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M30-21

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M30-21 [#permalink]

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Official 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.

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.


Answer: E
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Re: M30-21 [#permalink]

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New post 02 Dec 2014, 21:34
when -5<x< 5

Why do you multiply I X-5 I by -1 ? ? I am trying to understand the logic behind this calculation
When -5<x<5, then |x+5|=x+5 and |x-5|=-(x-5)=5-x.

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rsamant wrote:
when -5<x< 5

Why do you multiply I X-5 I by -1 ? ? I am trying to understand the logic behind this calculation
When -5<x<5, then |x+5|=x+5 and |x-5|=-(x-5)=5-x.


Absolute value properties:

When \(x \le 0\) then \(|x|=-x\), or more generally when \(\text{some expression} \le 0\) then \(|\text{some expression}| = -(\text{some expression})\). For example: \(|-5|=5=-(-5)\);

When \(x \ge 0\) then \(|x|=x\), or more generally when \(\text{some expression} \ge 0\) then \(|\text{some expression}| = \text{some expression}\). For example: \(|5|=5\).


Therefore, when -5 < x < 5, then x + 5 > 0 and x - 5 < 0, thus |x + 5| = x + 5 and |x - 5| = -(x - 5) = 5 - x.

Theory on Abolute Values: math-absolute-value-modulus-86462.html
Absolute value tips: absolute-value-tips-and-hints-175002.html

DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hard set on Abolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html


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Re: M30-21 [#permalink]

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New post 17 Dec 2014, 22:00
Can you please explain that in condition -5<x<5 , how we calulated that which side should have +ve and which side should have -ve.
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Re: M30-21 [#permalink]

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New post 20 Nov 2015, 08:52
Hello Bunuel,

Could you please explain the solution in some other way??

Thanks

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Re: M30-21 [#permalink]

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New post 02 Oct 2016, 14:14
Bunuel, I fell for the trap and assumed x was an integer as well.

If you can add, "x does not have to be an integer" in the explanation, I think it will help.

Thanks for a great question!

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Re: M30-21 [#permalink]

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New post 23 Oct 2016, 06:27
Bunuel wrote:
Official 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.

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.


Answer: E



Dear Bunuel,

Are the ranges correctly defined , shouldn't the ranges be

For critical points [-5,5]

the solution range would be

1) x< -5
2) -5 <= x < 5
3) x>=5

Thanks

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Re: M30-21 [#permalink]

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New post 23 Oct 2016, 06:41
rt1601 wrote:
Bunuel wrote:
Official 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.

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.


Answer: E



Dear Bunuel,

Are the ranges correctly defined , shouldn't the ranges be

For critical points [-5,5]

the solution range would be

1) x< -5
2) -5 <= x < 5
3) x>=5

Thanks


You have the same ranges - it does not matter in which range you include = sign.
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Re: M30-21 [#permalink]

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New post 13 Jan 2017, 08:50
I always get a bit lost with regards to >= versus >.

Could someone please clarify the following: if I have |X-5| which of the following is correct:

X<=-5 --> -X + 5

X<5 --> -X + 5

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Re: M30-21 [#permalink]

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New post 19 Jul 2017, 21:42
hi brunel
when x>5, y = 10
when x<5, y = -10
when -5<x<5, y= 2x...These 3 steps are clear to me.But my concern is as per your solution you have taken the following range:

Therefore for this range −10<(y=2x)<10.- still not clear with this part as why we are taking value from -10 to 10.What is the logic behind this

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sidagar wrote:
hi brunel
when x>5, y = 10
when x<5, y = -10
when -5<x<5, y= 2x...These 3 steps are clear to me.But my concern is as per your solution you have taken the following range:

Therefore for this range −10<(y=2x)<10.- still not clear with this part as why we are taking value from -10 to 10.What is the logic behind this


For \(-5 < x < 5\) range and we got \(y=2x\). Multiply \(-5 < x < 5\) by 2 to get \(-10 < 2x < 10\). Substitute 2x with y to get \(-10 < y< 10\).
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Re M30-21 [#permalink]

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New post 26 Aug 2017, 07:55
I think this is a poor-quality question and I don't agree with the explanation. possible mistake? in the range of -5<x<5, according to the explanation, y can get 19 values.
however, if we assign x= 4,3,2,1,0,-1,-2,-3,-4 it results return exactly 9 options. respectively y=2x as mentioned.

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New post 26 Aug 2017, 07:59
gleshem wrote:
I think this is a poor-quality question and I don't agree with the explanation. possible mistake? in the range of -5<x<5, according to the explanation, y can get 19 values.
however, if we assign x= 4,3,2,1,0,-1,-2,-3,-4 it results return exactly 9 options. respectively y=2x as mentioned.


You should read the question more carefully. We are not told that x is an integer so there are more integer values of y possible when x is not integer.
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M30-21 [#permalink]

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New post 29 Aug 2017, 15:36
This is a min/max and count of set problem in disguise.

For convenience, let |x+5| = a and |x-5| = b, so
y = a-b, where a≥0 and b≥0 (modulus' result can't be less than 0)

y is minimized when a=0, and maximized when b=0, so set a and b to these numbers:
a = |x+5| = 0 --> x1=-5
b = |x-5| = 0 --> x2=5

Sub x1 & x2 into the original question for the min/max of y
Min(y) = |-5+5| - |-5-5| = -10
Max(y) = |5+5| - |5-5| = 10

Thus the subset of y containing only integers has 10 - (-10) + 1 = 21 members in it

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Re M30-21 [#permalink]

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New post 18 Nov 2017, 01:29
I think this is a high-quality question and I agree with explanation. How do we get to know about the range which we have to check?

Like
x<-5
-5<x<5
and
x>5

Please explain @bunnel

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New post 18 Nov 2017, 01:54
jasanisanket24 wrote:
I think this is a high-quality question and I agree with explanation. How do we get to know about the range which we have to check?

Like
x<-5
-5<x<5
and
x>5

Please explain @bunnel


This is called key point/check point/transition point approach and is used to solve absolute value questions. Please check the links below:

10. Absolute Value



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.
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Re: M30-21   [#permalink] 18 Nov 2017, 01:54
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