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?
ratios / fraction problem
 GMATGuruNY
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Percentage of wheat in A = 2/5 = 40%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 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%)8.7531.25%1.25B(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:
X = 40% ryegrass, Y = 25% ryegrass, the mixture = 30% ryegrass.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%)1030%5Y(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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no need to mystify the weighted averages as invented alligation methodsachindia 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?
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)
B2A=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+11A)=162=14
enjoy pure math & algebra rather invented rules
Success doesn't come overnight!
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
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
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
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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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
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
 GMATGuruNY
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Let F = the first solution, L = the latter solution, and M = the resulting mixture of the two solutions.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
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 544540 L
Step 4: Calculate the distances between the numerators.
F 54945540 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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@Mitch : Thanks a ton for demystifying mixtures!GMATGuruNY wrote:Percentage of wheat in A = 2/5 = 40%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 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%)8.7531.25%1.25B(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.
@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%)Ra8.7568.75%1.25Rb(70%)
A/B = 1.25/8.75 = 1/7"
 sahilchaudhary
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Excellent new method Mitch.GMATGuruNY wrote:Percentage of wheat in A = 2/5 = 40%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 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%)8.7531.25%1.25B(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:
X = 40% ryegrass, Y = 25% ryegrass, the mixture = 30% ryegrass.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%)1030%5Y(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.
Sahil Chaudhary
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Hi Mitch,GMATGuruNY wrote:Let F = the first solution, L = the latter solution, and M = the resulting mixture of the two solutions.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
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 544540 L
Step 4: Calculate the distances between the numerators.
F 54945540 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.
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.