 2009
Posts: 47183

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16 Sep 2014, 00:22
1
1
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.

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Manager
Status: PLAY HARD OR GO HOME
Joined: 25 Feb 2014
Posts: 163
Location: India
Concentration: General Management, Finance
Schools: Mannheim
GMAT 1: 560 Q46 V22
GPA: 3.1

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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.

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

$$\frac{A1}{A2}=\frac{11.5}{1}$$
_________________

ITS NOT OVER , UNTIL I WIN ! I CAN, AND I WILL .PERIOD.

Manager
Status: PLAY HARD OR GO HOME
Joined: 25 Feb 2014
Posts: 163
Location: India
Concentration: General Management, Finance
Schools: Mannheim
GMAT 1: 560 Q46 V22
GPA: 3.1

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02 Nov 2014, 07:27
Hii Bunuel,
_________________

ITS NOT OVER , UNTIL I WIN ! I CAN, AND I WILL .PERIOD.

Math Expert
Joined: 02 Sep 2009
Posts: 47183

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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.

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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Joined: 05 Jan 2015
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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.

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

$$\frac{A1}{A2}=\frac{11.5}{1}$$
Manager
Joined: 16 Feb 2016
Posts: 53
Concentration: Other, Other

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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.

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.
Manager
Joined: 14 Oct 2012
Posts: 177

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01 Apr 2017, 08:28
My 2 cents:
>> !!!

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Intern
Joined: 29 May 2017
Posts: 3

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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.
Math Expert
Joined: 02 Sep 2009
Posts: 47183

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17 Aug 2017, 06:20
1
vaibhav1446 wrote:
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.

Edited. Hope it's more clear now.
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Concentration: Marketing, Entrepreneurship
Schools: McCombs
GMAT 1: 630 Q40 V40
GPA: 2.99
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25 Dec 2017, 14:28
You guys are really smart! I pause the exam and still get the wrong answer. Half the time, I can't understand the question! I've taken all the relevant classes. But the classes are taught in a linear fashion: a+b+c=d. But the CAT presents the questions as: xa^3+yb^2-d=zc. I should just stay retired...=( It is very frustrating to work on a problem for 15 minutes (even with a calculator) and still get the wrong answer!...This is akin to the medieval "trial by morsel!"
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Re: M04-08 &nbs [#permalink] 25 Dec 2017, 14:28
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# M04-08

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