Question

An alloy of gold, sliver and bronze contains 90% bronze, 7% gold and 3% silver. a second alloy of bronze and sliver only is melted with the first and the mixture contains 85% of bronze, 5% of gold and 10% of silver. find the percentage of bronze in the second alloy.

Answer 1

Explanation:

Mixture 1 bronze + Mixture 2 bronze = Mixture 3 bronze

0.9x + ay = 0.85(x + y)

0.9x - 0.85x = 0.85y - ay

0.05x = (0.85 - a)y

sub eq (1)

(0.85 - a)y = 0.05 * 5y/2

a = 1.45/2 = 0.725

a is 72.5%

Answer 2

Answer:

72.5%

Explanation:

Let A and B the first and second alloy weights respectively

x=% of bronze in alloy B

We know that B contains no GOLD.

.07A=.05(A+B)

B=.4A

substituting,

.9A+.4Ax=.85(A+.4A)

x=.725=72.5%