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For how many integer values of x, is x – 8 + 5 – x > x + 7? [#permalink]
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09 Jan 2017, 07:05
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For how many integer values of x, is x – 8 + 5 – x > x + 7? [#permalink]
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09 Jan 2017, 09:18
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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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 = 5x x + 7= x7 putting respective values in x – 8 + 5 – x > x + 7 x+8+5x> x7 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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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 8x +x5 > x7==> x>10 so x will 8, 9 two values
IInd case 8x +x5 > x+7 ==> x<4 so x will 5, 6 and 7 three values
IIIrd case 5 to 8 8x +5x > 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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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 = 5x x + 7= x7 putting respective values in x – 8 + 5 – x > x + 7 x+8+5x> x7 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 = 5x



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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, 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 8x +x5> x7==> x>10 so x will 8, 9 two values
IInd case 8x +x5 > x+7 ==> x<4 so x will 5, 6 and 7 three values
IIIrd case 5 to 8 8x +5x > x+7 ==> x<2 so no value similarly x>8 no value
so answer wil be C hi mbaprep2016seems 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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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 8x +x5> x7==> x>10 so x will 8, 9 two values
IInd case 8x +x5 > x+7 ==> x<4 so x will 5, 6 and 7 three values
IIIrd case 5 to 8 8x +5x > x+7 ==> x<2 so no value similarly x>8 no value
so answer wil be C hi mbaprep2016seems 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 5x changes to x5 , 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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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 8x +x5> x7==> x>10 so x will 8, 9 two values
IInd case 8x +x5 > x+7 ==> x<4 so x will 5, 6 and 7 three values
IIIrd case 5 to 8 8x +5x > x+7 ==> x<2 so no value similarly x>8 no value
so answer wil be C hi mbaprep2016seems 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 5x changes to x5 , is there any thing m doing wrong if x<7 then putting any value of x in 5x 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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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



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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, 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 mbaprep2016 Kudos is the way for thanking



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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, 17:07
Case 1: x8+5x > 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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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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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 (x8)(5x)>x(+7) > x<20 (partially valid since portions will be less than 7) 2) 7<x<5 (x8)+(5x)>x+7>x<2 (partially valid since values are within 7<x<5) 3) 5<x<8 (x8)(5x)>x+7 >x<4 (not valid since not within 5<x<8) 4) x>8 (x8)  (5x)>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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06 Nov 2017, 04:24
Even if you let the mode sign as it is and take the absolute values
x8+5x>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]
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26 Dec 2017, 20:17
Can anyone give me more detail explaination please I still dont understand how to solve this



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Re: For how many integer values of x, is x – 8 + 5 – x > x + 7? [#permalink]
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26 Dec 2017, 23:12
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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 STEPBYSTEP 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. Answer: E. 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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26 Dec 2017, 23:30
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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 STEPBYSTEP 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. Answer: E. Hope it's clear. Similar questions to practice: https://gmatclub.com/forum/forhowmany ... 33947.htmlhttps://gmatclub.com/forum/forhowmany ... 33945.htmlhttps://gmatclub.com/forum/ifxisani ... 46499.htmlhttps://gmatclub.com/forum/howmanypos ... 28782.htmlhttps://gmatclub.com/forum/howmanyint ... 10427.htmlhttps://gmatclub.com/forum/howmanyint ... 36656.htmlhttps://gmatclub.com/forum/ifyx5x5 ... 73626.htmlhttps://gmatclub.com/forum/forhowmany ... 31928.htmlhttps://gmatclub.com/forum/forhowmany ... 31929.htmlhttps://gmatclub.com/forum/howmanyval ... 52859.htmlhttps://gmatclub.com/forum/forhowmany ... 31930.htmlhttps://gmatclub.com/forum/forhowmany ... 03766.htmlhttps://gmatclub.com/forum/forwhatval ... 48561.htmlhttps://gmatclub.com/forum/howmanydif ... 75948.htmlhttps://gmatclub.com/forum/howmanyroo ... 79379.htmlhttps://gmatclub.com/forum/howmanyint ... 88429.htmlhttps://gmatclub.com/forum/ify2x2x ... 98566.htmlhttps://gmatclub.com/forum/howmanypos ... 12418.htmlhttps://gmatclub.com/forum/ifxisany ... 27794.html9. Inequalities For more check Ultimate GMAT Quantitative Megathread 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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12 Jan 2018, 14:46
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. Final Answer: GMAT assassins aren't born, they're made, Rich
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