BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

ABsolute Value Trap

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Senior | Next Rank: 100 Posts
Posts: 44
Joined: Sun Jun 13, 2010 3:07 am
Location: Bahrain
Thanked: 5 times
Followed by:3 members
GMAT Score:720

ABsolute Value Trap

by jainpiyushjain » Sun Mar 31, 2013 6:31 am
Question courtesy Manhattan GMAT CAT

Is x > 0?

(1) |x + 3| = 4x - 3

(2) |x + 1| = 2x - 1

OA to be posted lated

My question is [spoiler] when should we plug the values back in the absolute value equation to ensure the solution? I got this question wrong because I failed to plug the derived value of x back into the original equation and test them. Is it the case always or am I missing some information in the question? [/spoiler]

Thank you for the help
Top
Source: — Data Sufficiency |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Sun Mar 31, 2013 6:52 am
jainpiyushjain wrote:Question courtesy Manhattan GMAT CAT

Is x > 0?

(1) |x + 3| = 4x - 3

(2) |x + 1| = 2x - 1
One option here is to solve each equation for x, but this isn't really necessary.
We know that |something| > 0

Target question: Is x > 0?

Statement 1: |x + 3| = 4x - 3
Since 4x - 3 equals |some expression|, we know that 4x - 3 > 0
Let's solve this inequality for x.
4x - 3 > 0
4x > 3
x > 3/4
If x > 3/4, then x > 0
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: |x + 1| = 2x - 1
Since 2x - 1 equals |some expression|, we know that 2x - 1 > 0
Solve 2x - 1 > 0 for x
2x > 1
x > 1/2
If x > 1/2, then x > 0
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Sun Mar 31, 2013 7:00 am
jainpiyushjain wrote:
My question is when should we plug the values back in the absolute value equation to ensure the solution? I got this question wrong because I failed to plug the derived value of x back into the original equation and test them. Is it the case always or am I missing some information in the question?
Always plug in the results (to ensure the solution does not have extraneous roots) when solving equations involving absolute value.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
User avatar
Senior | Next Rank: 100 Posts
Posts: 44
Joined: Sun Jun 13, 2010 3:07 am
Location: Bahrain
Thanked: 5 times
Followed by:3 members
GMAT Score:720

by jainpiyushjain » Sun Mar 31, 2013 7:02 am
Brent@GMATPrepNow wrote: Always plug in the results (to ensure the solution does not have extraneous roots) when solving equations involving absolute value.

Cheers,
Brent
Noted & thank you
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Sun Mar 31, 2013 7:12 am
jainpiyushjain wrote:Question courtesy Manhattan GMAT CAT

Is x > 0?

(1) |x + 3| = 4x - 3

(2) |x + 1| = 2x - 1
Here's the algebraic approach.

Target question: Is x > 0?

Statement 1: |x + 3| = 4x - 3
We'll use the following rule: If |A| = B then A = B or A = -B
Here, we get: We get:
x + 3 = 4x - 3 or x + 3 = -(4x - 3)
Solve each equation to get: x = 2 or x = 0
IMPORTANT: Plug in both solutions to confirm that there are no extraneous roots.

x = 2
|x + 3| = 4x - 3
|2 + 3| = 4(2) - 3
5 = 5
Perfect, x = 2 is a solution

x = 0
|x + 3| = 4x - 3
|0 + 3| = 4(0) - 3
3 = -3
No good. So x = 0 is an extraneous solution.

Since we now know that x = 2, we can be certain that x > 0
So, statement 1 is SUFFICIENT


Statement 2: |x + 1| = 2x - 1
x + 1 = 2x - 1 or x + 1 = -(2x - 1)
Solve each equation to get: x = 2 or x = 0
Once again, when we plug the two values into the equation, we see that x = 0 is an extraneous root and that x = 2 works.
Since we now know that x = 2, we can be certain that x > 0
So, statement 2 is SUFFICIENT

Answer: D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Senior | Next Rank: 100 Posts
Posts: 33
Joined: Sat Dec 01, 2012 9:37 am
Thanked: 6 times

by mohit_1607 » Sun Mar 31, 2013 7:14 am
jainpiyushjain wrote:Question courtesy Manhattan GMAT CAT

Is x > 0?

(1) |x + 3| = 4x - 3

(2) |x + 1| = 2x - 1

OA to be posted lated

My question is [spoiler] when should we plug the values back in the absolute value equation to ensure the solution? I got this question wrong because I failed to plug the derived value of x back into the original equation and test them. Is it the case always or am I missing some information in the question? [/spoiler]

Thank you for the help
Statement 1:
|x + 3| = 4x - 3
Since, |x + 3| >= 0 therefore, 4x - 3 >= 0 or x >= 3/4
Hence, x > 0

Similarly Statement 2:
|x + 1| = 2x - 1
Since, |x + 1| >= 0 therefore, 2x - 1 >= 0 or x >= 1/2
Hence, x > 0

From each statement alone, we can check if x > 0 therefore answer should be
Option D

I don't think that we need to put the derived value of x back into the original equation.
We simply need to calculate the value of x from each equation and check if the condition is satisfying.
Polish your skills till you achieve perfection.
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 511
Joined: Wed Aug 11, 2010 9:47 am
Location: Delhi, India
Thanked: 344 times
Followed by:86 members

by Anju@Gurome » Sun Mar 31, 2013 7:16 am
jainpiyushjain wrote:...when should we plug the values back in the absolute value equation to ensure the solution? I got this question wrong because I failed to plug the derived value of x back into the original equation and test them. Is it the case always or am I missing some information in the question?
Although this problem does not require solving absolute value equations as pointed by Brent, I'm writing this post to answer your query.

