12 Days of Christmas GMAT Competition 2025 - Day 5: a, b, and c are

rahumangal

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19 Dec 2025, 14:55

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Bunuel

I- take a=1 b=-2 c= -3
range = 1+3=4
sd od {a,b,c} >sd {lal, lbl,lcl}
median =-2
possible
II- let a=1 b=-1 c=-3
product = 3 = prime
possible
III- let a=1 b=-1 c=-3
Mode of the moulus terms{1,1,3}= 1
possible
Ans-E

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19 Dec 2025, 15:19

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for std of {a, b, c} to be greater than {|a|, |b|, |c|}, the first set should have both negative and positive values. If it has only non negative values or only non positive values, then both sets will be equal.

Let a be the smallest and c be the largest. Range is 4. To have a negative and positive value, they need to cross over the origin. So potential options

{-3, -2 , 1}
{-3, -1, 1}
{-3, 0, 1}

{-2, -1, 2}
{-2, 0, 2}
{-2, 1, 2}

{-1, 0, 3}
{-1, 1, 3}
{-1, 2, 3}

Notice that (set 1, set 9), (set 2, set 8), (set 3, set 7), (set 4, set 6) differ by multiplication of -1. Why this matters ?
if a set has std = d, then set *k has std of d *|k|. So in this case all those similar sets will have same std.

Overall we need to check,
1. {-3, -2, 1}
2. {-3, -1, 1}
3. {-3, 0, 1}
4. {-2, -1, 2}

we need to ensure std of {a,b,c} > {|a|, |b|, |c|}
std{-3, -2, 1} => mean is -1.667, diffs = {-1.33, 0.33, 2.666}, squares = {~1, 0 , ~4} => sum = ~5
std{|-3|, |-2|, |1|} => mean is 2, diffs = {1, 0, -1}, squares = {1, 0, 1} => sum = 2
this works and so does the symmetric one {-1, 2, 3}

check {-3, -1, 1}
mean is -1, diffs = {-2, 0, 2}, squares = {4, 0, 4} => sum = 8
for {|-3|, |-1|, |1|} => means is 5/3 =1.667, diffs = {1.33, -0.66, -0.66}, squares = {} it will be less than 8
this works and so does the symmetric one {-1, 1, 3}

check for {-3, 0, 1}
mean is -0.66, diffs = {2.333, 0.66, 1.66}, squares = {~4, 0, ~1} => sum = ~5
for {|-3|, |0|, |1|} => mean is 1.66, diffs = {1.33, -1.66, -0.66}, squares = {~1, ~1, ~0} => sum = ~2

Actually, I just got an idea that, we can just check the ranges to see
{-3, -2, 1} => range = 4
{|-3|, |-2|, |1|} => range = 2
so yes.

based on the principle that both measure spreads and so if range{set A} > range{set B}, then std{set A} > std{setB}

So all the sets actually work.

Now
1. Possible. Take set 1 => {-3, -2, 1}
2. Possible. Take set 2 => {-3, -1, 1}
3. Possible. Take set 2 => mode of {|-3|, |-1|, |1|} is 1

Ans: E

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19 Dec 2025, 17:56

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Suppose, a,b,c are (-3,-2,1) which meets the S.D of (a,b,c) is greater than S.D of Mod of (a,b.c)

The median is -2, so, (1) could be true
The product is prime number is also true, example (-2,-1,1)
The mode of Mod of (a,b,c) is 1 ==> also true

Answer: E

Bunuel

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19 Dec 2025, 19:38

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a, b, and c are distinct non-zero integers, and the standard deviation of {a, b, c} is greater than the standard deviation of {|a|, |b|, |c|}. If the range of {a, b, c} is 4, which of the following could be true?

I. The median of {a, b, c} is -2
II. The product of a, b, and c is a prime number
III. The mode of {|a|, |b|, |c|} is 1

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III

Quick solution :-

Let numbers be -3 , -1 , 1

Mean= -1
So SD = 2

but SD of l-3l l-1l & l1l is less than 2
Also range is 1-(-3) =4
So numbers -3,-1,& 1 stands valid.

option 2 stands valid
option 3 stands valid.

answer choices does not have II & III only
So Option E is correct

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19 Dec 2025, 19:39

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Since asking for could be true even a single set satisfying option would suffice. let a,b,c be -3, -2 & 1 respectively. After taking mod of a, b, c spread reduces hence standard deviation will be less than origional nos.
Option 1 satisfied, option 3 satisfied.
For any prime no only two disting factors would be there 1 & the no itself so option two would never be true for three non Zero, distinct integers. Ans-D

Bunuel

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19 Dec 2025, 21:25

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Good question , and all statistics concepts are covered in this. Given a ,b ,c are distinct non-zero integers and the standard deviation of {a,b,c} is greater than the standard deviation of {|a| , |b| , |c|}. The range of {a,b,c} is also provided as 4. We need to find which of the following options could be/may be true.

