Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Does GMAT RC seem like an uphill battle? e-GMAT is conducting a free webinar to help you learn reading strategies that can enable you to solve 700+ level RC questions with at least 90% accuracy in less than 10 days.

Enter The Economist GMAT Tutor’s Brightest Minds competition – it’s completely free! All you have to do is take our online GMAT simulation test and put your mind to the test. Are you ready? This competition closes on December 13th.

The integers x and y are both positive, the remainder when x is divide
[#permalink]

Show Tags

Updated on: 11 Oct 2017, 04:34

2

20

00:00

A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

68% (02:31) correct 32% (02:39) wrong based on 340 sessions

HideShow timer Statistics

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

I got this right but it took me 2 minutes. I wonder if there is a quick way to solve this question without plugging numbers to X and Y. I wrote down like 6 numbers with remainder of 7 and 3 and found each of the numbers from the answer choices manually.

I also tried the other approach X=12Q+7, Y=12R+3 and calculated 2X+Y but couldn't understand what to do next. thanks

Re: The integers x and y are both positive, the remainder when x is divide
[#permalink]

Show Tags

07 Jan 2013, 04:34

10

9

roygush 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 got this right but it took me 2 minutes. I wonder if there is a quick way to solve this question without plugging numbers to X and Y. I wrote down like 6 numbers with remainder of 7 and 3 and found each of the numbers from the answer choices manually.

I also tried the other approach X=12Q+7, Y=12R+3 and calculated 2X+Y but couldn't understand what to do next. thanks

The remainder when x is divided by 12 is 7: \(x=12q+7\) --> \(2x=24q+14\). The remainder when y is divided by 12 is 3: \(y=12p+3\).

\(2x+y=(24q+14)+(12p+3)=24q+12p+12+5=12(2q+p+1)+5=(multiple \ of \ 12)+5\). Only D is not a multiple of 12 plus 5.

Re: The integers x and y are both positive, the remainder when x is divide
[#permalink]

Show Tags

06 Jan 2013, 01:50

3

roygush 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 got this right but it took me 2 minutes. I wonder if there is a quick way to solve this question without plugging numbers to X and Y. I wrote down like 6 numbers with remainder of 7 and 3 and found each of the numbers from the answer choices manually.

I also tried the other approach X=12Q+7, Y=12R+3 and calculated 2X+Y but couldn't understand what to do next. thanks

There are two statements in the question: 1) The remainder when x is divided by 12 is 7. This can be written as \(x=12I + 7\), where I is an integer. 2) The remainder when y is divided by 12 is 3. This can be written as \(y=12J + 3\), where J is an integer.

One thing to note here is that the maximum remainder that these two equations can generate is 11. So \(2x\) will not be equal to \(12I+14\) but will be equal to \(12I+2\).

Therefore \(2x + y = 12(I+J) + 5\) or any multiple of 12 + 5. 51 doesn't follows such pattern. Hence is the answer.
_________________

Re: The integers x and y are both positive, the remainder when x is divide
[#permalink]

Show Tags

07 Jan 2013, 07:04

Bunuel wrote:

roygush 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 got this right but it took me 2 minutes. I wonder if there is a quick way to solve this question without plugging numbers to X and Y. I wrote down like 6 numbers with remainder of 7 and 3 and found each of the numbers from the answer choices manually.

I also tried the other approach X=12Q+7, Y=12R+3 and calculated 2X+Y but couldn't understand what to do next. thanks

The remainder when x is divided by 12 is 7: \(x=12q+7\) --> \(2x=24q+14\). The remainder when y is divided by 12 is 3: \(y=12p+3\).

\(2x+y=(24q+14)+(12p+3)=24q+12p+12+5=12(2q+p+1)+5=(multiple \ of \ 12)+5\). Only D is not a multiple of 12 plus 5.

Answer: D.

bunuel, can we apply this method on any remainders problem? write the two equations and then work from there?
_________________

Re: The integers x and y are both positive, the remainder when x is divide
[#permalink]

Show Tags

07 Jan 2013, 09:44

roygush wrote:

Bunuel wrote:

roygush 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 got this right but it took me 2 minutes. I wonder if there is a quick way to solve this question without plugging numbers to X and Y. I wrote down like 6 numbers with remainder of 7 and 3 and found each of the numbers from the answer choices manually.

