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# M30-21

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Math Expert
Joined: 02 Sep 2009
Posts: 53066

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13 Oct 2014, 05:55
4
19
00:00

Difficulty:

95% (hard)

Question Stats:

38% (01:07) correct 63% (01:47) wrong based on 104 sessions

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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

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Math Expert
Joined: 02 Sep 2009
Posts: 53066

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13 Oct 2014, 05:57
3
10
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 $$y=|x+5|−|x−5|:$$

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

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Joined: 30 Jun 2012
Posts: 13

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02 Dec 2014, 20:34
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.
Math Expert
Joined: 02 Sep 2009
Posts: 53066

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03 Dec 2014, 03:16
2
2
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

Hope it helps.
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Joined: 20 Jan 2014
Posts: 142
Location: India
Concentration: Technology, Marketing

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17 Dec 2014, 21:00
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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Posts: 181
Location: India
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GMAT Date: 11-23-2015
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20 Nov 2015, 07:52
Hello Bunuel,

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

Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 53066

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21 Nov 2015, 01:47
WillGetIt wrote:
Hello Bunuel,

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

Thanks

Check here: if-y-x-5-x-5-then-y-can-take-how-many-integer-173626.html
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Current Student
Joined: 28 Aug 2016
Posts: 88
Concentration: Strategy, General Management

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02 Oct 2016, 13:14
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!
Intern
Joined: 12 Jul 2013
Posts: 7

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23 Oct 2016, 05:27
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 $$y=|x+5|−|x−5|:$$

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

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
Math Expert
Joined: 02 Sep 2009
Posts: 53066

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23 Oct 2016, 05:41
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 $$y=|x+5|−|x−5|:$$

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

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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Intern
Joined: 17 Aug 2016
Posts: 48

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13 Jan 2017, 07:50
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
Intern
Joined: 04 Aug 2014
Posts: 29
GMAT 1: 620 Q44 V31
GMAT 2: 620 Q47 V28
GPA: 3.2

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19 Jul 2017, 20:42
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
Math Expert
Joined: 02 Sep 2009
Posts: 53066

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19 Jul 2017, 20:52
2
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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Joined: 12 Aug 2017
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26 Aug 2017, 06:55
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.
Math Expert
Joined: 02 Sep 2009
Posts: 53066

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26 Aug 2017, 06:59
1
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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Joined: 19 Jun 2017
Posts: 11

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29 Aug 2017, 14:36
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
Intern
Joined: 17 Feb 2017
Posts: 6
Location: India
GMAT 1: 670 Q49 V32
GPA: 3.3

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18 Nov 2017, 00:29
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

Math Expert
Joined: 02 Sep 2009
Posts: 53066

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18 Nov 2017, 00:54
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

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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Joined: 08 Dec 2017
Posts: 1
GMAT 1: 700 Q48 V37

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04 Feb 2018, 19:17
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. How exactly did y=2x equation have its values between -10 and 10? There is nothing mentioned.
Math Expert
Joined: 02 Sep 2009
Posts: 53066

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04 Feb 2018, 19:26
rk85 wrote:
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. How exactly did y=2x equation have its values between -10 and 10? There is nothing mentioned.

This is 700+ question but I assure you it's 100% correct. You can check other solutions here: http://gmatclub.com/forum/if-y-x-5-x-5- ... 73626.html
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Re: M30-21   [#permalink] 04 Feb 2018, 19:26

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# M30-21

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