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Ordered pairs

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Source: — Data Sufficiency |
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by rdxb » Wed Oct 01, 2008 11:54 pm
My attempt was as follows, I took all values of y starting from 12and -12 (largest absolute given), and placed those values for which x was an integer.

x y
0 12
12 -12
1 10
11 -10
2 8
10 -8
3 6
9 -6
4 4
8 -4
5 2
7 -2
6 0

I realized there was a pattern which was following, so I din make the whole table, just realized it would be 13.....
but here , I have made the whole table...

Lets hope the answer is 13 ! :?
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by Mani_mba » Thu Oct 02, 2008 12:57 am
I would approach like this..

-12 <= y <= 12

For x to be an integer, Y must be even.

So out of 25 values of y, only 13 are even ( -12,-10 ......., 0 , 2, 4 , .... 10,12)

So there are 13 possible pairs that both x and y are integers.
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by farooq » Fri Jan 08, 2010 10:10 am
Mani_mba wrote:I would approach like this..

-12 <= y <= 12

For x to be an integer, Y must be even.

So out of 25 values of y, only 13 are even ( -12,-10 ......., 0 , 2, 4 , .... 10,12)

So there are 13 possible pairs that both x and y are integers.
Nice solution.

Very real to GMAT type question.
Regards,
Farooq Farooqui.
London. UK

It is your Attitude, not your Aptitude, that determines your Altitude.
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by mehravikas » Mon Jan 11, 2010 7:20 pm
Really good approach :-)
Mani_mba wrote:I would approach like this..

-12 <= y <= 12

For x to be an integer, Y must be even.

So out of 25 values of y, only 13 are even ( -12,-10 ......., 0 , 2, 4 , .... 10,12)

So there are 13 possible pairs that both x and y are integers.
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by kg » Wed Jan 20, 2010 7:50 pm
2x+y = 12, or x = 6- y/2, that implies, even y will give integer x
mody<=12, or 12<=y<=12, or y ranges from -12,-10,-8,.....0,............12
its an AP, -12+(n-1)2 = 12, that implies, n = 13 and hence 13 pairs.
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by gokyada » Thu Mar 10, 2011 8:59 pm
Given 2x+y=12 and |y|<=12, it means -12 <= y <= 12. For this range, x is between 12 and 0, inclusive. There are only 13 values for x. Hence I can have 13 ordered pairs.
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