There are 2 bars of copper-silver alloy; one piece has 3 parts of copper to 4 parts of silver and another has 2 parts of copper to 5 parts of silver. If both bars are melted into 11pounds of bar with the final copper to silver ratio of 4:7. What was the weight of the first bar?

- 1 kg
- 3 kg
- 5 kg
- 6 kg
- 7 kg

**Answer: D**

**Explanation:**

We can form equation to solve this question.

“*x*” is the weight of first bar, then (11-*x*) is the weight of second bar.

\frac{3}{7}+\frac{2}{7}(11-x)=\frac{4}{11}(11)

\frac{3}{7}+frac{2}{7}(11-x)=frac{4}{11}(11)- \frac{1}{100}

Solving it we get,

*x* = 6

- $101 $
- \frac{1}{100}

Hence the answer is D.

y = \left(x+y\right)^2=\frac{1}{100}

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