Tricky Modulus question

[Mo2men]

If y=|x+5|−|x−5|, then y can take how many integer values?

A. 5
B. 10
C. 11
D. 20
E. 21

[spoiler]OA: E[/spoiler]

[GMATGuruNY]

If y=|x+5|−|x−5|, then y can take how many integer values?

A. 5
B. 10
C. 11
D. 20
E. 21

The CRITICAL POINTS are where the expressions inside the absolute values are equal to 0.

x+5 = 0 when x=-5.
Substituting x=-5 into y = |x+5|−|x−5|, we get:
y = |-5+5| - |-5-5| = 0-10 = -10.

x-5 = 0 when x=5.
Substituting x=5 into y = |x+5|−|x−5|, we get:
y = |5+5| - |5-5| = 10-0 = 10.

The resulting values in blue indicate the range for y.
Thus, y can be equal to any integer value between -10 and 10, inclusive, implying that there are 21 integer options for y.

The correct answer is E.

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Hi Mo2men,

GMAT questions are often built around a pattern of some kind (and sometimes more than one pattern). If you don't immediately see the pattern, then you might have to do a bit of work to figure out what the pattern is (the good news is that the work is rarely all that difficult, but you might have to do a lot of little calculations).

From the answer choices, we know that there are at least 5, but no more than 21, possible INTEGER values for Y. How hard would it be to find them all?

We can TEST VALUES here. Try TESTing X=0, X=1, X=2, etc. and note the results. What types of patterns are you noticing. What if you use negative values for X? What if you use non-integer values for X (such as 1/2, 3/2, etc.)?

On a certain level, the GMAT will reward you for being a 'worker', so don't be afraid of playing around with a prompt. You'd be amazed how quickly you can figure most of these questions out.

GMAT assassins aren't born, they're made,
Rich

[Matt@VeritasPrep]

To make our lives easier, let's consider three cases:

Case I:
x - 5 < 0 < x + 5

Case II:
0 < x - 5 < x + 5

Case III:
x - 5 < x + 5 < 0

In the first case, we have |x - 5| = -(x - 5) = 5 - x, so |x + 5| - |x - 5| = (x + 5) - (5 - x) = 2x.
Since x - 5 < 0 < x + 5, we have -5 < x < 5. Either x or 2x must be an integer, so our solutions are

x = -4.5, -4, -3.5, -3, ..., 3, 3.5, 4, 4.5

which gives 19 possibilities.

In the second case, we have |x - 5| = x - 5, so |x + 5| - |x - 5| = x + 5 - (x - 5) = 10. This is one more possibility.

In the third case, we have |x + 5| - |x - 5| = -(x + 5) - -(x - 5) = -x -5 - (5 - x) = -10. This is one more possibility.

In all, we have 19 + 1 + 1 => 21 possibilities, so we're set.

Tough Q!