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Allegation

by sixpointer » Thu Nov 04, 2010 10:08 am
A mixture of 125 gallons of wine and water contains 20% water .How much water must be added to the mixture in order to increase the percentage of water to 25% of the new mixture ?????
a 10 gals
b 8.5 gals
c 8 gals
d 6.66 gals
e 8.33 gals

Please use allegation approach
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by limestone » Thu Nov 04, 2010 10:27 am
Hi,

Because we'll add more water into the original mixture, then the no. of gallons of wine is unchanged.

In the original 125 gallons of mixture, 20% is water, thus 80% is wine. Hence, no. of gallons of wine:

125 * 80% = 100 gallons

In the new mixture, water makes up 25%, thus 75% is wine. As no. of gallons of wine is unchanged, 100 gallons = 75% the new mixture volume.

The total volume of the new mixture is : 100 / 75% = 100/ 0.75 = 133.3 gallons.

No. of gallons of added water = 133.3 - 125 = 8.3

Pick E.
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by sixpointer » Thu Nov 04, 2010 10:40 am
limestone wrote:Hi,

Because we'll add more water into the original mixture, then the no. of gallons of wine is unchanged.

In the original 125 gallons of mixture, 20% is water, thus 80% is wine. Hence, no. of gallons of wine:

125 * 80% = 100 gallons

In the new mixture, water makes up 25%, thus 75% is wine. As no. of gallons of wine is unchanged, 100 gallons = 75% the new mixture volume.

The total volume of the new mixture is : 100 / 75% = 100/ 0.75 = 133.3 gallons.

No. of gallons of added water = 133.3 - 125 = 8.3




Pick E.

Hi I know this method but I want to learn allegation method :)
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by limestone » Thu Nov 04, 2010 11:08 am
I'm not sure about Alligation Method. But I think it should be in this way:

The original mixture 125 Gallons at 20% water ( A )
Added water Y Gallons at 100% water. (B)
New mixture X Gallons at 25% water ( The Result)

Each gallon of pure water added is surplus of 75% water compared to the desired result 25%.
Each gallons of A is missing 5% of water from the desired result.

So no. of gallons of the original mixture (A) that one gallon of pure water can help to increase water degree to 25%: 75%/5% = 15

So number of gallons of pure waters need to add to help increase all 125 gallons of mixture A to 25% water: 125/15 = 8.33 Gallons.

Hope it helps. Correct me if I'm wrong.
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by sixpointer » Thu Nov 04, 2010 11:44 am
limestone wrote:I'm not sure about Alligation Method. But I think it should be in this way:

The original mixture 125 Gallons at 20% water ( A )
Added water Y Gallons at 100% water. (B)
New mixture X Gallons at 25% water ( The Result)

Each gallon of pure water added is surplus of 75% water compared to the desired result 25%.
Each gallons of A is missing 5% of water from the desired result.

So no. of gallons of the original mixture (A) that one gallon of pure water can help to increase water degree to 25%: 75%/5% = 15

So number of gallons of pure waters need to add to help increase all 125 gallons of mixture A to 25% water: 125/15 = 8.33 Gallons.

Hope it helps. Correct me if I'm wrong.
Please explain the bold part . My mind is working like a tubelight :(
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by GMATGuruNY » Thu Nov 04, 2010 12:09 pm
sixpointer wrote:A mixture of 125 gallons of wine and water contains 20% water .How much water must be added to the mixture in order to increase the percentage of water to 25% of the new mixture ?????
a 10 gals
b 8.5 gals
c 8 gals
d 6.66 gals
e 8.33 gals

Please use allegation approach
Here's the alligation method.

When combining a lower percentage (L) with a higher percentage (H) to achieve a goal percentage (G):

The proportion needed of L = H-G (the positive difference between the higher percentage and the goal percentage)
The proportion needed of H = G-L (the positive difference between the lower percentage and the goal percentage)

In the problem above:
L = 20 (the original mixture)
H = 100 (the pure water)
G = 25 (the goal percentage of the final mixture)

Thus:
Proportion needed of L = 100-25 = 75
Proportion needed of H = 25-20 = 5
L:H = 75:5 = 15:1

Since the amount of the original mixture (the lower percentage L) is not changing from 125, we could set up the following proportion:

125/x = 15/1
x = 125/15 = 8.33

The correct answer is E.

Hope this helps!
Last edited by GMATGuruNY on Thu Nov 04, 2010 12:14 pm, edited 1 time in total.
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by limestone » Thu Nov 04, 2010 12:11 pm
1 Gallon of pure water has a surplus of 75%, while one gallons of A is lack of 5%.
So 1 gallons of pure water can compensate for 15 gallons of A. Or say in another way:

1 gallon of pure water + 15 gallons of A = 16 gallons of mixture that have 25% water.
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by Ian Stewart » Thu Nov 04, 2010 1:34 pm
sixpointer wrote:A mixture of 125 gallons of wine and water contains 20% water .How much water must be added to the mixture in order to increase the percentage of water to 25% of the new mixture ?????
a 10 gals
b 8.5 gals
c 8 gals
d 6.66 gals
e 8.33 gals

Please use allegation approach
You're combining a 20% mixture with a 100% mixture (pure water) to get a 25% mixture. You can use alligation algebraically here, as was done in a post above. I find it easier to visualize alligation on a number line. We can plot the three averages:

