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:

Given that a, b, c, and d are different nonzero digits and [#permalink]

Show Tags

31 Jan 2012, 17:51

2

This post received KUDOS

21

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

39% (01:42) correct
61% (02:26) wrong based on 226 sessions

HideShow timer Statistics

Given that a, b, c, and d are different nonzero digits and that 10d + 11c < 100 – a, which of the following could not be a solution to the addition problem below?

abdc + dbca

(A) 3689 (B) 6887 (C) 8581 (D) 9459 (E) 16091

This is the original question and its beyond my head to solve this. can someone please help and try to explain the concept of this?

Also , can you please tell how to solve if the question says COULD BE a solution to the addition problem?

Given that a, b, c, and d are different nonzero digits and that 10d + 11c < 100 – a, which of the following could not be a solution to the addition problem below? abdc + dbca (A) 3689 (B) 6887 (C) 8581 (D) 9459 (E) 16091

This is the original question and its beyond my head to solve this. can someone please help and try to explain the concept of this?

Also , can you please tell how to solve if the question says COULD BE a solution to the addition problem?

10d + 11c < 100 – a --> 10d+11c+a<100 --> (10d+c)+(10c+a)<100. (10d+c) is the way of writing two-digit integer dc and (10c+a) is the way of writing two-digit integer ca. Look at the sum:

abdc +dbca -----

Now, as two-digit integer dc + two-digit integer ca is less than 100, then there won't be carried over 1 to the hundreds place and as b+b=2b=even then the hundreds digit of the given sum must be even too. Thus 8581 could not be the sum of abdc+dbca (for any valid digits of a, b, c, and d).

Answer: C.

As for "could be true" questions: try our new "Must or Could be True Questions" tag - search.php?search_id=tag&tag_id=193 to learn more about this type of questions.

Bunuel - thanks for this buddy. Its almost clear apart from how did you get

Now, as two-digit integer dc + two-digit integer ca is less than 100 then there won't be carried over 1 to the hundreds place and as b+b must be ecen then the hundreds digit of the given sum must be even too _________________

Bunuel - thanks for this buddy. Its almost clear apart from how did you get

Now, as two-digit integer dc + two-digit integer ca is less than 100 then there won't be carried over 1 to the hundreds place and as b+b must be ecen then the hundreds digit of the given sum must be even too

Two-digit integer dc + two-digit integer ca is less than 100: suppose they are 12 and 23: ab12 +db23 ----- XY35

So, there is no carried over 1 to hundreds place or to b+b. Next, b+b=2b=even so the hundreds digit of the given sum, Y in our example, must be even too.

Re: Given that a, b, c, and d are different nonzero digits and [#permalink]

Show Tags

31 Jan 2012, 21:08

enigma123 wrote:

Given that a, b, c, and d are different nonzero digits and that 10d + 11c < 100 – a, which of the following could not be a solution to the addition problem below? abdc + dbca (A) 3689 (B) 6887 (C) 8581 (D) 9459 (E) 16091

This is the original question and its beyond my head to solve this. can someone please help and try to explain the concept of this?

Also , can you please tell how to solve if the question says COULD BE a solution to the addition problem?

Given 10d+11c+a<100

now we have to find the sum of -> 1000a+100b+10d+c+1000d+100b+10c+a

we simplify it further to get -> 1000a+1000d+200b+(11c+10d+a) we know that the value in bracket has to be less than 100 now take the options - let us take C which is 8581 we can break it to 8000+500+81 -> since each of a,b,c and d is an integer so we cannot find a value of b to get the sum of 500 Thus C is the answer

No that's not correct. The hundreds digit will be even if there is no carried over 1 from the sum of the tens digits. So you should check this first. Please refer to the complete solution above.

No that's not correct. The hundreds digit will be even if there is no carried over 1 from the sum of the tens digits. So you should check this first. Please refer to the complete solution above.

Re: Given that a, b, c, and d are different nonzero digits and [#permalink]

Show Tags

03 Jun 2013, 08:16

enigma123 wrote:

Given that a, b, c, and d are different nonzero digits and that 10d + 11c < 100 – a, which of the following could not be a solution to the addition problem below?

abdc + dbca

(A) 3689 (B) 6887 (C) 8581 (D) 9459 (E) 16091

This is the original question and its beyond my head to solve this. can someone please help and try to explain the concept of this?

