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

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

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New post 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"

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

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New post 09 Jul 2015, 22:17
ankushbagwale wrote:
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"

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,


Yes, when you are adding the two terms, there will no limit on the values y can take. Keep increasing x and y will keep increasing.

When x = 10, y = 20
When x = 20, y = 30
and so on...

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

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New post 31 Aug 2016, 01:42
I'd like to know if i did this correctly.

y=|x+5|−|x−5|

1. y = (x+5) - (x - 5) = 10
2. y = -(x+5) - (-(x-5)) = -10

Thus, -10 <= y <= 10

10 - (-10) = 20 + including integer zero = 21 integers
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Re: If y = |x + 5|-|x - 5|, then y can take how many integer  [#permalink]

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New post 02 Sep 2016, 00:42
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?
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Re: If y = |x + 5|-|x - 5|, then y can take how many integer  [#permalink]

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New post 02 Sep 2016, 01:35
La1yaMalhotra wrote:
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?



Check Bunuel's solution here: if-y-x-5-x-5-then-y-can-take-how-many-integer-173626.html#p1378885

In the range x <= -5, you have 1 solution.
In the range -x < x < 5, you have 19 solutions.
In the range x >= 5, you have 1 solution.

Total 21 solutions.
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If y = |x + 5|-|x - 5|, then y can take how many integer  [#permalink]

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New post 12 Mar 2017, 20:12
1
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.


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

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New post 13 Mar 2017, 00:26
ziyuen wrote:
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.


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?


Yes, in this case y can take infinitely many integer values. You can see this directly from the graph: when x<-5 or when x>5, the value of y increases along with increase in absolute value of x.
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Re: If y = |x + 5|-|x - 5|, then y can take how many integer  [#permalink]

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New post 22 May 2017, 09:28
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.

Try NEW absolute value DS question.


Hi Bunuel, in many a question I have seen this graph view for abs questions. Could you please explain how to draw a graph from the abs values equations.
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Re: If y = |x + 5|-|x - 5|, then y can take how many integer  [#permalink]

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New post 09 Jul 2017, 17:24
Can we say that in order to get the number of integer values that Y can take, we can take test cases where we will see the that maximum Y can be is 10 and minimum it can be is -10.
If we test cases such as x=5, -5, and x=6 , -6 we will see the the graph will not expand nor will it converge.

Therefore the range is 20 but we need to count zero as an integer therefore, the integer values will be 21.

had it been how many integer values will x take in such a way that y will be different, then we would have 11integer values including zero.
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Re: If y = |x + 5|-|x - 5|, then y can take how many integer  [#permalink]

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New post 29 Jul 2017, 14:04
VeritasPrepKarishma wrote:
Bunuel wrote:


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

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

Kudos for a correct solution.



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.



hi mam

Thank you for your wonderful solution to the problem...

Anyway, do you believe - on the basis of number line know-how you shared here - this problem can be solved by solving only "-10<= y <=10" ..?

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

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New post 30 Jul 2017, 06:24
VeritasPrepKarishma wrote:
Bunuel wrote:


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

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

Kudos for a correct solution.



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.



hi

the distance between 2 points is -10 NOT +10 why ...?
is this because the key points for |x+5| and for |x-5| are -5 and +5 respectively ..
please correct me if I am missing something ...

also, if we see the number line we can easily find that y=-9 as long as x= -4.5, but if we want to interact this relationship with the equation given, can it be seen as under ..?

x+5

= -4.5 + 5 = 0.5
and

x - 5

= -4.5 - 5 = -9.5

thus

|x+5| - |x-5|
= .5 - 9.5 = -9
please correct me if I am missing something...

can such problems as this one, however, be solved by recognizing that the number of integer values y can take will be the values that fall within -10<=y<=10 ....?
please correct me if I am missing something ....

anyway, please shed some light on me, giving some lessons on number lines so that I can solve with the help of number lines virtually any problem pertaining to absolute modulus ...

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

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New post 01 Aug 2017, 00:52
ssislam wrote:
VeritasPrepKarishma wrote:
Bunuel wrote:


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

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

Kudos for a correct solution.



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.



hi

the distance between 2 points is -10 NOT +10 why ...?
is this because the key points for |x+5| and for |x-5| are -5 and +5 respectively ..
please correct me if I am missing something ...

also, if we see the number line we can easily find that y=-9 as long as x= -4.5, but if we want to interact this relationship with the equation given, can it be seen as under ..?

x+5

= -4.5 + 5 = 0.5
and

x - 5

= -4.5 - 5 = -9.5

thus

|x+5| - |x-5|
= .5 - 9.5 = -9
please correct me if I am missing something...

can such problems as this one, however, be solved by recognizing that the number of integer values y can take will be the values that fall within -10<=y<=10 ....?
please correct me if I am missing something ....

anyway, please shed some light on me, giving some lessons on number lines so that I can solve with the help of number lines virtually any problem pertaining to absolute modulus ...

thanks in advance ...


