12 Days of Christmas GMAT Competition 2025 - Day 8: A museum recorded

gemministorm

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25 Dec 2025, 00:13

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Given: of 12 - min - 27 and all distinct -> listing 1 possibility -> {27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38} since there is not upper cap, there can be many possibilities.
need to find the upper cap -?
A -> range could be at-most 12 -> from our possibilities range is 11 ok if we give 12 as range then upper cap will be 39 -> 2 possible answers hence not sufficient.
B -> median - 33 -> even numbers and distinct hence 6th and 7th has to be 32 and 34 for the median to be 33. but still upper cap can be any value since no cap -> not sufficient.
combining A and B -> From A 2 possibilities - 1 gets eliminated -> only possible one is with the upper cap 39 -> hence both combined is sufficient.

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25 Dec 2025, 00:42

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Bunuel

The museum had visitors on each of the consecutive 12 days.

They are mentioned that the number of visitor count were all different.

The smallest daily visitor count =27

Largest daily visitor count = ?

Statement 1:

The range of the visitor counts for the 12 days was at most 12.

If the smallest visitor count =27

Range = Largest - smallest

Largest - Smallest <=12

Largest <= 12 + Smallest

Largest <= 39

So, the 12 consecutive numbers are : 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38.

Largest number can be either 38 or 39.

Hence, Insufficient.

Statement 2:

The median visitor count is 33

The smallest is 27, so the value of the 6th and 7th day will be 32 and 34 respectively.

The Largest value can be either 39, or any other higher integer.

Hence, Insufficient

Combining both Statements 1 and 2, we get

The 6th and 7th value is 32 and 34.

Leaving the value 33, pushes the max value to be 39 instead of 38.

Thus, maximum value = 39.

Hence, Sufficient

Option C

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25 Dec 2025, 01:02

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min = 27, different numbers on each 12 consecutive days
(1) range at most 39: max = 27 + 12 <= 39 => insufficient
(2) median = 33, median is the average of 6th and 7th days => we have 6th day is 32, 7th days is 34, max >= 34+5= 39 (different numbers but still not know each day)
(1) + (2): max = 39. C

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25 Dec 2025, 01:08

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Both statement together are sufficient but neither alone is sufficient.

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25 Dec 2025, 01:38

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Bunuel

Given :

Museum have different number of visitors for each 12 days

smallest daily visitor count is 27

What is largest daily visitor count

Statement 1 : X - 27 <= 12
X < = 39

So X could be 39 or 38 ; we don't know exact number so Insufficient

Statement 2 : We can't conclude anything by this ---- Insufficient

Statement 1 + 2 : ok so if median is 33 then the largest number will be 39 not 38 ( because 33 is on position 7 there) ----- Sufficient

Answer - C

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25 Dec 2025, 01:39

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Dmin = 27 , need to find Dmax

(1) Range <=12

Dmax-Dmin <=12

Dmax-27 <=12

Dmax<=39

Now if we place the data starting from 27 in ascending order , Dmax can be 38 or 39. So, statement 1 is not sufficient. Keeping in mid that all the data are unique as stated in question.So when starting the order need to place first with common difference of 1.

(2) Median is 33. It means d6 and d7 observation sum will lead to 66. Bot Dmax still can be anything . Statement 2 is not sufficient.

Now combining both statements , an unique order will be received since d6 <33 . So

d1 =27 d2 = 28 d3 = 29 d4 = 30 d5 = 31 d6 =32 d7 =34 d8 =35 d9 = 36 d10 =37 d11 = 38 d12 = 39. If you try to manipulate this order all numbers will not be accommodated to fit the range. And range needs to be max value to fit all unique data.

So C is the answer.

Bunuel

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25 Dec 2025, 02:00

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(1) This means the largest value can at most be 39.
If we can check, 27, 28, 29, ....., 38, 38 is the largest if we consider consecutive numbers. It could also be 39. So not sufficient.

(2) Since we have an even number of numbers, 33 is the average of the 6th and 7th value.
So 6th and 7th values can be (33, 33), (32, 34), (31, 35), (30, 36). We can eliminate (33, 33) as all are distinct.
Now, let's see if the remaining are possible,
27, 28, 29, 30, 31, ......(taking the minimum for the first 5 values), so the minimum value of 6th place is 32. So only (32, 34) is possible.
So the list is like this: 27, 28, 29, 30, 31, 32, 34, _, _, _, _, _. The largest could be anything here. So not sufficient.