Also , can you please tell how to solve if the question says COULD BE a solution to the addition problem?

Bunuel's solution is always the perfect one, but I just wanted to share what I did. Pretty much the same as above, though...

10d + 11c + a = 10d + 10c + c + a < 100

this means, 10d + 10c <= 90 (since it cannot be equal to hundred)

then, 10(d+c) <= 90

thus, d + c < 10 and also, c + a < 10

This concludes that hundreds digit needs to be even because there is no carry over.

Re: Given that a, b, c, and d are different nonzero digits and [#permalink]

Show Tags

27 Jun 2015, 23:50

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: Given that a, b, c, and d are different nonzero digits and [#permalink]

Show Tags

01 Jan 2017, 14:44

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: Given that a, b, c, and d are different nonzero digits and [#permalink]

Show Tags

03 Mar 2017, 10:08

Bunuel wrote:

enigma123 wrote:

Given that a, b, c, and d are different nonzero digits and that 10d + 11c < 100 – a, which of the following could not be a solution to the addition problem below? abdc + dbca (A) 3689 (B) 6887 (C) 8581 (D) 9459 (E) 16091

This is the original question and its beyond my head to solve this. can someone please help and try to explain the concept of this?

Also , can you please tell how to solve if the question says COULD BE a solution to the addition problem?

10d + 11c < 100 – a --> 10d+11c+a<100 --> (10d+c)+(10c+a)<100. (10d+c) is the way of writing two-digit integer dc and (10c+a) is the way of writing two-digit integer ca. Look at the sum:

abdc +dbca -----

Now, as two-digit integer dc + two-digit integer ca is less than 100, then there won't be carried over 1 to the hundreds place and as b+b=2b=even then the hundreds digit of the given sum must be even too. Thus 8581 could not be the sum of abdc+dbca (for any valid digits of a, b, c, and d).

Answer: C.

As for "could be true" questions: [b]try our new "Must or Could be True Questions" to learn more about this type of questions.

Hope it helps.

I adopted a different approach and was able to eliminate a different option.

The numeric form of abdc + dbca, lets call it sum S, is (1001a + 1010d + 200b + 11c).

From the the other given condition 10d+11c<100-a implies 11c <100-a-10d.

When we substitute for 11c in S, we get 1001a + 1010d + 200b + 11c < 1001a + 1010d + 200b + (100-a-10d) => S < 1000a + 1000d + 200b + 100

The minumum value of S (given that a,b,c,d are digits i.e. positive DIFFERENT integers) will be 3700.

Given that a, b, c, and d are different nonzero digits and [#permalink]

Show Tags

19 Aug 2017, 07:11

enigma123 wrote:

Given that a, b, c, and d are different nonzero digits and that 10d + 11c < 100 – a, which of the following could not be a solution to the addition problem below?

abdc + dbca

(A) 3689 (B) 6887 (C) 8581 (D) 9459 (E) 16091

This is the original question and its beyond my head to solve this. can someone please help and try to explain the concept of this?

Also , can you please tell how to solve if the question says COULD BE a solution to the addition problem?

This is one of the easiest methods;

abdc + dbca

Clearly unit digit of sum will depend upon (c+a) and tenth digit will depend upon (d+c); We have 10d + 11c < 100 – a or we can say 10d+11c+a< 100 or 10d+10c+c+a< 100 or 10(d+c)+ (c+a)< 100 Now just go through the options;

1. 3689 so (c+a)=9 and we have 10(d+c)+ (c+a)< 100..so (d+c)< 9...here it is 8 (tenth digit) ..so possible 2. 6887 so (c+a)=7 or 17 and we have 10(d+c)+ (c+a)< 100..so (d+c)<9 or 8...here it is 8 (tenth digit) ..so possible 3. 8581 so (c+a)=11..it cant be 1 as a, b, c, and d are different nonzero digits and we have 10(d+c)+ (c+a)< 100..so (d+c)< 8...here it is 8 (tenth digit) ..which can not be possible....So option C..

Give kudos if you find this solution useful. :)
_________________

Please hit kudos button below if you found my post helpful..TIA

Version 8.1 of the WordPress for Android app is now available, with some great enhancements to publishing: background media uploading. Adding images to a post or page? Now...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...