GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 15 Dec 2018, 19:37

### GMAT Club Daily Prep

#### 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

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

## Events & Promotions

###### Events & Promotions in December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• ### $450 Tuition Credit & Official CAT Packs FREE December 15, 2018 December 15, 2018 10:00 PM PST 11:00 PM PST Get the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) • ### FREE Quant Workshop by e-GMAT! December 16, 2018 December 16, 2018 07:00 AM PST 09:00 AM PST Get personalized insights on how to achieve your Target Quant Score. # The integers x and y are both positive, the remainder when x is divide  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Manager Joined: 01 Sep 2012 Posts: 117 The integers x and y are both positive, the remainder when x is divide [#permalink] ### Show Tags Updated on: 11 Oct 2017, 03:34 2 15 00:00 Difficulty: 55% (hard) Question Stats: 68% (02:29) correct 32% (02:43) wrong based on 382 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 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 _________________ If my answer helped, dont forget KUDOS! IMPOSSIBLE IS NOTHING Originally posted by roygush on 05 Jan 2013, 13:17. Last edited by Bunuel on 11 Oct 2017, 03:34, edited 3 times in total. Edited the question. ##### Most Helpful Expert Reply Math Expert Joined: 02 Sep 2009 Posts: 51218 Re: The integers x and y are both positive, the remainder when x is divide [#permalink] ### Show Tags 07 Jan 2013, 03:34 8 6 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. _________________ ##### General Discussion VP Status: Been a long time guys... Joined: 03 Feb 2011 Posts: 1131 Location: United States (NY) Concentration: Finance, Marketing GPA: 3.75 Re: The integers x and y are both positive, the remainder when x is divide [#permalink] ### Show Tags 06 Jan 2013, 00:50 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. _________________ Manager Joined: 01 Sep 2012 Posts: 117 Re: The integers x and y are both positive, the remainder when x is divide [#permalink] ### Show Tags 07 Jan 2013, 06:04 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? _________________ If my answer helped, dont forget KUDOS! IMPOSSIBLE IS NOTHING Math Expert Joined: 02 Sep 2009 Posts: 51218 Re: The integers x and y are both positive, the remainder when x is divide [#permalink] ### Show Tags 07 Jan 2013, 08:44 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. For more check GMAT Math Book chapter on remainders: remainders-144665.html Hope it helps. _________________ Manager Joined: 06 Jul 2011 Posts: 95 Re: The integers x and y are both positive, the remainder when x is divide [#permalink] ### Show Tags 25 Jun 2015, 05:30 x= 12q+7 => 2x = 24q+14 y = 12p+3 2x+y = 24q+14+12p+3 => 12(p+2q)+17 Any answer choice from which 17 is subtracted and it doesn't turn out to be a multiple of 12 is the answer. => 51-17 = 34 (not a multiple of 12). (D) is the answer. Senior Manager Status: Math is psycho-logical Joined: 07 Apr 2014 Posts: 417 Location: Netherlands GMAT Date: 02-11-2015 WE: Psychology and Counseling (Other) Re: The integers x and y are both positive, the remainder when x is divide [#permalink] ### Show Tags 30 Jul 2015, 08:04 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... Intern Joined: 08 Nov 2014 Posts: 2 Re: The integers x and y are both positive, the remainder when x is divide [#permalink] ### Show Tags 30 Jul 2015, 08:17 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. Ans - D Intern Joined: 08 Nov 2014 Posts: 2 Re: The integers x and y are both positive, the remainder when x is divide [#permalink] ### Show Tags 30 Jul 2015, 08:20 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. EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 13095 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: The integers x and y are both positive, the remainder when x is divide [#permalink] ### Show Tags 30 Jul 2015, 14:27 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. Thus, 51 is the option that is NOT possible… Final Answer: GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****

Intern
Joined: 19 May 2016
Posts: 29
Location: United States
Concentration: Strategy, Human Resources
GMAT 1: 710 Q46 V41
GMAT 2: 730 Q49 V41
GMAT 3: 680 Q46 V37
WE: Operations (Manufacturing)
Re: The integers x and y are both positive, the remainder when x is divide  [#permalink]

### Show Tags

15 Jul 2016, 18:36
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.

Goodluck!
CEO
Joined: 11 Sep 2015
Posts: 3238
The integers x and y are both positive, the remainder when x is divide  [#permalink]

### Show Tags

19 Apr 2018, 15:15
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

Cheers,
Brent

RELATED VIDEO FROM OUR COURSE

_________________

Test confidently with gmatprepnow.com

Director
Status: Tutor - BrushMyQuant
Joined: 05 Apr 2011
Posts: 610
Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
GPA: 3
WE: Information Technology (Computer Software)
Re: The integers x and y are both positive, the remainder when x is divide  [#permalink]

### Show Tags

28 Jun 2018, 07:08
1
Top Contributor
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.

_________________

Ankit

Check my Tutoring Site -> Brush My Quant

GMAT Quant Tutor
How to start GMAT preparations?
How to Improve Quant Score?
Gmatclub Topic Tags
Check out my GMAT debrief

How to Solve :
Statistics || Reflection of a line || Remainder Problems || Inequalities

Math Expert
Joined: 02 Sep 2009
Posts: 51218
Re: The integers x and y are both positive, the remainder when x is divide  [#permalink]

### Show Tags

28 Jun 2018, 07:38
BrushMyQuant 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.

___________________
Done. Thank you.
_________________
Re: The integers x and y are both positive, the remainder when x is divide &nbs [#permalink] 28 Jun 2018, 07:38
Display posts from previous: Sort by