Last visit was: 19 Nov 2025, 07:52

It is currently 19 Nov 2025, 07:52

Toolkit

Practice Tests

Quantitative Questions

Problem Solving (PS)

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Go to My Error Log Learn more

Request Expert Reply

Confirm Cancel

Nov 20
UCR MBA Impact - Live Q&A with UCR AdCom & Student
07:30 AM PST
-
08:30 AM PST
Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
Nov 22
GMAT RC Masterclass
11:00 AM IST
-
01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment
Nov 23
Conquer Algebra on the GMAT Focus Edition
11:00 AM IST
-
01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease.
Nov 25
Try OnDemand GMAT Masterclass
10:00 AM EST
-
11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.

Back to Forum

Create Topic

GMAT Club Olympics 2025 (Day 5): If a < b < c, and c^4 < b^4 < a^4

1 2 3 4 5 6

UfuomaOh

Joined: 14 Sep 2023

Last visit: 17 Nov 2025

Posts: 83

Own Kudos:

Given Kudos: 14

Products:

Posts: 83

Kudos: 50

05 Jul 2025, 00:56

Kudos

Bookmarks

a<b<c and c^4<b^4<a^4

The second expression is raised to power 4 which means that C to the power 4 and B to the power 4 as well as C to the power 4 are all positive.

However when raised to the power of 1 the order of magnitude reverses. So it can only mean that a is less than 0 and C is closest to zero than either a or b. In other words a, b and c are negative.

Option I abc> 0

It means negative x negative x negative is positive. However, this is not true. So this leaves this option as incorrect

Option II a3b5c7<0

given that a, b and c are negative and the power of a, b and c is odd. It means
a3, b5 and c7 are all negative. So negative x negative x negative is negative which is less than 0. so this option is true

Option III
a2b4c6= 0

For this to be true, it means that either a , b or c is 0. But since a<b<c and
c4<b4<a4. It means a cannot be zero, as the order of magnitude reverses for negative numbers.

Lets say a= -16 b=-4 and c=-2

Then of course c4<b4<a4. a cannot be zero because then a would be greater than b and c as we have established that a, b and c are negative numbers. This is also apparent for b. However c is the largest of the value and is closer to zero. so if we assume a=-16, b=-4 and c =0. Then a<b<c and C4<b4<a4. And it is possible that a2b4c6= 0 because -16*-4*0=0

We do not have option II and III. and given that option II is more likely, the answer is thus option II. which is B

Krishnajain1103

Joined: 27 Apr 2025

Last visit: 17 Nov 2025

Posts: 38

Own Kudos:

Given Kudos: 12

Posts: 38

Kudos: 2

05 Jul 2025, 01:08

Kudos

Bookmarks

the answer is B. The other options are wrong . only the second option holds true

Brownieeee

Joined: 24 Nov 2024

Last visit: 04 Aug 2025

Posts: 49

Own Kudos:

Given Kudos: 17

Products:

Posts: 49

Kudos: 31

05 Jul 2025, 02:03

Kudos

Bookmarks

Given: a < b < c and c4 < b4 < a4
All three numbers must be negative!
Because if they were positive, a < b < c would mean a4 < b4 < c4, but we're told c4 < b4 < a4.

Example: a = -6, b = -4, c = -2

Check: -6 < -4 < -2
Check: (-2)4 < (-4)4 < (-6)4 = 16 < 256 < 1296

Now test each option:
I. abc > 0?
negative × negative × negative = negative So abc < 0.
FALSE

II. a3b5c7 < 0?
negative × negative × negative = negative < 0.
TRUE

III. a2b4c6 = 0?
positive × positive × positive = positive ≠ 0.
FALSE

Only II works.
Answer: B

Dipan0506

Joined: 24 May 2021

Last visit: 17 Nov 2025

Posts: 72

Own Kudos:

Given Kudos: 3

Products:

Posts: 72

Kudos: 14

05 Jul 2025, 02:04

Kudos

Bookmarks

If a<b<c
�
<
�
<
�
, and c4<b4<a4

radhikamehta

Joined: 31 Jan 2023

Last visit: 12 Nov 2025

Posts: 9

Own Kudos:

