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Re: For how many integer values of x, is x – 8 + 5 – x > x + 7?
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08 Feb 2019, 09:47
Solve the question for x=3. You'll see that the answer is wrong...



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Re: For how many integer values of x, is x – 8 + 5 – x > x + 7?
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08 Feb 2019, 12:21
Hi merttural11, You're correct that X=3 is NOT a solution to the given inequality. The correct answer does NOT state that EVERY value of X is a solution though  it states that the number of solutions is INFINITE  because as X gets larger and larger (think about X=1000, X=1001, X=1002, etc.), we will clearly end up with an unlimited number of solutions . GMAT assassins aren't born, they're made, Rich
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Re: For how many integer values of x, is x – 8 + 5 – x > x + 7?
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07 Mar 2019, 08:47
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. Perhaps, I m missing something Please help me out if we take X=20 then the equation " lx8l+l5xl > lx+7l " yields 12 + 15 > 27 , which is not true; hence, E is incorrect.



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Re: For how many integer values of x, is x – 8 + 5 – x > x + 7?
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07 Mar 2019, 09:04
Rupesh1Nonly wrote: 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. Perhaps, I m missing something Please help me out if we take X=20 then the equation " lx8l+l5xl > lx+7l " yields 12 + 15 > 27 , which is not true; hence, E is incorrect. Check the red parts in the solution. For ALL value of x which are LESS than 2 or MORE than 20, x – 8 + 5 – x > x + 7 is true. It wont' hold true for x = 20 (only for x < 2 and x > 20). Since there are infinitely many integers LESS than 2 or MORE than 20, then the answer is E. Hope it's clear.
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Re: For how many integer values of x, is x – 8 + 5 – x > x + 7?
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07 Mar 2019, 11:33
Rupesh1Nonly wrote: 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. Perhaps, I m missing something Please help me out if we take X=20 then the equation " lx8l+l5xl > lx+7l " yields 12 + 15 > 27 , which is not true; hence, E is incorrect. Hi Rupesh1Nonly, I think that you might be confusing "an infinite number of values" with "every number possible is a correct value" You've proven that not every value fits the given inequality. Consider what happens when X becomes really large though (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. GMAT assassins aren't born, they're made, Rich
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Re: For how many integer values of x, is x – 8 + 5 – x > x + 7?
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12 Mar 2019, 05:52
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. Bunuel how do you assign the sign to the different modulus .Kindly shade more light i always get confused on what should be negative and what should be positive




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