Last visit was: 21 Apr 2026, 02:14

It is currently 21 Apr 2026, 02:14

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

Apr 16
Sneak Peek: $200 Off GMAT Prep
12:00 PM EDT
-
11:59 PM EDT
You’re first to know: Save $200 on GMAT prep starting tomorrow, 4/13. Get 6 official practice tests and study with real questions. Be ready to save!
Apr 26
Score High on the GMAT with Target Test Prep
10:00 AM EDT
-
11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 27
Try OnDemand GMAT Masterclass
11:00 AM EDT
-
12:00 PM EDT
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
Apr 28
Join - GMAT Day!
08:00 AM PDT
-
11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.

Back to Forum

Create Topic

12 Days of Christmas GMAT Competition 2025 - Day 6: For how many

1 2 3

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 21 Apr 2026

Posts: 109,715

Own Kudos:

810,342

[5]

Given Kudos: 105,795

Products:

Expert

Expert reply

Posts: 109,715

Kudos: 810,342

[5]

22 Dec 2025, 09:00

Kudos

Bookmarks

00:00

Start Timer

Pause Timer

Resume Timer

Show Answer

Hide

Show

History

Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

85% (hard)

Question Stats:

68% (02:00) correct

32% (02:44) wrong

based on 203 sessions

History

Date

Time

Result

Not Attempted Yet

For how many integer values of k is k^4 + 2k < 2k^3 + k^2?

A. 0
B. 1
C. 2
D. 4
E. More than 4

M46-16

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

Show Answer

Official Answer

Most Helpful Reply

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 21 Apr 2026

Posts: 109,715

Own Kudos:

810,342

Given Kudos: 105,795

Products:

Expert

Expert reply

Posts: 109,715

Kudos: 810,342

22 Dec 2025, 10:00

Kudos

Bookmarks

Bunuel

For how many integer values of k is k^4 + 2k < 2k^3 + k^2?

A. 0
B. 1
C. 2
D. 4
E. More than 4

GMAT Club Official Explanation:

k^4 + 2k < 2k^3 + k^2

k^4 - 2k^3 - k^2 + 2k < 0

k(k^3 - 2k^2 - k + 2) < 0

k(k^2(k - 2) - (k - 2)) < 0

k(k - 2)(k^2 - 1) < 0

k(k - 2)(k - 1)(k + 1) < 0

The transition points are $k = -1$, $k = 0$, $k = 1$, and $k = 2$ (these are the values of k at which a factor changes sign). This gives us five ranges:

$k < -1$
$-1 < k < 0$
$0 < k < 1$
$1 < k < 2$
$k > 2$

Next, test an extreme value for k: if k is a large enough number, say 100, then all factors will be positive, resulting in a positive value for the whole expression. Therefore, when k > 2, the expression is positive. Now, here’s the trick: since the expression is positive in the 5th range, it will be negative in the 4th range, positive again in the 3rd range, negative in the 2nd range, and positive in the 1st range, following the pattern: +, -, +, -, +.

Thus, the expression is negative for $-1 < k < 0$ and $1 < k < 2$.

Since k must be an integer, there are no integer values satisfying the inequality.

Answer: A.

General Discussion

adityamntr

Joined: 15 Jul 2023

Last visit: 21 Feb 2026

Posts: 111

Own Kudos:

Given Kudos: 13

Location: India

Concentration: General Management, Strategy

Schools: Booth '27 Fuqua '27 Kellog '27

Posts: 111

Kudos: 81

22 Dec 2025, 09:06

Kudos

Bookmarks

rewrite it as
k^4 - 2k^3 < k^2 - 2k
= k^2 (K^2 -2k) < K^2 - 2K
only possible when K^2 < 1
so no intger values will make it true

ManifestDreamMBA

Joined: 17 Sep 2024

Last visit: 21 Feb 2026

Posts: 1,387

Own Kudos:

897

[1]

Given Kudos: 243

Posts: 1,387

Kudos: 897

[1]

22 Dec 2025, 09:26

Kudos

Bookmarks

k^4 + 2k < 2k^3 + k^2
k^4 - k^2 < 2k^3 - 2k
k^2 (k^2 - 1) < 2k (k^2 - 1)
(k^2 - 1) (k^2 - 2k) < 0
(k+1) (k-1) (k-2) k < 0
(k-2) (k-1) k (k+1)< 0

k>2 is always positive
-1<=k<=2 is 0
k< -1 is always positive

There's no scenario where this is a negative value

Answer A

Bunuel

For how many integer values of k is k^4 + 2k < 2k^3 + k^2?

