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 350,000 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

4

This post received KUDOS

Expert's post

3

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

2

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

gmatclubot

Re: In the correctly worked addition problem above, A, B, C, D
[#permalink]
01 Aug 2014, 15:35

How the growth of emerging markets will strain global finance : Emerging economies need access to capital (i.e., finance) in order to fund the projects necessary for...

One question I get a lot from prospective students is what to do in the summer before the MBA program. Like a lot of folks from non traditional backgrounds...