ratios / fraction problem

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ratios / fraction problem

by sachindia » Thu Jul 05, 2012 11:33 pm
Two tins A and B contain mixtures of wheat and rice. In A, the weights of wheat and rice are in
the ratio 2 : 3 and in B they are in the ratio 3 : 7. What quantities must be taken from A and B to
form a mixture containing 5 kg of wheat and 11 kg of rice?
Regards,
Sach

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by GMATGuruNY » Fri Jul 06, 2012 1:44 am
sachindia wrote:Two tins A and B contain mixtures of wheat and rice. In A, the weights of wheat and rice are in
the ratio 2 : 3 and in B they are in the ratio 3 : 7. What quantities must be taken from A and B to
form a mixture containing 5 kg of wheat and 11 kg of rice?

1. 3kgs from A, 13 kg from B
2. 8kgs from A, 8 kg from B
3. 2kgs from A, 14 kg from B
4. 6kgs from A, 10 kg from B
Percentage of wheat in A = 2/5 = 40%
Percentage of wheat in B = 3/10 = 30%.
Percentage of wheat in the mixture = 5/16 = 31.25%.

The following approach is called alligation.
It's a very good way to handle MIXTURE PROBLEMS.

Step 1: Plot the 3 percentages on a number line, with the two starting percentages (40% and 30%) on the ends and the goal percentage (31.25%) in the middle.
A(40%)--------------------31.25%--------B(30%)

Step 2: Calculate the distances between the percentages.
A(40%)----------8.75---------31.25%----1.25----B(30%)

Step 3: Determine the ratio in the mixture.
The ratio of A to B in the mixture is the RECIPROCAL of the distances in red.
A : B = 1.25 : 8.75 = 1:7.

Only answer choice C has the same ratio:
2:14 = 1:7.

The correct answer is C.

Here's another:
Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 percent fescue. If a mixture of X and Y contains 30 percent ryegrass, what percent of the weight of the mixture is X ?

(A) 10%
(B) 33.1/3%
(C) 40%
(D) 50%
(E) 66.2/3%
X = 40% ryegrass, Y = 25% ryegrass, the mixture = 30% ryegrass.

X(40%)--------10--------30%-----5-----Y(25%)

X:Y = 5:10 = 1:2.
Since 1+2 = 3, of every 3 liters, 1 liter must be X and 2 liters must be Y, implying that the fraction of X in the mixture = 1/3 = 33 1/3%.

The correct answer is B.
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by jaiswalamrita » Mon Jul 09, 2012 12:21 pm
wonderful method Mitch..many thanks

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by kjallow » Wed Sep 19, 2012 5:24 am
GMATGuruNY - You are truly a GMAT Guru!! I have been struggling with mixture problems for a while. many thanks for this simple, easy to follow approach.

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by [email protected] » Sun Nov 04, 2012 9:42 am
Thanks Mitch for showing such a wonderful approach...

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by pemdas » Sun Nov 04, 2012 12:55 pm
sachindia wrote:Two tins A and B contain mixtures of wheat and rice. In A, the weights of wheat and rice are in the ratio 2 : 3 and in B they are in the ratio 3 : 7. What quantities must be taken from A and B to form a mixture containing 5 kg of wheat and 11 kg of rice?
no need to mystify the weighted averages as invented alligation method

A*(2/5)+B*(3/10)=5
A*(3/5)+B*(7/10)=11

solve for A and B to find the required,

4A+3B=50 (multiply by 2, 8a+6b=100)
6A+7B=110 (subtract the expression in parentheses above, 6A+7B=110 less 8A+6B=100)

B-2A=10 and B=2A+10
4A+3B=50 <> 4A+3(2A+10)=50 <> 4A+6A+30=50, A=20/10=2

Now, either solve for B or find B from (5+11-A)=16-2=14

enjoy pure math & algebra rather invented rules
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by TG_GMAT » Mon Mar 18, 2013 7:57 am
Hi Mitch,

I have a question on the alligation approach. Isnt the question asking for the ratio of wheat:rice. From the alligation method in your solution, are we not getting the ratio of wheat from A to wheat from B? What about rice?

