tonebeeze wrote:152.

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


x = (12-y)/2

For x to be an integer, (12-y) must be even.
Thus y must be even.
Since y lies between -12 and +12 (both inclusive) there are a total of 13 even numbers possible.
(-12,-10,-8,-6,-4,-2,0,2,4,6,8,10,12)

Thus there can be 13 possible solutions such that both x and y are integers.

OA = 13

tonebeeze wrote:152.

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

OA = 13

GMATGuruNY wrote:

tonebeeze wrote:152.


The formula for counting even integers is:

(Biggest - Smallest)/2 + 1

Hi. I just want to say that that formula works only to integers that begins and ends with even numbers.
If the set begins with odd numbers then it will be a little bit different

tonebeeze wrote:152.

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

OA = 13

Night reader wrote:

tonebeeze wrote:152.

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

OA = 13

Hi tonebeeze, legs up and hands down

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

Cool - now |12-2x| <=12 and
a) 12-2x <=12
b) 2x-12 <=12

a) x >= 0
b) x>=12

Can I ask you a question how we know that our interval starts at 0 and goes through exactly 12? - you must answer we know for sure ... how about going through 13 too - you may answer (dunnow, may be, no, yes, feel tired )
in fact x>= 12 and x can be 12 1/2 too so cancel the values beyond 12

0,1,2,3,4,5,6,7,8,9,10,11,12 => all in all we got 13 integers, correct?

bomond wrote:

GMATGuruNY wrote:

tonebeeze wrote:152.


The formula for counting even integers is:

(Biggest - Smallest)/2 + 1

Hi. I just want to say that that formula works only to integers that begins and ends with even numbers.
If the set begins with odd numbers then it will be a little bit different

Correct! When I'm counting even integers, I begin and end the set with even values and then apply the formula. So to count the even integers between 27 and 53, I would apply the formula to the range of integers from 28 to 52:

Number of even integers between 28 and 52 inclusive = (52-28)/2 + 1 = 13.

I use the same process to count odd integers: I begin and end the set with odd values and then apply the formula. To count the odd integers between -12 and 12, I would apply the formula to the range of integers from -11 to 11:

Number of odd integers between -11 and 11 inclusive = 11-(-11)/2 + 1 = 12.

ankur.agrawal wrote:

Night reader wrote:

tonebeeze wrote:152.

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

OA = 13

Hi tonebeeze, legs up and hands down

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

Cool - now |12-2x| <=12 and
a) 12-2x <=12
b) 2x-12 <=12

a) x >= 0
b) x>=12

Can I ask you a question how we know that our interval starts at 0 and goes through exactly 12? - you must answer we know for sure ... how about going through 13 too - you may answer (dunnow, may be, no, yes, feel tired )
in fact x>= 12 and x can be 12 1/2 too so cancel the values beyond 12

0,1,2,3,4,5,6,7,8,9,10,11,12 => all in all we got 13 integers, correct?

Hey 2x-12<=12 implies 2x<=24 i.e. dividing by two both sides gives us x<=12.

How do u get x>=12. Plz clarify.

The rule as i know is multiplying or dividing by a negative no flips the sign