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# For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?

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Math Expert
Joined: 02 Sep 2009
Posts: 43867
For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]

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09 Jan 2017, 06:05
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For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?

(A) 1

(B) 3

(C) 5

(D) 7

(E) Infinite
[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Aug 2009
Posts: 5660
For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]

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09 Jan 2017, 08:18
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Bunuel wrote:
For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?

(A) 1

(B) 3

(C) 5

(D) 7

(E) Infinite

Hi,

If you look at the left side...
When x is between 5 and 8, ans will be constant 3 .. this is less than the right hand side.

But as we move up above 8 OR below 5, increase in left hand side is more than the right hand side..
As there is no restriction on value of x, after a certain point both above 8 and below 5, LHS will be more than RHS..

So infinite values..
E

NOTE although we can easily find at what values of x , the LHS becomes greater than RHS..
This is not required here..
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

BANGALORE/-

Director
Joined: 05 Mar 2015
Posts: 962
Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]

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09 Jan 2017, 10:42
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Bunuel wrote:
For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?

(A) 1

(B) 3

(C) 5

(D) 7

(E) Infinite

just check the first assumption interval when x<-7
|x – 8| = -x+8
|5 – x| = 5-x
|x + 7|= -x-7
putting respective values in |x – 8| + |5 – x| > |x + 7|
-x+8+5-x> -x-7
x<20
but we have assumption x<-7
thus any value of x<-7 is satisfying the inequality
infinite values

Ans E
Manager
Joined: 29 May 2016
Posts: 130
For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]

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09 Jan 2017, 10:47
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i think we 4 cases here
x<-7 all modes negative
x between -7 to +5 in which two modes negative and one positive
x between 5 to 8 , i which two are posotive and one negative
x>8 all modes +ve

when x<-7
8-x +x-5 > -x-7==> x>-10 so x will -8, -9 two values

IInd case
8-x +x-5 > x+7 ==> x<-4 so x will -5, -6 and -7 three values

IIIrd case 5 to 8
8-x +5-x > x+7 ==> x<2 so no value
similarly x>8 no value

Manager
Joined: 29 May 2016
Posts: 130
For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]

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09 Jan 2017, 10:50
rohit8865 wrote:
Bunuel wrote:
For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?

(A) 1

(B) 3

(C) 5

(D) 7

(E) Infinite

just check the first assumption interval when x<-7
|x – 8| = -x+8
|5 – x| = 5-x
|x + 7|= -x-7
putting respective values in |x – 8| + |5 – x| > |x + 7|
-x+8+5-x> -x-7
x<20
but we have assumption x<-7
thus any value of x<-7 is satisfying the inequality
infinite values

Ans E

when x<-7
|x – 8| = -x+8 understood
|5 – x| = 5-x
Director
Joined: 05 Mar 2015
Posts: 962
Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]

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09 Jan 2017, 10:54
mbaprep2016 wrote:
i think we 4 cases here
x<-7 all modes negative
x between -7 to +5 in which two modes negative and one positive
x between 5 to 8 , i which two are posotive and one negative
x>8 all modes +ve

when x<-7
8-x +x-5> -x-7==> x>-10 so x will -8, -9 two values

IInd case
8-x +x-5 > x+7 ==> x<-4 so x will -5, -6 and -7 three values

IIIrd case 5 to 8
8-x +5-x > x+7 ==> x<2 so no value
similarly x>8 no value

hi mbaprep2016

seems to reconsider in highlighted part above in your calculation
Manager
Joined: 29 May 2016
Posts: 130
Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]

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09 Jan 2017, 11:00
rohit8865 wrote:
mbaprep2016 wrote:
i think we 4 cases here
x<-7 all modes negative
x between -7 to +5 in which two modes negative and one positive
x between 5 to 8 , i which two are posotive and one negative
x>8 all modes +ve

when x<-7
8-x +x-5> -x-7==> x>-10 so x will -8, -9 two values

IInd case
8-x +x-5 > x+7 ==> x<-4 so x will -5, -6 and -7 three values

IIIrd case 5 to 8
8-x +5-x > x+7 ==> x<2 so no value
similarly x>8 no value

hi mbaprep2016

seems to reconsider in highlighted part above in your calculation

I am asking the same thing ...

x has three points -7 , +5 and +8

when we are saying x<-7 so |5-x| changes to x-5 , is there any thing m doing wrong
Director
Joined: 05 Mar 2015
Posts: 962
Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]

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09 Jan 2017, 11:04
1
KUDOS
mbaprep2016 wrote:
rohit8865 wrote:
mbaprep2016 wrote:
i think we 4 cases here
x<-7 all modes negative
x between -7 to +5 in which two modes negative and one positive
x between 5 to 8 , i which two are posotive and one negative
x>8 all modes +ve

when x<-7
8-x +x-5> -x-7==> x>-10 so x will -8, -9 two values

IInd case
8-x +x-5 > x+7 ==> x<-4 so x will -5, -6 and -7 three values

IIIrd case 5 to 8
8-x +5-x > x+7 ==> x<2 so no value
similarly x>8 no value

hi mbaprep2016

seems to reconsider in highlighted part above in your calculation

I am asking the same thing ...