[1]

Given Kudos: 8

Location: India

GMAT Focus 1: 565 Q80 V80 DI74 Verified score

GMAT Focus 2: 615 Q86 V79 DI76 Verified score

GMAT Focus 3: 655 Q85 V81 DI81 Verified score

Posts: 9

Kudos: 10

[1]

05 Jul 2025, 02:59

Kudos

Bookmarks

Bunuel

If $a 0$

II. $a^3b^5c^7 < 0$

III. $a^2b^4c^6 = 0$

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

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

Given, a<b<c & c^4<b^4<a^4

Since it is a "could be true" question, we can check with only 1 solution if the options come out to be true.

I. $abc > 0$

Let a=(-3), b=(-2), c=1
-3<-2<1 & 1^4<(-2)^4<(-3)^4

Further abc = (-3)*(-2)*(1) = 6 > 1. Hence, this could be true.

II. $a^3b^5c^7 < 0$

Let a=(-3), b=(-2), c=(-1) [a<b<c & c^4<b^4<a^4]

a^3*b^5*c^7 = (-3)^3*(-2)^5*(-1)^7 = -64 < 0. Hence, this could be true.

III. $a^2b^4c^6 = 0$

Let a=(-3), b=(-2), c=0 [a<b<c & c^4<b^4<a^4]

a^2b^4c^6 = (-3)^2*(-2)^4*(0)^6 = 0. Hence, this could be true.

Answer: E. I, II, and III.

twinkle2311

Joined: 05 Nov 2021

Last visit: 18 Nov 2025

Posts: 150

Own Kudos:

167

[1]

Given Kudos: 10

Location: India

Concentration: Finance, Real Estate

Schools: ISB '26 Wharton '26

GPA: 9.041

Schools: ISB '26 Wharton '26

Posts: 150

Kudos: 167

[1]

05 Jul 2025, 03:23

Kudos

Bookmarks

We have a |b| > |c|
So a and b have to be negative (they are the smallest numbers but have the biggest sizes). c can be -ve, 0, or +ve.
Let's try some values
• a = –3, b = –2, c = 1. Then abc = (–3)(–2)(1) = 6 > 0, so I can happen.
• a = –3, b = –2, c = –1. Then (a^3*b^5*c^7) = (–3)^3 * (–2)^5 * (–1)^7 = –864 < 0, so II can happen.
• a = –3, b = –2, c = 0. Then (a^2*b^4*c^6) = (–3)^2 * (–2)^4 * (0)^6 = 0, so III can happen.
All three statements are possible, so the answer is E.

SaKVSF16

Joined: 31 May 2024

Last visit: 18 Nov 2025

Posts: 86

Own Kudos:

[1]

Given Kudos: 41

Products:

Posts: 86

Kudos: 79

[1]

05 Jul 2025, 04:11

Kudos

Bookmarks

Constraints : $a 0$

Suppose, $a= -3, b = -2, c = 1$, then $abc = -3*-2*1= 6 > 0$.

then $-3 <-2 < 1 $
and $(-3)^4 = 81, (-2)^4 = 16, 1^4 = 1 ---> 1^4 < (-2)^4 < (-3)^4 $

This satisfies the constraints given to us. So we have an example where $abc>0$ and the constraints are followed.

II. $a^3b^5c^7 < 0$

Suppose,$ a= -3, b = -2, c = -1$, then $a^3b^5c^7 < 0$ will be less than 0 since we have three negative numbers raised to odd powers so -ve * -ve*-ve will give a -ve number ehich is <0
also $-3 <-2 < -1 $
and $(-3)^4 = 81, (-2)^4 = 16, (-1)^4 = 1 ---> (-1)^4 < (-2)^4 < (-3)^4 $

This satisfies the constraints given to us. So we have an example where $a^3b^5c^7 < 0$ and the constraints are followed.

