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

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Bunuel

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19 Dec 2025, 10:00

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Bunuel

GMAT Club Official Explanation:

Since the standard deviation of {a, b, c} is greater than that of {|a|, |b|, |c|}, {a, b, c} must contain both positive and negative integers. This is because if all numbers were positive or all negative, the standard deviations of {a, b, c} and {|a|, |b|, |c|} would be the same.

This fact, together with the range being 4, implies that the least and greatest integers in the set can be (-3, 1), (-2, 2), or (-1, 3). So we can have the following sets:

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

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

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

Let's evaluate the options:

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

This could be true if we have the set {-3, -2, 1}.

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

This could be true if we have the set {-3, -1, 1}.

III. The mode of {|a|, |b|, |c|} is 1

This could be true if we have the set {-3, -1, 1} or {-1, 1, 3}.

Thus, all three options could be true.

Answer: E.

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paragw

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Updated on: 20 Dec 2025, 23:43

Originally posted by paragw on 19 Dec 2025, 09:11.
Last edited by paragw on 20 Dec 2025, 23:43, edited 1 time in total.

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Let speeds be v1 and v2, lengths be l1 and l2.

Passing the man:
l1/v1 = 40
=> l1=40v1

l2/v2 = 25
=> l2=25v2

passing each other:
(l1+l2 ) / (v1+v2) = 30
Substitute:
(40v1+25v2)/(v1+v2) = 30
40v1 + 25v2 = 30v1+30v2
10v1 = 5v2
v1/v2 = 1:2

Answer: C

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19 Dec 2025, 09:15

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all the cases are possible. below are exampls.

one thing clear i that at least 1 of the numbers shoudl be negative, for theier to be difference in sdev for with and without mod

1) -3 -2 1
2) -3 -1 1
3) -3 -1 1

hece all can be true

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19 Dec 2025, 09:35

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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

a,b,c are distinct non zero integers
(a,b,c) SD is > (lal,lbl,lcl)
range of a,b,c is 4
possible numbers
(-3,-2,-1) ; ( -2,1,2); ( -1,1,3) ; ( -3,-1,1)
#1
median is -2
true at ( -3,-2,-1) sufficient
#2
product of a,b,c is prime
(-3,-1,1) sufficient
#3
mode is of lal,lbl,lcl is 1
(-1,1,3)
sufficient

all conditions 1,2,3 are could be true

OPTION E is correct

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19 Dec 2025, 09:36

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Bunuel

The standard deviation of {a,b,c} is greater than {|a|,|b|,|c|}. This can only happen if taking the absolute value pulls the number closer, i.e., the original set must contain both positive and negative (sign cancellation increase spread).
Give, a,b,c are distinct; Range=4

Statement 1: Hit & Trial
Try: {-3,-2,1}
Satisfies every condition - POSSIBLE.

Statement 2: Hit & Trial
A prime number has exactly two factors, so the product must be (+-)p; That makes two of the numbers to be (+-)1
{-1,1,p}
No clear solution - NOT POSSIBLE.

Statement 3: Hit & Trial
Try: {-1,1,3}
Satisfies every condition - POSSIBLE.

Hence, OPTION D (I & III ONLY)

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19 Dec 2025, 09:39

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$a \neq b \neq c \neq 0$

IF sd of {|a|,|b|,|c|} < sd of {a,b,c} that means some values of a,b,c are negative

Let the sequence be a<b<c
c = a+4

I: -4,-2,0 satisfies all the given constraints and satisfies the median being -2 too. But c = 0, which isn't allowed.
Considering other possibility -3,-2,1, makes this work. Possible

II: For a,b,c to be a prime we need:
a) One of them being the prime and other numbers as 1,1 or -1,-1, which isn't possible as all integers are distinct.
b) -Negative of prime and a -1, say p,-1,1 (where p < 0)
p = 1-4 => p = -3 works -3,-1,1
Possible

III: From above -3,1,1 the other set with abs becomes 3,1,1 with mode 1. Possible

answer E

Bunuel

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19 Dec 2025, 09:40

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as the standard deviation of the set with absolute values is lower, that means, atleast one out of the a, b, c should be negative.
statement I can be true for -> -1, -2, 3
statement II can be true for -> -3,-1,1
statement III can be true for -> -1,1,3
so all can be true - E

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19 Dec 2025, 10:05

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choose(-3,-1,1)
range1-(-3)=4
(a,b,c)=(-3,-1,1)larger than(|a|,|b|,|c|)=(3,1,1)
sig({a,b,c)}>sig({|a|,|b|,|c|})

I. Median is -2: No
II.product abc is prime yes(-3,-1,1)
abc=(-3)(-1)(1)=3
3 is prime
III.Mode (|a|,|b|,|c|) is 1 Yes for(-3,-1,1) mode is 1
E.I,II, and III

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19 Dec 2025, 10:10

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a, b & c are distinct non-zero integers
SD of {a,b,c} > SD of {|a|, |b|, |c|}
Mean a, b & c have mix of positive and negative integers.