x has three points -7 , +5 and +8

when we are saying x<-7 so |5-x| changes to x-5 , is there any thing m doing wrong

if x<-7 then putting any value of x in |5-x| will be |5-(-x)| = |5+x| >0 thus after removing modulus only 5+x remains

hope it helps!!
Manager
Joined: 29 May 2016
Posts: 130
Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]

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09 Jan 2017, 11:19
rohit8865 , honestly you have given a shock to me ; i never deal with mode this way ..... i always multiply complete mode with +ve or -ve sign ... bases on x only , like the one i did for |x – 8|
but what you saying make sense ....

thanks i will check the logic again and will bother you in case any more doubt
Director
Joined: 05 Mar 2015
Posts: 962
Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]

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09 Jan 2017, 11:24
mbaprep2016 wrote:
rohit8865 , honestly you have given a shock to me ; i never deal with mode this way ..... i always multiply complete mode with +ve or -ve sign ... bases on x only , like the one i did for |x – 8|
but what you saying make sense ....

thanks i will check the logic again and will bother you in case any more doubt

mbaprep2016
Kudos is the way for thanking
Senior Manager
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Posts: 272
Location: United States
Concentration: Entrepreneurship, Marketing
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Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]

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09 Jan 2017, 16:07
Case 1:
x-8+5-x > x + 7
x < - 10
Case 2 :
-x + 8 - 5 + x > x + 7
x < - 4
Infinite
E
Is my approach right?
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I welcome analysis on my posts and kudo +1 if helpful. It helps me to improve my craft.Thank you

Director
Joined: 23 Jan 2013
Posts: 601
Schools: Cambridge'16
For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]

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26 Feb 2017, 20:14
1
KUDOS
Number line is a good alternative

--------- -7 ------------------- 5 ----------- 8 ------------->

we need distance from X to 5 plus distance from X to 8 to be more than X to -7.

Solution is all number <2 and >20

E
Intern
Joined: 09 Mar 2017
Posts: 42
For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]

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01 Sep 2017, 10:46
1
KUDOS
Bunuel wrote:
For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?

(A) 1

(B) 3

(C) 5

(D) 7

(E) Infinite

Bunuel- can you post an explanation for this problem when you get a second please? Thanks!

I split open all the modulus but that was very time consuming and I don't fully get the answer (even though I got it right). Here's my logic:

1) x<-7

-(x-8)(5-x)>-x(+7) --> x<20 (partially valid since portions will be less than -7)

2) -7<x<5

-(x-8)+(5-x)>x+7-->x<2 (partially valid since values are within -7<x<5)

3) 5<x<8

-(x-8)-(5-x)>x+7 -->x<-4 (not valid since not within 5<x<8)

4) x>8

(x-8) - (5-x)>x+7 -->x>20 (valid since all values >8).

Thus since x>20, then there are infinite solutions.

Is that reasoning and method correct?
Intern
Joined: 28 Aug 2016
Posts: 16
Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]

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06 Nov 2017, 03:24
Even if you let the mode sign as it is and take the absolute values

x-8+5-x>x+7
= -3>x+7
therefore, -10>x
x= -11,-12...........-3000 etc etc, infinite values in negative
Manager
Joined: 17 Sep 2017
Posts: 61
Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]

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26 Dec 2017, 19:17
Can anyone give me more detail explaination please I still dont understand how to solve this
Math Expert
Joined: 02 Sep 2009
Posts: 43867
Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]

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26 Dec 2017, 22:12
2
KUDOS
Expert's post
4
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lichting wrote:
Can anyone give me more detail explaination please I still dont understand how to solve this

For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?

(A) 1

(B) 3

(C) 5

(D) 7

(E) Infinite

STEP-BY-STEP DETAILED SOLUTION:

There are three transition points at -7, 5 and 8 (transition points are points for which expression in modulus is equal to 0). So, four ranges to check:

1. $$x < -7$$

For this range:
x - 8 is negative, so |x - 8| = -(x - 8).
5 - x is positive, so |5 - x| = 5 - x.
x + 7 is negative, so |x + 7| = -(x + 7).

Thus, |x – 8| + |5 – x| > |x + 7| becomes -(x - 8) + 5 - x > -(x + 7), which in turn is x < 20. Since, we consider x < -7 range, then finally for it we get x < -7.

2. $$-7 \leq x \leq 5$$

For this range:
x - 8 is negative, so |x - 8| = -(x - 8).
5 - x is positive, so |5 - x| = 5 - x.
x + 7 is positive, so |x + 7| = x + 7.

