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 2007:30 AM PST
-08:30 AM PST
Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
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 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.
27 May 2018, 05:31
Kudos
Bookmarks
D
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:
23% (03:07) correct
77%
(03:05)
wrong
based on 473
sessions
History
Date
Time
Result
Not Attempted Yet
A three digit number A with hundreds digit a, tens digit b and units digit c is written in reverse order to form another three digit number B. If B > A , is B + A > 1200?
(1) B - A is divisible by 8
(2) A is a multiple of 5
(1) B - A is divisible by 8
(2) A is a multiple of 5
06 Jun 2018, 22:13
Kudos
Bookmarks
GMATSkilled
A is 100a+10b+c
B is 100c+10b+a
Let's see the statements
1) B-A is div by 8
So (100c+10b+a)-(100a+10b+c)=99(c-a) is div by 8...
Since C and a are digits, ways the equation is satisfied
a) c is 9 and a is 1 , number becomes 9b1 and 1b9
Even you take b as max possible 9, sum becomes 991+199=1190, <1200
b) c is 8 and a is 0, but then A becomes two digit number so not possible..
So number can be only 9b1 and 1b9, and answer will be NO irrespective of value of B
Sufficient
2) A is a multiple of 5..
So C can be either 5 or 0 but 0 is not possible as the other number B becomes two digit number.
Hence ab5 and 5ba where 5ba>ab5 but the larger number has to be > 1200/2 or 600.. hence Ans is NO
Sufficient
D
27 May 2018, 06:50
Kudos
Bookmarks
Statement I alone is not sufficient to answer the question, since it gives us multiple possibilities.
Let's take statement II:
Since A is a multiple of 5, it should end with 5 or 0, since the reverse order number should also be a three digit number, the number cannot end with 0. Hence, A is of the form ab5. Then B should be 5ba, then B+A=5ba + ab5. Since B>A, a can take values (5,4,3,2,1), in any case A+B will always be <1200. Hence statement II alone is sufficient to answer.
IMO, option B.
Thanks and regards,
Sharan Salem
Let's take statement II:
Since A is a multiple of 5, it should end with 5 or 0, since the reverse order number should also be a three digit number, the number cannot end with 0. Hence, A is of the form ab5. Then B should be 5ba, then B+A=5ba + ab5. Since B>A, a can take values (5,4,3,2,1), in any case A+B will always be <1200. Hence statement II alone is sufficient to answer.
IMO, option B.
Thanks and regards,
Sharan Salem