We will create cases for each options and will check whether it's coming as truth. If it is then it may be true.

I . Let's take a = -3 , b= -2 , c =1. This makes range as 4 and all are distinct integers. |a| = 3 , |b|= 2 , |c|=1 . SD of |a| , |b| , |c| is also smaller than a ,b ,c as deviation is less. The median of a ,b ,c is also coming as -2 . This shows that I. may be true.

II. Let's take a = -3 , b = -1 , c = 1 . This makes range as 4 and all are distinct integers. |a| = 3 , |b| = 1 , |c| = 1 . SD of |a| , |b| , |c| is smaller than a , b, , c . The product of a ,b ,c, is 3 which is a prime number. So , II. could be true.

III. Take previous set of integers . Mode of |a| , |b| , |c| is 1. So III. could be true.

So all options could be true. So answer is E.

Bunuel

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19 Dec 2025, 22:18

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distinct non-zero integers.... and SD of {a, b, c} > SD of {|a|, |b|, |c|} -> to be something has to do with -ve integers [SD -> spread -> if entirely +ve or entirely -ve will have same spread hence must have some +ve and some -ve]
for simplicity assume a < b < c (since they are variable, order does not matter)

Given - range 4 = c - a => possible values of c,a = {3, -1}, {2, -2}, {1, -3}
acc to question Important: could be true [not must be true]

(correct) I -> median-> -2 -> possible in one of our values -> a,b,c => -3, -2, 1
II -> since -ve -> prime won't be possible
III -> mode = 1 -> max times 1 => given distinct numbers hence not possible.

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19 Dec 2025, 22:26

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I. Take {-3,-2,1}. The std dev of this set is still greater than {3,2,1}. TRUE
II. For the product to be a prime number, there should be two out of three numbers take the value 1. But a,b,c are distinct non-zero integers, so this is impossible. FALSE
III. Take {-3,-2,1} again. This satisfies 2 conditions and the mode of {3,2,1} is 1.

Answer: D

Bunuel

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19 Dec 2025, 22:33

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Given :
1. a,b,c are distinct non zero numbers
2. SD of {a,b,c} > SD of {|a| , |b|, |c|}
3. The range of {a,b,c} is 4
Statement 1:
If median is -2, then the number would look like:
{-3,-2,1} range is 1-(-3) = 4 and {|3|,|2|,|1|} --> Not possible

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19 Dec 2025, 23:16

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Given Conditions:
1) a, b, c are distinct & none are 0.
2) Range is 4 which is the difference of largest and smallest number.
3) standard deviation of {a,b,c} is greater tha {|a|, |b|, |c|} which only occurs if set of a,b,c contains both positive and negative numbers.
Given the range of 4, the possible values of a,b,c can be {-3,b,1}, {-1,b,3} and {-2,b,2} where I am assuming a is smallest and c is largest value.

Now, check for I,II and III statements.
I. Median of {a,b,c} is -2 which can be possible with set 1 {-3,b,1} if we take b as -2.
II. The product of a,b,c is prime number which is also possibel with set {-3,b,1} if we consider b=-1. (-3)*(-1)*(1) = 3 Prime no.
III. Mode of |a|,|b|,|c| is 1. This is also possible if with sets {-3,-1,1} and {-1,1,3}.

All three statements could be true. (E)

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19 Dec 2025, 23:37

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This is a could be true question. So all we need to do is prove the statements for one set of values.

Given the median of modulus of the values are lower than the actual values, we can deduce that the lowest number in the set is negative. Also the range should be 4

Statement 1 for a set (-3,-2,1) satisfies our constraints and the median is -2. Therefore, possible

Statement 2 for a set (-1, 1, 3) satisfies our constraints and the product is -3 and is prime. Therefore, possible

Statement 3 for the same set (-1,1,3) when taken under modulus gives the mode as 1. Therefore, possible

Option E is the answer.

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20 Dec 2025, 00:18

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Option 1 Median is -2
{-3,-2,1} Range=4
SD=2.236
SD of{3,2,}=0.816
So true

Option 2 The product of a,b,c is prime no.
{-3,-1,1} Range=4
SD=1.63
{3,1,1}
SD=0.94
So option 2 is also true.

