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Ratios and Mixtures

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Ratios and Mixtures

by Shridharvk » Tue Aug 10, 2010 6:42 am
The ratio, by volume, of acid to base to water in a certain solution is 4:15:20. The solution is altered so that the
ratio of acid to base is 3:5 and the ratio of acid to water remains same. If the solution initially contained 30mm
of base, what is the minimum amount of water that could be added in the second phase?

A) 18 B) 36 C) 50 D) 60 E) 90

Can someone help me in solving this question please?
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by GMATGuruNY » Tue Aug 10, 2010 7:11 am
Shridharvk wrote:The ratio, by volume, of acid to base to water in a certain solution is 4:15:20. The solution is altered so that the
ratio of acid to base is 3:5 and the ratio of acid to water remains same. If the solution initially contained 30mm
of base, what is the minimum amount of water that could be added in the second phase?

A) 18 B) 36 C) 50 D) 60 E) 90

Can someone help me in solving this question please?
Original A:B:W = 4:15:20.
B = 30, so actual values of A, B and W are all doubled:
A = 8
B = 30
W = 40

New ratio of A:B = 3:5
Unchanged ratio of A:W = 4:20 = 1:5 = 3:15 (Since A is common to each ratio, we want A to be represented by the same number in each ratio so that the ratios can be combined.)
So combining the ratios, we get a new ratio of A:B:W = 3:5:15.

We want to minimize the amount of water added, so we should keep B=30.
A:B:W = 3:5:15 = 18:30:90. (Every value in 3:5:15 is multiplied by 6 so that B=30.)
So new A=18.
B=30.
new W=90.
So new W - old W = 90-40=50.

The correct answer is C.
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by rahul goyal » Tue Aug 10, 2010 9:38 pm
GMATGuruNY wrote:
Shridharvk wrote:The ratio, by volume, of acid to base to water in a certain solution is 4:15:20. The solution is altered so that the
ratio of acid to base is 3:5 and the ratio of acid to water remains same. If the solution initially contained 30mm
of base, what is the minimum amount of water that could be added in the second phase?

A) 18 B) 36 C) 50 D) 60 E) 90

Can someone help me in solving this question please?
Original A:B:W = 4:15:20.
B = 30, so actual values of A, B and W are all doubled:
A = 8
B = 30
W = 40

New ratio of A:B = 3:5
Unchanged ratio of A:W = 4:20 = 1:5 = 3:15 (Since A is common to each ratio, we want A to be represented by the same number in each ratio so that the ratios can be combined.)
So combining the ratios, we get a new ratio of A:B:W = 3:5:15.

We want to minimize the amount of water added, so we should keep B=30.
A:B:W = 3:5:15 = 18:30:90. (Every value in 3:5:15 is multiplied by 6 so that B=30.)
So new A=18.
B=30.
new W=90.
So new W - old W = 90-40=50.

The correct answer is C.
Thank you verymuch GMATGuruNY .But i want little bit clear here,
why we should keep B=30.
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by aarati » Tue Aug 10, 2010 10:28 pm
GMATGuruNY wrote: Original A:B:W = 4:15:20.
B = 30, so actual values of A, B and W are all doubled:
A = 8
B = 30
W = 40

New ratio of A:B = 3:5
Unchanged ratio of A:W = 4:20 = 1:5 = 3:15 (Since A is common to each ratio, we want A to be represented by the same number in each ratio so that the ratios can be combined.)
So combining the ratios, we get a new ratio of A:B:W = 3:5:15.

We want to minimize the amount of water added, so we should keep B=30.
A:B:W = 3:5:15 = 18:30:90. (Every value in 3:5:15 is multiplied by 6 so that B=30.)
So new A=18.
B=30.
new W=90.
So new W - old W = 90-40=50.

The correct answer is C.
thank you.. but i am totally confused..can u explain me little bit clear .....
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by gmatmachoman » Wed Aug 11, 2010 12:40 am
rahul goyal wrote: , so we should keep B=30.

The correct answer is C.
Thank you verymuch GMATGuruNY .But i want little bit clear here,
why we should keep B=30.[/quote]


It is stated in the stem that BASE (B) is " If the solution initially contained 30mm
of base" . So B is 30
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by GMATGuruNY » Wed Aug 11, 2010 2:35 am
rahul goyal wrote:
GMATGuruNY wrote:
Shridharvk wrote:The ratio, by volume, of acid to base to water in a certain solution is 4:15:20. The solution is altered so that the
ratio of acid to base is 3:5 and the ratio of acid to water remains same. If the solution initially contained 30mm
of base, what is the minimum amount of water that could be added in the second phase?

A) 18 B) 36 C) 50 D) 60 E) 90

Can someone help me in solving this question please?
Original A:B:W = 4:15:20.
B = 30, so actual values of A, B and W are all doubled:
A = 8
B = 30
W = 40

New ratio of A:B = 3:5
Unchanged ratio of A:W = 4:20 = 1:5 = 3:15 (Since A is common to each ratio, we want A to be represented by the same number in each ratio so that the ratios can be combined.)
So combining the ratios, we get a new ratio of A:B:W = 3:5:15.