Range of {a, b, c} = 4

Which of the following could be true?

I. The median of {a, b, c} is -2
Let {a,b,c} = {-3,-2,1}
Range = 1 - (-3) = 4
SD of {-3,-2,1} > SD of {3,2,1}
COULD BE TRUE

II. The product of a,b & c is a prime number
Let {a,b,c} = {-3,-1,1}
Product of a,b & c = 3
Range = 1 -(-3) = 4
SD of {-3,-1,1} > SD of {3,1,1}
COULD BE TRUE

III. The mode of {|a|,|b|,|c|} = 1
Let {a,b,c} = {-3,-1,1}
Range = 1-(-3) = 4
SD of {-3,-1,1} > SD of {3,1,1}
The mode of {|-3|,|-1|,|1|} = 1
COULD BE TRUE

IMO E

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19 Dec 2025, 10:11

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And. E

Given, SD of {a, b, c}> SD of {|a|, |b|, |c|}. Hence atleast 1 of a,b,c is -ve but not all. If all are -ve then , SD of {a, b, c}= SD of {|a|, |b|, |c|}. And atleast 1 is postive as well.
Given Range is 4 and all are distinct non zero
Two possible sets of range are [-3,1][-1,3]
I. Median -2 . Possible solution (-3,-2,1)
II. Product is prime no. (-3,-1,1) & (-1,1,3) Product= 3
III. Mode = 1 , (-3,-1,1) & (-1,1,3) in | | is 1 .

Hence all three are possible.

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19 Dec 2025, 10:18

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Bunuel

Lets consider the statements:

Statement 1:
Lets consider a set with range 4 and -2
{-3, -2, 1}
Range = 1 - (-3) = 4
median = -2
absolute values = {1, 2, 3}, so SD decreases after absolute values
Statement 1 could be true

Statement 2:
lets consider: {-1, 1, 3}
product = -3 (prime)
range = 3 - (-1) = 4
absolute values = {1, 1, 3}, here the absolute values are not less spread out than the original, in fact the SD does not decrease
Statement 2 is not true

Statement 3:
lets consider: {-1, 1, 3}
absolute values: {1, 1, 3} -> mode = 1
range = 3 - (-1) = 4
SD decreases after absolute values
Statement 3 could be true

IMO Option D

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19 Dec 2025, 10:49

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COULD be true not MUST be true. So even if we find one possible case for each option, that option would be the answer.

I. Let $a = -3$, $b = -2$, $c = 1$

$=>$ median $= -2$; mean $= \frac{(-3-2+1)}{3} = -1.3$; range $= 1-(-3) = 4$

$|a| = 3$, $|b| = 2$, $|c| = 1$; This set is clearly less spaced out so possible.

II. Let $a = -3$, $b = -1$, $c = 1$

range = 4; |a| = 3, |b| = 1, |c| = 1; Less spaced out than a,b,c.

Product of a,b, and c $= -3*-1*1 = 3$ so possible.

III. Same case as above.

|a| = 3, |b| = 1, |c| = 1. Mode = 1. Possible.

The answer is E.

Bunuel

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19 Dec 2025, 11:10

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I think ans is D (1 & 3)
1 could be true when a,b,c are -3,-2,1
2. Could work when a,b,c = -3,-1,1
3. could be true if a,b,c = -2,1,2

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19 Dec 2025, 11:12

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At least one of the integers has to be negative. So, as range is 4, maximum value 3 & minimum is -3

let a=-3,b=-2,c=1
then 3,2,1 has smaller SD than a,b,c. .. Median=-2. Statement I POSSIBLE

let a=-3,b=-1,c=1
then 1,1,3 has smaller SD than a,b,c. Now, a*b*c=3.. A prime number.. Statement II Possible
Again 1,1,3 has mode as 1. So Statement III Possible.

Ans E

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19 Dec 2025, 11:18

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Bunuel

If the stddev is smaller when absolute value is taken, means that some portion of a, b and c are negative. We're looking for what may be true, so is potentially possible.

Possible sets with range 4: {-2, _, 2}, {-3, _, 1}, and {-1, _, 3}

Going through options.

I. You could slot in -2 into the second set above. True.
II. We could have a set that's {-3, -1, 1} which multiples to 3. True.
III. You can just take the example {-3, -1, 1} again for this. True.

Therefore answer is E!

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19 Dec 2025, 11:38

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The correct answer will be I,II & III only. That is E

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19 Dec 2025, 11:58

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Given the difference in SD when absolute value is considered we can safely conclude that some or one of the numbers a, b and c is negative
The range is the highest minus the least of the 4
Could be true means we just need one working example
Statement 1 we could have something like -3, -2, 1 can work hence could be true
Statement 2 We could have something like -1, 1, 3 which works too hence could be true
Statement 3 For this still -1, 1, 3 can work so all could be true
Ans E

Bunuel