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
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Re: In the correctly worked addition problem above, A, B, C, D [#permalink]
12 Mar 2012, 12:22
7
This post received KUDOS
Expert's post
10
This post was BOOKMARKED
In the correctly worked addition problem above, A, B, C, D, E, F, and G are distinct digits. What is the sum of E, F, and G ?
AD BD CD --- EFG
Notice that E can be only 1 or 2 (no sum of 3 two-digit numbers can give number more than 297).
(1) A, B, and C are consecutive odd integers. 3 cases are possible:
(i) A, B, and C are 1, 3, and 5 (it doesn't matter which is which) --> 1+3+5=9 then E (hundreds digit) can only be 1, which is not possible since we are told that the digits are distinct and we already have 1 (A, B, or C);
(ii) A, B, and C are 3, 5, and 7 (it doesn't matter which is which) --> 3+5+7=15 then E (hundreds digit) can only be 1. So, D can be 0, 2, 4, 6, 8, or 9. After trial and error we can get that only D=4 will give all distinct digits: 34 54 74 --- 162
E+F+G=9.
(iii) A, B, and C are 5, 7, and 9 (it doesn't matter which is which) --> 5+7+9=21 then E (hundreds digit) can only be 2. So, D can be 1, 3, 4, 6, or 8. After trial and error we can get that only D=8 will give all distinct digits: 58 78 98 --- 234
E+F+G=9.
So, as you can see in both valid cases (ii and iii) the sum of E, F, and G is 9. Sufficient.
(2) E = 2. After some trial and error you can find that several numbers can be found which will give different values for the sum of E, F, and G, for example: 58+78+98=234 and 38+78+98=214. Not sufficient.
Answer: A.
P.S. Though not very hard this question is not likely to appear on the GMAT because of long and boring math. _________________
Re: In the correctly worked addition problem above, A, B, C, D [#permalink]
12 Mar 2012, 23:12
3
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
carcass wrote:
In the correctly worked addition problem above, A, B, C, D, E, F, and G are distinct digits. What is the sum of E, F, and G ?
(1) A, B, and C are consecutive odd integers
(2) E = 2
I concur with Bunuel here. You are not likely to see this, at least not in DS format. You could possibly see something similar in PS format and it will be based on logic, not hit and trial. Hit and trial makes it long, repetitive and cumbersome, things GMAT doesn't mess with. You will have a starting point and there will be a reason why an alphabet will stand for a particular digit. _________________
Re: In the correctly worked addition problem above, A, B, C, D [#permalink]
17 Oct 2013, 22:09
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
Re: In the correctly worked addition problem above, A, B, C, D [#permalink]
18 Oct 2013, 08:34
Bunuel wrote:
In the correctly worked addition problem above, A, B, C, D, E, F, and G are distinct digits. What is the sum of E, F, and G ?
AD BD CD --- EFG
Notice that E can be only 1 or 2 (no sum of 3 two-digit numbers can give number more than 297).
(1) A, B, and C are consecutive odd integers. 3 cases are possible:
(i) A, B, and C are 1, 3, and 5 (it doesn't matter which is which) --> 1+3+5=9 then E (hundreds digit) can only be 1, which is not possible since we are told that the digits are distinct and we already have 1 (A, B, or C);
(ii) A, B, and C are 3, 5, and 7 (it doesn't matter which is which) --> 3+5+7=15 then E (hundreds digit) can only be 1. So, D can be 0, 2, 4, 6, 8, or 9. After trial and error we can get that only D=4 will give all distinct digits: 34 54 74 --- 162
E+F+G=9.
(iii) A, B, and C are 5, 7, and 9 (it doesn't matter which is which) --> 5+7+9=21 then E (hundreds digit) can only be 2. So, D can be 1, 3, 4, 6, or 8. After trial and error we can get that only D=8 will give all distinct digits: 58 78 98 --- 234
E+F+G=9.
So, as you can see in both valid cases (ii and iii) the sum of E, F, and G is 9. Sufficient.
(2) E = 2. After some trial and error you can find that several numbers can be found which will give different values for the sum of E, F, and G, for example: 58+78+98=234 and 38+78+98=214. Not sufficient.
Answer: A.
P.S. Though not very hard this question is not likely to appear on the GMAT because of long and boring math.
