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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

WOW!! I read your explanation for about 15 minutes to understand it... Nothing against the way you have written... But my basics are weak and reading your explanation and understanding it took some time!!

So the key in putting the values for x is find a break point i.e. At what point y will not be equal to 7. In this case x < -3 and x> 4 y is not equal to 7.

The 2nd thing (to understand and freshen up on my basics) is that the moment value of x < -3 and x > 4, you removed the equation in between the mod signs (given below) and changed the sign and solved the equation.

o If x < -3, y is simplified to:
y = -(x+3) + 4-x
<=> y = 1 - 2*x

o If x > 4, y is simplified to:
y = (x+3) + -(4-x)
<=> y = 2*x - 1

How to figure those spl. points? Should I just have in mind that the numbers that add up to zero are the spl. points?

y=|x+3|+|4-x|

Here, |-3+3|+|4-(-3)|=|0|+|7|=7 and |4+3|+|4-4|=|7|+|0|=7

Thanks Fig

Yes... We need to look for each value of x that makes flip the sign in each absolute value.

For instances o in |x - a| + |x - b| + |x - c| + |x - d| >>> the special points are a, b, c, and d.
o in |x - a|*|x - b| or |(x - a)(x - b)| >>> the special points are a and b

Re: Abs DS from OG [#permalink]
14 Jan 2007, 19:25

Sumithra wrote:

If x and y are integers and y=|x+3|+|4-x| does y=7?

(1)x<4 (2)x>-3

Any basic idea to be kept in mind instead of trying few values?

By plotting the equation on the xy plane one can also arrive at the answer. For the range [-3 , 4]: y = x+3 + 4-x = 7. An essential thing to remember is that, if 2 lines with slopes m and -m (same absolute value, opposite signs) are added, the result is a horizontal line; in this case, the line y = 7. Hope the diagram is illustrative.

Attachments

abs val.jpg [ 7.25 KiB | Viewed 4603 times ]

Last edited by Andr359 on 15 Jan 2007, 08:48, edited 1 time in total.

Re: Abs DS from OG [#permalink]
15 Jan 2007, 17:27

I think the simplest approach would be that since the question has ranges, it needs upper and lower boundary. and 1 & 2 together have upper and lower boundary in this case.

if 2) was x <-3, then answer for this would be E.

Sumithra wrote:

If x and y are integers and y=|x+3|+|4-x| does y=7?

1)x<4 2)x>-3

Any basic idea to be kept in mind instead of trying few values?

Re: Abs DS from OG [#permalink]
15 Jun 2009, 11:32

Since we are talking about absolute values with a + sign between them its pretty obvious that 1) and 2) can never be sufficient by themselves.

What is left is to test if -3<x<4. I did this by plugging in three numbers (just to be sure) in that range. It took me under 60 seconds to solve this one using that method.

Re: Abs DS from OG [#permalink]
14 May 2010, 19:06

@ Fig I've browsed through most of the problems on Absolute Value, and IMHO, you are the most facile & astute solver of questions on Modulus there is. _________________

In a Normal Distribution, only the Average'Stand Out'

Re: If x and y are integers and y=|x+3|+|4-x| does y=7? 1)x<4 [#permalink]
21 Oct 2012, 05:40

Sumithra wrote:

If x and y are integers and y=|x+3|+|4-x| does y=7?

1)x<4 2)x>-3

Any basic idea to be kept in mind instead of trying few values?

Use the meaning of absolute value: |x - a| is the distance between x and a on the number line.

|4 - x| = |x - 4| is the distance between x and 4. |x + 3| = |x - (-3)| is the distance between x and -3.

Since the distance between -3 and 4 is 7 (=4 - (-3)), y = 7 for every x between -3 and 4. Draw the number line, it is easy to understand. Also, we can immediately deduce that neither (1) nor (2) alone is sufficient.

Therefore, answer C. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Wow...I'm still reeling from my HBS admit . Thank you once again to everyone who has helped me through this process. Every year, USNews releases their rankings of...

I had an interesting conversation with a friend this morning, and I realized I need to add a last word on the series of posts on my application process. Five key words:...