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08 Aug 2022, 07:14
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E
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:
79% (03:09) correct
21%
(03:01)
wrong
based on 19
sessions
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Date
Time
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Not Attempted Yet
In a class of 10 students, the average marks of the top 6 students is 20. The average marks of the bottom 5 students is 6. The overall average of the class is 14. What can be the maximum marks obtained by the topper, if each student gets distinct marks?
A. 44
B. 48
C. 52
D. 56
E. 60
(adapted from gmatfree)
A. 44
B. 48
C. 52
D. 56
E. 60
(adapted from gmatfree)
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
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01 Sep 2022, 14:52
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I worked out the following formula for finding the solution in just one step
6 * 20 - [(6 * 20 + 5 * 6 - 10 * 14) * 5 + (4 * 5 / 2)] = 60 --> "E"
The explanation is as follows:
There is obviously a double counting, since we have 10 students but we are given the top 6 and bottom 5 (so 11 students...)
As a consequence, the average mark of the sixth student has been double counted
To work its value out we sum the total marks of the first 6 students and those of the bottom 5 and subtract the total marks of the 10 students:
6 * 20 + 5 * 6 - 10 * 14 = 10
As a result, we now know that the average mark of the 6th student is 10
Since we are required to maximize the average mark of the 1st student, we have to minimize those of the 2nd, 3rd, 4th and 5th student
On the top of that, the average marks have all to be distinct (though not explicitly stated, we could very well assume that this "distinctiveness" must be represented by - at least - a "plus one: +1" or a multiple thereof; that is to say, the set of the average marks must be composed by natural numbers separated by at least one unit)
In order to properly account for this "distinctiveness", we multiply 10 by 5 and we add the result of the "Gauss's Trick" for 4 students (4 * 5 / 2), a process that corresponds to having
6th Student's Average Mark = 10
5th SAM = 11
4th SAM = 12
3rd SAM = 13
2nd SAM = 14
Summing all these average marks and subtracting them from the total average marks of the first 6 students (6 * 20 = 120), we obtain that the maximum average mark of the 1st student that satisfies all the given constraints is 60
One possible set would therefore be
3-4-6-7-10-11-12-13-14-60
Posted from my mobile device
6 * 20 - [(6 * 20 + 5 * 6 - 10 * 14) * 5 + (4 * 5 / 2)] = 60 --> "E"
The explanation is as follows:
There is obviously a double counting, since we have 10 students but we are given the top 6 and bottom 5 (so 11 students...)
As a consequence, the average mark of the sixth student has been double counted
To work its value out we sum the total marks of the first 6 students and those of the bottom 5 and subtract the total marks of the 10 students:
6 * 20 + 5 * 6 - 10 * 14 = 10
As a result, we now know that the average mark of the 6th student is 10
Since we are required to maximize the average mark of the 1st student, we have to minimize those of the 2nd, 3rd, 4th and 5th student
On the top of that, the average marks have all to be distinct (though not explicitly stated, we could very well assume that this "distinctiveness" must be represented by - at least - a "plus one: +1" or a multiple thereof; that is to say, the set of the average marks must be composed by natural numbers separated by at least one unit)
In order to properly account for this "distinctiveness", we multiply 10 by 5 and we add the result of the "Gauss's Trick" for 4 students (4 * 5 / 2), a process that corresponds to having
6th Student's Average Mark = 10
5th SAM = 11
4th SAM = 12
3rd SAM = 13
2nd SAM = 14
Summing all these average marks and subtracting them from the total average marks of the first 6 students (6 * 20 = 120), we obtain that the maximum average mark of the 1st student that satisfies all the given constraints is 60
One possible set would therefore be
3-4-6-7-10-11-12-13-14-60
Posted from my mobile device
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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