Please help!
Thanks

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by TG_GMAT » Mon Mar 18, 2013 7:58 am
Hi Mitch,

I have a question on the alligation approach. Isnt the question asking for the ratio of wheat:rice. From the alligation method in your solution, are we not getting the ratio of wheat from A to wheat from B? What about rice? The 1:7 is for A(wheat):B(wheat), correct?

Please help!
Thanks

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by vonsumit » Tue Mar 19, 2013 7:46 am
unique but effective method

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by aaggar7 » Thu Apr 18, 2013 1:44 am
Hi Mitch,

Can u pls explain the steps for the below question using the same approach that u explained earlier

A mixture contains wine and water in the ratio 3:2.Another contains in the ratio 4:5.
How many gallons of latter to be mixed with 3 gallons of first so that resulting mix contains
equal qty of wine and water

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by GMATGuruNY » Thu Apr 18, 2013 11:45 am
aaggar7 wrote:Hi Mitch,

Can u pls explain the steps for the below question using the same approach that u explained earlier

A mixture contains wine and water in the ratio 3:2.Another contains in the ratio 4:5.
How many gallons of latter to be mixed with 3 gallons of first so that resulting mix contains
equal qty of wine and water
Let F = the first solution, L = the latter solution, and M = the resulting mixture of the two solutions.
Alligation can be performed only with percentages or fractions.

Step 1: Convert the ratios to FRACTIONS.
F:
Since wine:water = 3:2, and 3+2=5, wine/total = 3/5.
L:
Since wine:water = 4:5, and 4+5=9, wine/total = 4/9.
M:
Since wine:water = 1:1, and 1+1=2, wine/total= 1/2.

Step 2: Put the fractions over a COMMON DENOMINATOR.

F = 3/5 = 54/90.
L = 4/9 = 40/90.
M = 1/2 = 45/90.

Step 3: Plot the 3 numerators on a number line, with the two starting numerators (54 and 40) on the ends and the goal numerator (45) in the middle.
F 54-----------45-----------40 L

Step 4: Calculate the distances between the numerators.
F 54-----9-----45-----5-----40 L

Step 5: Determine the ratio in the mixture.
The ratio of F to L in the mixture is the RECIPROCAL of the distances in red.
F/L = 5/9.

Since F/L = 5/9, and the actual volume of F = 3 liters, we get:
5/9 = 3/L
5L = 27
L = 27/5 = 5.4 liters.
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by ygdrasil24 » Thu May 23, 2013 5:28 am
We should get the same answer when solving for rice ?

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by mevicks » Thu Sep 19, 2013 7:16 am
GMATGuruNY wrote:
sachindia wrote:Two tins A and B contain mixtures of wheat and rice. In A, the weights of wheat and rice are in
the ratio 2 : 3 and in B they are in the ratio 3 : 7. What quantities must be taken from A and B to
form a mixture containing 5 kg of wheat and 11 kg of rice?

1. 3kgs from A, 13 kg from B
2. 8kgs from A, 8 kg from B
3. 2kgs from A, 14 kg from B
4. 6kgs from A, 10 kg from B
Percentage of wheat in A = 2/5 = 40%
Percentage of wheat in B = 3/10 = 30%.
Percentage of wheat in the mixture = 5/16 = 31.25%.

The following approach is called alligation.
It's a very good way to handle MIXTURE PROBLEMS.

Step 1: Plot the 3 percentages on a number line, with the two starting percentages (40% and 30%) on the ends and the goal percentage (31.25%) in the middle.
A(40%)--------------------31.25%--------B(30%)

Step 2: Calculate the distances between the percentages.
A(40%)----------8.75---------31.25%----1.25----B(30%)

Step 3: Determine the ratio in the mixture.
The ratio of A to B in the mixture is the RECIPROCAL of the distances in red.
A : B = 1.25 : 8.75 = 1:7.

Only answer choice C has the same ratio:
2:14 = 1:7.

The correct answer is C.
@Mitch : Thanks a ton for demystifying mixtures!