You only have to plug the values back in the absolute value equation to ensure the solution only if you blindly follow algebraic approach.

But if you solve the absolute value problems algebraically using the definition of absolute values, i.e. the proper algebraic approach, you don't have plug back the result in the original equation.

For example, |x + 3| = 4x - 3
  • Now, if x ≥ -3, |x + 3| = (x + 3) ---> (x + 3) = 4x - 3 ---> x = 2 --> This is fine with our initial assumption of x ≥ -3
    And, if x < -3, |x + 3| = -(x + 3) ---> -(x + 3) = 4x - 3 ---> x = 0 --> A contradiction to our initial assumption of x < -3
Hence, only feasible solution is x = 2.
Anju Agarwal
Quant Expert, Gurome

Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.

§ GMAT with Gurome § Admissions with Gurome § Career Advising with Gurome §
Top
Senior | Next Rank: 100 Posts
Posts: 38
Joined: Mon Jun 21, 2010 12:37 pm

by subhakam » Thu Sep 05, 2013 4:58 pm
Anju - I did not understand this following equation/explanation that you stated for the algebraic route |x + 3| = 4x - 3 (question when (x+3) is positive)
Now, if x ≥ -3, |x + 3| = (x + 3)
Why should x be greater than or equal to 3 , it should be x greater than 3
this is because as i understand absolute values |x + 3| when it is positive case means (x+3)> 0 . This is because anything positive means > 0 . if it is = 0 that means it is a non negative (but not positive) value
So when we talk about absolute value do we always consider equal to or greater than zero or just plainly greater than zero?
Many thanks - appreciate the help - your answer will really help me understand this conceptually
Thanks
Subhakam
Top
Senior | Next Rank: 100 Posts
Posts: 38
Joined: Mon Jun 21, 2010 12:37 pm

by subhakam » Thu Sep 05, 2013 5:02 pm
Brent@GMATPrepNow wrote:
jainpiyushjain wrote:Question courtesy Manhattan GMAT CAT

Is x > 0?

(1) |x + 3| = 4x - 3

(2) |x + 1| = 2x - 1
Here's the algebraic approach.

Target question: Is x > 0?

Statement 1: |x + 3| = 4x - 3
We'll use the following rule: If |A| = B then A = B or A = -B
Here, we get: We get:
x + 3 = 4x - 3 or x + 3 = -(4x - 3)
Solve each equation to get: x = 2 or x = 0
IMPORTANT: Plug in both solutions to confirm that there are no extraneous roots.

x = 2
|x + 3| = 4x - 3
|2 + 3| = 4(2) - 3
5 = 5
Perfect, x = 2 is a solution

x = 0
|x + 3| = 4x - 3
|0 + 3| = 4(0) - 3
3 = -3
No good. So x = 0 is an extraneous solution.

Since we now know that x = 2, we can be certain that x > 0
So, statement 1 is SUFFICIENT


Statement 2: |x + 1| = 2x - 1
x + 1 = 2x - 1 or x + 1 = -(2x - 1)
Solve each equation to get: x = 2 or x = 0
Once again, when we plug the two values into the equation, we see that x = 0 is an extraneous root and that x = 2 works.
Since we now know that x = 2, we can be certain that x > 0
So, statement 2 is SUFFICIENT

Answer: D

Cheers,
Brent
Brent - another question for you. As you stated
If |A| = B then A = B or A = -B
What happens if |A| = |B|
Thanks
Subhakam
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Thu Sep 05, 2013 8:26 pm
subhakam wrote:
Brent - another question for you. As you stated
If |A| = B then A = B or A = -B
What happens if |A| = |B|
Thanks
Subhakam
Good question.

If |A| = |B|, then A = B or A = -B
You can test it with some values for A and B to confirm it.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Senior | Next Rank: 100 Posts
Posts: 38
Joined: Mon Jun 21, 2010 12:37 pm

by subhakam » Fri Sep 06, 2013 5:52 am
Thanks Brent- much appreciated
Subhakam
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 234
Joined: Tue Jul 16, 2013 9:00 am
Location: West Virginia
Thanked: 9 times

by Java_85 » Sun Sep 08, 2013 9:52 am
IMO D it is.
using (1) x is either 0 or 2 ==> x>0 not Valid
using (2) x is either 0 or 2 ==> x>0 not Valid
Top
Post new topic Post Reply
• Page 1 of 1