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

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by NandishSS » Tue Aug 30, 2016 6:34 am
If y=|x+5|−|x−5| then y can take how many integer values?

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

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by GMATGuruNY » Tue Aug 30, 2016 6:40 am
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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by NandishSS » Tue Aug 30, 2016 6:50 am
GMATGuruNY wrote: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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Is it applicable for all these type of problems?
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by GMATGuruNY » Tue Aug 30, 2016 7:12 am
NandishSS wrote:
GMATGuruNY wrote: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.
You are too fast too furious :-) Just Awesome

Is it applicable for all these type of problems?
y = |x ± a| - |x ± b|.
Given this expression, we can determine the range for y by plugging in the two critical points and solving.

Another example:
y = |x-2| - |x+10|.

Here, the critical points are 2 and and -10.
Plugging x=2 into y = |x-2| - |x+10|, we get:
y = |2-2| - |2+10| = -12.
Plugging x=-10 into y = |x-2| - |x+10|, we get:
y = |-10-2| - |-10+10| = 12.
Thus:
-12 ≤ y ≤ 12.
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by [email protected] » Tue Aug 30, 2016 10:59 am
Hi NandishSS,

Every question that you'll face on the GMAT is based on some combination of rules/formulas/patterns. In some cases, you'll immediately be able to spot the pattern or define the necessary formula/rules that you'll need to use to get to the correct answer. If none of that is immediately obvious to you, then you have to be ready to do some 'brute force' work to define what the pattern is (as opposed to just staring at the screen and hoping that something will come to you).

Here, we're asked how many integer values are possible from the given equation. From the answer choices, we know that there are at least five integers, but no more than 21 integers. Given the rather simple math involved, how long would it actually take you to find them all? And how quickly would you be able to spot the patterns involved? To start, try plugging in X=0, X=1, X=2. What pattern do you notice? What if you try negative integers? The prompt does NOT state that X has to be an integer, so what else could you TEST that would make Y an integer? With a few examples, you could probably define the pattern and then stop working.

Much of the math that you'll have to do to get through the Quant section isn't too tough, which is why the Quant section isn't really a 'math test.' It's a critical thinking test that requires lots of little math steps. You should expect to 'play around' with a bunch of questions on Test Day though; you'll be amazed how quickly you can solve many of them once you just get in the habit of working to find the patterns involved.

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by Matt@VeritasPrep » Thu Sep 01, 2016 5:27 pm
You could also think about this conceptually. |a - b| is the distance between a and b, and |a + b| is the distance between a and -b. So

|x + 5| = distance between x and -5

|x - 5| = distance between x and 5

We want the difference in those distances, or |x + 5| - |x - 5|. Since the difference between 5 and -5 is 10, the difference between |x + 5| and |x - 5| can be anything from -10 to 10, but nothing beyond that. (This isn't simply true for integers: we can get any real number from -10 to 10, but no real number outside of that range.)

To see why, imagine x at either end of the range. If x = -5, then we have 0 - 10, or -10. If x = 5, then we have 10 - 0, or 10. With 5 > x > -5, we can get any value between -10 and 10. If x > 5, then the difference simply becomes the difference between 5 and -5, or 10, and if x < -5, then the difference is simply -5 - 5, or -10, since the only relevant distance now is the distance from -5 to 5.

There are 21 integers in the set {-10, -9, ..., 0, ..., 9, 10}, so that's our pick.
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