Events & Promotions
|
|
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.
May 1609:00 PM PDT
-10:00 PM PDT
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!
May 1706:30 AM PDT
-08:30 AM PDT
We know Strengthen and Weaken questions account for more than 50% of the CR questions on the GMAT. With CR becoming even more important on GMAT Focus, it's time you strengthen your weaknesses with an approach that improves your solving time and accuracy!
May 1711:00 AM IST
-01:00 PM IST
Attend this session to learn how to Deconstruct arguments, Pre-Think Author’s Assumptions, and apply Negation Test.
May 1806:30 AM PDT
-08:30 AM PDT
Join Piyush Beriwala in this webinar as he guides you through a smart study plan to achieve the 99th %ile on the GMAT Focus. Learn the best strategies and tips to master the test and boost your score.
Jun 1006:00 AM PDT
-05:30 PM PDT
Register for the GMAT Club Virtual MBA Spotlight fair, the biggest MBA fair of the year. You will have a chance to hear Admissions directors from almost every Top 20 program speak, network with peers, and conveniently attend morning or afternoon sessions.
19 May 2020, 05:09
Kudos
Bookmarks
First two
bars: 0.98 Third
bar: 0.92
bars: 0.98 Third
bar: 0.92
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
69% (02:40) correct
31%
(03:02)
wrong
based on 1713
sessions
History
Date
Time
Result
Not Attempted Yet
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
18 Sep 2021, 05:37
Kudos
Bookmarks
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
15 Jun 2024, 09:01
Kudos
Bookmarks
parkhydel
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.
Simply imagine the scale of weighted averages.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
All 3 bars have equal weight. First two bars have same finesse. It simply means that the weight given to the finesse of the first 2 bars is twice the weight given to the finesse of the third bar. So the average finesse of .96 will be at a distance of 1:2 from the finesse of first two bars to the finesse of the third bar.
Hence 0.98 could be the finesse of first two bars and .92 the finesses of the third bar.
0.98 - - 0.96 - - - - 0.92
Their distance from the average will be in the ratio 1:2.
ANSWER: 0.98, 0.92
Video of weighted averages: https://www.youtube.com/watch?v=_GOAU7moZ2Q
