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Re: An alloy of copper and zinc contains copper and zinc in the ratio 5 : [#permalink]
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simple mixtures problem.

5x + y = 3x + 7y (since quantity of Cu and Zn in resultant mixture is equal)

2x = 6y
x = 3y

Ans. = D
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Re: An alloy of copper and zinc contains copper and zinc in the ratio 5 : [#permalink]
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Let Alloy_1 be x units, and let Alloy_2 be y units.

So, fraction of copper in Alloy_1 = 5x/8, and fraction of zinc in Alloy_1 = 3x/8.

Similarly, fraction of copper in Alloy_2 = y/8, and fraction of zinc in Alloy_2 = 7y/8.

Mixing them, we get
copper = 5x/8 + y/8; zinc = 3x/8 + 7y/8.

So, 5x+y = 3x+7y
-> x = 3y
-> x/y = 3/1.

So, they must be mixed in the ratio 3 : 1. Ans (D).
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Re: An alloy of copper and zinc contains copper and zinc in the ratio 5 : [#permalink]
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Putting together an equation with the numbers - 5x+y/ 3x+7y = 1

Solving it out - 2x=6y

Hence 3/1


Option D
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Re: An alloy of copper and zinc contains copper and zinc in the ratio 5 : [#permalink]
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Bunuel 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


Ans: D

Solution: While the ratio of Cu and Zn in both the mixture is given but not the quantity.
so we assume that M1 is x, and M2 is y.
now we know in M1= 5x/8 Cu, 3x/8 Zn
in M2 y/8 Cu and 7y/8

now we want to make the total quantity of Cu and Zn equal
use the following data and make equation

5x/8 + y/8 = 3x/8 + 7y/8
it gives x = 3y
so the ratio must be x:y = 3y:y = 3:1
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Re: An alloy of copper and zinc contains copper and zinc in the ratio 5 : [#permalink]
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let ratio of 1st alloy and 2nd alloy in 3rd mixture=k:1
5k+1=3k+7
k=3
ans 3:1 (D)
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Re: An alloy of copper and zinc contains copper and zinc in the ratio 5 : [#permalink]
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

let x=alloy 1
y=alloy 2
(5x/8)+y/8=4(x+y)/8
x/y=3
3:1
D

Originally posted by gracie on 23 Jan 2017, 10:44.
Last edited by gracie on 08 Mar 2018, 23:04, edited 1 time in total.
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Re: An alloy of copper and zinc contains copper and zinc in the ratio 5 : [#permalink]
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Hi All,

Beyond the algebra that can be used here, 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 a match
Answer B: 7:3 gives us (35C + 21Z) + (3C + 21Z) --> 38C + 42Z --> NOT a match
Answer C: 5:3 gives us (25C + 15Z) + (3C + 21Z) --> 28C + 36Z --> NOT a match
Answer D: 3:1 gives us (15C + 9Z) + (1C + 7Z) --> 16C + 16Z --> This IS a match
Answer E: 4:3 gives us (20C + 12Z) + (3C + 21Z) --> 23C + 33Z --> NOT a match

Final Answer:

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Re: An alloy of copper and zinc contains copper and zinc in the ratio 5 : [#permalink]
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Bunuel 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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Re: An alloy of copper and zinc contains copper and zinc in the ratio 5 : [#permalink]
Bunuel 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

Kudos for a correct solution.


Hi VeritasKarishma ,

I was trying to solve this using weighted average and here is what I did:

The final mixture contains alloy in the ratio 1:1

\(\frac{W1}{W2}\) = \((\frac{5}{3}-1)/(1-\frac{1}{7})\)

\(\frac{W1}{W2} = \frac{7}{9}\)

Can you please tell where did I go wrong ?
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Re: An alloy of copper and zinc contains copper and zinc in the ratio 5 : [#permalink]
ShreyKapil08 wrote:
Bunuel 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

Kudos for a correct solution.


