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

 It is currently 14 Nov 2019, 08:56 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # The integers x and y are both positive, the remainder when x is divide

Author Message
TAGS:

### Hide Tags

Manager  Joined: 01 Sep 2012
Posts: 114
The integers x and y are both positive, the remainder when x is divide  [#permalink]

### Show Tags

2
20 00:00

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

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, 14:17.
Last edited by Bunuel on 11 Oct 2017, 04:34, edited 3 times in total.
Edited the question.
Math Expert V
Joined: 02 Sep 2009
Posts: 59039
Re: The integers x and y are both positive, the remainder when x is divide  [#permalink]

### Show Tags

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.

_________________
##### General Discussion
Director  Status: Been a long time guys...
Joined: 03 Feb 2011
Posts: 994
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

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: 114
Re: The integers x and y are both positive, the remainder when x is divide  [#permalink]

### Show Tags

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.

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 V
Joined: 02 Sep 2009
Posts: 59039
Re: The integers x and y are both positive, the remainder when x is divide  [#permalink]

### Show Tags

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.

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: 81
Re: The integers x and y are both positive, the remainder when x is divide  [#permalink]

### Show Tags

1
1
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).

Senior Manager  Status: Math is psycho-logical
Joined: 07 Apr 2014
Posts: 402
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

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

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

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 V
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15444
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

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…

GMAT assassins aren't born, they're made,
Rich
_________________
Current Student B
Joined: 19 May 2016
Posts: 28
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

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!
GMAT Club Legend  V
Joined: 12 Sep 2015
Posts: 4063
The integers x and y are both positive, the remainder when x is divide  [#permalink]

### Show Tags

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

_________________
GMAT Tutor G
Status: Tutor - BrushMyQuant
Joined: 05 Apr 2011
Posts: 622
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

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.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 59039
Re: The integers x and y are both positive, the remainder when x is divide  [#permalink]

### Show Tags

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.
_________________
Non-Human User Joined: 09 Sep 2013
Posts: 13581
Re: The integers x and y are both positive, the remainder when x is divide  [#permalink]

### Show Tags

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.
_________________ Re: The integers x and y are both positive, the remainder when x is divide   [#permalink] 02 Jul 2019, 04:20
Display posts from previous: Sort by

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