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Manager  Joined: 20 Mar 2005
Posts: 138
For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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47 00:00

Difficulty:   55% (hard)

Question Stats: 65% (01:51) correct 35% (02:07) wrong based on 1228 sessions

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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/for-how-many ... 68675.html

Originally posted by Balvinder on 24 May 2007, 05:07.
Last edited by Bunuel on 28 Nov 2017, 19:14, edited 3 times in total.
Renamed the topic, edited the question and added the OA.
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Re: For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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34
21
tonebeeze wrote:
152.

$$2x + y = 12$$
$$|y| \leq 12$$

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

The solution of $$|y| \leq 12$$ is straight forward.
$$-12 \leq y \leq 12$$
(If you are not comfortable with this, check out my blog post:
http://www.veritasprep.com/blog/2011/01/quarter-wit-quarter-wisdom-do-what-dumbledore-did/

If both x and y have to be integers, y should be an integer and hence can take any value from the set {-12, -11, -10 ... 10, 11, 12} i.e. any one of 25 values (these are 25 values -12 to -1 (12 values), 0, 1 to 12 (another 12 values)) 13 of them are even and 12 of them are odd.

$$2x + y = 12$$
Every time y is even, x will be integer. e.g. y = 12, x = 0 (because x = (12 - even)/2 will be an integer)
Every time y is odd, x will be non-integer e.g. y = 1, x = 5.5 (because x = (12 - odd)/2 will not be an integer)

Therefore, for 13 values, x and y both will be integers.
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Re: For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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4
40
$$2x + y = 12$$
$$|y| \leq 12$$

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
##### General Discussion
Intern  Joined: 20 Apr 2007
Posts: 12
Re: For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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Can you explain why the Y values go from -12 to 12? I a little new at this forum. I originally got 12 total possibilities.
Manager  Joined: 11 Jun 2006
Posts: 204
Re: For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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3
1
2x + y = 12

|y| <= 12 means that y can be anything between 12 and -12 inclusive. Absolute values always indicates a range of numbers... this is the easy way to think about abs. values.

Ok, now you've narrowed down the answer choices to 25 possible numbers... which doesn't help you with the answers given. Next you need to find a way of eliminating more answer choices...

Simplify 2x + y = 12 to

x + y/2 = 3

Now looking at that, you know y has to be an even number to yield an integer... so the initial pool of 25 numbers is now narrowed down to 13, hence the answer.
Intern  Joined: 18 Sep 2006
Posts: 45
Re: For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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This is a good question, coz I did not pay attention to the "(x,y) that will yield x and y to be integers" part of the question, so I was stuck with the answer being 25 and was stumped by the choices.
Good job you guys....i guess I should read the question clearly
Retired Moderator Joined: 20 Dec 2010
Posts: 1516
Re: For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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6
Sol:
|y| <= 12
Means;
-12<=y<=12

2x + y = 12
x = (12-y)/2

x will be integers for y=even; because even-even = even and even is always divisible by 2.

We need to find out how many even integers are there between -12 and 12

((12-(-12))/2)+1 = (24/2)+1 = 12+1 = 13

Ans: "D"
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Joined: 02 Sep 2009
Posts: 61396
Re: For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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7
1
Baten80 wrote:
2x + y = 12
|y| <= 12

152. 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

Given: $$-12\leq{y}\leq{12}$$ and $$2x+y=12$$ --> $$y=12-2x=2(6-x)=even$$, (as $$x$$ must be an integer). Now, there are 13 even numbers in the range from -12 to 12, inclusive each of which will give an integer value of $$x$$.

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Re: For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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2
-12<= y <=12
gives 0<=x <=12

thus 13 values in total.
Retired Moderator Joined: 20 Dec 2010
Posts: 1516
Re: For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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10
2
Balvinder 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

|y| <= 12
-12<=y<=12

2x + y = 12
x=(12-y)/2

To find number of integer pairs, we just need to find even number of y's, because even y will make "(12-y)" even as well and only even numbers divide by 2 evenly to give an integer.

e.g.
x=(12-y)/2; for y=1; x=(12-1)/2=11/2=5.5(Not an integer because y is odd)
x=(12-y)/2; for y=0; x=(12-0)/2=12/2=6(An integer because y is even)

Thus, if we find the number of even y's, we should be good.

