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.
11 Aug 2021, 04:17
Kudos
Bookmarks
Nik711
The two red portions contradict each other:
Whereas C=5 in the first red statement, C=9 in the second.
rvgmat12
Joined: 19 Oct 2014
Last visit: 15 Nov 2025
Posts: 356
Own Kudos:
Given Kudos: 189
Location: United Arab Emirates
Schools: Wharton '23 Booth '23 INSEAD Jan'21 Intake
Products:
29 Mar 2022, 04:29
Kudos
Bookmarks
Official Explanation:
First, note that A, B, C, and D have to be digits between 1 and 9 (and the problem does not prevent some letters from having the same value).
The two rightmost columns both contain C + B = D. From that information, you can deduce that B + C < 9. (If B + C were 10 or more, then the rightmost column would “carry over” into the next column, making the tens digit into D + 1 rather than D.)
Next, A + B = C. A larger value of C, then, will reduce B (because B + C < 9) and therefore increase A (because A + B = C).
Since the questions asks to maximize the product for A and B, consider only the cases in which B + C = 9. Further, since A + B = C, you also know that B must be less than C. (Not sure why? Test a couple of real numbers to figure it out.) That leaves only a few cases, so write them out:
If B = 1 and C = 8, then A = 7. In this case, the product of A and B is 7.
If B = 2 and C = 7, then A = 5. In this case, the product of A and B is 10.
If B = 3 and C = 6, then A = 3. In this case, the product of A and B is 9.
If B = 4 and C = 5, then A = 1. In this case, the product of A and B is 4.
That’s all of the possible cases. The largest possible product is 10.
The correct answer is B.
First, note that A, B, C, and D have to be digits between 1 and 9 (and the problem does not prevent some letters from having the same value).
The two rightmost columns both contain C + B = D. From that information, you can deduce that B + C < 9. (If B + C were 10 or more, then the rightmost column would “carry over” into the next column, making the tens digit into D + 1 rather than D.)
Next, A + B = C. A larger value of C, then, will reduce B (because B + C < 9) and therefore increase A (because A + B = C).
Since the questions asks to maximize the product for A and B, consider only the cases in which B + C = 9. Further, since A + B = C, you also know that B must be less than C. (Not sure why? Test a couple of real numbers to figure it out.) That leaves only a few cases, so write them out:
If B = 1 and C = 8, then A = 7. In this case, the product of A and B is 7.
If B = 2 and C = 7, then A = 5. In this case, the product of A and B is 10.
If B = 3 and C = 6, then A = 3. In this case, the product of A and B is 9.
If B = 4 and C = 5, then A = 1. In this case, the product of A and B is 4.
That’s all of the possible cases. The largest possible product is 10.
The correct answer is B.
21 Aug 2023, 15:55
Kudos
Bookmarks
C+B = D =< 9
A+B = C
Therefore,
2B+A =<9
2(1) + 7 =< 9 -> 1*7 = 7
2(2) + 5 =< 9 -> 2*5 = 10
2(3) + 3 =< 9 -> 3*3 = 9
2(4) + 1 =< 9 -> 4*1 = 4
Can't go further since 2(5) > 9
A+B = C
Therefore,
2B+A =<9
2(1) + 7 =< 9 -> 1*7 = 7
2(2) + 5 =< 9 -> 2*5 = 10
2(3) + 3 =< 9 -> 3*3 = 9
2(4) + 1 =< 9 -> 4*1 = 4
Can't go further since 2(5) > 9
