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 1308:30 AM PDT
-09:30 AM PDT
In Episode 4 of our GMAT Ninja CR series, we tackle the most intimidating CR question type: Boldface & "Legalese" questions. If you've ever stared at an answer choice that reads, "The first is a consideration introduced to counter a position that...
May 1212:00 PM EDT
-11:59 PM EDT
Make the most of your break with the most realistic GMAT™ prep. Take up to $700 off select products.- May 14
08:30 AM PDT
-09:30 AM PDT
Most GMAT test-takers are intimidated by the hardest GMAT Verbal questions. In this session, Target Test Prep GMAT instructor Erika Tyler-John, a 100th percentile GMAT scorer, will show you how top scorers break down challenging Verbal questions.. - Jun 10
06:00 AM PDT
-06:15 PM PDT
Register for the GMAT Club Virtual MBA Spotlight Fair – the world’s premier event for serious MBA candidates. This is your chance to hear directly from Admissions Directors at nearly every Top 30 MBA program..
19 Mar 2024, 03:04
Kudos
Bookmarks
Athos, Porthos, Aramis, and D'Artagnan are dividing n gold coins they received from the queen. Athos gets 25% of all the coins, then Porthos gets 25% of the remaining coins, then Aramis gets 25% of the remaining coins, then D'Artagnan gets 25% of the remaining coins. Still, there are some coins left, which they divide equally among four of them. If n is the least number of coins the queen could have given them, then how many coins did D'Artagnan get?
A. 81
B. 189
C. 324
D. 405
E. 1,024
A. 81
B. 189
C. 324
D. 405
E. 1,024
19 Mar 2024, 03:04
Kudos
Bookmarks
Official Solution:
Athos, Porthos, Aramis, and D'Artagnan are dividing n gold coins they received from the queen. Athos gets 25% of all the coins, then Porthos gets 25% of the remaining coins, then Aramis gets 25% of the remaining coins, then D'Artagnan gets 25% of the remaining coins. Still, there are some coins left, which they divide equally among four of them. If n is the least number of coins the queen could have given them, then how many coins did D'Artagnan get?
A. 81
B. 189
C. 324
D. 405
E. 1,024
After each distribution, \(\frac{3}{4}\)th of the coins remain to be distributed to the next person. For instance:
Athos receives \(\frac{1}{4}n\) coins, leaving \(\frac{3}{4}n\) coins for distribution.
Then, Aramis receives \(\frac{1}{4}\)th of that remaining amount, leaving \(\frac{3}{4}(\frac{3}{4}n)=(\frac{3}{4})^2n\) coins to distribute.
Next, Porthos receives \(\frac{1}{4}\)th of the remaining amount, leaving \(\frac{3}{4}((\frac{3}{4})^2n)=(\frac{3}{4})^3n\) coins to distribute.
Subsequently, D'Artagnan receives \(\frac{1}{4}\)th of the remaining amount, which is \(\frac{1}{4}(\frac{3}{4})^3n\) coins, leaving \(\frac{3}{4}((\frac{3}{4})^3n)=(\frac{3}{4})^4n\) coins to distribute.
In the final distribution, D'Artagnan also receives \(\frac{1}{4}\)th of the remaining amount, an additional \(\frac{1}{4}(\frac{3}{4})^4n\) coins, making his total receipt \(\frac{1}{4}(\frac{3}{4})^3n + \frac{1}{4}(\frac{3}{4})^4n=(\frac{3^3}{4^4} + \frac{3^4}{4^5})n=\frac{189}{1,024}n\).
Given that \(\frac{189}{1,024}n\) must be an integer, the minimum value of \(n\) is 1,024. Therefore, the least number of coins D'Artagnan could have received is 189.
Answer: B
Athos, Porthos, Aramis, and D'Artagnan are dividing n gold coins they received from the queen. Athos gets 25% of all the coins, then Porthos gets 25% of the remaining coins, then Aramis gets 25% of the remaining coins, then D'Artagnan gets 25% of the remaining coins. Still, there are some coins left, which they divide equally among four of them. If n is the least number of coins the queen could have given them, then how many coins did D'Artagnan get?
A. 81
B. 189
C. 324
D. 405
E. 1,024
After each distribution, \(\frac{3}{4}\)th of the coins remain to be distributed to the next person. For instance:
Athos receives \(\frac{1}{4}n\) coins, leaving \(\frac{3}{4}n\) coins for distribution.
Then, Aramis receives \(\frac{1}{4}\)th of that remaining amount, leaving \(\frac{3}{4}(\frac{3}{4}n)=(\frac{3}{4})^2n\) coins to distribute.
Next, Porthos receives \(\frac{1}{4}\)th of the remaining amount, leaving \(\frac{3}{4}((\frac{3}{4})^2n)=(\frac{3}{4})^3n\) coins to distribute.
Subsequently, D'Artagnan receives \(\frac{1}{4}\)th of the remaining amount, which is \(\frac{1}{4}(\frac{3}{4})^3n\) coins, leaving \(\frac{3}{4}((\frac{3}{4})^3n)=(\frac{3}{4})^4n\) coins to distribute.
In the final distribution, D'Artagnan also receives \(\frac{1}{4}\)th of the remaining amount, an additional \(\frac{1}{4}(\frac{3}{4})^4n\) coins, making his total receipt \(\frac{1}{4}(\frac{3}{4})^3n + \frac{1}{4}(\frac{3}{4})^4n=(\frac{3^3}{4^4} + \frac{3^4}{4^5})n=\frac{189}{1,024}n\).
Given that \(\frac{189}{1,024}n\) must be an integer, the minimum value of \(n\) is 1,024. Therefore, the least number of coins D'Artagnan could have received is 189.
Answer: B
25 Nov 2024, 01:09
Kudos
Bookmarks
I think this is a high-quality question and I agree with explanation. IMO this feels like a 705+ question.
