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
Quant Review #152: Inequalities
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stormier
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Last edited by stormier on Fri Jan 14, 2011 4:45 pm, edited 1 time in total.
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GMATGuruNY
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Since 2x = even integer, y must also be an even integer.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
Since |y|≤12, y can be any even integer from -12 to 12.
Here is the formula for counting even integers between one even integer and another:
(Biggest - Smallest)/2 + 1
Thus, the number of even integers from -12 to 12 = (12-(-12))/2 + 1 = 13.
The correct answer is D.
Last edited by GMATGuruNY on Sat Jan 15, 2011 3:55 am, edited 1 time in total.
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GMATGuruNY wrote:Hi. I just want to say that that formula works only to integers that begins and ends with even numbers.tonebeeze wrote:152.
The formula for counting even integers is:
(Biggest - Smallest)/2 + 1
If the set begins with odd numbers then it will be a little bit different
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Night reader
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Hi tonebeeze, legs up and hands downtonebeeze 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
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) deleted (x>=12) sorry mechanical mistake 2x-12<=12, x<=12
solution interval is below
0,1,2,3,4,5,6,7,8,9,10,11,12 => all in all we got 13 integers
Last edited by Night reader on Sat Jan 15, 2011 12:39 pm, edited 1 time in total.
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ankur.agrawal
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Hey 2x-12<=12 implies 2x<=24 i.e. dividing by two both sides gives us x<=12.Night reader wrote:Hi tonebeeze, legs up and hands downtonebeeze 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
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 questionhow 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 tooso 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?
How do u get x>=12. Plz clarify.
The rule as i know is multiplying or dividing by a negative no flips the sign
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GMATGuruNY
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bomond wrote: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:GMATGuruNY wrote:Hi. I just want to say that that formula works only to integers that begins and ends with even numbers.tonebeeze wrote:152.
The formula for counting even integers is:
(Biggest - Smallest)/2 + 1
If the set begins with odd numbers then it will be a little bit different
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.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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Night reader
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Ankur thanks for your note. I made mechanical mistake there.ankur.agrawal wrote:Hey 2x-12<=12 implies 2x<=24 i.e. dividing by two both sides gives us x<=12.Night reader wrote:Hi tonebeeze, legs up and hands downtonebeeze 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
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
in fact x>= 12 and x can be 12 1/2 too
0,1,2,3,4,5,6,7,8,9,10,11,12 => all in all we got 13 integers, correct?
How do u get x>=12. Plz clarify.
The rule as i know is multiplying or dividing by a negative no flips the sign
