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by davidlee05 » Sat Jan 05, 2013 7:10 pm
I'm not sure if this question is considered easy or hard, but it was the latter for me when I tried. Can someone please explain in detail on how to answer this question?

This question was from a practice Kaplan test prep exam.

Thanks in advance for your help!

The integers x and y are both positive, the remainder when x is divided by 12 is 7, and the remainder when y is divided by 12 is 3. Each of the following is a possible value of EXCEPT

A. 125
B. 101
C. 77
D. 51
E. 41
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by GMATGuruNY » Sat Jan 05, 2013 8:23 pm
davidlee05 wrote:The integers x and y are both positive, the remainder when x is divided by 12 is 7, and the remainder when y is divided by 12 is 3. Each of the following is a possible value of 2x+y EXCEPT

A. 125
B. 101
C. 77
D. 51
E. 41
I believe that the question stem intends to ask for possible values of the expression in red.
The solution below is based upon this assumption.

The remainder when x is divided by 12 is 7.
In other words, x is 7 more than a multiple of 12:
x = 12a + 7.

The remainder when y is divided by 12 is 3.
In other words, y is 3 more than a multiple of 12:
y = 12b + 3.

Thus:
2x+y =

= 2(12a + 7) + (12b + 3)

= 24a + 14 + 12b + 3

= 24a + 12b + 12 + 5

= 12(2a + b + 1) + 5

= (multiple of 12) + 5.

In other words, 2x + y is 5 more than a multiple of 12.
Thus, when 5 is subtracted from a viable answer choice, the result must be a multiple of 12:

A. 125 - 5 = 120.
B. 101 - 5 = 96.
C. 77 - 5 = 72.
D. 51 - 5 = 46.
E. 41 - 5 = 36.

Only D does not yield a multiple of 12.

The correct answer is D.
Last edited by GMATGuruNY on Sun Jan 06, 2013 5:21 am, edited 1 time in total.
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by buzzdeepak » Sun Jan 06, 2013 4:29 am
x/12 => R7 (Remainder 7)
Therefore, 2x / 12 => R14 = R(12+2) = R2

y/12 => R3

Now, 2x+y=> R2+R3 = R5, i.e, we need a number which when divided by 12 yields a remainder of 5.

D is the only one which doesn't yield R5, hence the answer
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by bnpetteway » Tue Jan 08, 2013 5:36 pm
@GuruNY, is there a simpler and faster way or is this it? Also would you be able to plug in the answers into the equation in the question to get your answer?
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by GMATGuruNY » Tue Jan 08, 2013 6:21 pm
bnpetteway wrote:@GuruNY, is there a simpler and faster way or is this it? Also would you be able to plug in the answers into the equation in the question to get your answer?
An alternate approach would be to make a list of possible values for 2x and y and then see which answer choices could be the value of 2x+y.
The integers x and y are both positive, the remainder when x is divided by 12 is 7, and the remainder when y is divided by 12 is 3. Each of the following is a possible value of 2x+y EXCEPT

A. 125
B. 101
C. 77
D. 51
E. 41
The remainder when x is divided by 12 is 7.
x = 12a + 7 = 7, 19, 31, 43, 55, 67...
Thus:
2x = 14, 38, 62, 86, 110...

The remainder when y is divided by 12 is 3.
y = 12b + 3 = 3, 15, 27, 39, 51, 63...

Now see which answer choices could be the value of 2x+y.

A: 125
If 2x = 110 and y = 15, 2x+y = 110+15 = 125.

B: 101
If 2x = 86 and y = 15, 2x+y = 86+15 = 101.

C: 77
If 2x = 62 and y = 15, 2x+y = 62+15 = 77.

E: 41
If 2x = 38 and y = 3, 2x+y = 38+3 = 41.

Since A, B, C and E are all possible values of 2x+y, the correct answer is D.
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by davidlee05 » Wed Jan 09, 2013 9:47 am
Thank you for the replies/help!
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