Difference between A and B is taken to be A - B in GMAT (At some other places, we take difference to mean Greater - Smaller)
Distance from a point is always positive.

For any point to the left of -5, say x = -6,
Distance from -5 = 1
Distance from 5 = 11

Difference between "distance from -5" and "distance from 5" = 1 - 11 = -10

Yes, if x = -4.5, then y = -9. We need integer values of y which we get. This is to show that y will take values between -10 and 10. That, x could be a decimal (which is acceptable) but we might still get an integer value for y.
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Re: If y = |x + 5|-|x - 5|, then y can take how many integer  [#permalink]

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New post 30 Sep 2018, 08:51
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1
Bunuel wrote:
If \(y=|x+5|-|x-5|\), then \(y\) can take how many integer values?

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


For this question, it's useful to know that |a - b| represents the DISTANCE from point a to point b on the number line.
For example, |3 - 10| = the DISTANCE from 3 to 10 on the number line.
Since |3 - 10| = |-7| = 7, we know that 7 is the distance from 3 to 10 on the number line

So, in this case, |x - 5| = the distance from x to 5
Likewise, since x + 5 = x - (-5), we know that |x + 5| = |x - (-5)|
So, |x + 5| = the distance from x to -5

The equation y = |x + 5| - |x - 5| has two critical points. These are x-values that MINIMIZE the value of |x + 5| and MINIMIZE the value of |x - 5|
First, |x + 5| is minimized when x = -5. That is, when x = -5, |x + 5| = |(-5) + 5| = |0| = 0
Next, |x - 5| is minimized when x = 5. That is, when x = 5, |x - 5| = |5 - 5| = |0| = 0

Let's add these critical points ( x = 5 and x = -5) to the number line.
Image

Notice that these critical points divide the number line into 3 regions.
Let's see what happens to the value of y when x lies in each region.

Let's start with region 1
Image
When x lies in region 1 (e.g., x = -7), notice that the blue bar represents the value of |x - 5|
When x lies in region 1 (e.g., x = -7), notice that the red bar represents the value of |x + 5| (aka |x - (-5)|
Since the distance between -5 and 5 is 10, we can see that the length of the blue bar must be 10 units LONGER than the red bar
So, it must be true that |x + 5| - |x - 5| = -10

Now let's generalize.
For ANY value of x in region 1, the length of the blue bar will be 10 units LONGER than the red bar.
So, for ANY value of x in region 1, |x + 5| - |x - 5| = -10

---------------------------------------------------------------------
Now let's see what's going on in region 3
Image
When x lies in region 3 (e.g., x = 8), notice that the blue bar represents the value of |x - 5|
When x lies in region 3 (e.g., x = 8), notice that the red bar represents the value of |x + 5| (aka |x - (-5)|
Since the distance between -5 and 5 is 10, we can see that the length of the red bar must be 10 units LONGER than the blue bar
So, it must be true that |x + 5| - |x - 5| = 10

To generalize, we can say that, for ANY value of x in region 3, |x + 5| - |x - 5| = 10

IMPORTANT ASIDE: At this point, we've shown that |x + 5| - |x - 5| can equal 10 and -10
---------------------------------------------------------------------
Now onto region 2
Let's see what happens when x = -4
Image
If x = -4, then |x + 5| (the red bar) = 1 and |x - 5| = 9
So, |x + 5| - |x - 5| = 1 - 9 = -8

Now let's see what happens when x = 3.5
[img]https://i.imgur.com/1SyK8Ii.png[/img
If x = 3.5, then |x + 5| = 8.5 and |x - 5| = 1.5
So, |x + 5| - |x - 5| = 8.5 - 1.5 = 7
---------------------------------------------------------------------

CONCLUSION: At this point, we've shown that |x + 5| - |x - 5| can equal 10, -10, and all values BETWEEN 10 and -10

There are 21 INTEGERS, from -10 to 10 inclusive.
So, y can have 21 different integer values

Answer: E

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

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New post 30 Sep 2018, 21:33
I attempted this by taking the maximum and minimum values of y.
The minimum value is possible when |x+5| is least and |x-5| Is maximum. The least possible value of an absolute is 0.
Hence
X+5 = 0
X=-5

Therefore, y = 0 - 10
Y = -10

Similarly , maximum value of y is when |x+5| is maximum and |x-5| is least. The least values of an absolute is 0.
Therefore
y = 10 + 0
y = 10.


