Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Mar 2305:30 AM PDT
-07:30 AM PDT
Achieving a high GMAT score while balancing a hectic work life is challenging, but with the right strategy, it's absolutely possible. Discover the ultimate GMAT study strategy designed exclusively for working professionals...
Mar 1910:00 AM EDT
-11:59 PM EDT
Get a massive $250 off the TTP OnDemand GMAT course by using the coupon code FLASH25 at checkout. If you prefer learning through engaging video lessons, TTP OnDemand GMAT is exactly what you need.
Mar 2610:00 AM PDT
-11:00 AM PDT
Confused about how to get the most from your GMAT mock tests? In this video, learn the perfect strategy to analyze GMAT mocks effectively and boost your score. We break down the common mistakes GMAT aspirants make when using mock tests
Mar 2610:00 AM EDT
-11:59 PM EDT
The Target Test Prep GMAT Flash Sale is LIVE! Get 25% off our game-changing course and save up to $400 today! Use code FLASH25 at checkout. This limited-time deal won’t last long, so grab your discount now!- Mar 27
09:30 AM PDT
-10:30 AM PDT
Preparing for the GMAT? Avoid these 5 major mistakes that can lower your score and hurt your MBA dreams. In this expert panel discussion, 4 top GMAT coaches - GMAT Ninja, Jeff Miller from TTP, Rida Shafiq from eGMAT, and Chris Gentry from Manhattan Prep - Mar 27
10:00 AM PDT
-11:00 AM PDT
What happens after getting an MBA? Is it worth the investment? To find out, I interviewed 10 GMAT Club professionals who have completed their MBAs and built careers in consulting, finance, tech, and entrepreneurship. Their answers might surprise you! 
Mar 3011:00 AM IST
-01:00 PM IST
Attend this session to evaluate your current skill level, learn process skills, and solve through tough Arithmetic questions.
02 Mar 2012, 10:36
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
65% (02:05) correct
35%
(01:54)
wrong
based on 2448
sessions
History
Date
Time
Result
Not Attempted Yet
If x, y, and d are integers and d is odd, are both x and y divisible by d ?
(1) x + y is divisible by d.
(2) x − y is divisible by d.
DS44402.01
(1) x + y is divisible by d.
(2) x − y is divisible by d.
DS44402.01
02 Mar 2012, 11:04
Kudos
Bookmarks
If x, y and d are integers and d is odd, are both x and y divisible by d?
Question asks whether both x and y are multiples of d.
(1) x+y is divisible by d --> if \(x=y=d=1\) then the answer is YES but if \(x=2\), \(y=4\) and \(d=3\) then the answer is NO. Not sufficient.
(2) x-y is divisible by d --> if \(x=y=d=1\) then the answer is YES but if \(x=4\), \(y=1\) and \(d=3\) then the answer is NO. Not sufficient.
(1)+(2) From (1) \(x+y=dq\) and from (2) \(x-y=dp\) --> sum those two equations: \(2x=d(q+p)\) --> \(x=\frac{d(q+p)}{2}\). Now, since \(x\) is an integer and \(d\) is an odd number then \(\frac{q+p}{2}=integer\) (d/2 cannot be an integer because d is odd) --> \(x=d*integer\) --> \(x\) is a multiple of \(d\). From (1) {multiple of d}+y={multiple of d} --> hence \(y\) must also be a multiple of \(d\). Sufficient.
Answer: C.
Below might help to understand this concept better.
If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=6\) and \(b=9\), both divisible by 3 ---> \(a+b=15\) and \(a-b=-3\), again both divisible by 3.
If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 ---> \(a+b=11\) and \(a-b=1\), neither is divisible by 3.
If integers \(a\) and \(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=5\) and \(b=4\), neither is divisible by 3 ---> \(a+b=9\), is divisible by 3 and \(a-b=1\), is not divisible by 3;
OR: \(a=6\) and \(b=3\), neither is divisible by 5 ---> \(a+b=9\) and \(a-b=3\), neither is divisible by 5;
OR: \(a=2\) and \(b=2\), neither is divisible by 4 ---> \(a+b=4\) and \(a-b=0\), both are divisible by 4.
Hope it's clear.
Question asks whether both x and y are multiples of d.
(1) x+y is divisible by d --> if \(x=y=d=1\) then the answer is YES but if \(x=2\), \(y=4\) and \(d=3\) then the answer is NO. Not sufficient.
(2) x-y is divisible by d --> if \(x=y=d=1\) then the answer is YES but if \(x=4\), \(y=1\) and \(d=3\) then the answer is NO. Not sufficient.
(1)+(2) From (1) \(x+y=dq\) and from (2) \(x-y=dp\) --> sum those two equations: \(2x=d(q+p)\) --> \(x=\frac{d(q+p)}{2}\). Now, since \(x\) is an integer and \(d\) is an odd number then \(\frac{q+p}{2}=integer\) (d/2 cannot be an integer because d is odd) --> \(x=d*integer\) --> \(x\) is a multiple of \(d\). From (1) {multiple of d}+y={multiple of d} --> hence \(y\) must also be a multiple of \(d\). Sufficient.
Answer: C.
Below might help to understand this concept better.
If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=6\) and \(b=9\), both divisible by 3 ---> \(a+b=15\) and \(a-b=-3\), again both divisible by 3.
If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 ---> \(a+b=11\) and \(a-b=1\), neither is divisible by 3.
If integers \(a\) and \(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=5\) and \(b=4\), neither is divisible by 3 ---> \(a+b=9\), is divisible by 3 and \(a-b=1\), is not divisible by 3;
OR: \(a=6\) and \(b=3\), neither is divisible by 5 ---> \(a+b=9\) and \(a-b=3\), neither is divisible by 5;
OR: \(a=2\) and \(b=2\), neither is divisible by 4 ---> \(a+b=4\) and \(a-b=0\), both are divisible by 4.
Hope it's clear.
General Discussion
02 Mar 2012, 11:22
Kudos
Bookmarks
Nice explanation Bunuel - i got it down to C & E and made an educated guess because I couldnt get any example to not work given both conditions..
the part of your explanation that took me a few reads to understand was the
x = D(Q+p)/2 and since d is an odd number (Q+P)/2 must be an integer..
Then i realized since D is odd it couldnt cancel out the 2 below the line to get x to equal an integer..
the part of your explanation that took me a few reads to understand was the
x = D(Q+p)/2 and since d is an odd number (Q+P)/2 must be an integer..
Then i realized since D is odd it couldnt cancel out the 2 below the line to get x to equal an integer..