A. 0
B. 1
C. 2
D. 4
E. More than 4

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

Kinshook

Major Poster

Joined: 03 Jun 2019

Last visit: 20 Apr 2026

Posts: 5,985

Own Kudos:

5,855

[1]

Given Kudos: 163

Location: India

Schools: HBS '26 CBS '26 Wharton '26

GMAT 1: 690 Q50 V34 Verified score

WE:Engineering (Transportation)

Products:

Schools: HBS '26 CBS '26 Wharton '26

GMAT 1: 690 Q50 V34 Verified score

Posts: 5,985

Kudos: 5,855

[1]

22 Dec 2025, 09:37

Kudos

Bookmarks

$k^4 - 2k^3 - k^2 + 2k < 0$
$k^3(k-2) - (k-2) <0$
$(k^3-1)(k-2) < 0$
$k(k-1)(k+1)(k-2) <0$

Using wavy line method: -

1 < k < 2
-1 < k < 0

Since there are no integers in these intervals, there are no values of k possible.

IMO A

vikramadityaa

Joined: 28 Jul 2025

Last visit: 23 Dec 2025

Posts: 55

Own Kudos:

[1]

Given Kudos: 1

Posts: 55

Kudos: 41

[1]

22 Dec 2025, 09:39

Kudos

Bookmarks

Bunuel

For how many integer values of k is k^4 + 2k < 2k^3 + k^2?

A. 0
B. 1
C. 2
D. 4
E. More than 4

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

k^4+2k<2k^3+k^2
k^4+2k-2k^3-k^2<0
k(k^3-2k^2-k+2)<0
Factor the cubic:
k^3-2k^2-k+2
(k^2-1)(k-2)
(k-1)(k+1)(k-2)
k(k-1)(k+1)(k-2)<0
Critical Ponits: -1;0;1;2
Check Intervals: Negative Possibilities; (-1,0); (1,2)
No integer value is possible in both cases.
Hence, OPTION A.

Archit3110

Major Poster

Joined: 18 Aug 2017

Last visit: 20 Apr 2026

Posts: 8,625

Own Kudos:

5,190

[1]

Given Kudos: 243

Status:You learn more from failure than from success.

Location: India

Concentration: Sustainability, Marketing

GMAT Focus 1: 545 Q79 V79 DI73 Verified score

GMAT Focus 2: 645 Q83 V82 DI81 Verified score

GPA: 4

WE:Marketing (Energy)

Products:

GMAT Focus 2: 645 Q83 V82 DI81 Verified score

Posts: 8,625

Kudos: 5,190

[1]

22 Dec 2025, 09:55

Kudos

Bookmarks

For how many integer values of k is k^4 + 2k < 2k^3 + k^2?

A. 0
B. 1
C. 2
D. 4
E. More than 4

given equation can be simplified

k^4+2k-2k^3-k^2<0

k(k^3-2k^2-k+2)<0

test with k = -2,-1,0,1,2 , 3

we get no integer values where value is meeting the equation given ...

option A ; 0 is correct

forestmayank

Joined: 05 Nov 2025

Last visit: 31 Mar 2026

Posts: 103

Own Kudos:

[1]

Given Kudos: 6

Posts: 103

Kudos: 87

[1]

22 Dec 2025, 09:55

Kudos

Bookmarks

Moving everything to LHS:
K^4 - 2K^3 - K^2 + 2K < 0
K (K^3 - 2K^2 - K + 2) < 0
K [K^2 (K - 2) - 1 (K - 2)] < 0
K [(K^2 - 1) (K - 2)] < 0
K (K - 1)(K + 1)(K - 2) < 0

From the above, K < 0, 1, -1, 2
The valid integer value intervals from the above values of K will be
K < -1
Taking sample value -2 for test check in the equation we get
(-2)(-2-1)(-2+1)(-2-2) < 0
(-2)(-3)(-1)(-4) < 0
24 < 0 which is not valid. Therefore, negative integers are not possible.

Similarly, checking for K < 2, the inequality doesn't hold true.
Hence there are no integer values which satisfy the equation.

Answer 0, Option A

remdelectus

Joined: 01 Sep 2025

Last visit: 21 Apr 2026

Posts: 58

Own Kudos:

[1]

Given Kudos: 3

Posts: 58

Kudos: 48

[1]

22 Dec 2025, 10:04

Kudos

Bookmarks

K^4+2K<2K^3+K^2
K^4-2K^3-K^2+2K<0
K(K-2)(K-1)(K+1)<0
roots K= -1,0,1,2
no integer k satisfying the inequality

=0

Ayeka

Joined: 26 May 2024

Last visit: 20 Apr 2026

Posts: 528

Own Kudos:

402

[1]

Given Kudos: 158

Location: India

Schools: ISB

GMAT Focus 1: 645 Q82 V83 DI80 Verified score

GPA: 4.2

Schools: ISB

GMAT Focus 1: 645 Q82 V83 DI80 Verified score

Posts: 528

Kudos: 402

[1]

22 Dec 2025, 10:07

Kudos

Bookmarks

k^4 -2k^3 -k^2 +2k <0
(k+1)(k-1)(k)(k-2) <0
This equation results in 0 when k=-1,1,0,2 . The equation should be less than 0, so these four numbers do not count. No integer value of strictly follows the above equation...

