SOLUTION

If \(y=|x+5|-|x-5|\), then \(y\) can take how many integer values?

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

When \(x\leq{-5}\), then \(|x+5|=-(x+5)=-x-5\) and \(|x-5|=-(x-5)=5-x\).
Hence in this case \(y=|x+5|-|x-5|=-x-5-(5-x)=-10\).
1 integer value of \(y\) for this range.

When \(-5<x<5\), then \(|x+5|=x+5\) and \(|x-5|=-(x-5)=5-x\).
Hence in this case \(y=|x+5|-|x-5|=x+5-(5-x)=2x\).
Therefore for this range \(-10<(y=2x)<10\).
19 integer values of \(y\) for this range (from -9 to 9, inclusive).

When \(x\geq{5}\), then \(|x+5|=x+5\) and \(|x-5|=x-5\).
Hence in this case \(y=|x+5|-|x-5|=x+5-(x-5)=10\).
1 integer value of \(y\) for this range.

Total = 1 + 19 + 1 = 21.

Answer: E.

If anyone interested here is a graph of \(y=|x+5|-|x-5|\):



As you can see y is a continuous function from -10 to 10, inclusive.

Try NEW absolute value DS question.


_________________

If \(y=|x+5|-|x-5|\), then \(y\) can take how many integer values?

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

Sol :\(y=|x+5|-|x-5|\), then \(y\)

We see that there is a following range of values

x<-5
-5< x<5
x>5

Now for any value of x<-5 the expression will give the value -10...
Consider x=-6 so we have |-1|-|-11|= 1-11=-10
Consider x=-10, we have |-5|-|-15|= 5-15=-10
Similarly when x>5 the expression will give value of 10

Consider x=7, |12|-|2|=10
X=100,|105|-|95|=10

So we have 2 values of y possible -10, 10

Consider the range -5<x<5...At x=-5 and 5 value is -10 and 10 respectively

Consider x=-4, |-1|-|-9|=-8
x=-3,|-2|-|-8|=-6
x=-2, |-3|-|-7|=-4
Similarly x=-1 will give you different value

Like wise x=0 will give you one value and value of x from 1 to 4 you get different values of y.

So total possible values -10,10,0, + 4 values -1 to-4 and +4 values for range of x 1 to 4= 11

Ans C
_________________

Use the number line for mod questions:

___________-5 ______________0______________5___________

You want the values of y which is the difference between "distance from -5" and "distance from 5".

Anywhere on the left of -5,

<------------------------------------------------
<-------------
___________-5 ______________0______________5___________

The difference between the two distances will always be -10 (taking difference to mean what it does for GMAT)

In between -5 and 5, can y can the value of -9? Sure. Move 0.5 to the right of -5. At x = -4.5, distance from -5 will be .5 and distance from 5 will be 9.5. So y = 0.5 - 9.5 = -9
Note that x needn't be an integer. Only y needs to be an integer.

Similarly, at various points between -5 and 5, y will take all integer values from -9 to 9.

To the right of 5, the difference between the two distances will always be 10.

So values that y can take: -10, -9, ... 0 ... 9, 10 i.e. a total of 21 values.
_________________

Here's a visual way of thinking through this question.

First of all, we see that the given expression for y contains two modulus expressions: |x + 5| and |x - 5|

What do these expressions mean?

|x-5| represents the distance of point x from the point 5 on the number line.

|x+5| can be written as |x - (-5)|. So, |x+5| represents the distance of point x from the point -5 on the number line.

So, let's plot the points 5 and -5 on the number line, and consider the different possible values of x.

Case 1: x < -5

Depicting this case on the number line:



It's clear that in this case, |x-(-5)| - |x-5| = -(distance between points - 5 and 5 on the number line) = -10. The minus sign here comes because the bigger distance (|x-5|) is subtracted from the smaller distance (|x+5|)

Case 2: -5 < = x <= 5

Depicting this case on the number line:



Let's say x = -5. In that case, it's easy to see that |x-(-5)| - |x-5| = -10

If x = -4.5, then |x-(-5)| - |x-5| = 0.5 - 9.5 = -9
.
.
.

If x = 5, then |x-(-5)| - |x-5| = 10

Thus, we see that in Case 2, all integral values between -10 and 10, inclusive, are possible for y.

Case 3: x > 5

Depicting this case on the number line:



It's easy to see that in this case, |x-(-5)| - |x-5| = +10

Combining all cases

We see that the following integral values of y are possible: -10, -9, -8 . . . 8, 9, 10

That is, 21 values in all.

Hope this visual way of thinking through this and other similar absolute value questions was helpful for you!

Japinder
_________________

Dear erikvm

The way to read the equation '|x+b| = c is: "the distance of x from the point -b on the number line is c'. So 2 values of x are possible: one that is c units to the left of -b and the second that is c units to the right of -b.



As you can see, '-b' is indeed the mid-point between the two possible values of x. But note that I didn't start my analysis of |x+b| = c from this result. I started from the fact that |x+b| represents the distance of x from the point -b on the number line.

Let's now come to the question at hand:

Start your analysis by considering that |x+5| represents the distance of x from -5 and |x-5| represents the distance of x from 5.

Try solving the question from here on. Go through the solution I've posted above, and if something is still not clear, I'll be happy to help!

