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.

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]
31 Jan 2012, 16:51

3

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

37% (02:50) correct
63% (02:24) wrong based on 70 sessions

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]
31 Jan 2012, 20:08

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]
03 Jun 2013, 07:16

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.