@ygdrasil24 : Yes, one can work with the rice ratios and still end up with the final A to B ratio as 1 : 7

Consider the following:


Mixture A
(W)40%__________________60%(R)


Mixture B
(W)30%__________________70%(R)


Final Mixture
(5KGS)31.25%_____________68.75%(11KGS)



(60%)Ra----------8.75-----------68.75%----1.25-----Rb(70%)

A/B = 1.25/8.75 = 1/7"

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by sahilchaudhary » Mon Dec 09, 2013 5:02 am
GMATGuruNY wrote:
sachindia wrote:Two tins A and B contain mixtures of wheat and rice. In A, the weights of wheat and rice are in
the ratio 2 : 3 and in B they are in the ratio 3 : 7. What quantities must be taken from A and B to
form a mixture containing 5 kg of wheat and 11 kg of rice?

1. 3kgs from A, 13 kg from B
2. 8kgs from A, 8 kg from B
3. 2kgs from A, 14 kg from B
4. 6kgs from A, 10 kg from B
Percentage of wheat in A = 2/5 = 40%
Percentage of wheat in B = 3/10 = 30%.
Percentage of wheat in the mixture = 5/16 = 31.25%.

The following approach is called alligation.
It's a very good way to handle MIXTURE PROBLEMS.

Step 1: Plot the 3 percentages on a number line, with the two starting percentages (40% and 30%) on the ends and the goal percentage (31.25%) in the middle.
A(40%)--------------------31.25%--------B(30%)

Step 2: Calculate the distances between the percentages.
A(40%)----------8.75---------31.25%----1.25----B(30%)

Step 3: Determine the ratio in the mixture.
The ratio of A to B in the mixture is the RECIPROCAL of the distances in red.
A : B = 1.25 : 8.75 = 1:7.

Only answer choice C has the same ratio:
2:14 = 1:7.

The correct answer is C.

Here's another:
Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 percent fescue. If a mixture of X and Y contains 30 percent ryegrass, what percent of the weight of the mixture is X ?

(A) 10%
(B) 33.1/3%
(C) 40%
(D) 50%
(E) 66.2/3%
X = 40% ryegrass, Y = 25% ryegrass, the mixture = 30% ryegrass.

X(40%)--------10--------30%-----5-----Y(25%)

X:Y = 5:10 = 1:2.
Since 1+2 = 3, of every 3 liters, 1 liter must be X and 2 liters must be Y, implying that the fraction of X in the mixture = 1/3 = 33 1/3%.

The correct answer is B.
Excellent new method Mitch.
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by Poisson » Mon Aug 29, 2016 5:31 pm
GMATGuruNY wrote:
aaggar7 wrote:Hi Mitch,

Can u pls explain the steps for the below question using the same approach that u explained earlier

A mixture contains wine and water in the ratio 3:2.Another contains in the ratio 4:5.
How many gallons of latter to be mixed with 3 gallons of first so that resulting mix contains
equal qty of wine and water
Let F = the first solution, L = the latter solution, and M = the resulting mixture of the two solutions.
Alligation can be performed only with percentages or fractions.

Step 1: Convert the ratios to FRACTIONS.
F:
Since wine:water = 3:2, and 3+2=5, wine/total = 3/5.
L:
Since wine:water = 4:5, and 4+5=9, wine/total = 4/9.
M:
Since wine:water = 1:1, and 1+1=2, wine/total= 1/2.

Step 2: Put the fractions over a COMMON DENOMINATOR.

F = 3/5 = 54/90.
L = 4/9 = 40/90.
M = 1/2 = 45/90.

Step 3: Plot the 3 numerators on a number line, with the two starting numerators (54 and 40) on the ends and the goal numerator (45) in the middle.
F 54-----------45-----------40 L

Step 4: Calculate the distances between the numerators.
F 54-----9-----45-----5-----40 L

Step 5: Determine the ratio in the mixture.
The ratio of F to L in the mixture is the RECIPROCAL of the distances in red.
F/L = 5/9.

Since F/L = 5/9, and the actual volume of F = 3 liters, we get:
5/9 = 3/L
5L = 27
L = 27/5 = 5.4 liters.
Hi Mitch,

Could you please explain how to apply this approach to the following question? Each time I try it, I get an incorrect answer.

Mix A is 35% red and 65 % blue.
Mix B is 50% red and 50% blue.
If you begin with 20 pounds of mix A, how many pounds of mix B do you need to end up with a combined mixture that is 40% red and 60% blue?

A. 5
B. 10
C. 15
D. 20
E. 25

I keep ending up with an answer of 5 once I take the reciprocal of the distances.

Thanks very much.