Last visit was: 28 Apr 2026, 03:13

It is currently 28 Apr 2026, 03:13

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 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

For how many integer values of x, is |x – 6| > |3x + 6|?

1 2

KarishmaB

Tutor

Joined: 16 Oct 2010

Last visit: 27 Apr 2026

Posts: 16,447

Own Kudos:

79,434

[1]

Given Kudos: 485

Location: Pune, India

Expert

Expert reply

Posts: 16,447

Kudos: 79,434

[1]

05 Jun 2020, 05:43

Kudos

Bookmarks

Kritisood

VeritasKarishma

Bunuel

For how many integer values of x, is |x – 6| > |3x + 6|?

(A) 1

(B) 3

(C) 5

(D) 7

(E) Infinite

Think in terms of the definition of absolute values:

|x - 6| > 3*|x + 2|

Distance from 6 is greater than 3 times the distance from -2

-------- (-2) -------- (0) ------------------------------ (6) -------------

At point 0, the distance from 6 will be equal to 3 times the distance from -2. So on the left of 0, the distance from 6 will be greater.

The distance between -2 and 6 is 8. Distance from 6 will be equal to 3 times the distance from -2 at a point 4 units to the left of -2 i.e. -6. To its left, distance from 6 will be less. So between -6 to 0, distance from 6 is greater than 3 times the distance from -2. We have 5 integer values lying between these two.

Answer (C)

Hi VeritasKarishma i was trying to solve this problem by finding the critical points. the critical points I arrived at are -2 and 6. and I'm not able to find the answer through this. the critical points as mentioned by JeffTargetTestPrep are -6 and 0.

could you help me in where I am going wrong in deducing the critical points?

-- to elaborate how I got the critical points
x-6=0
x=6

and 3x+6=0
x=-2

https://gmatclub.com/forum/is-x-1-1-x-1 ... 82478.html
in this question the statement was |x + 1| = 2|x - 1| hence critical points mentioned are -1 and 1; using the same logic i arrived at the critical points here.

JeffTargetTestPrep VeritasKarishma i am really confused about this. could you please assist?

The transition points are -2 and 6 as you derived.
But note that the question has a negative between the terms:

|x – 6| - |3x + 6| > 0

If you are trying to use the same method that we do when the terms are added, you will not get the answer. Here, you need to use the method I have discussed above.

robinbjoern

Joined: 22 Mar 2020

Last visit: 13 Aug 2021

Posts: 25

Own Kudos:

Given Kudos: 102

Posts: 25

Kudos: 11

02 Jul 2020, 07:19

Kudos

Bookmarks

I don't understand following problem:

When I raise both sides to 2 and get RHS to LHS I get the following inequality

(x-6)^2-(3x+6)^2>0
(x-6+3x+6)(x-6-3x-6)>0
(4x)(-2X-12)>0

So values for X will be 0 and -6

BUUT

Number line is

....+++.....-6..........---.............0...........+++..........

And as LHS should be BIGGER than 0 I could take infinitive values for X

Where am I wrong?

GMAT0010

Joined: 17 Sep 2019

Last visit: 08 Dec 2022

Posts: 104

Own Kudos:

Given Kudos: 516

GMAT 1: 710 Q49 V38 Verified score

GMAT 2: 680 Q49 V33 Verified score

Posts: 104

Kudos: 56

06 Jul 2020, 01:50

Kudos

Bookmarks

chetan2u

Bunuel

For how many integer values of x, is |x – 6| > |3x + 6|?

(A) 1

(B) 3

(C) 5

(D) 7

(E) Infinite

Hi sobby,
The actual way would be...

Since both sides are positive, square both sides
\((x-6)^2>(3x+6)^2..........x^2-12x+36>9x^2+36x+36.........8x^2+48x<0\)
8x(x+6)<0.....

For 8x(x+6) to be NEGATIVE, one of 8x and x+6 will be negative and other positive..

