I31-13

Question

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.

Bunuel · Answer

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

hitakshim · Answer

I did not quite understand the solution. I do not understand, why even consider 38?
It is stated in first that the range was 12 at most.. taking 39 that way is easy, all of the middle numbers can be different you can omit any one. Why consider 38 to be the largest at all?

Bunuel · Answer

I'm not really following what you are trying to say there, but 39 is only the  maximum possible value  from statement (1), not a required value, so  38 is still possible  and must be considered.

SkyBulb · Answer

I did not quite understand the solution. How can range be 12 in case 38 is largest and 27 is lowest? Its a bit confusing for 1st statement.

Bunuel · Answer

The range does not have to be exactly 12. Statement (1) says the range is at most 12, meaning 12 or less. So if the smallest count is 27 and the largest is 38, the range is 38 - 27 = 11, which still satisfies statement (1). If the largest is 39, the range is 39 - 27 = 12, which also satisfies statement (1). That is why statement (1) alone is not sufficient.

janvibatra.50 · Answer

Hi, isn't Statement (2) sufficient by itself? 
 
We know the smallest visitor count is 27 and all 12 counts are distinct integers. 
 
From Statement (2), the median is 33. Since there are 12 observations, the median is the average of the 6th and 7th numbers, so the 6th and 7th numbers must add up to 66.   Since 66 is even, the two numbers must have the same parity (o/o or e/e). These numbers can neither be equal (all counts are distinct) nor be consecutive. Doesn't that force them to be 32 and 34?

If that's the case, then the first six numbers have to be 27, 28, 29, 30, 31, 32, and the remaining five numbers have to be 35, 36, 37, 38, and 39. That uniquely determines the largest visitor count as 39. 
 
Am I missing something here?

Bunuel · Answer

You are correct that the 6th and 7th counts must be 32 and 34.

However, the remaining five counts do  not  have to be 35, 36, 37, 38, and 39. They can be any five distinct integers greater than 34, such as 35, 36, 37, 38, and 100.

Thus, the largest count is not determined, so Statement (2) is not sufficient.

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