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How many values of x are there such that x is an integer and
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21 Nov 2019, 10:41
21 Nov 2019, 10:41
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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
Kudos for the right answer and solution.
A. One
B. Two
C. Three
D. Four
E. Five
Kudos for the right answer and solution.
Moderator
Joined: 02 Jan 2020
Status:GRE Quant Tutor
Posts: 1115
Given Kudos: 9
Location: India
Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
GPA: 2.8
WE:Engineering (Computer Software)
How many values of x are there such that x is an integer and
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07 Jun 2020, 03:43
07 Jun 2020, 03:43
\(|3x — 2| \) is an absolute value of 3x-2
Now, 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
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
So, there are 5 values of x which are possible
x = -1,0,1,2,3
So, Answer will be E
Hope it helps!
Watch the following video to learn how to Solve Inequality + Absolute value Problems
Now, 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
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
So, there are 5 values of x which are possible
x = -1,0,1,2,3
So, Answer will be E
Hope it helps!
Watch the following video to learn how to Solve Inequality + Absolute value Problems
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: How many values of x are there such that x is an integer and
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22 Feb 2021, 05:22
22 Feb 2021, 05:22
Carcass wrote:
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
Kudos for the right answer and solution.
A. One
B. Two
C. Three
D. Four
E. Five
Kudos for the right answer and solution.
---------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
