Absolute value question

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jfranco23: Senior | Next Rank: 100 Posts; Posts: 75; Joined: Fri Jan 16, 2009 2:00 pm; Thanked: 1 times; GMAT Score:550

by jfranco23 » Fri Mar 13, 2009 5:15 am

2x + y = 12
/y/<=12

For how many ordered pairs (x,y) that 2 solutions of the system above are x and y both integers?

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

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Source: Beat The GMAT — Problem Solving | Fri Mar 13, 2009 5:15 am

cramya: Legendary Member; Posts: 2467; Joined: Thu Aug 28, 2008 6:14 pmFollowed by: 11 members; Thanked: 331 times

by cramya » Fri Mar 13, 2009 6:24 am

2x+y=12

x=12-y/2

y has to be even satisfying |y|<=12

12,10,8,6,4,2,0,-2,-4,-6,-8,-10,-12

13 ordered pairs

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jfranco23: Senior | Next Rank: 100 Posts; Posts: 75; Joined: Fri Jan 16, 2009 2:00 pm; Thanked: 1 times; GMAT Score:550

by jfranco23 » Fri Mar 13, 2009 7:38 am

how did you get that conclusion?

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DanaJ: Site Admin; Posts: 2567; Joined: Thu Jan 01, 2009 10:05 amFollowed by: 550 members; Thanked: 712 times; GMAT Score:770

by DanaJ » Fri Mar 13, 2009 7:59 am

jfranco23: notice that you have the first condition:
2x + y = 12
Now, 2x is obviously even, and so is 12. This is why you have that:
(even number) + y = (even number), making y even as well.

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sureshbala: Master | Next Rank: 500 Posts; Posts: 319; Joined: Wed Feb 04, 2009 10:32 amFollowed by: 9 members; Location: Delhi; Thanked: 84 times

by sureshbala » Fri Mar 13, 2009 7:59 am

Given 2x+y = 12
and |y|<=12.
Hence -12<=y<=12.
So y can take 25 values from -12 to 12.

We have 2x= 12-y
i.e x = (12-y)/2= 6 - y/2
Hence x will be integer if y is divisible by 2.
From -12 to 12, there are 13 values which are divisible by 2.

Hence the answer is 13

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