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If y = |x-2| + |x| - |x+2| where x is an integer, then y can take how

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If y = |x-2| + |x| - |x+2| where x is an integer, then y can take how  [#permalink]

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New post 14 May 2017, 09:26
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If y = |x-2| + |x| - |x+2| where x is an integer, then y can take how many non-zero integral values between -10 and 10, exclusive?

A.10
B.11
C.12
D.13
E.14

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Re: If y = |x-2| + |x| - |x+2| where x is an integer, then y can take how  [#permalink]

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New post 15 May 2017, 00:45
I am not sure about how to do this... Can anybody please suggest a process?
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If y = |x-2| + |x| - |x+2| where x is an integer, then y can take how  [#permalink]

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New post 15 May 2017, 02:51
msk0657 wrote:
If y = |x-2| + |x| - |x+2| where x is an integer, then y can take how many non-zero integral values between -10 and 10, exclusive?

A.10
B.11
C.12
D.13
E.14


Start substituting values for the simplest approach to this question.

y = |x-2| + |x| - |x+2|

@x = 0, y = 0
@x = 1, y = -1
@x = 2, y = -2
@x = 3, y = -1
@x = 4, y = 0
@x = 5, y = 1
@x = 6, y = 2
@x = 7, y = 3
@x = 8, y = 4
@x = 9, y = 5
and so on so all positive values till 10

so total Integer values of x = 12
{-2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

Answer: Option C
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If y = |x-2| + |x| - |x+2| where x is an integer, then y can take how  [#permalink]

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New post 15 May 2017, 03:11
mine is A, only 10 values, after breaks the absolute sign, y can only takes range from -2 to 9, and except 0 => 10 values
-2 0 2
x-2 - 0 +
x - 0 +
x+2 - 0 +
-/x+2/ x+2 -x-2

x >=2 y = x -4 => 10 values
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Re: If y = |x-2| + |x| - |x+2| where x is an integer, then y can take how  [#permalink]

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New post 15 May 2017, 03:25
chesstitans wrote:
mine is A, only 10 values, after breaks the absolute sign, y can only takes range from -2 to 9, and except 0 => 10 values
-2 0 2
x-2 - 0 +
x - 0 +
x+2 - 0 +
-/x+2/ x+2 -x-2

x >=2 y = x -4 => 10 values


I see two mistakes here
1) -2 to 9 there are 12 values
2) y may be 0 @x=0
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Re: If y = |x-2| + |x| - |x+2| where x is an integer, then y can take how  [#permalink]

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New post 15 May 2017, 05:04
msk0657 wrote:
If y = |x-2| + |x| - |x+2| where x is an integer, then y can take how many non-zero integral values between -10 and 10, exclusive?

A.10
B.11
C.12
D.13
E.14


Use the number line to figure this out.


................. -5 ........ -4 ....... -3 ....... -2 ....... -1 ........ 0 ........ 1 ........ 2 ........ 3 ......... 4 ........ 5 ..............

We need the sum of "distance from 0" + "distance from 2" and from that we need to subtract the "distance from -2". This will be y.
A few things to note right away:
At 0, distance from 2 = distance from -2 and hence y = 0

At the right of 0, distance from -2 and distance from 0 effectively will always add up to -2. We will just keep adding distance from 2 to this.
At x = 1, this will give us y = -1.
At x = 2, this will give us y = -2.
At x = 3, this will give us y = -1.
and now the distance from 2 will keep increasing so we will get all values from 0 to infinity. In our desired range, we get 0 to 9.

At the left of 0, the distance from -2 will always get cancelled out by distance from 0 and hence y will always be positive. We have already got all positive values of y.

So all values of y from -2 to 9 are possible i.e. 12 values.

Answer (C)
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Re: If y = |x-2| + |x| - |x+2| where x is an integer, then y can take how  [#permalink]

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New post 15 May 2017, 05:54
Hi,

As explained above, the values y can take is from -2 to 9...
When x is 0, y=0...
At all values of x above 2, an increase of 1 in value of x, increase in x-2 and x will also be 1 each and x+2 will increase by 1..
Thus effect will be 1+1-1=1..
At x as 2, y=|2-2|+|2|-|2+2|=-2.. so y can take all values of -2 and above it..
So 12 values..

But be careful, inspite of doing everything correct, you will land up with wrong answer as we are looking for non-zero..
There are 12 values including 0..
Ans=12-1=11

B

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Re: If y = |x-2| + |x| - |x+2| where x is an integer, then y can take how  [#permalink]

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New post 15 May 2017, 08:11
msk0657 wrote:
If y = |x-2| + |x| - |x+2| where x is an integer, then y can take how many non-zero integral values between -10 and 10, exclusive?

A.10
B.11
C.12
D.13
E.14


If \(x<-2\) then \(y=(2-x)-x+(x+2)=4-x\)
\(x=-3 \implies y=7\)
\(x=-4 \implies y=8\)
If \(x\) decreases then \(y\) increases.

If \(-2 \leq x < 0\) then \(y=(2-x)-x-(x+2)=-3x\)
\(x=-2 \implies y=6\)
\(x=-1 \implies y=3\)

If \(0 \leq x < 2\) then \(y=(2-x)+x-(x+2)=-x\)
\(x=0 \implies y=0\)
\(x=1 \implies y=-1\)

If \(x \geq 2\) then \(y=(x-2)+x-(x+2)=x-4\)
\(x=2 \implies y=-2\)
\(x=3 \implies y=-1\)
\(x=4 \implies y=0\)
...
\(x=13 \implies y=9\)
\(x=14 \implies y=10\)
If \(x\) increases then \(y\) increases.

Based on cases above, we see that \(y\) could receive value from \(-2\) to positive infinitive. Hence, y could receive 13 values between -10 and 10 (from -10 to 10).

However, the question asks for non-zero integral values, hence we need to exclude 0 and 10. Hence the answer will be 11.
-2, -1, 1, 2, 3, 4, 5, 6, 7, 8, 9

The answer is B
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Re: If y = |x-2| + |x| - |x+2| where x is an integer, then y can take how  [#permalink]

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New post 16 May 2017, 04:52
GMATinsight wrote:
msk0657 wrote:
If y = |x-2| + |x| - |x+2| where x is an integer, then y can take how manynon-zero integral values between -10 and 10, exclusive?

A.10
B.11
C.12
D.13
E.14


Start substituting values for the simplest approach to this question.

y = |x-2| + |x| - |x+2|

@x = 0, y = 0
@x = 1, y = -1
@x = 2, y = -2
@x = 3, y = -1
@x = 4, y = 0
@x = 5, y = 1
@x = 6, y = 2
@x = 7, y = 3
@x = 8, y = 4
@x = 9, y = 5
and so on so all positive values till 10

so total Integer values of x = 12
{-2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

Answer: Option C


The question asks fro Non-zero integer values.
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Re: If y = |x-2| + |x| - |x+2| where x is an integer, then y can take how  [#permalink]

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New post 17 May 2017, 10:55
Bunuel plz can u explain this qn...
i m not very confident on how the sign changes for range related values as above soln does.

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Re: If y = |x-2| + |x| - |x+2| where x is an integer, then y can take how &nbs [#permalink] 17 May 2017, 10:55
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