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mixture

This topic has 5 expert replies and 4 member replies
vipulgoyal Master | Next Rank: 500 Posts Default Avatar
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mixture

Post Sun Oct 19, 2014 11:20 pm
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?

A. 144
B. 122.5
C. 105.10
D. 72
E. 134

plz explain with alligation approach only

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Post Mon Oct 20, 2014 11:32 am
Hi vipulgoyal,

You asked about the allegation method (which Mitch has explained, but this question can be solved using the traditional weighted average formula. I'm curious about the source of this question because the ratios and answer choices are a bit more "complicated-looking" than a typical GMAT question on this subject would be.

With the given ratios of milk to water….

Mix A: 2:5
Mix B: 5:4
90 gallons of B
And the goal of a 40% milk mixture, we can set up the following equation:

[(2/7)A + (5/9)B]/(A+B) = .4

Now, substituting in the value for B gets us…

[(2/7)A + 50]/(A+90) = .4

From here, we have a 1-variable equation, that requires just a bit of algebra to solve…

(2/7)A + 50 = .4A + 36

The GMAT expects you the know that 1/7 = about .142 (it's actually a 6-digit "repeater", but you're normally not asked for that). Substituting in .142 for 1/7 gets us…

.284A + 50 = .4A + 36
14 = .116A
14/.116 = 14,000/116 = about 120

The only answer that's close is B

GMAT assassins aren't born, they're made,
Rich

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GMAT/MBA Expert

Post Mon Oct 20, 2014 8:05 pm
I'd like to point out that Rich is applying a formula for dealing with weighted averages. The formula goes like this:
Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...

For more information on weighted averages, you can watch this free GMAT Prep Now video: http://www.gmatprepnow.com/module/gmat-statistics?id=805

Cheers,
Brent

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Post Wed Oct 22, 2014 4:07 pm
vipulgoyal wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?

A. 144
B. 122.5
C. 105.10
D. 72
E. 134
When solving mixture questions, I find it useful to sketch the solutions with the ingredients SEPARATED.

First recognize that if mixture A has a milk to water ratio of 2:5, then the mixture is 2/7 milk.
Also recognize that if mixture B has a milk to water ratio of 5:4, then the mixture is 5/9 milk.

Start with 90 gallons of mixture B, which is 5/9 milk:
ttp://postimg.org/image/csi0pfanz/" target="_blank">
When we draw this with the ingredients separated, we see we have 50 gallons of milk in the mixture.

Next, we'll let x = the number of gallons of mixture A we need to add.
Since 2/7 of mixture A is milk, we know that (2/7)x = the volume of MILK in this mixture:
ttp://postimg.org/image/elkxdqvun/" target="_blank">

At this point, we can ADD the two solutions (PART BY PART) to get the following volumes:
ttp://postimg.org/image/bg0bnjd8f/" target="_blank">

Since the RESULTING mixture is 40% milk (i.e., 40/100 of the mixture is milk), we can write the following equation:
[50 + (2/7)x]/(90 + x) = 40/100
Simplify to get: [50 + (2/7)x]/(90 + x) = 2/5
Cross multiply to get: 5[50 + (2/7)x] = 2(90 + x)
Expand: 250 + (10/7)x = 180 + 2x
Subtract 180 from both sides to get: 70 + (10/7)x = 2x
Multiply both sides by 7 to get: 490 + 10x = 14x
Rearrange: 490 = 4x
Solve: x = 490/4 = 245/2 = 122.5

Answer: B

Cheers,
Brent

Here are some additional mixture questions to practice with:
http://www.beatthegmat.com/liters-of-mixture-x-in-the-50-mixture-t271387.html
http://www.beatthegmat.com/percentage-mixture-t268631.html
http://www.beatthegmat.com/rodrick-mixes-a-martini-that-has-a-volume-of-n-ounces-havi-t270387.html
http://www.beatthegmat.com/mixure-problem-quite-confusing-t261767.html
http://www.beatthegmat.com/mixture-ratio-problem-2-t191643.html

_________________
Brent Hanneson – Founder of GMATPrepNow.com
Use our video course along with Beat The GMAT's free 60-Day Study Guide

Check out the online reviews of our course
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vipulgoyal Master | Next Rank: 500 Posts Default Avatar
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Post Mon Oct 20, 2014 4:16 am
Thanks once again...