I also tried the other approach X=12Q+7, Y=12R+3 and calculated 2X+Y but couldn't understand what to do next. thanks

The remainder when x is divided by 12 is 7: \(x=12q+7\) --> \(2x=24q+14\). The remainder when y is divided by 12 is 3: \(y=12p+3\).

\(2x+y=(24q+14)+(12p+3)=24q+12p+12+5=12(2q+p+1)+5=(multiple \ of \ 12)+5\). Only D is not a multiple of 12 plus 5.

Answer: D.

bunuel, can we apply this method on any remainders problem? write the two equations and then work from there?

I wouldn't say any, but \(dividend=divisor*quotient+remainder\) formula is indeed very handy to deal with questions about remainders.

Re: The integers x and y are both positive, the remainder when x is divide
[#permalink]

Show Tags

30 Jul 2015, 09:04

Another way to do it is the following, which took me around 4 minutes (perhaps someone quick in calculations could do it faster, dunno...). So:

x=12q+7 = 19, 31, 43, 55, 67 etc. Doudling it: 38, 62, 126, 110, 134 etc. y=12z+3 = 15, 27, 39, 51, 63 etc.

At this point, we can already see that D is the answer, as D=51, which is a possible value of y alone. So, it is not possible that if we add x to 51 it will remain 51. So, ANS D

Do you think it is doable like this in under 2 minutes, even though it took me 4? I am asking as I am terribly slow in calculations. Performing calculations is a serious problem of mine in gmat...

Re: The integers x and y are both positive, the remainder when x is divide
[#permalink]

Show Tags

30 Jul 2015, 09:17

roygush 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 got this right but it took me 2 minutes. I wonder if there is a quick way to solve this question without plugging numbers to X and Y. I wrote down like 6 numbers with remainder of 7 and 3 and found each of the numbers from the answer choices manually.

I also tried the other approach X=12Q+7, Y=12R+3 and calculated 2X+Y but couldn't understand what to do next. thanks

Hi,

x/12 = 7 (remainder) and y/12 = 3 (remainder). Hence the least value of x has to be 7 and least value of y is 3. Therefore, 2x+y = 17. Hence, look for an answer choice that is a multiple of 17, which is 51.

Re: The integers x and y are both positive, the remainder when x is divide
[#permalink]

Show Tags

30 Jul 2015, 09:20

pacifist85 wrote:

Another way to do it is the following, which took me around 4 minutes (perhaps someone quick in calculations could do it faster, dunno...). So:

x=12q+7 = 19, 31, 43, 55, 67 etc. Doudling it: 38, 62, 126, 110, 134 etc. y=12z+3 = 15, 27, 39, 51, 63 etc.

At this point, we can already see that D is the answer, as D=51, which is a possible value of y alone. So, it is not possible that if we add x to 51 it will remain 51. So, ANS D

Do you think it is doable like this in under 2 minutes, even though it took me 4? I am asking as I am terribly slow in calculations. Performing calculations is a serious problem of mine in gmat...

Hi,

x/12 = 7 (remainder) and y/12 = 3 (remainder). Hence the least value of x has to be 7 and least value of y is 3. Therefore, 2x+y = 17. Hence, look for an answer choice that is a multiple of 17, which is 51.

Ans - D

The above method, if used, can be solved within 2 mins. I took 1.5 mins to do so.

Re: The integers x and y are both positive, the remainder when x is divide
[#permalink]

Show Tags

30 Jul 2015, 15:27

1

Hi All,

I'm a big fan of TESTing VALUES on this question along with a bit of 'brute force' math (pacifist85's approach showcases this tactic nicely). There is one detail worth noting and one aspect I would add to it: since this is an EXCEPT question, once you find the exception, you can stop working.

The detail in pacifist85s math is that Q and Z could both be 0, so X COULD = 7 (and by extension, 2X COULD = 14) and Y COULD = 3.

