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

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

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New post 09 Jan 2017, 07:05
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Kudos [?]: 132809 [0], given: 12378

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

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New post 09 Jan 2017, 09:18
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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..
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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

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

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New post 09 Jan 2017, 11:42
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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

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

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New post 09 Jan 2017, 11:47
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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

so answer wil be C

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

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New post 09 Jan 2017, 11:50
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



please help me in thi s
when x<-7
|x – 8| = -x+8 understood
|5 – x| = 5-x

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

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New post 09 Jan 2017, 11:54
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

so answer wil be C



hi mbaprep2016

seems to reconsider in highlighted part above in your calculation

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

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New post 09 Jan 2017, 12:00
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

so answer wil be C



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

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

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New post 09 Jan 2017, 12:04
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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

so answer wil be C



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!!

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

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New post 09 Jan 2017, 12:19
rohit8865 , honestly you have given a shock to me :shock: ; 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

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

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New post 09 Jan 2017, 12:24
mbaprep2016 wrote:
rohit8865 , honestly you have given a shock to me :shock: ; 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
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Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]

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New post 09 Jan 2017, 17:07
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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For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]

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New post 26 Feb 2017, 21:14
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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

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

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New post 01 Sep 2017, 11:46
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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



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?

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

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New post 06 Nov 2017, 04:24
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

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Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?   [#permalink] 06 Nov 2017, 04:24
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