M12-18

Official Solution:

A metal bar weighing 20 kilograms, consisting of tin and silver, has lost 2 kilograms of its weight when placed in water. If 10 kilograms of tin lose 1.375 kilograms in water and 5 kilograms of silver lose 0.375 kilograms in water, what is the proportion of tin to silver in the bar?

A. \(\frac{1}{4}\)
B. \(\frac{2}{5}\)
C. \(\frac{1}{2}\)
D. \(\frac{3}{5}\)
E. \(\frac{2}{3}\)


Let \(t\) be the amount of tin in the bar (in kilograms) and \(s\) be the amount of silver in the bar (in kilograms). Since 10 kilograms of tin lose 1.375 kilograms in water, \(t\) kilograms of tin will lose \(\frac{t}{10}*1.375=0.1375*t\) kilograms in water.

Since 5 kilograms of silver lose 0.375 kilograms in water, \(s\) kilograms of silver will lose \(\frac{s}{5}*0.375= 0.075*s\) kilograms in water.

The total weight lost by both tin and silver is \(0.1375*t + 0.075*s = 2\) kilograms.

Since \(t\) is equal to \(20 - s\), we can substitute this into the equation to get \(0.1375 *(20 - s) + 0.075 * s = 2\). Solving for \(s\), we find that \(s = 12\). Hence, \(t = 20 - s = 20 - 12 = 8\).

Finally, the ratio of tin to silver in the bar is calculated as \(\frac{t}{s} = \frac{8}{12} = \frac{2}{3}\).


Answer: E