Hi Bunuel ,
Just curious to know why A=1,B=3,C=5 and sum = 9 and the resulting number 96 cannot be the case. It says any digits so why EFG =096 cant be a possibility?? _________________
Re: In the correctly worked addition problem above, A, B, C, D [#permalink]
20 Oct 2013, 12:20
Expert's post
adg142000 wrote:
Bunuel wrote:
In the correctly worked addition problem above, A, B, C, D, E, F, and G are distinct digits. What is the sum of E, F, and G ?
AD BD CD --- EFG
Notice that E can be only 1 or 2 (no sum of 3 two-digit numbers can give number more than 297).
(1) A, B, and C are consecutive odd integers. 3 cases are possible:
(i) A, B, and C are 1, 3, and 5 (it doesn't matter which is which) --> 1+3+5=9 then E (hundreds digit) can only be 1, which is not possible since we are told that the digits are distinct and we already have 1 (A, B, or C);
(ii) A, B, and C are 3, 5, and 7 (it doesn't matter which is which) --> 3+5+7=15 then E (hundreds digit) can only be 1. So, D can be 0, 2, 4, 6, 8, or 9. After trial and error we can get that only D=4 will give all distinct digits: 34 54 74 --- 162
E+F+G=9.
(iii) A, B, and C are 5, 7, and 9 (it doesn't matter which is which) --> 5+7+9=21 then E (hundreds digit) can only be 2. So, D can be 1, 3, 4, 6, or 8. After trial and error we can get that only D=8 will give all distinct digits: 58 78 98 --- 234
E+F+G=9.
So, as you can see in both valid cases (ii and iii) the sum of E, F, and G is 9. Sufficient.
(2) E = 2. After some trial and error you can find that several numbers can be found which will give different values for the sum of E, F, and G, for example: 58+78+98=234 and 38+78+98=214. Not sufficient.
Answer: A.
P.S. Though not very hard this question is not likely to appear on the GMAT because of long and boring math.
Hi Bunuel ,
Just curious to know why A=1,B=3,C=5 and sum = 9 and the resulting number 96 cannot be the case. It says any digits so why EFG =096 cant be a possibility??
From the stem we can assume that E is not 0. _________________
Re: In the correctly worked addition problem above, A, B, C, D [#permalink]
20 Oct 2013, 19:27
Expert's post
adg142000 wrote:
Hi Bunuel ,
Just curious to know why A=1,B=3,C=5 and sum = 9 and the resulting number 96 cannot be the case. It says any digits so why EFG =096 cant be a possibility??
To add to what Bunuel said, the question stem tells us that when we add three 2 digit numbers, we get a 3 digit number. Had you obtained a 2 digit number as the sum, you would have wirtten the addition as
Re: In the correctly worked addition problem above, A, B, C, D [#permalink]
01 Aug 2014, 15:34
VeritasPrepKarishma wrote:
carcass wrote:
In the correctly worked addition problem above, A, B, C, D, E, F, and G are distinct digits. What is the sum of E, F, and G ?
(1) A, B, and C are consecutive odd integers
(2) E = 2
I concur with Bunuel here. You are not likely to see this, at least not in DS format. You could possibly see something similar in PS format and it will be based on logic, not hit and trial. Hit and trial makes it long, repetitive and cumbersome, things GMAT doesn't mess with. You will have a starting point and there will be a reason why an alphabet will stand for a particular digit.
That's a relief. Cause I was able to solve the problem but it took over 5 mins. _________________
......................................................................... +1 Kudos please, if you like my post
Re: In the correctly worked addition problem above, A, B, C, D [#permalink]
01 Aug 2014, 15:35
Bunuel wrote:
In the correctly worked addition problem above, A, B, C, D, E, F, and G are distinct digits. What is the sum of E, F, and G ?
AD BD CD --- EFG
Notice that E can be only 1 or 2 (no sum of 3 two-digit numbers can give number more than 297).
(1) A, B, and C are consecutive odd integers. 3 cases are possible:
(i) A, B, and C are 1, 3, and 5 (it doesn't matter which is which) --> 1+3+5=9 then E (hundreds digit) can only be 1, which is not possible since we are told that the digits are distinct and we already have 1 (A, B, or C);
(ii) A, B, and C are 3, 5, and 7 (it doesn't matter which is which) --> 3+5+7=15 then E (hundreds digit) can only be 1. So, D can be 0, 2, 4, 6, 8, or 9. After trial and error we can get that only D=4 will give all distinct digits: 34 54 74 --- 162
E+F+G=9.
(iii) A, B, and C are 5, 7, and 9 (it doesn't matter which is which) --> 5+7+9=21 then E (hundreds digit) can only be 2. So, D can be 1, 3, 4, 6, or 8. After trial and error we can get that only D=8 will give all distinct digits: 58 78 98 --- 234
E+F+G=9.
