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15 Jun 2020, 07:06
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
67% (02:50) correct
33%
(02:48)
wrong
based on 337
sessions
History
Date
Time
Result
Not Attempted Yet
AA
+
BB
+
CC
___
ABC
In the correctly worked addition problem above, A, B, and C are distinct digits. What is the sum of A, B and C ?
A. 12
B. 15
C. 17
D. 18
E. 19
Are You Up For the Challenge: 700 Level Questions
+
BB
+
CC
___
ABC
In the correctly worked addition problem above, A, B, and C are distinct digits. What is the sum of A, B and C ?
A. 12
B. 15
C. 17
D. 18
E. 19
Are You Up For the Challenge: 700 Level Questions
yashikaaggarwal
Senior Moderator - Masters Forum
Joined: 19 Jan 2020
Last visit: 29 Mar 2026
Posts: 3,089
Given Kudos: 1,510
Location: India
Schools: Cranfield (A) Warwick MiM "23 (M)
GPA: 4
WE:Analyst (Internet and New Media)
15 Jun 2020, 12:28
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Bookmarks
=> 11A+11B+11C = 100A+10B+C
=> B+10C = 89A
Nearest no. To 89 which is a multiple of 10 is 80
Therefore C = 8
=> B+80 = 89A
=> on comparing both sides
=> B = 9 & A = 1
Sum of A+B+C = 1+9+8 = 18
Answer is D
Posted from my mobile device
=> B+10C = 89A
Nearest no. To 89 which is a multiple of 10 is 80
Therefore C = 8
=> B+80 = 89A
=> on comparing both sides
=> B = 9 & A = 1
Sum of A+B+C = 1+9+8 = 18
Answer is D
Posted from my mobile device
09 May 2024, 17:12
Kudos
Bookmarks
AA
+
BB
+
CC
___
ABC
In the correctly worked addition problem above, A, B, and C are distinct digits. What is the sum of A, B and C ?
AA = 10A + 1A = 11A
BB = 10B + 1B = 11B
CC = 10C + 1C = 11C
11A + 11B + 11C = 100A + 10B + C
B + 10C = 89A
B + 10C is a multiple of 89.
B + 10C cannot be greater than 99 since B and C are digits. So, B + 10C = 89.
B must be 9.
C must be 8.
Thus, since B + 10C = 89A, A = 1
A + B + C = 1 + 9 + 8 = 18
A. 12
B. 15
C. 17
D. 18
E. 19
Correct answer: D
+
BB
+
CC
___
ABC
In the correctly worked addition problem above, A, B, and C are distinct digits. What is the sum of A, B and C ?
AA = 10A + 1A = 11A
BB = 10B + 1B = 11B
CC = 10C + 1C = 11C
11A + 11B + 11C = 100A + 10B + C
B + 10C = 89A
B + 10C is a multiple of 89.
B + 10C cannot be greater than 99 since B and C are digits. So, B + 10C = 89.
B must be 9.
C must be 8.
Thus, since B + 10C = 89A, A = 1
A + B + C = 1 + 9 + 8 = 18
A. 12
B. 15
C. 17
D. 18
E. 19
Correct answer: D
