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copper-zinc alloy

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by finance » Mon Jul 18, 2011 4:27 am
An alloy of copper and zinc contains copper and zinc in the ratio 5 : 3. Another alloy of copper and zinc contains copper and zinc in the ratio 1 : 7. In what ratio should the two alloys be mixed so that the resultant alloy contains equal proportions of copper and zinc?


A. 1 : 5

B. 7 : 3

C. 5 : 3

D. 3 : 1

E. 4 : 3
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by GMATGuruNY » Mon Jul 18, 2011 4:39 am
finance wrote:An alloy of copper and zinc contains copper and zinc in the ratio 5 : 3. Another alloy of copper and zinc contains copper and zinc in the ratio 1 : 7. In what ratio should the two alloys be mixed so that the resultant alloy contains equal proportions of copper and zinc?


A. 1 : 5

B. 7 : 3

C. 5 : 3

D. 3 : 1

E. 4 : 3
Fraction of copper is Alloy 1 = 5/8.
Fraction of copper in Alloy 2 = 1/8.
Desired fraction of copper in mixture = 4/8. (Since we want the mixture to be equal proportions copper and zinc.)

We can use alligation.
Alligation dictates the following:

The proportion of each element in the mixture is equal to the distance between the fraction attributed to the other element in the mixture and the fraction attributed to the final mixture.

Proportion of Alloy 1 = 4/8 - 1/8 = 3/8.
Proportion of Alloy 2 = 5/8 - 4/8 = 1/8.
Alloy 1:Alloy 2 = 3/8 : 1/8 = 3:1.

The correct answer is D.
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by finance » Mon Jul 18, 2011 5:03 am
Thank you for your answer GuruNY..

I solved as follows but I m not sure if its right or just by chance gave the right answer..

Let x be the amount of Alloy 1 and y the amount of Alloy 2.

Since zinc and copper will be in same proportions, then

5x+1y = 3x+7y
x = 3y
so alloy1/alloy2 = 3y/y = 3
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by amit2k9 » Mon Jul 18, 2011 7:00 am
you can use the allegation formula here,

Cheaper (B)/ Dearer (A) = (5/8-1/2)/ (1/2-1/8)

thus B/A = 1/2 hence

ratio is 3:1 for B/(A+B).
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by Revathir » Fri Aug 22, 2014 7:26 am
whats is solution (with explanation) for this ques?
In an alloy, zinc and copper are in the ratio 1 : 3. In the 2nd alloy the same elements are in the ratio 3 : 4. In what ratio should these two alloys be mixed to form a new alloy in which the 2 elements are 5 : 4?
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by GMATGuruNY » Fri Aug 22, 2014 8:19 am
Revathir wrote:whats is solution (with explanation) for this ques?
In an alloy, zinc and copper are in the ratio 1 : 3. In the 2nd alloy the same elements are in the ratio 3 : 4. In what ratio should these two alloys be mixed to form a new alloy in which the 2 elements are 5 : 4?
This mixture is not possible.

In the first alloy, zinc/total = 1/4 = 25%.
In the second alloy, zinc/total = 3/7 = less than 50%.
In the mixture of the two alloys, zinc/total = 5/9 = more than 50%.

While each of the alloys is LESS THAN 50% zinc, the mixture of the two alloys is MORE THAN 50% zinc.
Not possible.
For the mixture to be viable, the percentage of zinc in the mixture must be BETWEEN the percentage of zinc in the first alloy and the percentage of zinc in the second alloy.
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by Katy_ » Sun Aug 24, 2014 5:42 am
finance wrote:An alloy of copper and zinc contains copper and zinc in the ratio 5 : 3. Another alloy of copper and zinc contains copper and zinc in the ratio 1 : 7. In what ratio should the two alloys be mixed so that the resultant alloy contains equal proportions of copper and zinc?


A. 1 : 5

B. 7 : 3

C. 5 : 3

D. 3 : 1

E. 4 : 3
Alloy 1: 5 copper (5c)and 3 zinc (3z)

Alloy 2: 1c and 7z

Alloy 1 : Alloy 2 = m : n

copper = zinc <=> 5c x m + 1c x n = 3z x m + 7z x n

<=> 5m + n = 7n + 3m

<=> m = 3n

=> 3:1


Did I do it right?
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by [email protected] » Sun Aug 24, 2014 12:34 pm
Hi Katy_,

Yes, your approach is correct. Beyond the algebra that you used, there are often other approaches that are fairly straight-forward (and sometimes faster/easier than doing "math"). Since the answers provide 5 ratios for us, we can use them "against" the information in the prompt to find the one ratio that gives us an equal proportion of copper and zinc. Let's TEST THE ANSWERS:

The two ratios…
C:Z
5:3

and

C:Z
1:7

Answer A: 1:5 gives us (5C + 3Z) + (5C + 35Z) --> 10C and 38Z --> NOT EQUAL
Answer B: 7:3 gives us (35C + 21Z) + (3C + 21Z) --> 38C + 42Z --> NOT EQUAL
Answer C: 5:3 gives us (25C + 15Z) + (3C + 21Z) --> 28C + 36Z --> NOT EQUAL
Answer D: 3:1 gives us (15C + 9Z) + (1C + 7Z) --> 16C + 16Z --> EQUAL!!!
Answer E: 4:3 gives us (20C + 12Z) + (3C + 21Z) --> 23C + 33Z --> NOTE EQUAL

Final Answer: D

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by Jeff@TargetTestPrep » Mon Mar 12, 2018 3:40 pm
finance wrote:An alloy of copper and zinc contains copper and zinc in the ratio 5 : 3. Another alloy of copper and zinc contains copper and zinc in the ratio 1 : 7. In what ratio should the two alloys be mixed so that the resultant alloy contains equal proportions of copper and zinc?

A. 1 : 5

B. 7 : 3

C. 5 : 3

D. 3 : 1

E. 4 : 3
We can let the ratio of the first alloy of copper to zinc = 5x: 3x

We can let the ratio of the second alloy of copper to zinc = y : 7y

We can create the equation:

5x + y = 3x + 7y

2x = 6y

x = 3y

x/y = 3/1

Answer: D

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