Option 3 {|a|,|b|,|c|} Mode is 1
{-3,-1,1}, Range=4, SD=1.63
{3,1,1} Mode is 1, SD=0.94
So option 3 is also true

Answer is E

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20 Dec 2025, 00:39

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Bunuel

Given c = a + 4

Statement I
The median of {a, b, c} is -2

Consider Set {-3, -2, 1}
{|a|, |b|, |c|} = {1, 2, 3}

We need not calculate sd, whichever has sum of differences with mean high that has higher sd

{-3,-2,1}, mean = -4/3 = -1.333

| -3 + 1.333 | + |-2 + 1.333| + 2.333 = 4.666

{3,2,1} mean = 2
|3-2| + |1-2| = 2

So {-3, -2, 1} has higher sd than {3,2,1}
Could be true.
Eliminate B,C

Statement II: The product of a, b, and c is a prime number
{ -1, 1, 3} product is 3 which is a prime

{ -1, 1, 3}, mean = 1
sd = |-1-1| + |3-1| = 4

{1,1,3}, mean = 5/3 = 1.666

sd = |1-1.66| + |1-1.66| + |3-1.66| = 2.66

Could be true.

Eliminate A,D

You can choose answer here only,
Correct Answer: E

Statement III:
The mode of {|a|, |b|, |c|} is 1, same example above works for this.
Could be true.

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20 Dec 2025, 01:01

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The problem states that the Standard Deviation (the spread) of {a,b,c}> standard deviation of their absolute values {|a|, |b| , |c|}.
This Means for the spread to get smaller after taking absolute values some numbers must be negative .
Why ? If you take a negative number like -3 and a positive number like 1 , they are 4 units apart.If you take their absolute values , they become 3 and 1 , which are only 2 units apart..
Let's test a set of numbers : {-3,-1,1}

ST 1 : Median of {a,b,c } is -2
Set = { -3,-2,1}
Check Range = 1-(-3) = 4
SD spread becomes smaller when absolute = {3,2,1}
This could be true ✅

ST 2: The product of a,b,c is a prime number
Let's use our first set { -3,-1,1}
( -3)x(-1) x 1 = 3
This could be true ✅

ST 3 : The mode of {|a| , |b| , |c|} is 1
Absolute values of {|-3|,|-1|, |1|}={ 3,1,1} , Since 1 appears twice . Could be True ✅

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20 Dec 2025, 01:10

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Bunuel

a,b,c are DISTINCT NON ZERO integers.

Non zero integers include both positive as well as negative integers.

The standard deviation (S.D) of set (a,b,c) is GREATER THAN standard deviation (S.D) of set ( mod a, mod b, mod c).

If the set includes negative numbers, then the dispersion is more. S.D is also more.

This means we have negative numbers amongst a,b,c.

Given range of a,b,c = 4

Thus, c = a+4

The possible combinations of numbers with positive and negative numbers are:

Case 1: (-3,b,1)

Case 2: (-2,b,2)

Case 3: (-1,b,3)

we need to find which could be true ? NOT MUST BE TRUE, even a single case satisfying this criteria is acceptable.

I. The median of {a,b,c} is -2.

Out of the three cases, case 1 holds good.

(-3,-2,1) median = -2

S.D of (-3,-2,1) > S.D of (3,2,1)

Satisfies.

II. The product of a,b,c is prime number.

since, the set is a combination of positive and negative integers, odd number of negative integers will yield -ve value.

product which involves 2 is not considered. Because, 2 being the only even prime, any multiplication will yield an even number.

Case 2 is eliminated.

case 1: (-3,-1,1) OR case 3: (-1,1,3) is acceptable.

product equals 3 ( a prime number)

satisifes.

III. Mode of ( mod a, mod b, mod c) is 1.

Is there any single case which satisfies this criteria.

Mode is the frequency of repetition of integers.

case 1: (-3,-1,1) , when represented in mod we get (3,1,1)

case 3: (-1,1,3), when represented in mod we get (1,1,3)

Satisfies.

option E (I, II, and III)

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20 Dec 2025, 01:28

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Bunuel

Range is 4

Please note it is asking for "could" condition, so we need one example which is true

1 : median is -2

Let's take example set {-3,-2,1}

Range is 4

and S.D of {-3,-2,1} > S.D of { 3,2,1} ------ True

2 : Product is prime

Ex : { -3,-1,1}

Range is 4

and S.D is greater here too -------- True

3 : Mode is 1

We can use same Ex - { -3,-1,1} --- > {3, 1,1} ----- True

Answer is E

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20 Dec 2025, 05:07

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Bunuel