III. $a^2b^4c^6 = 0$

Suppose, $a= -3, b = -2, c = 0$, then $a^2b^4c^6= (-3)^2 * (-2)^4 * (0)^6 = 0 $

also $-3 <-2 < 0 $
and $(-3)^4 = 81, (-2)^4 = 16, 0^4 = 0 ---> [m] 0^4 < (-2)^4 < (-3)^4 $

This satisfies the constraints given to us. So we have an example where $a^2b^4c^6 = 0$ and the constraints are followed.

So all three statements can be satisfied using certain examples..hence answer is E , I, II, and III

SSWTtoKellogg

Joined: 06 Mar 2024

Last visit: 19 Nov 2025

Posts: 57

Own Kudos:

[1]

Given Kudos: 14

Location: India

Schools: Kellogg INSEAD Jan '26 Ross '26 Duke '26 Darden '26 Anderson '26 Tepper '26 Johnson '26 Kelley '26

GMAT Focus 1: 595 Q83 V78 DI77 Verified score

GMAT Focus 2: 645 Q87 V79 DI79 Verified score

GPA: 8.8

Products:

Schools: Kellogg INSEAD Jan '26 Ross '26 Duke '26 Darden '26 Anderson '26 Tepper '26 Johnson '26 Kelley '26

GMAT Focus 2: 645 Q87 V79 DI79 Verified score

Posts: 57

Kudos: 35

[1]

05 Jul 2025, 04:31

Kudos

Bookmarks

Try to draw a number line, keep a, b < 0 and keep the c flexible to check different options.

I. abc>0 : Let's take a=-4, b=-2, c=1 , it satisfies.

II. a3b5c7<0 : Let's take a=-4, b=-2, c=-1 , it satisfies.

III. a2b4c6=0 : Let's take a=-4, b=-2, c=0, it satisfies.

E

crimson_king

Joined: 21 Dec 2023

Last visit: 19 Nov 2025

Posts: 127

Own Kudos:

131

[1]

Given Kudos: 103

GRE 1: Q170 V170 Verified score

Posts: 127

Kudos: 131

[1]

05 Jul 2025, 04:48

Kudos

Bookmarks

If a,b & c are three numbers & if we assume c to be zero & a & b to be negative numbers then a^2*b^4*c^6 can be equal to 0.

A is equal to -6, B is equal to -5 & C is equal to -1. Hence a^3*b^5*c^7 will be negative

A is equal to -6 & B is equal to -5 & C is equal to 1. Hence a*b*c >0.

Hence all three conditions are possible & hence the correct answer is option (E)

Aishna1034

Joined: 21 Feb 2023

Last visit: 19 Nov 2025

Posts: 219

Own Kudos:

Given Kudos: 150

Products:

Posts: 219

Kudos: 64

05 Jul 2025, 05:04

Kudos

Bookmarks

The question asks which of these could be true. Here the relationship is reversed totally if exponents become even, which happens when a no. is negative. Keeping that in mind, statement 1- cant be true, 3 negative no.s cant result into something postive.
Statement 2- can be true , all powers are odd.
statement3- cant be true, as can never be equal to 0. Hence, B

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

GMAT2point0

Joined: 07 May 2025

Last visit: 18 Nov 2025

Posts: 31

Own Kudos:

[1]

Given Kudos: 96

Posts: 31

Kudos: 12

[1]

05 Jul 2025, 05:23

Kudos

Bookmarks

Let's go statement by statement and find if a set of values for a,b,c satisfy it.

i) abc > 0

if a=-10, b=-5, c=1
a<b<c & c^4 < b^4 < a^4
then abc=50, which is greater than zero

ii) a^3 * b^5 * c^7 < 0
if all a,b,c are less than -1, then this is true

iii) a^2 * b^4 * c^6 = 0
if a and b are negative integers, and c is zero, this is true

Rahilgaur

Joined: 24 Jun 2024

Last visit: 19 Nov 2025

Posts: 104

Own Kudos:

[1]

Given Kudos: 45

GMAT Focus 1: 575 Q81 V82 DI72 Verified score

Products:

GMAT Focus 1: 575 Q81 V82 DI72 Verified score

Posts: 104

Kudos: 74

[1]

05 Jul 2025, 05:29

Kudos

Bookmarks

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

For could be true, we need only one case to satisfy the given equation,

a<b<c (1st inequality) => c^4<b^4<a^4 (2nd inequality)

1. abc>0,
either a,b,c all should be positive ==> multiple set of numbers will satisfy the 1st inequality but not second one.

or any 2 out of a,b,c can be positive.. --> case 1 - a= -3, b=-2, c=1 --> 1<16<81 (satisfying second equation too )

- Could be true.