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GMAT/MBA Expert

Post Mon Oct 20, 2014 3:59 am
vipulgoyal wrote:
yes the ans is B
my approach, plz tell where i went wrong

2/5---2/3(40%milk&60%water)----5/4
2/3-2/5= 4/15
5/4-2/3= 7/12
ratio of mixture A:B:: 7/12:4/15
so if B is 4/15 A should be 7/12
therefore if B is 90, A should be 7/12*15/4*90 = 196.8(not an option)
Your solution attempts to perform alligation with RATIOS.
Allligation can be performed only with PART-TO-WHOLE relationships (percentages or fractions).
It cannot be performed with ratios.
In my solution above, I converted the ratios for solutions A and B to FRACTIONS.

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vipulgoyal Master | Next Rank: 500 Posts Default Avatar
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Post Mon Oct 20, 2014 3:54 am
thanks Mitch, but i still swamped where I went wrong ??

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vipulgoyal Master | Next Rank: 500 Posts Default Avatar
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Post Mon Oct 20, 2014 3:41 am
yes the ans is B
my approach, plz tell where i went wrong

2/5---2/3(40%milk&60%water)----5/4
2/3-2/5= 4/15
5/4-2/3= 7/12
ratio of mixture A:B:: 7/12:4/15
so if B is 4/15 A should be 7/12
therefore if B is 90, A should be 7/12*15/4*90 = 196.8(not an option)

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GMAT/MBA Expert

Post Mon Oct 20, 2014 3:22 am
vipulgoyal wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?

A. 144
B. 122.5
C. 105.10
D. 72
E. 134
The following approach is called ALLIGATION -- a great way to handle mixture problems.
For an explanation of alligation, check my 2 posts here:
http://www.beatthegmat.com/ratios-fraction-problem-t115365.html

Note:
Alligation can be performed only with percentages or fractions.
In the problem above, the given ratios must be converted.

Step 1: Convert the ratios to FRACTIONS.
A:
Since milk:water = 2:5, and 2+5=7, milk/total = 2/7.
B:
Since milk:water = 5:4, and 5+4=9, milk/total = 5/9.
Mixture:
milk/total = 40% = 2/5.

Step 2: Put the fractions over a COMMON DENOMINATOR.

A = 2/7 = (2*5*9)/(5*7*9) = 90/(5*7*9).
B = 5/9 = (5*5*7)/(5*7*9) = 175/(5*7*9).
Mixture = 2/5 = (2*7*9)/(5*7*9) = 126/(5*7*9).

Step 3: Plot the 3 numerators on a number line, with the numerators for A and B on the ends and the numerator for the mixture in the middle.
A 90------------126------------175 B

Step 4: Calculate the distances between the numerators.
A 90-----36-----126-----49-----175 B

Step 5: Determine the ratio in the mixture.
The required ratio of A to B is the RECIPROCAL of the distances in red.
A/B = 49/36.

Since A/B = 49/36, and B=90 gallons, we get:
49/36 = A/90
49/2 = A/5
2A = 5*49
2A = 245
A = 122.5.

The correct answer is B.

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Last edited by GMATGuruNY on Mon Oct 20, 2014 4:32 am; edited 1 time in total

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gmatcracker0123 Junior | Next Rank: 30 Posts Default Avatar
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Post Mon Oct 20, 2014 2:29 am
I used the method of alligation according to my understanding and used approxiamtions to reach to 121.5 as the answer. Could you confirm whether B is the OA so that i can either explain my approach or rework on it if my answer is incorrect?

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