Given that a, b, c, and d are different nonzero digits and that 10d +

Question

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 · Accepted Answer

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: ab dc +db ca ----- 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 8 5 81 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 - https://gmatclub.com/forum/search.php?se ... tag_id=193 to learn more about this type of questions. Hope it helps.

enigma123 · Answer

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 · Answer

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.

Hope it's clear.

enigma123 · Answer

Perfect!!!! What an explanation. Thanks a ton.

subhajeet · Answer

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

M3tm4n · Answer

Isn't it be easier just to look for b? The hundreds digit of the sum must be even. There is only one answer choice where the hundreds digit is odd.

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Bunuel · 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.

Hope it helps.

wfmd · Answer

Hey Bunuel,

most is clear, except for the starting point:

Why is 10d + 11c < 100 – a the same as (10d+c)+(10c+a)<100 ???
That's not clear for me... :-/

Bunuel · Answer

10d + 11c < 100 – a

Add a to both sides: 10d + 11c + a< 100;
10d + (10c + c) + a< 100;
(10d+c)+(10c+a)<100.

Hope it's clear.

Nixondutta · Answer

Best explanation ever. couldn't be more clear.

htieuan · Answer

My question is b is nonzero digit, how can 2b=0 in option E?

Posted from my mobile device

Bunuel · Answer

______________________
What if b = 5?

udaypratapsingh99 · Answer

Given:  10d + 11c < 100 – a,
 Inference:   Neither of 'd' and 'c' can be 9

Now,
 a b dc
 +d b ca

1) As above addition operation has unit b on hundreds place,  b+b must give an even number  unless a 1 is carried from the addition at the tens place. 
2) As neither of 'd' and 'c' can be 9, both (i) digit of sum at tens place to be 8 and (ii) carry over of 1 to hundreds place, cannot happen together.

Thus, option (C) 8  58  1 is not possible.

C is the Correct Answer

Kinshook · Answer

Asked: 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

abdc = 1000a + 100b + 10d + c
dbca = 1000d + 100b + 10c + a

Sum of last 2 digits = 10d + 11c + a < 100; There is no carry over to hundredth digit

Hundredth digit = 200b is even; therefore, abdc + dbca can not be 8581.
 
IMO C

DeepanshuGupta · Answer

I am all clear with the question only if I take a simple assumption that the no. are positive. But my point is that where is it written that they are positive? It is written that the numbers are non-zero. Now non-zero could be either positive or negative. Is it so that we have eliminated the possibility of numbers being negative by looking at the answer options. Kindly explain.

bumpbot · Answer

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