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Bunuel
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If y = |x-2| + |x| - |x+2| where x is an integer, then y can take how 30 Dec 2022, 12:35
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95% (hard)

Question Stats:

33% (02:48) correct 67% (03:11) wrong based on 158 sessions

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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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B
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gmatophobia
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If y = |x-2| + |x| - |x+2| where x is an integer, then y can take how 30 Dec 2022, 13:23
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Bunuel
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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We can find the possible value of y corresponding to the possible range of x.

y can take values from -2 to 9, hence can take 12 possible values.

However, as we need non zero values, we have to subtract 1.

12 - 1 = 11

Option B
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If y = |x-2| + |x| - |x+2| where x is an integer, then y can take how 12 Jan 2023, 23:43
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In the elegant solution by gmatphobia,

Just want to add that in zone 2, the equation will be y = –3x, and in zone 3, it will be y = –x.
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If y = |x-2| + |x| - |x+2| where x is an integer, then y can take how 21 Jan 2023, 11:32
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Is there any quicker way to solve this other than listing all the scenarios?
I needed at least 5 mins to solve this
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If y = |x-2| + |x| - |x+2| where x is an integer, then y can take how 07 Oct 2024, 09:42
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Tiffsootmoot8
Is there any quicker way to solve this other than listing all the scenarios?
I needed at least 5 mins to solve this
Show more
Unfortunately not. It's among the hardest question too.

But Hit and trial may work, however, not without knowing how absolute values work - some randomness is required.

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

Since we have two absolute entities looking opposite to each other, there must be a pattern that they follow.

Checking for |x-2| first, we can try extreme values - let's take
x = -10, it gives y = 14 (invalid)
x = -9, it gives y = 13 (again invalid, we can jump and check more)
x = -5, it gives y = 11
x = -4, it gives y = 10
x = -3, it gives y = 7
x = -2, it gives y = 6
x = 0, it gives y = 0 (eliminate)
x = 1, it gives y = -1
x = 2, it gives y = -2 (already know why it flips from here)
x = 3, it gives y = -1
x = 4, it gives y = 0 (again we know why)
x = 5, it gives y = 1
x = 6, it gives y = 2 (pattern here)
...
x = 10, it gives y = 6
x = 11, it gives y = 7
x = 12, it gives y = 8 (thus y values increase with x)
x = 13, it gives y = 9 (we can stop here)

From x = 5 to x = 13 we have 9 different values of y.
For x = 1 and x = 2 we have 2 more values of y


Total 11 values

Answer B.
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If y = |x-2| + |x| - |x+2| where x is an integer, then y can take how 04 Nov 2024, 05:10
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Tiffsootmoot8
Is there any quicker way to solve this other than listing all the scenarios?
I needed at least 5 mins to solve this
Show more
I think focusing on region 1 or region 4 might help since they are the only ones with maximum range for y. The middle 2 regions have a boundary on both sides but region 1 & 4 have limitations on only 1 side (ex, x≥2). So y=x-4 could offer the maximum no. of values that y can take (which is possible when x takes either 1 of 11 values.

I could be wildly wrong about this, but this seems plausible to whatever level of understanding I have currently
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