For 0 values of k, k^4 +2k < 2k^3 +k^2

A

vasu1104

Joined: 10 Feb 2023

Last visit: 21 Apr 2026

Posts: 388

Own Kudos:

233

[1]

Given Kudos: 664

Location: Canada

Products:

Posts: 388

Kudos: 233

[1]

22 Dec 2025, 10:28

Kudos

Bookmarks

k^4 + 2k < 2k^3 + k^2
k^4+2k-2k^3-k^2<0
k^2(k^2-1) - 2k(k^2-1) <0
k(k-2)(k-1) (k+1) <0

in this equality there are no integers available
ans is A

Bunuel

For how many integer values of k is k^4 + 2k < 2k^3 + k^2?

A. 0
B. 1
C. 2
D. 4
E. More than 4

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

dp1234

Joined: 15 Nov 2021

Last visit: 20 Apr 2026

Posts: 93

Own Kudos:

[1]

Given Kudos: 166

Products:

Posts: 93

Kudos: 66

[1]

22 Dec 2025, 10:31

Kudos

Bookmarks

Bunuel

For how many integer values of k is k^4 + 2k < 2k^3 + k^2?

A. 0
B. 1
C. 2
D. 4
E. More than 4

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

or kˆ4-2kˆ3-kˆ2+2k <0
or (kˆ2-2k)(Kˆ2-1) < 0

or (k-2)(k-1)k(k+1) <0

note, k can only be integer.
for k=0,1,2 LHS=0 ; For K=3,4,5..... -> LHS = +ve
for k=-1, LHS =0, For k= -2,-3,.... -> LHS = +ve

Thus answer is no value of k. Option A

750rest

Joined: 27 Jul 2022

Last visit: 21 Apr 2026

Posts: 46

Own Kudos:

[1]

Given Kudos: 1,126

Concentration: Marketing, Operations

Products:

Posts: 46

Kudos: 34

[1]

22 Dec 2025, 10:55

Kudos

Bookmarks

k^4 + 2k -2k^3 - k^2 = 0 --> By hit & trial one solution is 1 hence after factorisation we get (k-2)*(k-1)k(K+1) < 0 putting on number line & comparing we get solution as -1 < k < 0 & 1 < k < 2 . Hence no integer in both range. Answer - option A.

Bunuel

For how many integer values of k is k^4 + 2k < 2k^3 + k^2?

A. 0
B. 1
C. 2
D. 4
E. More than 4

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

harishg

Joined: 18 Dec 2018

Last visit: 09 Apr 2026

Posts: 176

Own Kudos:

174

[1]

Given Kudos: 31

GMAT Focus 1: 695 Q88 V84 DI81 Verified score

Products:

GMAT Focus 1: 695 Q88 V84 DI81 Verified score

Posts: 176

Kudos: 174

[1]

22 Dec 2025, 11:01

Kudos

Bookmarks

We can rewrite the given equation as

k^4- 2k^3- k^2+2k <0

k^4- k^2 -2k^3+2k<0

k^2(k^2-1)-2k(k^2-1)<0

k(k-2)(k+1)(k-1)<0

Using the wavy curve theory, we can say that 1<k<2 and -1<k<0 are regions which satisfy the above inequality. Since no integers can be formed from the constraints, there are zero values.

Therefore, Option A

sitrem

Joined: 19 Nov 2025

Last visit: 24 Feb 2026

Posts: 91

Own Kudos:

[1]

Given Kudos: 238

Posts: 91

Kudos: 84

[1]

22 Dec 2025, 11:08

Kudos

Bookmarks

to find the answer i simply solved the inequality and checked how many integers would respect the inequality.

k^4 + 2k < 2k^3 + k^2
k^4 + 2k - 2k^3 - 2k^2 < 0
k^3 * (k - 2) - k * (k - 2) < 0
k (k^2 - 1) (k - 2) < 0
k (k + 1) (k - 1) (k - 2) < 0

check the sign of the product in each interval created by the critical points 0, -1, +1, +2:

the equation is negative in -1 < k < 0 and in 1 < x < 2.

no integral is inside these intervals

Answer A

MANASH94

Joined: 25 Jun 2025

Last visit: 11 Apr 2026

Posts: 88

Own Kudos:

[1]

Given Kudos: 16

Location: India

Schools: IIM IIM ISB

GPA: 2.9

Schools: IIM IIM ISB

Posts: 88

Kudos: 63

[1]

22 Dec 2025, 11:08

Kudos

Bookmarks

Lets bring everything to the same side :
k^4 - 2k^3 - k^2 + 2k < 0....................(i)

k^3(k - 2) - k(k -2) < 0
(k^3 - k)(k - 2)<0
k(k^2 -1)(k-2)<0
(k)(k+1)(k-1)(k-2)<0.
k = -1, 0, 1, 2.

