Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2512:30 AM EDT
-01:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
Apr 2601:30 AM EDT
-02:30 AM EDT
Learn how Keshav, a Chartered Accountant, scored an impressive 705 on GMAT in just 30 days with GMATWhiz's expert guidance. In this video, he shares preparation tips and strategies that worked for him, including the mock, time management, and more.
Apr 2610:00 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 2711:00 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 2808:00 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.
May 0111:00 AM PDT
-12:00 PM PDT
Free- full-length test + 15 concept videos + 150+ short videos + study plan!
22 Nov 2019, 03:26
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
71% (01:15) correct
29%
(01:17)
wrong
based on 646
sessions
History
Date
Time
Result
Not Attempted Yet
How many values of x are there such that x is an integer and \(|3x — 2| < 8\)?
A. One
B. Two
C. Three
D. Four
E. Five
A. One
B. Two
C. Three
D. Four
E. Five
22 Nov 2019, 07:48
Kudos
Bookmarks
Bunuel
---------ASIDE---------------------
When solving inequalities involving ABSOLUTE VALUE, there are 2 things you need to know:
Rule #1: If |something| < k, then –k < something < k
Rule #2: If |something| > k, then EITHER something > k OR something < -k
Note: these rules assume that k is positive
------------------------------------
Take: \(|3x — 2| < 8\)
Applying Rule #1, we can write: \(-8 < 3x — 2 < 8\)
Add 2 to all sides to get: \(-6 < 3x < 10\)
Divide all sides by 3 to get: \(-2 < x < \frac{10}{3}\)
In other words: \(-2 < x < 3 \frac{1}{3}\)
So, the INTEGER values of \(x\) that satisfy the above inequality are: \(x = -1, 0, 1, 2\) and \(3\)
There are FIVE such values.
Answer: E
Cheers,
Brent
General Discussion
Updated on: 19 Jan 2025, 05:09
Originally posted by BrushMyQuant on 21 May 2022, 13:23.
Last edited by BrushMyQuant on 19 Jan 2025, 05:09, edited 2 times in total.
Last edited by BrushMyQuant on 19 Jan 2025, 05:09, edited 2 times in total.
Kudos
Bookmarks
↧↧↧ Detailed Video Solution to the Problem ↧↧↧
\(|3x — 2| \) is an absolute value of 3x-2
So, Answer will be E
Hope it helps!
Watch the following video to learn how to Solve Inequality + Absolute value Problems
\(|3x — 2| \) is an absolute value of 3x-2
| Whenever there is an absolute value we need to consider two cases | |
| -Case 1: Whatever is inside the modulus is >= 0 3x-2 >= 0 If the value inside the modulus is non-negative then the modulus can be removed directly As 3x-2 >= 0 which means 3x >=2 or \(x >= \frac{2}{3}\) so \(|3x — 2| \) = 3x - 2 => \(3x — 2 < 8\) => 3x < 8+2 => 3x < 10 => x < \(\frac{10}{3}\) Now we have two conditions for x \(\frac{2}{3} <= x < \frac{10}{3}\) => So, x = 1,2,3 is a SOLUTION | -Case 2: Whatever is inside the modulus is < 0 3x-2 < 0 If the value inside the modulus is negative then the modulus can be removed after multiplying the terms inside the modulus by -1 As 3x-2 >= 0 which means 3x < 2 or \(x < \frac{2}{3}\) so \(|3x — 2| \) = -(3x - 2) => \(-(3x — 2) < 8\) => 3x > 2 - 8 => 3x > -6 => x > \(\frac{-6}{3}\) => x > -2 Now we have two conditions for x \(-2 < x < \frac{2}{3}\) => So, x = -1,0 is a SOLUTION |
So, Answer will be E
Hope it helps!
Watch the following video to learn how to Solve Inequality + Absolute value Problems
