Gold is 19 times as heavy as water and copper is 9 times as heavy as water. In what ratio should gold and copper be mixed so that the mixture is 15 times as heavy as water?

Introduction / Context:
This is a standard mean-density (alligation) problem. We are given relative densities with respect to water and asked for the mixing ratio to achieve a target mean density for the alloy.

Given Data / Assumptions:

• Gold density ratio = 19.
• Copper density ratio = 9.
• Mixture target ratio = 15.

Concept / Approach:
Use weighted average: (19a + 9b) / (a + b) = 15, where a and b are amounts of gold and copper. Solve for a : b. Alternatively, apply alligation rule: parts are 15 − 9 and 19 − 15.

Step-by-Step Solution:
19a + 9b = 15a + 15b. (19 − 15)a = (15 − 9)b ⇒ 4a = 6b. a / b = 6 / 4 = 3 / 2. Therefore, gold : copper = 3 : 2.

Verification / Alternative check:
Alligation method: Difference from mean—gold: 19 − 15 = 4; copper: 15 − 9 = 6. Ratio is 6 : 4 = 3 : 2 for gold : copper (reverse placement), confirming the same result.

Why Other Options Are Wrong:
1 : 2 or 2 : 3 give a mean below or above 15. 19 : 135 is unrelated to the balanced mean.

Common Pitfalls:
Reversing the differences in alligation or taking arithmetic mean instead of weighted mean. Always use weighted average for mixture problems.

Final Answer:
3 : 2