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For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? 09 Jan 2017, 07:05
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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
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For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? 26 Dec 2017, 23:12
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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

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|.

Answer: E.

Hope it's clear.
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For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? 26 Dec 2017, 23:30
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Bunuel
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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|.

Answer: E.

Hope it's clear.
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9. Inequalities



For more check Ultimate GMAT Quantitative Megathread



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

(A) 1

(B) 3

(C) 5

(D) 7

(E) Infinite
Show more

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..
General Discussion
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For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? 09 Jan 2017, 11:42
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Bunuel
For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?

(A) 1

(B) 3

(C) 5

(D) 7

(E) Infinite
Show more

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|? 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|? 12 Jan 2018, 14:46
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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:
Show Spoiler
E

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For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? 07 Mar 2019, 08:47
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Bunuel
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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|.

Answer: E.

Hope it's clear.
Show more



Perhaps, I m missing something
Please help me out

if we take X=20

then the equation " lx-8l+l5-xl > lx+7l " yields 12 + 15 > 27 , which is not true; hence, E is incorrect.
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For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? 07 Mar 2019, 09:04
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Bunuel
lichting
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|.

Answer: E.

Hope it's clear.



Perhaps, I m missing something
Please help me out

if we take X=20

then the equation " lx-8l+l5-xl > lx+7l " yields 12 + 15 > 27 , which is not true; hence, E is incorrect.
Show more

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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For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? 07 Mar 2019, 11:33
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Bunuel
lichting
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|.

Answer: E.

Hope it's clear.



Perhaps, I m missing something
Please help me out

if we take X=20

then the equation " lx-8l+l5-xl > lx+7l " yields 12 + 15 > 27 , which is not true; hence, E is incorrect.
Show more

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.

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For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? 01 Jul 2020, 04:28
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Bunuel
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|.

Answer: E.

Hope it's clear.
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Hi Bunuel,

I have a question regarding the use of the "<=" or ">=" signs on the ranges.

How do you choose in which of them you will use it? Or it doesn´t matter?

Thank you in advance.
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For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? 01 Jul 2020, 04:36
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Bunuel
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|.

Answer: E.

Hope it's clear.

Hi Bunuel,

I have a question regarding the use of the "<=" or ">=" signs on the ranges.

How do you choose in which of them you will use it? Or it doesn´t matter?

Thank you in advance.
Show more

As long as you include it, it does not matter in which one you do.
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For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? 21 Jun 2021, 13:49
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Hello,
Is there a quicker way to solve this than to check the values of x in each internal?

Is there someway to arrive at the answer by looking at the check points?

Posted from my mobile device
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For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? 21 Jun 2021, 19:53
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Hi rahulms94,

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. By extension, the fastest way to answer it is NOT to look at the 'check points.'

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.

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For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? 22 Aug 2022, 20:06
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I have a question on the last range (where x>8) , how does x fall in the range. Doesn't it mean X>20 but not from 8 to 19?


Bunuel
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 [highlight]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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For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? 22 Aug 2022, 22:48
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ayushimani
I have a question on the last range (where x>8) , how does x fall in the range. Doesn't it mean X>20 but not from 8 to 19?
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8 is a critical point for the expression as x>8 will give some values in the modulus as positive and some as negative. Accordingly, you change the sign while opening the modulus.

Now, when you do that you realise x>20, which is in the range x>8, so our assumption on the start is correct. Only thing is that we get more specific range that matches with our initial assumption.

A simpler example would be
|x|=2
Take x>0, then x=2.
So, the value of x is 2 although we took range as x>0.
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For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? 03 Sep 2024, 03:39
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Bunuel , i approached this question by squaring both sides (LHS is a modulus additions and RHS is a mod, hence both sides are positive)
Upon simplifying, I ended up with
x^2 - 40x + 40 >0
Looking at the equation, I can see that this is true for all large values of x (say x =100)..
Is this a correct approach to mark the answer as infinite.
Is there any shortcoming in following this approach for similar questions ? I understand the explanation you've provided, but that would require me to open mods across 3-4 ranges of x and I fear that can lead to mistakes during the exam.
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For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? 03 Sep 2024, 07:35
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shaurya_gmat
Bunuel , i approached this question by squaring both sides (LHS is a modulus additions and RHS is a mod, hence both sides are positive)
Upon simplifying, I ended up with
x^2 - 40x + 40 >0
Looking at the equation, I can see that this is true for all large values of x (say x =100)..
Is this a correct approach to mark the answer as infinite.
Is there any shortcoming in following this approach for similar questions ? I understand the explanation you've provided, but that would require me to open mods across 3-4 ranges of x and I fear that can lead to mistakes during the exam.
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After squaring you'd end up with:

\(2x^2 - 26x + 2|5 - x|*|x - 8| + 89 > x^2 + 14x + 49\)

\(x^2 - 40x + 40 + 2|5 - x|*|x - 8| > 0\)

So, you won't get x^2 - 40x + 40 > 0, meaning that a single squaring won't eliminate the modulus signs.
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For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? 23 Oct 2024, 09:56
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Bunuel I used the Wavy Curve Method here, out of desperation and against my intuition, and got the right answer. I suspect this is out of pure luck since the problem is additive and not multiplicative.

The method you mention above makes sense to me, but I'm concerned that it may take too much time on the exam. Is this a problem that I should expect to see on exam day? Upon further reflection, though, I suppose the key here is to notice that one of the ranges extends to infinity. Once you reach this point, you can mark the correct answer. Maybe the best protocol is to start with the most extreme ranges, then.
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For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? 31 Oct 2025, 14:25
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