Hi VeritasKarishma ,

I was trying to solve this using weighted average and here is what I did:

The final mixture contains alloy in the ratio 1:1

\(\frac{W1}{W2}\) = \((\frac{5}{3}-1)/(1-\frac{1}{7})\)

\(\frac{W1}{W2} = \frac{7}{9}\)

Can you please tell where did I go wrong ?


I think I understand where I went wrong.

I should not use the ratio copper is to alloy as is. It should be "part" by "whole".

\(\frac{W1}{W2}\) = \((\frac{5}{8}-\frac{1}{2})/(\frac{1}{2}-\frac{1}{8})\)

\(\frac{W1}{W2} = \frac{1}{3}\)
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An alloy of copper and zinc contains copper and zinc in the ratio 5 : [#permalink]
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ShreyKapil08 wrote:
ShreyKapil08 wrote:
Bunuel 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

Kudos for a correct solution.


Hi VeritasKarishma ,

I was trying to solve this using weighted average and here is what I did:

The final mixture contains alloy in the ratio 1:1

\(\frac{W1}{W2}\) = \((\frac{5}{3}-1)/(1-\frac{1}{7})\)

\(\frac{W1}{W2} = \frac{7}{9}\)

Can you please tell where did I go wrong ?


I think I understand where I went wrong.

I should not use the ratio copper is to alloy as is. It should be "part" by "whole".

\(\frac{W1}{W2}\) = \((\frac{5}{8}-\frac{1}{2})/(\frac{1}{2}-\frac{1}{8})\)

\(\frac{W1}{W2} = \frac{1}{3}\)


Yes, think about the physical quantities you are dealing with - concentration (part/whole), volume, amount
Concentration of copper is 5/8 and 1/8 in the two. You need to mix them to get concentration of 1/2.
You need to find the ratio of the volumes in which you mix them.

Check here:

Weighted Avg and Mixtures:
https://anaprep.com/arithmetic-weighted-averages/
https://anaprep.com/arithmetic-mixtures/
and these videos:
https://www.youtube.com/watch?v=_GOAU7moZ2Q
https://www.youtube.com/watch?v=VdBl9Hw0HBg

Originally posted by KarishmaB on 16 Jun 2020, 22:22.
Last edited by KarishmaB on 21 Dec 2023, 08:16, edited 1 time in total.
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Re: An alloy of copper and zinc contains copper and zinc in the ratio 5 : [#permalink]
Bunuel 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

Kudos for a correct solution.

Amount of copper in 40 gm of mixture=20gm....(1:1, 1/2*40=20)
Amount of copper in alloy 1 in 40 gm of mixture=25 gm(5:3..5/8*40=25)
Amount of copper in alloy 1 in 40 gm of mixture=5 gm(1:7..1/8*40=5)
C2..............C...............C1
5................20.................25
C2/C1=(25-20)/(20-5)
=5/15
C1:C2=3:1
D:)
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Re: An alloy of copper and zinc contains copper and zinc in the ratio 5 : [#permalink]
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VeritasKarishma wrote:
Yes, think about the physical quantities you are dealing with - concentration (part/whole), volume, amount
Concentration of copper is 5/8 and 1/8 in the two. You need to mix them to get concentration of 1/2.
You need to find the ratio of the volumes in which you mix them.

Check here: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... -mixtures/


Thank you for confirming VeritasKarishma,

I did go through the blog you mentioned above and others in the series. I must say I could never approach the Weighted average and mixture problems the way I do now after reading those blogs. Thank you for sharing such an insightful write up.
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Re: An alloy of copper and zinc contains copper and zinc in the ratio 5 : [#permalink]
Consider either copper or zinc. I am gonna consider copper here. There are two alloys with ratio 5:3 and 1:7.To make the ratio equal means it should be 4:4. So I am gonna use the quick and simplest way to calculate the ratio.
5 1
4
3 : 1

Posted from my mobile device
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Re: An alloy of copper and zinc contains copper and zinc in the ratio 5 : [#permalink]
Bunuel 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

Kudos for a correct solution.


Going through the options is the easy approach.

3*(5:3)+1*(1:7)=15:9+1:7=16:16

And that's what we need.

Answer D.
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