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problem from mixtures and alligation

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by readsid » Fri Dec 11, 2009 6:33 am
Q1) Two tins A and B contain mixtures of wheat and rice. In A the weights of wheat and rice are in the ration 2:3 and in B they are in the ratio of 3:7. what quantities must be taken from A and B to form a mixture containing 5kg of wheat and 11 kgs 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

how do i apply the alligation rule ...i keep getting a -ve answear... ??
Last edited by readsid on Fri Dec 11, 2009 1:05 pm, edited 1 time in total.
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by jasz » Fri Dec 11, 2009 6:58 am
IMO C

Tin A contains wheat and rice in ratio 2:3. Let's suppose concrete values are 2x and 3x for wheat and rice respectively.
Similarly, Tin B has ratios 3:7. Therefore, concrete values are 3y and 7y.

Since we need to have 5kg wheat and 11kg rice:
2x+3y = 5
3x+7y = 11
Solving for x,y we get x=2/5 and y=7/5

Mixture to be taken from Tin A = 2x+3x = 5x = 2kgs
Mixture to be taken from Tin B = 3y+7y = 10y = 14kgs
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by readsid » Fri Dec 11, 2009 12:59 pm
thanks jasz .... but how do i use the alligation method...on the above problem ??
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by sharchandran » Mon Jun 27, 2011 8:57 pm
Hello,

For allegation , I guess you can use the following method.

Let X be the quatity of mixture taken from tin A.
Then automatically 16 - X will be the quantity taken from Tin B. ( As 5:11 has a total of 16 parts )

Now we know that we need 5kg of wheat and 11 kg of rice.

So in the first tin A --> Wheat is 2/5. So we shall take 2/5X of mixture from tin A
In Tin B, wheat is 3/10. So lets take 3/10 ( 16 - X )

Totally , we get
2/5X = 3(16-X)/10 = 5.

Solve, we get the answer of X as 2.
16-X will be 14 :)

Hope this helps.

Regards and All the best.
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by GMATGuruNY » Mon Jun 27, 2011 11:47 pm
readsid wrote:Q1) Two tins A and B contain mixtures of wheat and rice. In A the weights of wheat and rice are in the ration 2:3 and in B they are in the ratio of 3:7. what quantities must be taken from A and B to form a mixture containing 5kg of wheat and 11 kgs 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

how do i apply the alligation rule ...i keep getting a -ve answear... ??
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%.

Alligation dictates the following:

The proportion of each element in the mixture = the distance between the percentage attributed to the other element in the mixture and the percentage attributed to the mixture.

Thus:
Proportion of A in the mixture = |30-31.25| = 1.25.
Proportion of B in the mixture = |40-31.25| = 8.75.

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.
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by amit2k9 » Tue Jun 28, 2011 3:20 am
the equations are
(2/5)A + (3/10)B = 5
(3/5)A + (7/10)B = 11
solving gives
(1/10)A = 2/10. thus C
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by cvsmech » Tue Jun 28, 2011 10:39 am
weighted average question.

2x/5+3y/10=5
3x/5+7y/10=11

solve for x, y
X = 2kg and y = 14 kg.

Check: Weight of rice and wheat should be same as weight taken from mixture x and y

5 + 11 = x + y =16
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by stuffstuff » Tue Jun 28, 2011 2:44 pm
You can also solve it using a simpler method of choosing values and plugging them in.
You know that you need 5 kg of wheat and 11 kg of rice.

we know that,
mixture A is 40% [2/(2+3)] wheat.
mixture B is 30% [3/(3+7)] wheat.

test out the options to see which produces 5 kg of wheat.

1) (3kg * .4) + (13kg * .3) = 5.1
2) (8kg * .4) + (8kg * .3) = 5.6
3) (2kg * .4) + (14kg * .3) = 5
4) (6kg * .4) + (10kg * .3) = 5.4


without even having to address the rice situation, you've eliminated all but option 3.
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