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Bunuel
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20 students took the test. The arithmetic mean of the scores was 4.75 24 Jan 2022, 05:45
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

55% (hard)

Question Stats:

69% (02:20) correct 31% (02:31) wrong based on 257 sessions

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20 students took the test. The arithmetic mean of the scores was 4.75 and the range was 2. If the only possible scores were {2,3,4,5}, what is the greatest possible number of excellent scores (5)?

(A) 19
(B) 17
(C) 14
(D) 13
(E )12


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20 students took the test. The arithmetic mean of the scores was 4.75 31 Jan 2022, 18:02
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Bunuel
20 students took the test. The arithmetic mean of the scores was 4.75 and the range was 2. If the only possible scores were {2,3,4,5}, what is the greatest possible number of excellent scores (5)?

(A) 19
(B) 17
(C) 14
(D) 13
(E )12

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Breaking Down the Info:

Since the range was 2, and 5 is definitely one of the scores, our smallest score would be 3.

The sum of all the score is \(4.75*20 = (5 - 0.25)*20 = 100 - 5 = 95\).

If we had all 3's, the total score would only be 60. We would need to "distribute" the rest of the 35. We can give 34 = 2*17 = 17 increments of 2's away to have 17 of 5's.

Answer: B
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20 students took the test. The arithmetic mean of the scores was 4.75 15 Jan 2025, 10:37
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Bunuel
20 students took the test. The arithmetic mean of the scores was 4.75 and the range was 2. If the only possible scores were {2,3,4,5}, what is the greatest possible number of excellent scores (5)?

(A) 19
(B) 17
(C) 14
(D) 13
(E )12


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(3x + 4y + 5z)/20 = 4.75
3x + 4y + 5z = 95 --- (1)

x + y + z = 20
3x + 3y + 3z = 60 --- (2)

From (1) and (2)

y + 2z = 35
z = (35 - y)/2
For z to be an integer, y needs to be 1, which makes z = 17

IMO: B
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