(1) + (2) Largest can at most be 39.
From (2): 27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 38, 39 (taking the minimum for the remaining 5 values), we can see 39 is the minimum and also the maximum value for the largest number. So the largest dialy visitor count is 39. Sufficient.

Option C.

Bunuel

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25 Dec 2025, 02:19

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Bunuel

considering the sorted set of visitor count

s1) max can be 38 or 39
s2) 6th + 7th = 66,

7th element can only be 34,8th can be >= 35
12th element >= 39

using s1 and s2
we get 12th element 39

hene answer is C

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25 Dec 2025, 04:16

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Assessing each statement:

Statement 1: Taking the most range given lowest was 27 => Highest can be 39. All counts were different. 27,28,29,30,31,32,33,34,35,36,37,38 can also be a solution. Hence, insufficient.
Statement 2: Median count was 33. This does not cap the final value, and it can go up to any number. Not sufficient.

Statement 1 & 2 together =>
Median cannot be 33 in the case listed in Statement 1. It can only be 33 when this is the case => 27,28,29,30,31,32,34,35,36,37,38,39 => Highest is 39. Both together sufficient.

Answer => C

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25 Dec 2025, 04:36

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Vistor count was different everyday and smallest daily visitor count=27

(1) range is atmost 12
Smallest=27
Range<= 27+12 =39
Smallest is 27, then the possibilities are 27,28,29,30,31,32,33,34,35,36,37,38 or 27,28,29,30,31,32,33,34,35,36,37,39
Largest can be 38 or 39
Insufficient

(2) median is was 33
Knowing median doesnt help us know the largest daily visitor count
Insufficient

(1)&(2) we still can’t uniquely identify the largest value, it could be 38 or 39
Insufficient

E

redandme21

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25 Dec 2025, 04:42

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all different
smallest=27
largest?

(1)
range<=12
largest<=27+12=39

But between 27 and 39 there are 39-27+1=13 numbers.
We must choose a subset of 12 elements from a set of 13 elements.
If the subset contains 39 -> largest=39
If the subset doesn't contain 39 -> it must contain 38 -> largest=38

Insufficient

(2)
median=33
As 12 is even, then the set must contain 32 and 34 (and not 33).
The first numbers must be: 27, 28, 29, 30, 31, 32, 34.
But the other 5 numbers can be any different value and the largest can be 100 or 200, for example.

Insufficient

(1) + (2)
As the list of numbers must not contain 33 (as median is 33) and range<=12, then largest must be 39.

Sufficient

IMO C

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25 Dec 2025, 05:01

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(1)
If the number of visitors for each of 12 consecutive days is:
27 28 29 30 31 32 33 34 35 36 37 38 range=11<=12
then largest is 38

But if it is:
27 28 29 30 31 32 33 35 36 37 38 39 range=12<=12
then largest is 39

Condition insufficient

(2)
If the number of visitors for each of 12 consecutive days is:
27 28 29 30 31 32 34 35 36 37 38 39 median=33
then largest is 39

But if it is:
27 28 29 30 31 32 34 35 36 37 38 50 median=33
then largest is 50

Condition insufficient

(1)+(2)
The only possiblity is that the number of visitors for each of 12 consecutive days is:
27 28 29 30 31 32 34 35 36 37 38 39 range=12<=12 and median=33

largest number of visitors=39

Conditions sufficient

Answer C

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25 Dec 2025, 05:08

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12 days
Visitor counts are different
The smallest count was 27
Largest = ?

(1) Range is at most 12
Since each count is different, this tells that the max would be 39
Sufficient.

(2) The median is 33. Max could be any number. Can't exactly say. Not Sufficient.

(A) is the answer

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25 Dec 2025, 05:09

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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.
Range is almost= 12
Largest-27≤12
Largest≤ 39
Since there are 12 different integers and smallest is 27, there are two possibilities for largest daily visitor count 38 or 39. It is insufficient.

