M03-30

Official Solution:

There are two bars made of gold-silver alloy. The first bar contains 2 parts of gold and 3 parts of silver by weight, while the second contains 3 parts of gold and 7 parts of silver by weight. If both bars are melted and combined to form a single 8-kilogram bar with a gold-to-silver ratio of 5:11 by weight, what was the weight of the first bar?

A. 1 kilogram
B. 3 kilograms
C. 5 kilograms
D. 6 kilograms
E. 7 kilograms


Let's assume that the weight of the first bar is \(x\) kilograms and the weight of the second bar is \(8-x\) kilograms. To determine the value of \(x\), we must construct an equation that we can solve for \(x\). One possible method is to equate the weights of gold in the first and second bars to the weight of gold in the final mixture.

The weight of gold in the first bar is \(\frac{2}{2+3}*x\) kilograms, and the weight of gold in the second bar is \(\frac{3}{3+7}*(8-x)\) kilograms. By adding these two quantities and setting the result equal to the weight of gold in the final bar, which is \(\frac{5}{5+11} * 8 = 2.5\) kilograms, we get the equation \(\frac{2}{5}*x + \frac{3}{10}*(8-x)=2.5\).

To simplify the equation, we multiply both sides by 10, giving us \(4x + 3*(8-x)=25\). Solving for \(x\), which represents the weight of the first bar, we find that \(x=1\) kilogram.


Answer: A