Now between the minimum and maximum value of -10 and 10; their are 21 values.
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If y = |x + 5|-|x - 5|, then y can take how many integer  [#permalink]

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New post 19 Dec 2018, 09:30
Bunuel wrote:

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

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

\(?\,\,\,:\,\,\,\# \,\,\,{\rm{integer}}\,\,{\rm{values}}\,\,{\rm{for}}\,\,\,y = \left| {x + 5} \right| - \left| {x - 5} \right|\)


\(\left( {\rm{i}} \right)\,\,\,x \le - 5\,\,\,\,\, \Rightarrow \,\,\,\,\,y = - \left( {x + 5} \right) - \left( {5 - x} \right) = - 10\,\,\,\,\,\, \Rightarrow \,\,\,\,\,1\,\,{\rm{integer}}\)


\(\left( {{\rm{ii}}} \right)\,\,\, - 5 < x \le 5\,\,\,\,\, \Rightarrow \,\,\,\,\,y = \left( {x + 5} \right) - \left( {5 - x} \right) = 2x\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{other}}\,\,10 \cdot 2\,\,{\rm{integers}}\,\,\,\left( * \right)\,\,\,\)

\(\left( * \right)\,\,\,x\,\,{\mathop{\rm int}} \,\,\,\left( { - 4 \le x \le 5\,\,\, \Rightarrow \,\,\,10\,\,{\rm{options}}} \right)\,\,\,\,{\rm{or}}\,\,\,\,\,\left\{ \matrix{
\,x \ne {\mathop{\rm int}} \hfill \cr
\,x = {{{\mathop{\rm int}} } \over 2} \hfill \cr} \right.\,\,\,\,\,\,\,\,\left( { - 9 \le \mathop{\rm int}\,{\rm{odd}}\,\, \le 9\,\,\,\,\, \Rightarrow \,\,\,10\,\,{\rm{options}}} \right)\)


\(\left( {{\rm{iii}}} \right)\,\,\,x > 5\,\,\,\,\, \Rightarrow \,\,\,\,\,y = \left( {x + 5} \right) - \left( {x - 5} \right) = 10\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{already}}\,\,{\rm{obtained}}\,\,{\rm{in}}\,\,\left( {{\rm{ii}}} \right)\)



\(? = 1 + 20 = 21\)



This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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If y = |x + 5|-|x - 5|, then y can take how many integer  [#permalink]

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New post 19 Dec 2018, 10:36
Bunuel wrote:

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

Image

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


Exactly, Bunuel!

This gives us (now explicitly) an immediate alternate solution:


Image


There are 21 integers between -10 and 10, both of them included (see blue interval).

Regards,
Fabio.
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If y = |x + 5|-|x - 5|, then y can take how many integer  [#permalink]

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New post 28 Feb 2019, 19:20
x<-5,
y=-10-------------------> 1 value

-5<=x<5
y=2x
-10<=2x<10-------------------> 20 value

x=>5
y=10-----------> 1 value

Total= 22

Please someone tell me where I am going wrong.
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Re: If y = |x + 5|-|x - 5|, then y can take how many integer  [#permalink]

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New post 15 May 2019, 16:12
Isn't this a GMATClub test question? OR is it actually sourced from GMAT prep?
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Re: If y = |x + 5|-|x - 5|, then y can take how many integer  [#permalink]

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New post 17 May 2019, 03:30
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 \(y=|x+5|-|x-5|\):

Image

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

Try NEW absolute value DS question.

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



Hi,

May i know how you take select the ranges for "x" ,as per my understanding we used to solve the absolute value questions using 2 conditions i.e. x>0 and x<0.
please also explain in which type of questions we need to select different ranges
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If y = |x + 5|-|x - 5|, then y can take how many integer  [#permalink]

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New post 31 Aug 2019, 05:26
If y=|x+5|−|x−5| then y can take how many integer values?
We need to find integer values of y. x can be an integer or non-integer.


if you enter in x as -5,-4.5,-4,-3.5,-3,-2.5,-2,-1.5,-1,-0.5, 0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5 in y=|x+5|−|x−5| then you will get 21 different values of y. Beyond -5 and 5 you will get integer values same as the ones you get by entering the above 21 numbers.

Hence 21.
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If y = |x + 5|-|x - 5|, then y can take how many integer   [#permalink] 31 Aug 2019, 05:26

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