|
|
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.
10 Dec 2019, 05:47
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
57% (02:15) correct
43%
(02:35)
wrong
based on 199
sessions
History
Date
Time
Result
Not Attempted Yet
In an intercollegiate tournament, Rosebud college won 70% of the first 80 games played during the first session of the year. In order for the college to win at least 80% of the total games played during the entire year, how many more it will have to play in the remaining sessions of the year? (Note: The college wins all the games in the remaining sessions)
I. 35
II. 40
III. 50
A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III
I. 35
II. 40
III. 50
A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III
10 Dec 2019, 23:53
Kudos
Bookmarks
This is a question where good old Algebra works very well. The most important key word in this question is ‘at least 80%’. Keep in mind that the overall percentage of wins need not be an integer value, as long as it is more than or equal to 80%.
Rosebud college won 70% of the first 80 games. This means that they won 56 of the first 80 games, whilst losing the rest.
Let Rosebud play ‘x’ games for the remaining sessions of the year. In this case,
Total number of games played during the entire year = 80 + x.
Total number of wins during the entire year = 56 + x ( since the question says that the college wins all the games in the remaining sessions).
Since the college has to win at least 80% of the total games played during the year,
56 + x ≥ \(\frac{4}{5}\) (80 +x) {Note that the fraction \(\frac{4}{5}\) is the equivalent of 80% and we are using the ≥ since this symbol represents 'at least'}. Solving the above inequality, we get x≥40.
Therefore, the possible values for x are 40 or 50. Statement I is not a possible value for x.
The correct answer option is D.
Note that answer option B is a trap answer for all of you who missed the ‘at least 80%’ part of the question stem and went ahead and found out 80% of the total games. This is a very common mistake that can happen in this question, so make sure that you re-read the question just to make sure that your answer matches the constraints.
Hope that helps!
Rosebud college won 70% of the first 80 games. This means that they won 56 of the first 80 games, whilst losing the rest.
Let Rosebud play ‘x’ games for the remaining sessions of the year. In this case,
Total number of games played during the entire year = 80 + x.
Total number of wins during the entire year = 56 + x ( since the question says that the college wins all the games in the remaining sessions).
Since the college has to win at least 80% of the total games played during the year,
56 + x ≥ \(\frac{4}{5}\) (80 +x) {Note that the fraction \(\frac{4}{5}\) is the equivalent of 80% and we are using the ≥ since this symbol represents 'at least'}. Solving the above inequality, we get x≥40.
Therefore, the possible values for x are 40 or 50. Statement I is not a possible value for x.
The correct answer option is D.
Note that answer option B is a trap answer for all of you who missed the ‘at least 80%’ part of the question stem and went ahead and found out 80% of the total games. This is a very common mistake that can happen in this question, so make sure that you re-read the question just to make sure that your answer matches the constraints.
Hope that helps!
General Discussion
10 Dec 2019, 23:10
Kudos
Bookmarks
Bunuel
Let the minimum games required be x, so total games played = 80+x..
Thus, 80% of (80+x)=70% of 80 +x......\(\frac{80}{100}*(80+x)=\frac{70}{100}*80+x................6400+80x=5600+100x........800=20x...x=40\)
So 40 and 50 will ensure at least 80% wins
D