Thus, |x – 8| + |5 – x| > |x + 7| becomes -(x - 8) + 5 - x > x + 7, which in turn is x < 2. Since, we consider $$-7 \leq x \leq 5$$ range, then finally for it we get $$-7 \leq x < 2$$.

3. $$5 < x < 8$$

For this range:
x - 8 is negative, so |x - 8| = -(x - 8).
5 - x is negative, so |5 - x| = -(5 - x).
x + 7 is positive, so |x + 7| = x + 7.

Thus, |x – 8| + |5 – x| > |x + 7| becomes -(x - 8) -(5 - x) > x + 7, which in turn is x < -4. Since, we consider $$5 < x < 8$$ range and x < -4 is out of it, then for this range the inequality does not have a solution (no x from this range satisfied the inequality).

3. $$x \geq 8$$

For this range:
x - 8 is positive, so |x - 8| = x - 8.
5 - x is negative, so |5 - x| = -(5 - x).
x + 7 is positive, so |x + 7| = x + 7.

Thus, |x – 8| + |5 – x| > |x + 7| becomes x - 8 -(5 - x) > x + 7, which in turn is x > 20 (already entirely in the range we consider)

So, we got that |x – 8| + |5 – x| > |x + 7| is true for x < 2 (combined range from cases 1 and 2) and x > 20. Thus, infinitely many integer values of x satisfy |x – 8| + |5 – x| > |x + 7|.

Hope it's clear.
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Math Expert
Joined: 02 Sep 2009
Posts: 43867
Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]

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26 Dec 2017, 22:30
1
KUDOS
Expert's post
1
This post was
BOOKMARKED
Bunuel wrote:
lichting wrote:
Can anyone give me more detail explaination please I still dont understand how to solve this

For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?

(A) 1

(B) 3

(C) 5

(D) 7

(E) Infinite

STEP-BY-STEP DETAILED SOLUTION:

There are three transition points at -7, 5 and 8 (transition points are points for which expression in modulus is equal to 0). So, four ranges to check:

1. $$x < -7$$

For this range:
x - 8 is negative, so |x - 8| = -(x - 8).
5 - x is positive, so |5 - x| = 5 - x.
x + 7 is negative, so |x + 7| = -(x + 7).

Thus, |x – 8| + |5 – x| > |x + 7| becomes -(x - 8) + 5 - x > -(x + 7), which in turn is x < 20. Since, we consider x < -7 range, then finally for it we get x < -7.

2. $$-7 \leq x \leq 5$$

For this range:
x - 8 is negative, so |x - 8| = -(x - 8).
5 - x is positive, so |5 - x| = 5 - x.
x + 7 is positive, so |x + 7| = x + 7.

Thus, |x – 8| + |5 – x| > |x + 7| becomes -(x - 8) + 5 - x > x + 7, which in turn is x < 2. Since, we consider $$-7 \leq x \leq 5$$ range, then finally for it we get $$-7 \leq x < 2$$.

3. $$5 < x < 8$$

For this range:
x - 8 is negative, so |x - 8| = -(x - 8).
5 - x is negative, so |5 - x| = -(5 - x).
x + 7 is positive, so |x + 7| = x + 7.

Thus, |x – 8| + |5 – x| > |x + 7| becomes -(x - 8) -(5 - x) > x + 7, which in turn is x < -4. Since, we consider $$5 < x < 8$$ range and x < -4 is out of it, then for this range the inequality does not have a solution (no x from this range satisfied the inequality).

3. $$x \geq 8$$

For this range:
x - 8 is positive, so |x - 8| = x - 8.
5 - x is negative, so |5 - x| = -(5 - x).
x + 7 is positive, so |x + 7| = x + 7.

Thus, |x – 8| + |5 – x| > |x + 7| becomes x - 8 -(5 - x) > x + 7, which in turn is x > 20 (already entirely in the range we consider)

So, we got that |x – 8| + |5 – x| > |x + 7| is true for x < 2 (combined range from cases 1 and 2) and x > 20. Thus, infinitely many integer values of x satisfy |x – 8| + |5 – x| > |x + 7|.

Hope it's clear.

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Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]

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12 Jan 2018, 13:46
Hi All,

This is a great 'concept' question, meaning that if you recognize the concepts involved, you don't have to do much math to get to the correct answer.

To start, notice that we're adding two absolute values that must sum to a total that is greater than a third individual absolute value. Consider what happens when X becomes really large (for example, X = 1,000). The differences in the values of the individual absolute values becomes negligible, meaning that we're essentially ending up with 1000 + 1000 > 1000. This will occur for every integer value of X at higher and higher values - thus, there's really no reason to try to count them all up - there's an infinite set of solutions.

[Reveal] Spoiler:
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Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?   [#permalink] 12 Jan 2018, 13:46
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