A and B are Two Aloys of Gold and Copper Prepared By Mixing Metals in GMAT Problem Solving

Question: A and B are two alloys of gold and copper prepared by mixing metals in proportions 5: 3 and 5: 11 respectively. If equal quantities of this alloys are melted to a third alloy C, what would be the proportion of gold and copper in the alloys thus formed?

1. 25: 33
2. 33: 25
3. 15: 17
4. 17: 15
5. 13: 15

“A and B are two alloys of gold and copper prepared by mixing metals in”- is a topic of the GMAT Quantitative reasoning section of GMAT. The question has been taken from the book “Mathematics Simplified.” To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

Solution and Explanation:
Approach Solution 1:

The given condition states that two alloys A and B of gold and copper which are mixed together have equal proportions respectively including 5:3 and 5: 11. If another alloy is meted, say C and all of these A and B had to be involved in it, gold and copper quantities would be required.
This problem resolves the question by finding the quantity required for gold and copper to become the third alloy.
The given condition states that alloy A has proportion 5:3, and copper has a proportion of 5:11.
Adding the alloy A’s proportion, it can be found that it is 8.
Total quantity of alloy B which is equal to 16
Finding the LCM of 8 and 16, it can be evaluated that it is 16.
so 8 units of Alloy A will have 5 units of gold and 3 units of copper.
Hence, 16 units of A will have 10 units of gold and 6 units of copper.
This simplicity implies that Alloy B has proportion 5:11, so 16 units of Alloy B will have 5 units of gold and 11 units of copper.
Accordingly, the quantity of A is supposed to be added in quantity of C = 10 + 5
The quantity of A and B can be equal to A + B = (6 + 11)
Mixing 16 units of Alloy A with 16 units of Alloy B will give me 15 units of gold and 17 units of copper,
Hence, the ratio of the proportion of the mixing metals, it can be evaluated that (10 + 5) : (6 + 11). This equals to 15:17

Correct Answer: C

Approach Solution 2:
Total quantities of Alloy A=5+3=8
Total quantity of Alloy B=5+11=16
LCM of 8 and 16 is 16.
Then we have quantity of A in C as (10+6)
We have quantity of A and B as (5+11)
Hence the ratio is (10+5):(6+11)⟹15:17.

Correct Answer: C

Approach Solution 3:
Given:

The ratio of gold and silver in the alloy A= 5: 3
The ratio of gold and silver in the alloy B= 5: 11

Calculation:

Let 1kg of A and 1 kg of B taken to form 2 kg of C
=> Gold in 1 kg of A = 5/8
=> Copper in 1 kg of A = 3/8 
=> Gold in 1 kg of B = 5/16 
=> Copper in kg of B = 11/16
=> Gold in 2 kg of C = (5/8) + (5/16) = 15/16
=> Copper in 2 kg of C = (3/8) + (11/16) = 17/16 

Now, 
=> Ratio of gold to copper in C = (15/16) : (17/16) = 15:17 

Therefore, the ratio of gold and silver in the alloy C is 15 :17. 

Correct Answer: C

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