So, as you can see in both valid cases (ii and iii) the sum of E, F, and G is 9. Sufficient.
(2) E = 2. After some trial and error you can find that several numbers can be found which will give different values for the sum of E, F, and G, for example: 58+78+98=234 and 38+78+98=214. Not sufficient.
Answer: A.
P.S. Though not very hard this question is not likely to appear on the GMAT because of long and boring math.
how long does it take when you did your error and trial? It took me almost 7 minutes _________________
......................................................................... +1 Kudos please, if you like my post
Re: In the correctly worked addition problem above, A, B, C, D [#permalink]
06 Jul 2015, 20:35
Another version of the same numbers could be 16 +36 + 56 = 108 => 9
Bunuel wrote:
In the correctly worked addition problem above, A, B, C, D, E, F, and G are distinct digits. What is the sum of E, F, and G ?
AD BD CD --- EFG
Notice that E can be only 1 or 2 (no sum of 3 two-digit numbers can give number more than 297).
(1) A, B, and C are consecutive odd integers. 3 cases are possible:
(i) A, B, and C are 1, 3, and 5 (it doesn't matter which is which) --> 1+3+5=9 then E (hundreds digit) can only be 1, which is not possible since we are told that the digits are distinct and we already have 1 (A, B, or C);
(ii) A, B, and C are 3, 5, and 7 (it doesn't matter which is which) --> 3+5+7=15 then E (hundreds digit) can only be 1. So, D can be 0, 2, 4, 6, 8, or 9. After trial and error we can get that only D=4 will give all distinct digits: 34 54 74 --- 162
E+F+G=9.
(iii) A, B, and C are 5, 7, and 9 (it doesn't matter which is which) --> 5+7+9=21 then E (hundreds digit) can only be 2. So, D can be 1, 3, 4, 6, or 8. After trial and error we can get that only D=8 will give all distinct digits: 58 78 98 --- 234
E+F+G=9.
So, as you can see in both valid cases (ii and iii) the sum of E, F, and G is 9. Sufficient.
(2) E = 2. After some trial and error you can find that several numbers can be found which will give different values for the sum of E, F, and G, for example: 58+78+98=234 and 38+78+98=214. Not sufficient.
Answer: A.
P.S. Though not very hard this question is not likely to appear on the GMAT because of long and boring math.
Re: In the correctly worked addition problem above, A, B, C, D [#permalink]
07 Jul 2015, 00:07
1
This post received KUDOS
Expert's post
kelvind13 wrote:
Another version of the same numbers could be 16 +36 + 56 = 108 => 9
Bunuel wrote:
In the correctly worked addition problem above, A, B, C, D, E, F, and G are distinct digits. What is the sum of E, F, and G ?
AD BD CD --- EFG
Notice that E can be only 1 or 2 (no sum of 3 two-digit numbers can give number more than 297).
(1) A, B, and C are consecutive odd integers. 3 cases are possible:
(i) A, B, and C are 1, 3, and 5 (it doesn't matter which is which) --> 1+3+5=9 then E (hundreds digit) can only be 1, which is not possible since we are told that the digits are distinct and we already have 1 (A, B, or C);
(ii) A, B, and C are 3, 5, and 7 (it doesn't matter which is which) --> 3+5+7=15 then E (hundreds digit) can only be 1. So, D can be 0, 2, 4, 6, 8, or 9. After trial and error we can get that only D=4 will give all distinct digits: 34 54 74 --- 162
E+F+G=9.
(iii) A, B, and C are 5, 7, and 9 (it doesn't matter which is which) --> 5+7+9=21 then E (hundreds digit) can only be 2. So, D can be 1, 3, 4, 6, or 8. After trial and error we can get that only D=8 will give all distinct digits: 58 78 98 --- 234
E+F+G=9.
So, as you can see in both valid cases (ii and iii) the sum of E, F, and G is 9. Sufficient.
(2) E = 2. After some trial and error you can find that several numbers can be found which will give different values for the sum of E, F, and G, for example: 58+78+98=234 and 38+78+98=214. Not sufficient.
Answer: A.
P.S. Though not very hard this question is not likely to appear on the GMAT because of long and boring math.
No, this does not work: A and E must be distinct. please refer to the highlighted part. _________________
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Wow! MBA life is hectic indeed. Time flies by. It is hard to keep track of the time. Last week was high intense training Yeah, Finance, Accounting, Marketing, Economics...