Now using the critical points and, if we do a number line analysis then we get the following:
The intervals are:
k < -1, -1<k<0, 0< k< 1, 1< k < 2, k > 2.
If we take any values and put it in (i) then we get the signs as: + , -, +, - and +. For this we just need to replace any value of k that satisfies the equation.
Based on the signs we see that only two conditions satisfy the requirement:
(0,1) and (1,2) where the signs are negative.
However if we count the number of solutions here in these two intervals we do not have a single integer that satisfy the condition.
Hence the correct ans should be A: 0.

Bunuel

For how many integer values of k is k^4 + 2k < 2k^3 + k^2?

A. 0
B. 1
C. 2
D. 4
E. More than 4

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

AviNFC

Joined: 31 May 2023

Last visit: 10 Apr 2026

Posts: 306

Own Kudos:

366

[1]

Given Kudos: 5

Posts: 306

Kudos: 366

[1]

22 Dec 2025, 11:10

Kudos

Bookmarks

$k^4 - 2k^3 - k^2 +2k$ < 0
$k(k-1)(k-2)(k+1)$ <0

if we plot on number line, the expression gives negative value for 1<k<2 & -<k<0 .. These regions don't contain any integers.
So, answer zero

Ans A

asdf611111

Joined: 18 Jun 2025

Last visit: 21 Apr 2026

Posts: 5

Own Kudos:

[1]

Posts: 5

Kudos: 5

[1]

22 Dec 2025, 11:15

Kudos

Bookmarks

on solving equation taking variables on one side of the equation final equation would be
K(k-2)(k-1)(k+1) < 0
and according to the question since K belongs to an integer
hence crititcal points of the equation are only 0,1,2 and - 1
and considering only integers after 2 and values than -1 would not staify the above inequality
hence 0 is the answer to the question

Dereno

Joined: 22 May 2020

Last visit: 21 Apr 2026

Posts: 1,398

Own Kudos:

1,372

[1]

Given Kudos: 425

Products:

Posts: 1,398

Kudos: 1,372

[1]

22 Dec 2025, 11:23

Kudos

Bookmarks

Bunuel

For how many integer values of k is k^4 + 2k < 2k^3 + k^2?

A. 0
B. 1
C. 2
D. 4
E. More than 4

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

Let’s rearrange the equation

k^4 +2k - 2*k^3 - k^2 <0

k^3 (k-2) -k*(k-2) <0

(k^3 -k) (k-2) <0

k (k^2 -1) (k-2) <0

k * (k+1) * (k-1) * (k-2) < 0

The critical points are : -1, 0, 1, 2

Let’s use the wavy graph concept.

—————(-1)———-(0)———(1)———-(2)—————

Since the question is about <0 , we look for negative regions in the graph.

We don’t need to stress so much on the critical points as we don’t have an equal sign in the inequality.

From negative infinity to -1, the sign is positive,

Between -1 and 0, the sign is negative.

Between 0 and 1, the sign is positive.

Between 1 and 2, the sign is negative.

Between 2 and positive infinity , the sign is positive.

There exists no integer values, between the mentioned critical points , which has the outcome of negative.

Option A : 0

arnab24

Joined: 16 Jan 2024

Last visit: 25 Feb 2026

Posts: 96

Own Kudos:

[1]

Given Kudos: 7

Location: India

Schools: ISB '26

GPA: 8.80

Products:

Schools: ISB '26

Posts: 96

Kudos: 81

[1]

22 Dec 2025, 11:30

Kudos

Bookmarks

K^4+2k < 2k^3 +k^2

k^4+2k-2k^3-k^2 <0

k[k^3-2k^2-k+2] <0

k[k^2(k-2)-(k-2)] <0

k[(k^2-1)(k-2)] <0

k(k+1)(k-1)(k-2)<0

applying wavy-curvy method,

k - (-1,0) U (1,2)

0 integer values of k possible ,

So , Answer is A

Bunuel

For how many integer values of k is k^4 + 2k < 2k^3 + k^2?

A. 0
B. 1
C. 2
D. 4
E. More than 4

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more