I would start with answer E because it's smallest (so it would have the least number of possible sums that could equal it).

41 = 38 + 3. It's possible, so it's NOT what we're looking for.

51 though…using the possible values of 2X…

2X = 14; Y would have to be 37 (which is NOT possible). 2X = 38; Y would have to be 13 (which is NOT possible). 2X = 62; this is already TOO BIG.

Re: The integers x and y are both positive, the remainder when x is divide
[#permalink]

Show Tags

15 Jul 2016, 19:36

roygush 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 got this right but it took me 2 minutes. I wonder if there is a quick way to solve this question without plugging numbers to X and Y. I wrote down like 6 numbers with remainder of 7 and 3 and found each of the numbers from the answer choices manually.

I also tried the other approach X=12Q+7, Y=12R+3 and calculated 2X+Y but couldn't understand what to do next. thanks

Just for others who may be interested, the first time I did this problem it took me almost 5 minutes, but I got it right. However, upon review, I figured out that since you understand the concept, all you have to do is divide the options by 12, and the right answer will give you a different remainder from the others. Hence, all you need to do is divide at least 3 options by 12, until one gives you a different remainder than the rest.

The integers x and y are both positive, the remainder when x is divide
[#permalink]

Show Tags

19 Apr 2018, 16:15

Top Contributor

roygush 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

One approach is to test values

IMPORTANT: When it comes to remainders, we have a nice rule that says: If N divided by D, leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc. For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.

Okay, onto the question....

The remainder when x is divided by 12 is 7 Possible values of x = 7, 19, 31, 43, 55, ... So, possible values of 2x are: 14, 38, 62, 86, 110, ...

The remainder when y is divided by 12 is 3 Possible values of y are: 3, 15, 28, 40, 52, ...

Each of the following is a possible value of 2x + y EXCEPT

A) 125 This equals 110 + 15. ELIMINATE A B) 101 This equals 86 + 15. ELIMINATE B C) 77 This equals 62 + 15. ELIMINATE C D) 51 E) 41 This equals 38 + 3. ELIMINATE E

Re: The integers x and y are both positive, the remainder when x is divide
[#permalink]

Show Tags

28 Jun 2018, 08:08

1

Top Contributor

Please move to PS

Bunuel wrote:

roygush 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 got this right but it took me 2 minutes. I wonder if there is a quick way to solve this question without plugging numbers to X and Y. I wrote down like 6 numbers with remainder of 7 and 3 and found each of the numbers from the answer choices manually.

I also tried the other approach X=12Q+7, Y=12R+3 and calculated 2X+Y but couldn't understand what to do next. thanks

The remainder when x is divided by 12 is 7: \(x=12q+7\) --> \(2x=24q+14\). The remainder when y is divided by 12 is 3: \(y=12p+3\).

\(2x+y=(24q+14)+(12p+3)=24q+12p+12+5=12(2q+p+1)+5=(multiple \ of \ 12)+5\). Only D is not a multiple of 12 plus 5.

Re: The integers x and y are both positive, the remainder when x is divide
[#permalink]

Show Tags

28 Jun 2018, 08:38

BrushMyQuant wrote:

Please move to PS

Bunuel wrote:

roygush 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 got this right but it took me 2 minutes. I wonder if there is a quick way to solve this question without plugging numbers to X and Y. I wrote down like 6 numbers with remainder of 7 and 3 and found each of the numbers from the answer choices manually.

I also tried the other approach X=12Q+7, Y=12R+3 and calculated 2X+Y but couldn't understand what to do next. thanks

The remainder when x is divided by 12 is 7: \(x=12q+7\) --> \(2x=24q+14\). The remainder when y is divided by 12 is 3: \(y=12p+3\).

\(2x+y=(24q+14)+(12p+3)=24q+12p+12+5=12(2q+p+1)+5=(multiple \ of \ 12)+5\). Only D is not a multiple of 12 plus 5.

Re: The integers x and y are both positive, the remainder when x is divide
[#permalink]

Show Tags

02 Jul 2019, 04:20

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________