--20---25----------------------------100---

First, notice we aren't adding much pure water here; the overall average is much closer to 20% than to 100%. Now the ratio of the distances from each average to the overall average (i.e. the distances from the two extreme numbers to the middle number) must be equal to the ratio of the amounts of our two components. The distances are 25-20 = 5, and 100-25 = 75, so we must mix our two components in a 75 to 5, or 15 to 1 ratio. Since we have 125 gallons of the 20% mixture, we must have 125/15 = 8 1/3 gallons of pure water.
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by sixpointer » Thu Nov 04, 2010 9:54 pm
Thankyou seniors :)
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by kamalakarthi » Sat Mar 19, 2011 12:57 pm
Mitch,

How do we identify h = 100. Also how do we equate 125/x = 15/1 , I can understand the part how we got 15:1 but not able to understand the other side of the equation and how can we equate it.
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by GMATGuruNY » Sun Mar 20, 2011 2:30 pm
kamalakarthi wrote:Mitch,

How do we identify h = 100. Also how do we equate 125/x = 15/1 , I can understand the part how we got 15:1 but not able to understand the other side of the equation and how can we equate it.
The percentage of water in the original mixture is 20%
The percentage of water in the pure water to be added is 100%.
The percentage water in the final mixture is 25%.

Using alligation:
Proportion needed of the original mixture = 100-25 = 75.
Proportion needed of the pure water = 25-20 = 5.

Thus, in the final mixture:
Original:Water = 75:5 = 15:1.

Since no part of the original mixture is being removed, the entire 125 gallons of the original mixture will remain in the final mixture. We need to determine only how much water must be added to the 125 gallons to yield a 15:1 ratio of original:water. Thus, we can set up the following proportion:

125/W = 15/1.
W = 125/15 = 8.33.
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by ankurmit » Sat May 07, 2011 10:41 pm
--thanks Mitch
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by shayam » Sat Sep 17, 2011 11:23 am
Good afternoon People!!

Few days back, I had practiced few problems on weighted average /allegation and I thought I had won!!
But when i tried to think about allegation today, i just didn't manage it out.
I am hopelessly back and I need some good help.

In weighted average problem or mixtures, what if we are given three components instead of two .

For eg: say a Solution of 2% , 10% and 15% wine in a mixture of water and wine and the weighted average is given as 5% wine

How do we find the ratio of individual components by using allegation(H-G/ G-L) approach..
or How do we deduce the weights in this case ??

If we are given only two say 2% , 10% wine in a mixture of water and wine with WA as 5%
I can happily derive weights as 5/8 and 3/8

What can i do when I have more than 2 ?

What am i failing to understand here?? Would appreciate your help..!!!
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by GMATGuruNY » Sat Sep 17, 2011 12:41 pm
shayam wrote:Good afternoon People!!

Few days back, I had practiced few problems on weighted average /allegation and I thought I had won!!
But when i tried to think about allegation today, i just didn't manage it out.
I am hopelessly back and I need some good help.

In weighted average problem or mixtures, what if we are given three components instead of two .

For eg: say a Solution of 2% , 10% and 15% wine in a mixture of water and wine and the weighted average is given as 5% wine

How do we find the ratio of individual components by using allegation(H-G/ G-L) approach..
or How do we deduce the weights in this case ??

If we are given only two say 2% , 10% wine in a mixture of water and wine with WA as 5%
I can happily derive weights as 5/8 and 3/8

What can i do when I have more than 2 ?

What am i failing to understand here?? Would appreciate your help..!!!
When more than 2 elements are being combined, an infinite number of ratios are possible.

Let x = the 2% wine solution, let y = the 10% wine solution, and let z = the 15% wine solution.
These 3 solutions are being mixed to form a solution of 5% wine.

To make the math easier, let's use integer values to represent the amounts of wine.
Amount of wine in x = 2x.
Amount of wine in y = 10y.
Amount of wine in z = 15z.
Amount of wine in the mixture = 5(x+y+x).

Since the total amount of wine in x,y and z is equal to the total amount of wine in the mixture:
2x + 10y + 15z = 5(x+y+z)
2x + 10y + 15z = 5x + 5y + 5z
3x = 5y + 10z
x = (5y + 10z)/3.

The equation above tells us the value of x in terms of y and z.
It does not tell us the ratio of x:y:z.

If we plug y=2 and z=2 into the equation above, then x = 10.
Thus, a ratio of x:y:z = 10:2:2 = 5:1:1 will yield a solution that is 5% wine.

If we plug y=4 and z=1 into the equation above, then x=10.
Thus, a ratio of x:y:z = 10:4:1 will yield a solution that is 5% wine.

Since many ratios are possible, the ratio of x:y:z cannot be determined.
If the problem asks for one element IN TERMS OF the other 2 elements, we can use algebra as I did above.
For a non-algebraic approach, please check my solution here:

https://www.beatthegmat.com/mixtures-t83395.html
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by jzw » Sun Mar 18, 2012 6:50 am
GMATGuruNY wrote:
Here's the allegation method.

When combining a lower percentage (L) with a higher percentage (H) to achieve a goal percentage (G):

The proportion needed of L = H-G (the positive difference between the higher percentage and the goal percentage)
The proportion needed of H = G-L (the positive difference between the lower percentage and the goal percentage)
Mitch - been feverishly studying this. Can you please clarify if the following more pedantic definition is correct? My issue I think was understanding "the higher percentage of what, that which is there already or that which is to be added?"

The proportion needed of L = H-G (the positive difference between the higher percentage of the solution being added and the goal percentage)
The proportion needed of H = G-L (the positive difference between the lower percentage of the existing solution[/b) and the goal percentage)

Is the above correct?

Also - you made a statement below "Since the amount of the original mixture (the lower percentage L) is not changing from 125..." How do we know this? Ie, if one is adding water to the mixture, won't the final volume increase from 125 gallons to - something more than this? Please advise. Thanks!
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