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Mixtures allegation

by vishal.pathak » Tue Mar 27, 2012 1:46 am
2 vessels contain mixture of spirit and water. 1st vessel s:w = 8:3, 2nd 5:1. 35 litre flask contains in the ratio 4:1. How many litres were taken from 1st vessel
[spoiler]Ans: 11[/spoiler]

Participation in soccer league is 10% higher this year. Participation of men increased by 5% and that of women increased by 20%. What is the % of female in the league now
a.1/3 b.4/11 c.2/5 d.4/9 e.1/2
[spoiler]OA E

In the 1st question we cannot apply the mixtures allegation unless the base is same i.e we have the ratio of spirit to the entire mixture but in 2nd we can apply the allegation although 5% is an increase on the number of BOYS last year, 20% is an increase on the number of GIRLS last year.

My question really is what should we check before applying the allegation[/spoiler]

Why am I not able to apply the allegation in the following question

A certain military vehicle can run on pure Fuel X, pure Fuel Y, or any mixture of X and Y. Fuel X costs $3 per gallon; the vehicle can go 20 miles on a gallon of Fuel X. In contrast, Fuel Y costs $5 per gallon, but the vehicle can go 40 miles on a gallon of Fuel Y. The vehicle can cover 7 and 1/7 miles for every dollar of fuel currently in its tank. What is the cost per gallon of the fuel mixture currently in the vehicle's tank?

(40/5 - 50/7)/(50/7 - 20/3) = 9/5 and [spoiler]the correct ans is the ratio of 1st fuel to 2nd = 3:1[/spoiler]


Regards,
Vishal
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by GMATGuruNY » Tue Mar 27, 2012 4:29 am
vishal.pathak wrote:
Why am I not able to apply the allegation in the following question

A certain military vehicle can run on pure Fuel X, pure Fuel Y, or any mixture of X and Y. Fuel X costs $3 per gallon; the vehicle can go 20 miles on a gallon of Fuel X. In contrast, Fuel Y costs $5 per gallon, but the vehicle can go 40 miles on a gallon of Fuel Y. The vehicle can cover 7 and 1/7 miles for every dollar of fuel currently in its tank. What is the cost per gallon of the fuel mixture currently in the vehicle's tank?

(40/5 - 50/7)/(50/7 - 20/3) = 9/5 and [spoiler]the correct ans is the ratio of 1st fuel to 2nd = 3:1[/spoiler]


Regards,
Vishal
Your solution is correct.
You determined that for every $9 spent on X, $5 must be spent on Y.
Since X costs $3 per gallon, $9 = 3 gallons.
Since Y costs $5 per gallon, $5 = 1 gallon.
Thus, the ratio of X:Y = 3:1.

To answer the question stem:
For every 3 gallons of X that is purchased, 1 gallon of Y must be purchased.
Thus, for every 4 gallons purchased, $9 is spent on X and $5 is spent on Y.
Cost per gallon = (9+5)/4 = $3.50.
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by vishal.pathak » Tue Mar 27, 2012 5:22 am
GMATGuruNY wrote:
vishal.pathak wrote:
Why am I not able to apply the allegation in the following question

A certain military vehicle can run on pure Fuel X, pure Fuel Y, or any mixture of X and Y. Fuel X costs $3 per gallon; the vehicle can go 20 miles on a gallon of Fuel X. In contrast, Fuel Y costs $5 per gallon, but the vehicle can go 40 miles on a gallon of Fuel Y. The vehicle can cover 7 and 1/7 miles for every dollar of fuel currently in its tank. What is the cost per gallon of the fuel mixture currently in the vehicle's tank?

(40/5 - 50/7)/(50/7 - 20/3) = 9/5 and [spoiler]the correct ans is the ratio of 1st fuel to 2nd = 3:1[/spoiler]


Regards,
Vishal
Your solution is correct.
You determined that for every $9 spent on X, $5 must be spent on Y.
Since X costs $3 per gallon, $9 = 3 gallons.
Since Y costs $5 per gallon, $5 = 1 gallon.
Thus, the ratio of X:Y = 3:1.

To answer the question stem:
For every 3 gallons of X that is purchased, 1 gallon of Y must be purchased.
Thus, for every 4 gallons purchased, $9 is spent on X and $5 is spent on Y.
Cost per gallon = (9+5)/4 = $3.50.
Hi Mitch,

Many many thanks for answering this.

Please help me understand how subtracting and dividing miles/dollar resulted in a ratio of dollars.

