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08 Jul 2025, 08:00
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D
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A class of seven students took a quiz, and each received a different integer score between 0 and 100. If the average (arithmetic mean) score was 30 and the median was 27, what is the minimum possible value for the highest score?
A. 34
B. 35
C. 36
D. 37
E. 38
A. 34
B. 35
C. 36
D. 37
E. 38
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08 Jul 2025, 08:30
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Bunuel
GMAT Club Official Explanation:
Say the seven scores are a, b, c, d, e, f, and g, where a < b < c < d < e < f < g (since all scores are different). We are asked to find the lowest possible value of g.
We are told the average of the seven scores is 30, so the total sum is:
a + b + c + d + e + f + g = 30 * 7 = 210
We are also told the median is 27, so d = 27:
a < b < c < (d = 27) < e < f < g
To minimize g, we need to maximize the other numbers. The maximum values of a, b, and c are 24, 25, and 26 respectively.
In this case: a + b + c + d = 24 + 25 + 26 + 27 = 102
Thus, e + f + g = 210 - 102 = 108.
If those three numbers were equal, we’d get e = f = g = 108/3 = 36. However, since they must be distinct, the least value g can take is when e = 35, f = 36, and g = 37.
The final distribution that yields the minimum possible value of g is: {24, 25, 26, 27, 35, 36, 37}.
Answer: D.
General Discussion
08 Jul 2025, 08:14
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A class of seven students took a quiz, and each received a different integer score between 0 and 100.
If the average (arithmetic mean) score was 30 and the median was 27, what is the minimum possible value for the highest score?
Let the scores = {x1, x2, x3, 27, x4, x5, x6}
x1 + x2 + x3 + 27 + x4 + x5 + x6 = 30*7 = 210
x1 + x2 + x3 + x4 + x5 + x6 = 210 - 27 = 183
To minimize x6 (highest score), we have to maximise all other score.
Let x3 = 26; x2 = 25; x1 = 24; x5 = x6 - 1 ; x4 = x6 - 2
24 + 25 + 26 + 27 + (x6-2) + (x6-1) + x6 = 210
x6 = 37
The minimum value of highest score = x6 = 37
IMO D
If the average (arithmetic mean) score was 30 and the median was 27, what is the minimum possible value for the highest score?
Let the scores = {x1, x2, x3, 27, x4, x5, x6}
x1 + x2 + x3 + 27 + x4 + x5 + x6 = 30*7 = 210
x1 + x2 + x3 + x4 + x5 + x6 = 210 - 27 = 183
To minimize x6 (highest score), we have to maximise all other score.
Let x3 = 26; x2 = 25; x1 = 24; x5 = x6 - 1 ; x4 = x6 - 2
24 + 25 + 26 + 27 + (x6-2) + (x6-1) + x6 = 210
x6 = 37
The minimum value of highest score = x6 = 37
IMO D
