M12-18

I have used a bit different approach.

This problem is the same as solution problems, though it is formulated in other way.

Let's translate given problem to conventional solution problem:

Let's denote taken weight of tin as X and taken weight of silver as Y.

1. Tin loses 1.375/10 = 0.1375 kg for 1 kg of its weight.
2. Silver loses 0.375/5 = 0.075 kg for 1 kg of its weight.
3. Metal bar loses 2/20 = 0.1 kg for 1 kg of its weight.

0.1375X + 0.075Y
--------------------- = 0.1;
X+Y

0.1375X + 0.075Y = 0.1 (X+Y)

0.0375X = 0.025Y

X/Y = 0.025/0.0375 = 2/3

You are welcome)

This question, like almost all weighted average questions, can be made simpler with allegation. For every 10 kg of tin, 1.375 kg dissolves in water. For every 10 kg of silver, .750 kg dissolves in water. We need a weighted average of tin and silver that will dissolve 1 kg per 10 kg of combined metal. Notice that 1 kg dissolving per 10 kg of combined metal is equivalent to 2 kg dissolving per 20 kg of combined metal:

ratio of tin:

\(1 - .750 = .250\)

ratio of silver:

\(1.375 - 1 = .375\)

\(\frac{.250}{.375}=\frac{2}{3}\)

I've attached a visual representation of allegation for those who are interested.

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

I think this is a high-quality question and I agree with explanation.

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.