2. a^3b^5c^7 < 0,
either all are negative or one of 3 is negative.

case 1. - a=-3, b=-2, c=-1 --> 1<16<81 -- could be true

3. a^2b^4c^6 = 0

any one is 0 --> a=0 not possible a^4 can't be highest

b=0 --> a=-4 , b=0 , c= 4 (can't be less than or equal to b)

c=0 ---> a=-4, b= -1, c=0 ---> 0<1<16 --could be true.
Hence E is the answer.

SaanjK26

Joined: 08 Oct 2022

Last visit: 18 Nov 2025

Posts: 77

Own Kudos:

Given Kudos: 69

Location: India

Posts: 77

Kudos: 63

05 Jul 2025, 05:44

Kudos

Bookmarks

We are given : a b means c is positive.
So only valid sign combination: a < 0 0; only

Statement II could be true.

Answer: B

LastHero

Joined: 15 Dec 2024

Last visit: 11 Nov 2025

Posts: 134

Own Kudos:

147

[1]

Given Kudos: 1

Posts: 134

Kudos: 147

[1]

05 Jul 2025, 06:03

Kudos

Bookmarks

This is a "could be" question.
The easiest way is to find examples that meet the requirements.

(a, b, c) = (-10, -5, 2) for I.
(a, b, c) = (-10, -5, -2) for II.
(a, b, c) = (-10, -5, 0) for III.

All the three options are possible.

The right answer is E

Fuyu

Joined: 13 Dec 2024

Last visit: 17 Nov 2025

Posts: 36

Own Kudos:

[1]

Given Kudos: 19

Products:

Posts: 36

Kudos: 12

[1]

05 Jul 2025, 06:30

Kudos

Bookmarks

Given:
a < b < c and c^4 < b^4 < a^4

there are three cases where this is true
1. All a, b and c are negative.
2. a and b are negative and c is positive, such that |c| < |b| < |a|
3. a and b are negative and c is 0.

I. abc > 0

Case 2 checks out for this. If c=+ve, a=-ve, b=-ve
(-ve)*(-ve)*(+ve)=(+ve) which is more than 0.
Can be true.

II. a^3b^5c^7 < 0

Case 1 checks out for this.
We know sign of variable with odd power remains as is.
So, (-ve)*(-ve)*(-ve)=(-ve) which is less than 0.
Can be true.

III. a^2b^4c^6 = 0

Case 3 checks out for this. If c=0
Zero raise to power of anything is 0. And anything multiplied by 0 is 0.
Can be true.

Answer: E

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

adityaprateek15

Joined: 26 May 2023

Last visit: 19 Nov 2025

Posts: 268

Own Kudos:

104

[1]

Given Kudos: 309

Location: India

Schools: Ross '26 Tepper '26 McCombs '26

GPA: 2.7

Products:

Schools: Ross '26 Tepper '26 McCombs '26

Posts: 268

Kudos: 104

[1]

05 Jul 2025, 06:37

Kudos

Bookmarks

For could be true questions, try to assume one set of variables that satisfies the given condition and can prove the statement to be true.

For St.1, if a= -4, b= -2 and c= 1

a<b<c and c^4<b^4<a^4

abc = -4*-2*1 = 8>0

St. 1 could be true

St.2, if a= -4, b= -3, c=-1

a<b<c and c^4<b^4<a^4

a^3b^5c^7<0
(-4)^3*(-3)^5*(-1)^7

You don't need to calculate here since all the values have odd power which means a^3b^5c^7<0

St.2 could be true

St.3 if a= -2, b= -1, c= 0

a^2=4, b^4=1, c^6=0

a^2b^4c^6=0

St.3 could be true

All statements could be true.

