Forum Home > GMAT > Quantitative > Problem Solving (PS)
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
The finesse of a gold bar is the weight of the gold present in the bar
[#permalink]
19 May 2020, 05:09
19 May 2020, 05:09
4
Kudos
36
Bookmarks
Show timer
00:00
Difficulty:
55%
(hard)
Question Stats:
73% (02:44) correct
27%
(03:03)
wrong
based on 597
sessions
The finesse of a gold bar is the weight of the gold present in the bar divided by the total weight of the bar. A jeweler has three gold bars. The three bars have identical weights, and the first two bars have identical finesse. If all three bars were melted and combined into one bar, the finesse of the resulting bar would be 0.96.
Select a finesse of the First two bars and a finesse of the Third bar that are jointly consistent with the given information. Make only two selections, one in each column.
Select a finesse of the First two bars and a finesse of the Third bar that are jointly consistent with the given information. Make only two selections, one in each column.
ID: 100426
|
First two bars |
Third bar |
|
| 0.92 | ||
| 0.93 | ||
| 0.94 | ||
| 0.98 | ||
| 0.99 |
ShowHide Answer
Official Answer
First two
bars: 0.98 Third
bar: 0.92
bars: 0.98 Third
bar: 0.92
GRE Forum Moderator
Joined: 02 Nov 2016
Posts: 14018
Given Kudos: 5865
GPA: 3.62
Re: The finesse of a gold bar is the weight of the gold present in the bar
[#permalink]
18 Sep 2021, 05:37
18 Sep 2021, 05:37
1
Kudos
2
Bookmarks
Expert Reply
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
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
Re: The finesse of a gold bar is the weight of the gold present in the bar
[#permalink]
14 Jun 2024, 19:37
14 Jun 2024, 19:37
1
Kudos
Expert Reply
Since all bars have identical weight and it speaks of average finesse, the question can be solved through weighted average method.
Let the weight be x each, so the ratio of those with higher finesse and lowers finesse will be (x+x):x or 2:1
Thus, the finesse too should be in the same ratio. The distance between the lower finesse and 0.96 should be two times that between higher and 0.96
So, one has to be higher: If it is next higher from 0.96, then 0.98, so (0.98-0.96)=1/2 * (0.96-y)….2*0.02 = 0.96-y
Or y=0.92
Let the weight be x each, so the ratio of those with higher finesse and lowers finesse will be (x+x):x or 2:1
Thus, the finesse too should be in the same ratio. The distance between the lower finesse and 0.96 should be two times that between higher and 0.96
So, one has to be higher: If it is next higher from 0.96, then 0.98, so (0.98-0.96)=1/2 * (0.96-y)….2*0.02 = 0.96-y
Or y=0.92
