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How many values can the integer p=|x+3|-|x-3| assume?

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Bunuel

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06 Jan 2017, 07:14

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Salvetor

What is the level of this question. Above 650?

Yes. You can check the difficulty level from the original post: how-many-values-can-the-integer-p-x-3-x-3-assume-152859.html#p1225434

Bunuel

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13 Feb 2017, 06:53

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Zarrolou

How many values can the integer \(p=|x+3|-|x-3|\) assume?

A)6
B)7
C)13
D)12
E)Infinite values

My own question, as always any feedback is appreciated
Click here for the OE.

This is a copy of the following GMAT Club question: https://gmatclub.com/forum/if-y-x-5-x-5 ... 73626.html

goforgmat

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06 Apr 2018, 06:37

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Zarrolou

Good job arpanpatnaik, vinaymimani!

Official explanation

The function \(|x+3|-|x-3|\) for values \(\geq{}3\) equals \(6\), and for values \(\leq{}-3\) equals \(-6\)
For the middle values it follows the equation \(2x\) (as the users above correctly say)

However there is a quicker way to get to the answer than counting the possible values.

Its upper limit is \(6\), its lower limit is \(-6\) and the function \(2x\) is monotonic and increasing (and continuous), so will assume all values between 6 and -6 included.
(This is not theory necessary for the GMAT, but if notice the fact that \(2x\) must pass for all values between 6 and -6, you can save time)

So the values that the integer p can assume are \(-6,-5,...,0,...,5,6\) TOT=\(13\)

The correct answer is C

For for clarity, below there is the graph of \(|x+3|-|x-3|\) that will make my explanation more clear.

Can you please explain how you drew the graph with two variables. sort of struggling with that.

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17 Sep 2019, 22:09

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missed the last step saying p is an integer... but this is a nice question!

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03 Nov 2019, 09:25

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Zarrolou

How many values can the integer \(p=|x+3|-|x-3|\) assume?

A)6
B)7
C)13
D)12
E)Infinite values

My own question, as always any feedback is appreciated
Click here for the OE.

Asked: How many values can the integer \(p=|x+3|-|x-3|\) assume?

Region 1: x<-3
p = -x-3 - (3-x) = -x - 3 - 3 + x = -6

Region 2: -3<=x<=3
p = x+3 - (3-x) = x + 3 - 3 +x = 2x
p can take integer values between -6 and 6
p = {-6,,-5,-4,-3,-2,-1,0,1,2,3,4,5,6}

Region 3: x>3
p = x+ 3 - (x-3) = x + 3 - x + 3 = 6

Combining results of all regions: -
p = {-6,,-5,-4,-3,-2,-1,0,1,2,3,4,5,6}
p takes 13 different values

IMO C

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04 Nov 2020, 04:54

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excuse me guys and Zarrolou, I don't think i know how to do this type of question. Where topic should i be look into?
many thanks

Jaya6

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11 Jan 2022, 08:26

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Hi,
why am I getting 4 scenarios?
x+3-x+3= 6
x+3+x-3= 2x
-x-3-x+3=-2x
-x-3+x-3= -6

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