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08 Jul 2025, 08:52
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Given:
Let the scores be: a1 < a2 < a3 < a4 < a5 < a6 < a7
Since median = 27, we have a4 = 27
Maximizing the lower scores (a1, a2, a3):
To make them as large as possible while staying below 27:
Maximizing the middle-high scores (a5, a6)
If the highest score is x, then to maximize the others:
Solving for x:
All scores must sum to 210:
24 + 25 + 26 + 27 + (x-2) + (x-1) + x = 210
102 + 3x - 3 = 210
99 + 3x = 210
3x = 111
x = 37
Check:
Scores are 24, 25, 26, 27, 35, 36, 37
- 7 students, all different integer scores
- Average = 30, so total points = 7 × 30 = 210
- Median = 27 (the middle score when arranged in order)
Let the scores be: a1 < a2 < a3 < a4 < a5 < a6 < a7
Since median = 27, we have a4 = 27
Maximizing the lower scores (a1, a2, a3):
To make them as large as possible while staying below 27:
- a3 = 26 (biggest integer less than 27)
- a2 = 25 (biggest integer less than 26)
- a1 = 24 (biggest integer less than 25)
Maximizing the middle-high scores (a5, a6)
If the highest score is x, then to maximize the others:
- a6 = x - 1 (biggest integer less than x)
- a5 = x - 2 (biggest integer less than x - 1)
Solving for x:
All scores must sum to 210:
24 + 25 + 26 + 27 + (x-2) + (x-1) + x = 210
102 + 3x - 3 = 210
99 + 3x = 210
3x = 111
x = 37
Check:
Scores are 24, 25, 26, 27, 35, 36, 37
- Sum = 210
- Average = 30
- Median = 27
08 Jul 2025, 08:55
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mean=30, n=7 , median is 27 which is the 4th position
total=30*7=210
for the maximum score to be the lowest as possible we will try to max the lower scores but all are different that means , 24,25,26,27 which makes a total of 102
now we are left with 210-102=108
108/3=36
So ,the 5th,6th,7th term can be 35,36,37
hence,37(D)
total=30*7=210
for the maximum score to be the lowest as possible we will try to max the lower scores but all are different that means , 24,25,26,27 which makes a total of 102
now we are left with 210-102=108
108/3=36
So ,the 5th,6th,7th term can be 35,36,37
hence,37(D)
MaxFabianKirchner
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Concentration: Finance, International Business
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Schools: ESSEC MiF '26 ESADE HEC Imperial LBS Oxford MFE '26 Sloan (MIT) NYU Stern NUS Bocconi IE'26
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08 Jul 2025, 09:00
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To find the minimum possible value for the highest score, all other scores must be as large as possible while adhering to the given constraints. The problem states there are seven different integer scores. The mean of 30 implies the total sum of the scores is 210 (30 * 7). The median of 27 means the fourth score in ascending order is 27.
Let the seven scores in ascending order be s1, s2, s3, 27, s5, s6, s7. To minimize s7, we must maximize the other scores. The three scores below the median must be distinct integers less than 27. The largest possible values for these scores are therefore 24, 25, and 26. The sum of these four known scores (24, 25, 26, and 27) is 102.
The sum of the three remaining scores (s5, s6, s7) must be 210 - 102 = 108. These three scores must be distinct integers greater than 27. To make the largest of these three (s7) as small as possible, the three values must be as close to each other as possible. The three consecutive integers that sum to 108 are 35, 36, and 37. These values satisfy all conditions.
Therefore, the minimum possible value for the highest score is 37.
Let the seven scores in ascending order be s1, s2, s3, 27, s5, s6, s7. To minimize s7, we must maximize the other scores. The three scores below the median must be distinct integers less than 27. The largest possible values for these scores are therefore 24, 25, and 26. The sum of these four known scores (24, 25, 26, and 27) is 102.
The sum of the three remaining scores (s5, s6, s7) must be 210 - 102 = 108. These three scores must be distinct integers greater than 27. To make the largest of these three (s7) as small as possible, the three values must be as close to each other as possible. The three consecutive integers that sum to 108 are 35, 36, and 37. These values satisfy all conditions.
Therefore, the minimum possible value for the highest score is 37.
