Official Explanation
According to the text, the finesse of a gold bar is equal to the weight of gold in the bar divided by the total weight of the bar. The text also indicates that the three gold bars have the same weight, that the first two bars have the same finesse, and that melting and combining the three bars would result in a gold bar with a finesse of 0.96. Since the first two bars have identical finesse and identical weight, the weight of the gold present in those bars must also be identical.
Let x equal the weight of the gold in each of the first two bars, let y equal the weight of the gold in the third bar, and let w be the total weight of each bar. Then, since the finesse of the combined bar is 0.96, it must be the case that
\(\frac{2x+y}{3w}=0.96\) or \(2(\frac{x}{w})+(\frac{y}{w})=(0.96)(3)=2.88\). But then if \(F_{1}=\frac{x}{w}\) is the finesse of each of the first two bars,
and \(F_{3}=\frac{y}{w}\) is the finesse of the third bar, then \(2F_{1}+F_{3}=2.88\). Among the available options, substituting 0.98
for \(F_{1}\) and 0.92 for \(F_{3}\) gives us \(2(0.98) + 0.92 = 1.96 + 0.92 = 2.88\).
No other pair of options yields the correct result.
Answer
Third bar: 0.92
First two bars: 0.98