(2) The median visitor count over the 12 days was 33.
Since the number of visitors are consecutive, median= (x6+x7)/2 = 33
x6+x7 =66
x6 and x7 can be 32,34 or 31,35, or 30,36 etc. alone it is insufficient

Both (1)&(2)
Largest vistor count is 38 or 39
Median is 33
If largest is 38,
Visitors= {27,28,29,30,31,32,33,34,35,36,37,38}
Median= (32+33)/2 = 32.5
Median is 33, largest can't be 38.
If largest visitor count is 39,
Visitors= {27,28,29,30,31,32,34,35,36,37,38,39}
Median= (32+34)/2 = 33 Correct
It is sufficient

C

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25 Dec 2025, 05:21

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(1)
There are 13 numbers between 27 and 27+12=39 inclusive.
Choosing 12 numbers from 13 makes that the largest number can be 38 or 39

Condition (1) is insufficient

(2)
12 days -> median = (n6+n7)/2 -> n6+n7 = 33*2=66
This means that 33 can not be one of the numbers.

Condition (2) is insufficient

(1)+(2)
If 33 can not be one of the numbers and range must be at most 12 then there is only one list of numbers
{27 28 29 30 31 32 34 35 36 37 38 39}
and the largest number is 39

Condition (1) and (2) are sufficient

The answer is C

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25 Dec 2025, 05:35

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(1)
Range <= 12
27+12=39, so largest daily visitor count <= 39
As all counts are different, there are 39-27+1=13 possible counts to assign to 12 days.
If 39 is one of the assigned counts then largest daily visitor count is 39.
If 39 isn't one of the assigned counts then largest daily visitor count is 38.

insufficient

(2)
Median=33
Median must be ((6th count)+(7th count))/2
No count can be 33 because all counts are different.
But largest daily visitor count has not a sigle answer.

insufficient

(1) and (2)
With the two conditions there are only 12 possible counts to assign to 12 days (33 is not possible).
largest daily visitor count is 39

sufficient

The correct answer is C

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25 Dec 2025, 06:05

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12 days visitor counts were recorded which were all different. smallest count=27
largest count?
1. range (visitor count) <=12
27,28,29,30,31,32,33,34,35,36,37,38, and 27,28,29,30,32,33,34,35,36,37,38,39
since both distributions satisfy all conditions so insufficient.
2. median (visitor count)=33
this option just tells that 6th and 7th counts are 32 and 34 respectively ,no restriction on the upper limit,---insufficient
together its clear that only 27,28,29,30,31,32,34,35,36,37,38,39 satisfies all conditions
hence C

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25 Dec 2025, 06:19

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it will be both required as the first option only gives 39 as max and it can fit 13 values, so there are multiple variables, but the 2nd one put together with first gives us median and we can determine the exact highest

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25 Dec 2025, 06:23

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Given:
Number of days: n = 12
Visitor counts: (x1, x2, ..., x12) (all distinct positive integers)
Sorted counts: x1 < x2 < .... < x(12)
Smallest count: x1 = 27
We are asked to find x12

Statement 1: The range of the visitor counts for the 12 days was at most 12. The range is defined as x12 - x1 (Max - Min)
Given x12 - 27 <= 12, we have x12 <= 39.
Since there are 12 distinct integers starting from 27, the minimum possible value for the largest integer is x1 + (n - 1) = 27 + 11 = 38.
Therefore, x12 could be 38 (if the counts are 27, 28, ...., 38) or 39 (if the common difference is of one in the sequence).
Since we have more than one possible value for x12, Statement 1 is not sufficient.

Statement 2. The median visitor count over the 12 days was 33.
For 12 values, the median is the average of the 6th and 7th terms: (x6 + x7)/2 = 33, so x6 + x7 = 66.
Since the values are distinct integers and x6 < x7, the possible pairs for (x6, x7) are (32, 34), (31, 35), (30, 36), ....
However, we need to fit 4 distinct integers between x1 = 27 and x6 (namely x2, x3, x4, x5).
If x6 = 32, the integers can be (27, 28, 29, 30, 31, 32).
If x6 = 31, there are only three integers (28, 29, 30) available to fill the four spots between 27 and 31. Which is not possible.
Thus, x6 must be 32, which forces x7 = 34.
Knowing x7 = 34, the remaining values x8, x9, x10, x11, x12 must be at least 35, 36, 37, 38, 39 respectively.

Since we now know that x12 >= 39, there is no upper threshold provided. So Statement 2 is not sufficient.

Combining both.
From Statement 2, we established that x7 = 34. To have 5 distinct integers greater than 34, the largest integer x12 must be at least 39.
From Statement 1, we found told x12 <=39.
The only integer that satisfies both conditions is x12= 39.
Both statements together are sufficient.

Ans should be C.

Bunuel