We want to minimize the amount of water added, so we should keep B=30.
A:B:W = 3:5:15 = 18:30:90. (Every value in 3:5:15 is multiplied by 6 so that B=30.)
So new A=18.
B=30.
new W=90.
So new W - old W = 90-40=50.

The correct answer is C.
Thank you verymuch GMATGuruNY .But i want little bit clear here,
why we should keep B=30.
In the original mixture, A=8, B=30, W=40.
The new ratio is A:B:W = 3:5:15.
We want to increase W by the smallest amount possible.
Since B is proportional to W, the smallest possible value of B will give us the smallest possible value of W.
We can't remove any base, so we can't decrease B. B=30 is the smallest possible value of B.
In the new ratio if B=30, then W=90. (B/W = 5/15 = 30/90).
So in the new ratio W=90 is the smallest possible value of W.

New W - old W = 90 - 40 = 50.

Does this help?
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by zaarathelab » Mon Oct 17, 2011 6:24 am
Mitch, I don't understand why we can't decrease the value of B beyond 30. When it says that the new ratio is 3:5:15, the new ratio could have been formed after decreasing their individual quantities.

The multiplying factor for the new ratio could be 1. In which case water could be 15mm. which is lower than 40mm!

I might be missing something in this question!
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by nandy1984 » Mon Oct 17, 2011 10:00 am
GMATGuruNY wrote:
rahul goyal wrote:
GMATGuruNY wrote:
Shridharvk wrote:The ratio, by volume, of acid to base to water in a certain solution is 4:15:20. The solution is altered so that the
ratio of acid to base is 3:5 and the ratio of acid to water remains same. If the solution initially contained 30mm
of base, what is the minimum amount of water that could be added in the second phase?

A) 18 B) 36 C) 50 D) 60 E) 90

Can someone help me in solving this question please?
Original A:B:W = 4:15:20.
B = 30, so actual values of A, B and W are all doubled:
A = 8
B = 30
W = 40

New ratio of A:B = 3:5
Unchanged ratio of A:W = 4:20 = 1:5 = 3:15 (Since A is common to each ratio, we want A to be represented by the same number in each ratio so that the ratios can be combined.)
So combining the ratios, we get a new ratio of A:B:W = 3:5:15.

We want to minimize the amount of water added, so we should keep B=30.
A:B:W = 3:5:15 = 18:30:90. (Every value in 3:5:15 is multiplied by 6 so that B=30.)
So new A=18.
B=30.
new W=90.
So new W - old W = 90-40=50.

The correct answer is C.
Thank you verymuch GMATGuruNY .But i want little bit clear here,
why we should keep B=30.
In the original mixture, A=8, B=30, W=40.
The new ratio is A:B:W = 3:5:15.
We want to increase W by the smallest amount possible.
Since B is proportional to W, the smallest possible value of B will give us the smallest possible value of W.
We can't remove any base, so we can't decrease B. B=30 is the smallest possible value of B.
In the new ratio if B=30, then W=90. (B/W = 5/15 = 30/90).
So in the new ratio W=90 is the smallest possible value of W.

New W - old W = 90 - 40 = 50.

Does this help?
GMATguruNY, couldnot understand why we need to take b =30 inorder to minimize water...Please explain it by using different values for A,B and W. or by explaining what happens to value of W if B is not chosen 30...Thanks....
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by GMATGuruNY » Mon Oct 17, 2011 11:14 am
zaarathelab wrote:Mitch, I don't understand why we can't decrease the value of B beyond 30. When it says that the new ratio is 3:5:15, the new ratio could have been formed after decreasing their individual quantities.

The multiplying factor for the new ratio could be 1. In which case water could be 15mm. which is lower than 40mm!

I might be missing something in this question!
My solution above reflects the intention of the question: that the amount of base cannot be reduced and must be at least 30mm.
If the amount of base can be reduced, then it is not necessary to add ANY water, and the question has no correct answer.

The writer of the question apparently assumed that it would not be possible to EXTRACT base from the solution and that the only way to yield the desired ratio would be to ADD ingredients.

To any chemists out there: is it possible to extract only base from a solution of acid, base and water?
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by zaarathelab » Tue Oct 18, 2011 1:16 am
GMATGuruNY wrote:
zaarathelab wrote:Mitch, I don't understand why we can't decrease the value of B beyond 30. When it says that the new ratio is 3:5:15, the new ratio could have been formed after decreasing their individual quantities.

The multiplying factor for the new ratio could be 1. In which case water could be 15mm. which is lower than 40mm!

I might be missing something in this question!
My solution above reflects the intention of the question: that the amount of base cannot be reduced and must be at least 30mm.
If the amount of base can be reduced, then it is not necessary to add ANY water, and the question has no correct answer.

The writer of the question apparently assumed that it would not be possible to EXTRACT base from the solution and that the only way to yield the desired ratio would be to ADD ingredients.

To any chemists out there: is it possible to extract only base from a solution of acid, base and water?
Mitch, your approach is right. These questions can be tricky at times. BTW this one is from GMAT paper tests.
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by sachindia » Wed Oct 17, 2012 7:29 pm
To any chemists out there: is it possible to extract only base from a solution of acid, base and water?
:) :)

nice sense of humor!
Regards,
Sach
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