Best Regards

Japinder
_________________

Actually, don't think of it from the mid-point perspective because then it is not useful in many circumstances. The actual logic of absolute value is this:
|x - a| gives you the "distance of x from point a on the number line"

You might want to check out this post to understand this perspective:
http://www.veritasprep.com/blog/2011/01 ... edore-did/

So y = |x+5|−|x−5| gives you the difference between "distance of x from -5" and "distance of x from 5"
So if x = 0, difference between "distance of x from -5" and "distance of x from 5" will be "5" - "5" = 0. This gives you y = 0.

Similarly, you can handle something like this: |x+5| + |x−5|
This is sum of "distance of x from -5" and "distance of x from 5".

Using this method, you can solve this question like this: if-y-x-5-x-5-then-y-can-take-how-many-integer-173626.html#p1379135
_________________

VeritasPrepKarishma

I am thinking the number of integer solutions to y= | x+5| + | x-5|

In this case we will have three cases:
1. When x <-5

here y can will have 2x values.

2. when -5<=x <= +5

then we have only one integral solution.

3. when x > 5

y will have 2x values

So essentially it will have infinite integral values.

Please let me know if my understanding is correct,

In the range -5<x<=5, there are only 11 integers. So, if we are solving for x using -10<=2x<10 and it gives us 20 integers,then only 11 of these are possible within the range -5<x<=5. Can someone please help me understand as to why we are going with 21 as the answer?

Dear Bunuel, If the question changed to \(y=|x+5|+|x-5|\).

When \(x\leq{-5}\), then \(|x+5|=-(x+5)=-x-5\) and \(|x-5|=-(x-5)=5-x\).
Hence in this case \(y=|x+5|+|x-5|=-x-5+(5-x)=-2x\).

When \(-5<x<5\), then \(|x+5|=x+5\) and \(|x-5|=-(x-5)=5-x\).
Hence in this case \(y=|x+5|+|x-5|=x+5+(5-x)=10\).
1 integer values of \(y\) for this range.

When \(x\geq{5}\), then \(|x+5|=x+5\) and \(|x-5|=x-5\).
Hence in this case \(y=|x+5|+|x-5|=x+5+(x-5)=2x\).

Does the answer equal to infinite?

_________________

There are a lot of good mathematical solutions here, but I'll be honest - even as somebody with a 790 GMAT, that's absolutely not what I'd do. I'm not confident in my ability to consistently draw correct graphs for complex absolute value functions like this one. Especially not within 2 minutes.

I would (and I did) actually just start plugging in numbers to see what happens. The biggest answer choice is 21, so there can't possibly be more than 21 values to count. That really isn't that many, if I work quickly. The numbers in the problem aren't tough either (just adding and subtracting 5). I might be able to find some patterns along the way that would speed it up as well. Plus, if I come up with an easy strategy for a problem that I know will eventually work, that means I can safely spend a little extra time on the problem. I know that I'll figure it out, so I might spend as much as 3 minutes on this one if I really had to.

Okay, let's try it:

x = 0: y = 5 - 5 = 0
x = 1: y = 6 - 4 = 2
x = 2: y = 7 - 3 = 4

Interesting - I notice that there's a pattern there. To save time, I'm going to guess that the values keep increasing until I hit x = 5, because |x + 5| will keep getting bigger, and |x - 5| will keep getting smaller. I'm going to assume that x = 3, x = 4, and x = 5 give me the values 6, 8, and 10.

If I go bigger than x = 5, what happens? How about x = 6?
|6 + 5| - |6 - 5| = 10 - interesting - that's a value I already got.
|7 + 5| - |7 - 5| = 10 - same value again. Boring. I'm going to stop testing bigger numbers.

Right now, I've got on my paper: 0, 2, 4, 6, 8, 10.

Now let's test negative numbers, since we're dealing with absolute values.

x = -1: |-1 + 5| - |-1 - 5| = 4 - 6 = -2
x = -2: |-2 + 5| - |-2 - 5| = 3 - 7 = -4

Okay, I'm going to guess that it's symmetrical, and that I can get the values -2, -4, -6, -8, and -10. If I had extra time, I could check that.

I'm almost done, but let me think just a little bit more. Is there any way I could possibly get a different value? Maybe I should try some fractions, like x = 1/2... just to see what happens.

x = 1/2: |1/2 + 5| - |1/2 - 5| = 5.5 - 4.5 = 1

Interesting! That gave me a value I didn't have before. How about x = 1/3?

x = 1/3: |1/3 + 5| - |1/3 - 5| = 5.333 - 4.6666 = not an integer

I'm thinking right now that I have to work with halves, since when I subtract, it has to come out to an integer. So x = 1/2 gives me a value of 1. x = 3/2 gives a value of 3, x = 5/2 gives a value of 5... I can probably also get -1, -3, -5, ... just by dealing with negative numbers instead.

Finally, how big can these numbers get? What about x = 9/2?

x = 9/2: |9/2 + 5| - |9/2 - 5| = 9.5 - 0.5 = 9

x = 11/2: |11/2 + 5| - |11/2 - 5| = 10.5 - 0.5 = 10

x = 13/2: |13/2 + 5| - |13/2 - 5| = 11.5-1.5 = 10

Okay, it gets 'stuck' at 10 again.

Looks like my values are:

0, 2, 4, 6, 8, 10
-2, -4, -6, -8, -10
1, 3, 5, 7, 9
-1, -3, -5, -7, -9

For a total of 21 values.

I did a lot of writing and a lot of arithmetic, but I didn't take that much time on this one. That's because I went with a plan I felt pretty confident about right from the beginning. I took the 'easy way out' - that's what you want to do on the GMAT!
_________________