-12<=y<=12
What is the first even number greater than or equal to -12?
-12
What is the last even number smaller than or equal to +12?
+12

$$Count=\frac{Last Even-First Even}{2}+1$$

$$Count=\frac{12-(-12)}{2}+1=12+1=13$$

Ans: "D"
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Re: For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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2
1
Number of ordered pairs = number of integers between 12 and 0 (both inclusive)
= 13
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Re: For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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Hi Karishma

Using the number properties this indeed is very convenient to solve. I was wondering can we substitute y = 12 - 2x in the inequality and solve for the possible values of x.
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Re: For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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ratinarace wrote:
Hi Karishma

Using the number properties this indeed is very convenient to solve. I was wondering can we substitute y = 12 - 2x in the inequality and solve for the possible values of x.

Certainly and it is quick too.

y = 12 - 2x
Whenever x is an integer, y will be an integer. So if we can solve for integral values of x, the number of values we get will be the number of solutions.

$$|y| \leq 12$$

$$|12 - 2x| \leq 12$$

$$|x - 6| \leq 6$$

From 6, x should be at a distance less than or equal to 6. So x will lie from 0 to 12 i.e. 13 values. (Check http://www.veritasprep.com/blog/2011/01 ... edore-did/ if this is not clear)

There are 13 solutions.
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Re: For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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Responding to a pm:
Changing the sign within the mod has no impact on anything outside the mod.

$$|6 - x| \leq 12$$ is same as
$$|x - 6| \leq 12$$

Think about it: Whether you write |x| or |-x|, it is the same.
|6| = |-6|

So for every value of x,
|x - 6| = |6 - x|
So you don't need to flip the inequality sign.

|x - 6| and - |x - 6| are of course different. If you change |x - 6| to - |x - 6|, you will need to flip the inequality sign.
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Re: For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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4
$$y=12-2x=2*(6-x).$$
Since $$|y| \leq 12 , -12 \leq y \leq 12$$ . Substituting for y from above, $$-6 \leq (6-x) \leq 6.$$. This reduces to $$x \geq 0$$ and $$x \leq 12.$$ Including 0 and 12 there are thus 13 integer solutions.
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Re: For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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2
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?

y=12-2x
|y|<=12
|12-2x| <= 12
12 - 2x <= 12
-2x <= 0
x>=0

-(12-2x) <= 12
-12+2x <= 12
2x <= 24
x<=12

13 solutions between 0 and 12 inclusive.
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Posts: 61396
Re: For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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2
2
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

Given: $$-12\leq{y}\leq{12}$$ and $$2x+y=12$$ --> $$y=12-2x=2(6-x)=even$$, (as $$x$$ must be an integer). Now, there are 13 even numbers in the range from -12 to 12, inclusive each of which will give an integer value of $$x$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: for-how-many-ordered-pairs-x-y-that-are-solutions-of-the-110687.html
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Re: For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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abs 12 = -12 all the way to 12, which is 25 integers including "0".

y= 2(6-x) = even.

So y = even. and y is equal the 25 number range. How many possible even numbers are in that range?

Answer is 13 possible even y numbers including zero "0".
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Re: For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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tonebeeze wrote:
$$2x + y = 12$$
$$|y| \leq 12$$

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

|y| <= 12 means range of y is -12 <= Y <= +12. which means Y can take any of the value in the set (-12, -11, -10......-1,0,1.....10,11,12).

now that we are given 2x + y = 12, y = 12 - 2x

we can include all the integer values for X as a solution for y = 12 - 2x as long as y falls in the above range mentioned. Such values of X are (0,1,2....12). 13 is the count for this set. Answer is D.
_________________ Re: For how many ordered pairs (x , y) that are solutions of the   [#permalink] 11 Apr 2016, 23:23

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