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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 \(y=|x+5|-|x-5|\):

Attachment:

WolframAlpha--yx5-x-5_--2014-07-08_0728.png [ 12.01 KiB | Viewed 13431 times ]

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

Try NEW absolute value DS question.
_________________

Re: If y = |x + 5|-|x - 5|, then y can take how many integer [#permalink]

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01 Jul 2014, 08:14

1

This post received KUDOS

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
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

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.
_________________

Re: If y = |x + 5|-|x - 5|, then y can take how many integer [#permalink]

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01 Jul 2014, 21:55

Plastelin wrote:

proposed solution was wrong. del.

You solution i presume was correct...No need to delete your post just because someone else did it differently. Your post will be helpful for other members in the forum as well as one can learn from different ways of attempting a question...

Good luck
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Re: If y = |x + 5|-|x - 5|, then y can take how many integer [#permalink]

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01 Jul 2014, 23:13

Just test the number. Alway start with one of the integer given in the question. Let start with -5 we get -10. Now test 5 gives us 10. Now test 0 gives us 0. Now test 6 gives us 10 or -6 gives us 9. So the pattern is as we increase beyond 6 or -6 value is constant at 10. So the total is 11 including 0 between -5 and 5.
_________________

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.

Kudos points given to correct solutions above.

Try NEW absolute value DS question.
_________________

Re: If y = |x + 5|-|x - 5|, then y can take how many integer [#permalink]

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12 Sep 2014, 13:58

Bunuel wrote:

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 \(y=|x+5|-|x-5|\):

Attachment:

WolframAlpha--yx5-x-5_--2014-07-08_0728.png

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

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 \(y=|x+5|-|x-5|\):

Attachment:

WolframAlpha--yx5-x-5_--2014-07-08_0728.png

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

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!

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!

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

Hey, thanks for trying to explain, but I still don't understand this logic :[

I do understand that we get two different midpoints, somewhere on the number line, but I don't understand how we can decide on their range. How do I know that the range for for the two values is 5 steps to the right/left of |x+5| and |x-5|

I have been taught that if we have a |x+b| = c. The midpoint is always -b.

I tried applying it here: so |x+5| gives us midpoint -5. and |x-5| gives us midpoint 5.

However - how do I know the actual range? I'm not actually given something like "|x+5| < 2" or whatever.

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"

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".

Amusing! Though on a serious note, try to understand the concept behind it and the HARD question will be bumped up to at least MEDIUM in no time!
_________________

WE: Project Management (Non-Profit and Government)

Re: If y = |x + 5|-|x - 5|, then y can take how many integer [#permalink]

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04 Jul 2015, 13:13

VeritasPrepKarishma wrote:

erikvm wrote:

I have been taught that if we have a |x+b| = c. The midpoint is always -b.

I tried applying it here: so |x+5| gives us midpoint -5. and |x-5| gives us midpoint 5.

However - how do I know the actual range? I'm not actually given something like "|x+5| < 2" or whatever.

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"

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".

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