Option E

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

harshnaicker

Joined: 13 May 2024

Last visit: 25 Sep 2025

Posts: 84

Own Kudos:

[1]

Given Kudos: 35

Posts: 84

Kudos: 60

[1]

05 Jul 2025, 06:52

Kudos

Bookmarks

Let us consider a=-3, b=-2, c=1. Then c^4 < b^4 < a^4.

Case I is possible.
For case III, consider a=-2, b=-1 and c=0. This would satisfy case III
At this point, we can directly choose option E since that is the only option which has both Case I and Case III.
But just to be sure, let us simulate case II as well.

a= -3, b=-2,c =-1
This will make case II possible.

Correct answer is option E

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

ODST117

Joined: 15 Aug 2024

Last visit: 29 Oct 2025

Posts: 173

Own Kudos:

Given Kudos: 149

Posts: 173

Kudos: 85

05 Jul 2025, 06:56

Kudos

Bookmarks

The only way $a<b<c $and $c^4<b^4<a^4$ can be true at the same time is when all three of the variables are negative.

(1). $abc > 0$ -> This can never be true. Product of three negatives cannot be greater than$ 0$.

(2). $a^3b^5c^7 < 0$ -> This has to be true. All variables are raised to the odd power, meaning they retain their sign (which is negative). Product of three negatives is negative.

(3). $a^2b^4c^6 = 0$ -> $a$, $b$, and $c$ are different values. Their product can never be $0$.

Answer - B

Gmat232323

Joined: 22 Sep 2022

Last visit: 18 Nov 2025

Posts: 86

Own Kudos:

[1]

Given Kudos: 143

Location: India

Schools: ISB '27

GPA: 3.04

Products:

Schools: ISB '27

Posts: 86

Kudos: 20

[1]

05 Jul 2025, 07:27

Kudos

Bookmarks

If $a<b<c$ but $c^4<b^4<a^4$ then there could be multiple possible cases. Let's try solving this problem by plugging in numbers:
i) $a=-2$, $b=-1$, $c=0$ then $a^4=16$, $b^4=1$, $c^4=0$
$a^2b^4c^6 =(-2)^2(-1)^4(0)^6= 0$ ; possible

ii) $a=-3$, $b=-2$, $c=1$ then $a^4=81$, $b^4=16$, $c^4=1$
$abc$=$(-3)(-2)(1)=6 > 0$ ; possible

iii) $a=-3$, $b=-2$, $c=-1$
$a^3b^5c^7$ = $(-3)^3(-2)^5(-1)^7$ = $(-27)(-32)(-1)$ = neg*neg*neg = neg < 0 ; possible

Since all three cases could be true, the answer is E.

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

RedYellow

Joined: 28 Jun 2025

Last visit: 09 Nov 2025

Posts: 80

Own Kudos:

[1]

Posts: 80

Kudos: 74

[1]

05 Jul 2025, 07:30

Kudos

Bookmarks

I. If a=-30, b=-20 and c=10 then c^4<b^4<a^4 and abc>0
II. If a=-30, b=-20 and c=-10 then c^4<b^4<a^4 and a^3b^5c^7<0
III. If a=-30, b=-20 and c=0 then c^4<b^4<a^4 and a^2b^4c^6=0

I, II, and III could be true.

Correct answer is E

1 2 3 4 5 6

Moderators:

Bunuel

Math Expert

105389 posts

Krunaal

Tuck School Moderator

805 posts

Prep Toolkit

Announcements

Master in Management, Finance, Business Analytics and more Master’s Programs!

Live Profile Evaluation Chat Session with ARINGO

Wednesday, November 19, 2025
09:00 am PDT | 10:30 pm IST
Location: GMAT Club Chat

Top GMAT Focus Debriefs

GMAT Club Olympics 2025 (Day 5): If a < b < c, and c^4 < b^4 < a^4

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

GMAT Club Olympics 2025 (Day 5): If a < b < c, and c^4 < b^4 < a^4

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards