There are two different alloys, A and B. Alloy A has copper and Iron

Question

There are two different alloys, A and B. Alloy A has copper and Iron in the ratio of 5:7, and Alloy B has copper and Iron in the ratio of 4:3. In what ratio should Alloy A and B be mixed such that percentage of iron and copper in the final mixture is equal.

A. 2:1
B. 2:3
C. 5:6
D. 6:7
E. 4:3

Show Answer

Official Answer

D

A. 2:1
B. 2:3
C. 5:6
D. 6:7
E. 4:3

CatNinja7 · Accepted Answer

Please see attachment for information

Akkiiii · Answer

I am getting the answer as A I,.e 2:1

Can anybody help with the answer and how you did it

The way I did it is 
C1:I1= 5x:7x
C2:I2= 4y:3y

5x+4y/ 7x+3y= 1:1

x:y= 2:1

AlieuK117 · Answer

I went about the solution using two paths)

Path A:
Alligation Method:

4/7 ..... 7/12
        1/2
1/12 ..... 1/14

This yeilds a ratio of 6:7 (This matches answer choice D but I am unsure if I used the method correctly)

Path B:
Alloy A) 5:7:12

Alloy B) 4:3:7

Then the combiantion would yield 
Combination) 12:7:1/2

I am unsure of how to procede from this point. May I get some help in solving this problem or breaking the question down further?

devdattakhoche · Answer

We are given: 
- Alloy A has Copper to Iron ratio =  \frac{5}{12}  
- Alloy B has Copper to Iron ratio =  \frac{4}{7}

We want the final mixture to have a 1:1 ratio of Copper and Iron, i.e., Copper fraction =  \frac{1}{2} .

Let the mixing ratio of Alloy A to Alloy B be x : y.

Using the weighted average method: 
 \frac{x \cdot \frac{5}{12} + y \cdot \frac{4}{7}}{x + y} = \frac{1}{2}

Multiply both sides by (x + y): 
 x \cdot \frac{5}{12} + y \cdot \frac{4}{7} = \frac{1}{2}(x + y)

Multiply both sides by 2: 
 2x \cdot \frac{5}{12} + 2y \cdot \frac{4}{7} = x + y

Simplify: 
 \frac{10}{12}x + \frac{8}{7}y = x + y  
 \frac{5}{6}x + \frac{8}{7}y = x + y

Multiply the entire equation by 42 (LCM of 6 and 7) to eliminate denominators: 
 35x + 48y = 42x + 42y

Simplify: 
 6y = 7x  
 \Rightarrow \frac{x}{y} = \frac{6}{7}

\boxed{6:7}

Feb2024 · Answer

Hello Experts  KarishmaB   Bunuel   chetan2u   gmatophobia   MartyMurray   IanStewart

I used the following approach and got the ratio for Alloy A: Alloy B as 7:6, and not 6:7.

Alloy A:  Copper: Iron :: 5:7  which is equivalent to  Copper: Iron :: 15:21

Alloy B:  Copper: Iron :: 4:3  which is equivalent to  Copper: Iron :: 28:21

Thus this way, from the above,  Quantity of Iron is achieved as 21 units .

To get the Quantity of Copper as 21 units, we can use weighted average method, which gives us  Alloy A: Alloy B  as  7:6.

Can you point out the fault?

KarishmaB · Answer

To use weighted averages, we work with the concentration of any one ingredient.

So A has 5/12th of copper and B has 4/7th of copper. Final mix should have 1/2 copper. 
Now use weighted averages on it and you will get 6:7

There are two different alloys, A and B. Alloy A has copper and Iron

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There are two different alloys, A and B. Alloy A has copper and Iron