Stuck on Alloy Mixing Question [Arithmetic and Algebra Session 4 HW]

GRE Quant Problem Solving
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Hi all. There’s something about this question that I cannot get out of my head.

I’ve watched Greg’s explanation several times and (spoilers), he comes to the conclusion that the answer is A – but in watching his explanation, I can’t help but think we’re approaching this question incorrectly. Here’s my reasoning – please help me see where I diverge from the correct course!

We have Alloys A and B. These alloys are made up of two basic elements. We’ll call these elements e1 and e2. Alloy A is a ratio of 5e1 : 3e2. Alloy B is a ratio of 1e1 : 2e2. Now, we’re told that there’s a new Alloy X, which is made by mixing alloys A and B in a ratio of 4:3, and we’re asked to solve for the composition of the two basic elements (e1 and e2) in Alloy X.

So, I take 4 parts Alloy A and 3 parts Alloy 3 and add them together:

4(5e1 / 3e2) + 3(1e1 / 2e2)

I multiply out the 4 and the 3 (interpreting them as 4/1 and 3/1, respectively):

(20e1/3e2) + (3e1/2e2)

I multiply the left figure by 2/2 and the right by 3/3 to get the same base:

(40e1/6e2) + (9e1/6e2)

and then add:

49e1/6e2

Which leads me to believe that when we mix 4 parts Alloy A with 3 parts Alloy B, we get Alloy X, which is 49 parts e1 to 6 parts e2.

In the video explanation, I feel like we didn’t solve for the composition of the basic elements. Obviously, my took me to a non-existent answer choice, so I’m in the wrong here, but I’m having trouble seeing my error. Can anyone point out my mistake?

Thank you!

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The problem with this that you are converting ratio to numbers, for instance you could also have
5e1, 3e2 for A but 2e1, 4e2 for B
This way you get (5+2)7e1 in total
and (3+4)7 e2 in total
e1:e2 = 7:7 = 1:1

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Thanks for the response!

That’s a good point – I understand that Alloy B could be 1e1, 2e2… 2e1, 4e2… (X)e1, (2X)e2… Any of those would work.

… but what’s confusing me is that the question states that Alloy X must be made with a 4:3 ratio of Alloy A to Alloy B.

If I did…
Alloy A - 5e1, 3e2 (5:3 ratio)
Alloy B - 2e1, 4e2 (1:2 ratio)

Isn’t this 1x Alloy A and 2x Alloy B? This is a 1:2 ratio, not a 4:3 mix required by the question.

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Ah my bad for 5e1, 3e2 for A but 2e1, 4e2 for B
The final mixture would be:
e1 = 20 + 6 = 26
e2 = 12 + 12 = 24
e1:e2 = 13:12
Which is different than what you get when B = 2e1 and 4e2
I was basically trying to demonstrate that the ratio alone of e1 and e2 cannot be used to solve in this manner