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Intern  B
Joined: 03 Feb 2019
Posts: 1
Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?  [#permalink]

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Solve the question for x=3. You'll see that the answer is wrong...
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?  [#permalink]

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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|?  [#permalink]

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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

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|.

Hope it's clear.

Perhaps, I m missing something

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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Joined: 02 Sep 2009
Posts: 58424
Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?  [#permalink]

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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

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|.

Hope it's clear.

Perhaps, I m missing something

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.

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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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?  [#permalink]

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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

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|.

Hope it's clear.

Perhaps, I m missing something

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.

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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Joined: 13 Sep 2018
Posts: 12
Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?  [#permalink]

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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

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|.

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|?   [#permalink] 12 Mar 2019, 05:52

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