Please throw some light on the 1st part of the question as well.
Are there some pre-conditions for applying this allegation. Why do we have to have the same base (spirit/(total liquid)) for the 1st problem and how are we able to solve the 2nd problem without making the base common. In the 2nd problem, 10% is the increase in the number of BOYS and 20% is the increase in the number of GIRLS

Please help

Regards,
Vishal
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by GMATGuruNY » Tue Mar 27, 2012 6:56 am
vishal.pathak wrote: Participation in soccer league is 10% higher this year. Participation of men increased by 5% and that of women increased by 20%. What is the % of female in the league now
a.1/3 b.4/11 c.2/5 d.4/9 e.1/2

Why do we have to have the same base (spirit/(total liquid)) for the 1st problem and how are we able to solve the 2nd problem without making the base common. In the 2nd problem, 10% is the increase in the number of BOYS and 20% is the increase in the number of GIRLS
OA E
Alligation must be performed with a common base.
Here, the common base is PEOPLE.

The league is composed of two groups: the MEN'S GROUP and the WOMEN'S GROUP.
The number of PEOPLE in the men's group increased by 5%.
The number of PEOPLE in the women's group increased by 20%.
The number of PEOPLE in the MIXTURE -- the men's group + the women's group -- increased by 10%.

Alligation dictates the following:
The proportion needed of each ingredient in the mixture is equal to the distance between the OTHER TWO PERCENTAGES attributed to the mixture.
Thus:
M:W = (20-10) : (10-5) = 10:5 = 2:1.
This represents the ratio of men to women BEFORE each group increased.

To determine the fraction of women AFTER each group increased:
Let the starting values be M=20 and W=10.
Increased by 5%, M = 21.
Increased by 20%, W = 12.
(new W)/(new total) = 12/(21+12) = 12/33 = 4/11.

The correct answer is B.

If the OA is E, it's incorrect.
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by vishal.pathak » Tue Mar 27, 2012 9:52 pm
GMATGuruNY wrote:
vishal.pathak wrote:
Why am I not able to apply the allegation in the following question

A certain military vehicle can run on pure Fuel X, pure Fuel Y, or any mixture of X and Y. Fuel X costs $3 per gallon; the vehicle can go 20 miles on a gallon of Fuel X. In contrast, Fuel Y costs $5 per gallon, but the vehicle can go 40 miles on a gallon of Fuel Y. The vehicle can cover 7 and 1/7 miles for every dollar of fuel currently in its tank. What is the cost per gallon of the fuel mixture currently in the vehicle's tank?

(40/5 - 50/7)/(50/7 - 20/3) = 9/5 and [spoiler]the correct ans is the ratio of 1st fuel to 2nd = 3:1[/spoiler]


Regards,
Vishal
Your solution is correct.
You determined that for every $9 spent on X, $5 must be spent on Y.
Since X costs $3 per gallon, $9 = 3 gallons.
Since Y costs $5 per gallon, $5 = 1 gallon.
Thus, the ratio of X:Y = 3:1.

To answer the question stem:
For every 3 gallons of X that is purchased, 1 gallon of Y must be purchased.
Thus, for every 4 gallons purchased, $9 is spent on X and $5 is spent on Y.
Cost per gallon = (9+5)/4 = $3.50.
Thanks a lot for your help, Mitch

Regards,
Vishal
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by krishna239455 » Wed Mar 28, 2012 8:41 am
i have tried to solve the first question by algebra.
Let 'x' be the common factor for ratio of Spirit and water in 1st vessel. Hence ratio will be 8x:3X and the volume will be 8x+3x=11x
Let 'y' be the common factor for ratio of spirit and water in 2nd vessel. Hence ratio will be 5y:y and the volume will be 5y+y=6y.
Now it is given that both the volumes were mixed to make 35 ltrs mix with 4:1 ratio. This gives me two equations:
11x+6y=35.....(1)
8x+5y/3x+y = 4/1.....(2)
Solving both the equations i get x=1. Hence volume in 1st vessel is 11*1=11
Tell me if my approach is wrong!!!
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by krishna239455 » Wed Mar 28, 2012 8:58 am
Dear Mitch
I have question on the second problem about Boys and Girls ratio.
You have considered allegation with "% increment" all across.
So the result should also be "% increment" (as in allegation the unit should remain same all across). That is the ratio 2:1 is the ratio of %increased boys to % increased girls.
Answer should be 1/3.
Tell me if i am wrong!![/u]
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by ronnie1985 » Wed Mar 28, 2012 9:37 am
First question is not clear
Second question must have asked for the ratio of mixture and teh method adopted by you is corect
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