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A museum recorded the number of visitors for each of 12 consecutive days. If the visitor counts were all different and the smallest daily visitor count was 27, what was the largest daily visitor count?
(1) The range of the visitor counts for the 12 days was at most 12.
(2) The median visitor count over the 12 days was 33.
(1) The range of the visitor counts for the 12 days was at most 12.
(2) The median visitor count over the 12 days was 33.
13 Jan 2026, 11:25
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Bookmarks
Official Solution:
A museum recorded the number of visitors for each of 12 consecutive days. If the visitor counts were all different and the smallest daily visitor count was 27, what was the largest daily visitor count?
(1) The range of the visitor counts for the 12 days was at most 12.
The range equals the largest count minus the smallest count. Since the smallest daily count was 27 and the range was at most 12, the largest count could be at most 39. Because the 12 daily counts were all different, they must be 12 of the 13 consecutive integers from 27 to 39.
If 39 is missing, the counts could be {27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38}, so the largest is 38.
If a different number is missing, the counts could include 39, such as {27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 38, 39}, so the largest is 39.
Not sufficient.
(2) The median visitor count over the 12 days was 33.
For 12 numbers, the median is the average of the 6th and 7th numbers when the counts are ordered from least to greatest. A median of 33 means the 6th and 7th counts sum to 66. This, however, does not determine the largest count. Different sets with smallest 27 and median 33 can have different largest values.
Not sufficient.
(1)+(2) From (1), the largest is either 38 or 39. If the largest were 38, the only set of 12 distinct integers from 27 to 38 would be {27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38}. Its median would be (32 + 33)/2 = 32.5, not 33, so the largest cannot be 38.
Therefore, the largest must be 39. With largest = 39, we must choose 12 numbers from 27 to 39. To have median 33, the 6th and 7th numbers must average to 33. The only set that works is {27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 38, 39} (omitting 33). This gives largest = 39.
Sufficient.
Answer: C
A museum recorded the number of visitors for each of 12 consecutive days. If the visitor counts were all different and the smallest daily visitor count was 27, what was the largest daily visitor count?
(1) The range of the visitor counts for the 12 days was at most 12.
The range equals the largest count minus the smallest count. Since the smallest daily count was 27 and the range was at most 12, the largest count could be at most 39. Because the 12 daily counts were all different, they must be 12 of the 13 consecutive integers from 27 to 39.
If 39 is missing, the counts could be {27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38}, so the largest is 38.
If a different number is missing, the counts could include 39, such as {27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 38, 39}, so the largest is 39.
Not sufficient.
(2) The median visitor count over the 12 days was 33.
For 12 numbers, the median is the average of the 6th and 7th numbers when the counts are ordered from least to greatest. A median of 33 means the 6th and 7th counts sum to 66. This, however, does not determine the largest count. Different sets with smallest 27 and median 33 can have different largest values.
Not sufficient.
(1)+(2) From (1), the largest is either 38 or 39. If the largest were 38, the only set of 12 distinct integers from 27 to 38 would be {27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38}. Its median would be (32 + 33)/2 = 32.5, not 33, so the largest cannot be 38.
Therefore, the largest must be 39. With largest = 39, we must choose 12 numbers from 27 to 39. To have median 33, the 6th and 7th numbers must average to 33. The only set that works is {27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 38, 39} (omitting 33). This gives largest = 39.
Sufficient.
Answer: C