If x is positive, 8x will be positive and x+6 will also be positive..
So x should be NEGATIVE..
Then 8x will be negative, and thus x+6 should be positive ..
For that \(x+6>0.....X>-6....\) But x<0.

So range becomes \(0>x>-6\)...
Values are \(-1,-2,-3,-4,-5\)
5 value
C

That's a nice approach,
I would like to add something too!
Here : x^2+6x<0 --> x^2+6x+9<9
Now take x^2+6x+9 = |x+3|^2 <9 .. after simplifying, we get -3 <x+3<3
= -6<x<0
5 integer values.
Hope this helps!

BritLee

Joined: 08 Oct 2021

Last visit: 20 Mar 2023

Posts: 5

Given Kudos: 38

Location: Canada

Posts: 5

Kudos: 0

05 Aug 2022, 06:27

Kudos

Bookmarks

JeffTargetTestPrep

Bunuel

For how many integer values of x, is \(|x – 6| > |3x + 6|\)?

(A) 1

(B) 3

(C) 5

(D) 7

(E) Infinite

To solve an inequality where the absolute value of an expression is greater than (or less than) the absolute value of another expression, we need to find the values that make the absolute values of the two expressions equal to each other. That is, to solve |x – 6| > |3x + 6|, we first need to solve |x – 6| = |3x + 6|. Recall that if |A| = |B|, then either A = B (Case 1) or A = -B (Case 2).

Case 1: When \(x - 6 = 3x + 6\):

\(x - 6 = 3x + 6\)

\(-2x = 12\)

\(x = -6\)

Case 2: When \(x - 6 = -(3x + 6)\):

\(x - 6 = -(3x + 6)\)

\(x - 6 = -3x - 6\)

\(4x = 0\)

\(x = 0\)

These two values, -6 and 0, are the critical values for determining the solution to the given inequality. Now when we place these two values on the number line, we will see that they partition the number line into 3 intervals:

\(x < -6\)

\(-6 < x < 0\)

\(x > 0\)

For each of these intervals, we will pick a value in that interval (for example, for x < -6, we can pick -7) and test whether it satisfies the given inequality. If it does, then EVERY number in that interval will satisfy the inequality. If it doesn’t, then NO numbers in that interval will satisfy the inequality. Recall that our given inequality is |x – 6| > |3x + 6|. Let’s begin testing the intervals.

1) For \(x < -6\), let’s pick x = -7:

|-7 – 6| > |3(-7) + 6| ?

\(|-13| > |-15|\)?

\(13 > 15\) → No!

Thus, no numbers in the interval \(x < -6\)will satisfy \(|x – 6| > |3x + 6|.\)

2) For \(-6 < x < 0\), let’s pick x = -1:

\(|-1 – 6| > |3(-1) + 6|\)?

\(|-7| > |3|\) ?

\(7 > 3\)→ Yes

Thus, every number in the interval \(-6 < x < 0\) will satisfy \(|x – 6| > |3x + 6|.\)

3) For \(x > 0\), let’s pick x = 1:

\(|1 – 6| > |3(1) + 6|\) ?

\(|-5| > |9|\) ?

\(5 > 9\) → No

Thus, no numbers in the interval \(x > 0\) will satisfy \(|x – 6| > |3x + 6|.\)As we can see, the only interval that satisfies \(|x – 6| > |3x + 6|\) is \(-6 < x < 0\). In other words, the solution to \(|x – 6| > |3x + 6|\) is \(6 < x < 0\). In this interval, there are 5 integers: -5, -4, -3, -2, and -1.

Answer: C

Hi, Jeff - thank you for your explanation. For Case 2, why do we not also negate A in finding out the value of x?

bumpbot

Non-Human User

Joined: 09 Sep 2013

Last visit: 04 Jan 2021

Posts: 38,986

Own Kudos:

1,119

Posts: 38,986

Kudos: 1,119

18 Dec 2025, 11:43

Kudos

Bookmarks

Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.

1 2

Moderators:

Bunuel

Math Expert

109948 posts

Krunaal

Tuck School Moderator

852 posts