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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

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..
_________________

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

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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

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

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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

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

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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

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

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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

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

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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

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

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09 Jan 2017, 12:04

1

This post received 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

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

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09 Jan 2017, 12:19

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

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

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09 Jan 2017, 12:24

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