Two alloys of gold and copper are prepared in the ratios 7 : 2 and 7 : 11 respectively. Equal masses of these alloys are melted to form a third alloy C. What is the ratio of gold to copper in alloy C?

Introduction / Context:
When mixing alloys, you must combine actual fractions of constituents, not the ratios directly. Since equal quantities (equal masses) of the two alloys are mixed, the overall fraction of a metal in the mixture is the simple average of the two alloys’ fractions for that metal.

Given Data / Assumptions:

• Alloy A: gold : copper = 7 : 2 ⇒ gold fraction = 7/9.
• Alloy B: gold : copper = 7 : 11 ⇒ gold fraction = 7/18.
• Equal masses of A and B are mixed to form C.

Concept / Approach:
For equal masses, the gold fraction in C equals the average of the gold fractions of A and B. The copper fraction is then 1 − (gold fraction). Finally, convert the fractions into a simplest whole-number ratio.

Step-by-Step Solution:
Gold fraction in A = 7/9Gold fraction in B = 7/18Gold fraction in C = [(7/9) + (7/18)] / 2 = (14/18 + 7/18) / 2 = (21/18) / 2 = 21/36 = 7/12Copper fraction in C = 1 − 7/12 = 5/12Hence, gold : copper = (7/12) : (5/12) = 7 : 5

Verification / Alternative check:
Assume 1 kg of A and 1 kg of B. Gold mass in C = 1*(7/9) + 1*(7/18) = 21/18 = 7/6 kg out of 2 kg total ⇒ fraction 7/12. Copper becomes 5/12. Ratio 7:5 confirmed.

Why Other Options Are Wrong:
They either invert the ratio or reflect averaging the ratios (incorrect) rather than the underlying fractions by mass.

Common Pitfalls:
Taking the mean of 7:2 and 7:11 directly; you must convert ratios to fractions first and then average because mixing is by mass.

Final Answer:
7 : 5