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2x + y = 12 |y| <= 12 For how many ordered pairs (x , y)

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Senior Manager
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2x + y = 12 |y| <= 12 For how many ordered pairs (x , y) [#permalink]

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New post 30 Aug 2007, 20:56
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

2x + y = 12
|y| <= 12

For how many ordered pairs (x , y) that are solutions of the system above are x and y both integers?
A. 7
B. 10
C. 12
D. 13
E. 14

P.S. please no answers like "i think it's B or C" :)
Manager
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Re: I have no clue [#permalink]

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New post 30 Aug 2007, 21:23
Sergey_is_cool wrote:
2x + y = 12
|y| <= 12

For how many ordered pairs (x , y) that are solutions of the system above are x and y both integers?
A. 7
B. 10
C. 12
D. 13
E. 14

P.S. please no answers like "i think it's B or C" :)


since |y| is 12 so y lies between -12 to 12. now if we substitute -12 in the equation we get x = 12 , if substitute -11 we get a fraction. again if we substitute y = -10 we get x = 11 if we substitute y = -9 we get x as a fraction. so since we are interested only in integers, there would half as many pairs as the range of Y which 24/2 = 12. so answer is C
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Re: I have no clue [#permalink]

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New post 30 Aug 2007, 21:25
Sergey_is_cool wrote:
2x + y = 12
|y| <= 12

For how many ordered pairs (x , y) that are solutions of the system above are x and y both integers?

A. 7
B. 10
C. 12
D. 13
E. 14

P.S. please no answers like "i think it's B or C" :)


D. 13 :-D

cus y can olny be even integer between -12 to 12 inncluding.
Senior Manager
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New post 30 Aug 2007, 21:32
when y= 12, x = 0
when y = -12, x = 12


so x can have values from 0 to 12 inclusive..

so (D)
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New post 30 Aug 2007, 21:38
sidbidus wrote:
when y= 12, x = 0
when y = -12, x = 12


so x can have values from 0 to 12 inclusive..

so (D)


yep .. my mistake.. missed the godamm zero in between... :-)
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New post 30 Aug 2007, 23:27
hello sergey ! nice to see you back in the forum.

see here:

http://www.gmatclub.com/forum/7/46071

:-D
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New post 30 Aug 2007, 23:59
x = (12-y)/2 = 6 - y/2

So we are going for even values of y. That's 12,10,8,6,4,2,0,-2,-4,-6,-8,-10,-12. Total 13 solutions.
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Re: I have no clue [#permalink]

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New post 01 Sep 2007, 00:27
Keeping in mind the second rule of the problem, start by trying 1, 0, and -1 for x (simple, quick numbers) and you notice that x>=0.

Now try to find the top end for x, I started with 6, too low. 12, that's as high as you can go for x, maxing out y.

12-(-1) = 13.

And just because I am not actually taking the test, and am kind OCD:

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

:)
Re: I have no clue   [#permalink] 01 Sep 2007, 00:27
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2x + y = 12 |y| <= 12 For how many ordered pairs (x , y)

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