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.
Nov 1912:30 PM EST
-01:30 PM EST
Learn how Keshav, a Chartered Accountant, scored an impressive 705 on GMAT in just 30 days with GMATWhiz's expert guidance. In this video, he shares preparation tips and strategies that worked for him, including the mock, time management, and more
Nov 2001:30 PM EST
-02:30 PM IST
Learn how Kamakshi achieved a GMAT 675 with an impressive 96th %ile in Data Insights. Discover the unique methods and exam strategies that helped her excel in DI along with other sections for a balanced and high score.
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2407:00 PM PST
-08:00 PM PST
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
08 Jan 2022, 12:50
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
38% (02:29) correct
62%
(02:38)
wrong
based on 87
sessions
History
Date
Time
Result
Not Attempted Yet
If a and b are positive integers greater than 1 and least common multiple of 5a and 5b is 60, what is the value of a?
(1) Greatest Common Divisor of 3a and 3b is 6.
(2) Twice of Greatest Common Divisor of 3a and 3b is 3b.
(1) Greatest Common Divisor of 3a and 3b is 6.
(2) Twice of Greatest Common Divisor of 3a and 3b is 3b.
Updated on: 09 Jan 2022, 08:20
Originally posted by ReedArnoldMPREP on 08 Jan 2022, 14:13.
Last edited by ReedArnoldMPREP on 09 Jan 2022, 08:20, edited 1 time in total.
Last edited by ReedArnoldMPREP on 09 Jan 2022, 08:20, edited 1 time in total.
Kudos
Bookmarks
Quote:
Okay, so I have divisibility here. The givens are that a,b are POS INT, and both > 1.
60 is the LCM of 5a and 5b... That means 5a and 5b are both factors of 60.
Least common multiple of 5a and 5b is 60. A little tricky... Let's think. 60 = 2^2 * 3 * 5 ... the 5 overlap is confusing me, let me just write the factors of 60:
1*60
2*30
3*20
4*15
5*12
6*10
5a and 5b must both be multiples of 5. So 5a and 5b are either 60, 30, 20, 15, or 10 (can't be '5' since a,b≠1).. Meaning a and b are either 12, 6, 4, 3, or 2.
But I know not every pair works (5a = 20 and 5b = 10, for instance, doesn't work since 20 is the LCM of 20 and 10, not 60).
Okay, I probably COULD keep going through possible cases here... I glance at the statements and they kind of bum me out so I am going to keep working through possible cases:
a and b could be: 12 and 6, 12 and 4, 12 and 3, or 12 and 2; 6 and 4, 4 and 3.
{NOTE: One of the easiest things to mix up on a problem like this is applying a constraint of '5a and 5b' to 'a and b.' a and b's least common multiple can be something other than 60. 5a and 5b are the ones constrainted with a LCM of 60, so I got these possible a and b pairs by thinking about 5a and 5b being: 60 and 30, 60 and 20, 60 and 15, 60 and 12, and 60 and 10; 30 and 20; and 20 and 15....Technically, I suppose '60 and 60' so a and b both 12 would also have to be considered).
Quote:
I had a momentary thought about this "Well I won't know which is a," ... but I'm not sure a and b won't be the *same* (...though it seems very unlikely since the only situation I can think that works would be a=b=12). Still, I'm cautious.
Multiplying my pairs of possible a and b and seeing which have a GCD of 6 tells me a and b could be: 12 and 2, or 6 and 4. But I do not know which is which.
Eliminate A and D
Quote:
Had to pause for a moment to understand what this meant. It tells me the GCD of 3a and 3b is *half* of 3b. But there's no divisor or a number between half that number in itself... So 3a *must* be that number [NOTE: see correction below]. That is 3a = (1/2)3b. Which means
a = (1/2)b
...Okay so I stopped here. Something wasn't clicking. The statements CANNOT contradict themselves and neither case that was possible in statement 1 was possible in statement 2. 12 and 6 are the only statements that work for statement 2, but 12 and 6 don't work for statement 1...
I think this question is poorly designed. Someone correct me if I'm missing something!
CORRECTION!
I made a bad assumption. It is true that one situation that follows statement 2's rule is when b = 2a. But that is not the *only* scenario that follows b's rule. It works whenever b has some prime factors of a *and then one extra two*.
So b=12 and a = 6 is a case that works, **but so does a = 6 and b = 4.**
Here's another fun part. In either case a = 6.
Great question! Tricky stuff.
Updated on: 09 Jan 2022, 09:13
Originally posted by gmatophobia on 09 Jan 2022, 02:06.
Last edited by gmatophobia on 09 Jan 2022, 09:13, edited 1 time in total.
Last edited by gmatophobia on 09 Jan 2022, 09:13, edited 1 time in total.
Kudos
Bookmarks
Official Explanation
Step 1: Analyse Question Stem
* a and b are positive integers greater than 1
* LCM(5a, 5b) = 60
* Thus, we can infer that LCM(a, b) = \(\frac{16}{5}\) = 12 = \(2^2 * 3^1\)
* Since a and b are > 1
* Some of the possible combinations of a and b are (4, 6), (3, 4), (2, 12), (6,12), and (12, 12)
We need to find the value of a.
Statement 1: GCD of 3a and 3b =6
Gcd(3a, 3b) =6
* Thus, the GCD(a, b) =\(\frac{6}{3}\) = 2
* We know that LCM of (a, b) = \(2^2 * 3^1\)
Thus, we can infer that possible combinations of (a, b) can be (4, 6), (6, 4), (2, 12) or (12, 2)
So, there can be multiple values of a.
Hence, statement 1 is not sufficient, we can eliminate Options A and D.
Statement 2: Twice of GCD of 3a and 3b is 3b.
2*GCD(3a, 3b) = 3b
GCD(a, b) = \(\frac{b}{2}\)
Let’s see which of the combination satisfy this condition.
download.png [ 6.32 KiB | Viewed 3444 times ]
The only cases that satisfy the condition are (6, 4) (6, 12)
And in both these case the value of a = 6.
Hence, statement 2 is sufficient, the correct answer is Option B.
Step 1: Analyse Question Stem
* a and b are positive integers greater than 1
* LCM(5a, 5b) = 60
* Thus, we can infer that LCM(a, b) = \(\frac{16}{5}\) = 12 = \(2^2 * 3^1\)
* Since a and b are > 1
* Some of the possible combinations of a and b are (4, 6), (3, 4), (2, 12), (6,12), and (12, 12)
We need to find the value of a.
Statement 1: GCD of 3a and 3b =6
Gcd(3a, 3b) =6
* Thus, the GCD(a, b) =\(\frac{6}{3}\) = 2
* We know that LCM of (a, b) = \(2^2 * 3^1\)
Thus, we can infer that possible combinations of (a, b) can be (4, 6), (6, 4), (2, 12) or (12, 2)
So, there can be multiple values of a.
Hence, statement 1 is not sufficient, we can eliminate Options A and D.
Statement 2: Twice of GCD of 3a and 3b is 3b.
2*GCD(3a, 3b) = 3b
GCD(a, b) = \(\frac{b}{2}\)
Let’s see which of the combination satisfy this condition.
Attachment:
download.png [ 6.32 KiB | Viewed 3444 times ]
The only cases that satisfy the condition are (6, 4) (6, 12)
And in both these case the value of a = 6.
Hence, statement 2 is sufficient, the correct answer is Option B.
