Please refer to the addition above. Here A, B, C, and D rep

I tried substituting some numbers - 83+38=121 which suffices the given equation so does 74+47. We have more than one solution for this problem so it can not be determined with the given information. Answer is E

For CDC
to find the hundreds digit,
even if we take AB = 98, which is the highest number we can take,
we get AB + BA = 98 + 89 = 187

Therefore, hundreds digit can only be equal to '1'.
So, units digit is also equal to 1.

CDC = 1D1

Now, to get units digit as 1 in CDC,
let us take B = 2
So, we can not take any digit other than 9 as A

So, AB = 92
BA = 29
Sum = AB + BA = 121 = CDC

A + B + C + D = 9+2+1+2 = 14

Choice A is the answer

C has to be 1 because of the digit.
A,B --> 29+92=121 (this is incorrect cuz D is becoming same as A)
38+83=121 (this could be correct combination since all the digits are unique)
74+47=121 (this could be correct combination since all the digits are unique)
56+65=121 (this could be correct combination since all the digits are unique)
So if we add the integers for each possible combination --> 3+8+1+2=14, 5+6+1+2=14, 7+4+1+2=14
So answer is A.

Solution :

given

(10A+B) + ( 10B+A) = 100 C + 10 D+ C

we get upon simplification :

11(A+B) = 101C+ 10 D

Now both sides need to be multiple of 11 . secondly maximum 11 ( A+B) = 11*18 = 198 so C can be at maximum 1. Zero not possible as as given distinct non-zero digits

So we apply divisibility rules by 11 - if CDC is divisible by 11 then C+C- D=0 so 2C = D

C=1 so , D=2

so 11 so Now we get 11(A+B) = 121

So A+B =11

so A+B +C+D = 11+ 1+2 = 14

So answer is A) 14

Answer: A
A, B between 1,9
Thus A<A+B<18

unit digit A+B = C ------- (1)
ten's digit A+B+(borrow from A+B) = CD (2)

from eqn 2 : C can be 1,2
Let C =1
combining it with eqn 1
then , A+B = 11
there fore A+B+ (borrow from A+B) =11+1 =12
A+B+C+D = 11+1+2 = 14

Let C =2
combine it with eqn 1
A+B >20
Not possible

Therefore option A

As the sum of two distinct digits cannot exceed 17, we have C = 1 necessary ( from C at hundreds place)
Also we have D = C+1 (from units & tens digit relation) , thus D =2.

Thus A+ B, will give 11, thus A+B+C+D = 11+1+2 = 14.

Assign numbers as per place i.e unit, tens, and hundred.
10A+B
10B+A
---------
100C+10D+C

So , 11A+11B = 101C+10D

As A, B, C, D are distinct positive integers

C=1 because if C=2 then sum will be 202+10D and 11A+11B will be less than (88+99).
So last digit is 1

Now substitute numbers in A&B to get last digit 1.
(2,9) (3,8) (4,7) etc

11A+11B = 121, so C=1 and D=2 and A+B =11.

Thus A+B+C+D =14

Given B + A = C
and A + B + carryover = C D

The maximum value of A and B could be 9 + 9 = 18
and the carry over can be 1.
hence A + B + 1 = D and carry over 1
Hence C = 1
Therefore the value of 1 can be from 11 => which is 6 + 5 or 7 + 4 or 8 + 3
in any case A + B = 11
C = 1
and D = A + B + 1 = > 11 + 1 => 12 => 2
hence A + B + C + D = 11 + 1 + 2 => 14
Answer is A = 14

Since two single digit numbers can only result in a carry over of maximum 1 (as biggest single digit numbers are 9 and 8 and 9+8=17, so carry over will be the tens digit, i.e. 1)=> C has to be 1, as C is the carry over of A+B (tens)+ Carry over of B+A (ones digit, whcih again can be max 1).

Hence C =1 (as C is non zero)

Hence ones digit of B+A (ones) also has to be 1, which is possible only if A+B is 11=> to visualise - case of A and B being unordered pair- 2,9
3,8
4,7
5,6
Hence D =2
Hence A+B+C+D=11+1+2=14.

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A, B, C, D are distinct non-zero integers.
9+9=18 (max)

thus, C can only be 1 and no other integer.
6+5 = 11
7+4 = 11
8 + 3 = 11
9 + 2 = 11

A + B = 11 which makes C = 1
When 1 is carried forward, D becomes 2

So, A + B + C + D = 11 + 1 + 2 = 14 (A)

From the sum of two number image=>.
10A + B + 10B + A = 100C + 10D + C
11(A+B) = 101C + 10D ------ (i)

Also, B + A = 10 + C --------(ii)
Since, A+B gives C and B+A gives D, that means there is one carry over from the units digit that has given the difference of 1(Maximum that can get carry over in a sum of single digits is 1).

Therefore, D=C+1 ----------(iii)

Put (iii) and (ii) in (i),

11(10 + C) = 101C + 10(C + 1)
110 + 11C=101C + 10C +10
100C=100
C=1

So, D=2
A+B= 11
A+B+C+D = 11+1+2= 14

Answer - Option 'A'

AB+BA = CDC
since these are digits
11A+11B = CDC
11(A+B) = CDC
now a+b can not exceed 18
so max value can be 198
but c is at hundreds and units place
so 11(A+B) = 1D1
only 121 is such multiple of 11
11(a+b) = 121
a+b = 11
thus 11+1+2= 14
A is the answer

answer is A=14
83+83=121
total of two, the two-digit number is a three-digit number
and the same digit at hundred and unit digit that show that it should start and end with 1
playing with a number for a certain time will eventually give you the right answer.

65+56=121
A=6, B=5, C=1, D=2
Therefore, A+B+C+D= 6+5+1+2=14

Answer:A

14 is the answer
A=4
B=7
C=1
D=2

How frequently are these type of questions were asked in GMAT?

OA is A,
sum of A+ B+ C+D =14

Explanation: from the figure in the right most column ,C = unit digit of B + A
in the middle column D = unit digit of A + B
As C and D are distinctive, so it means the sum of A+ B > 9 whereas there will be some carry for the middle column ( carry + A+B = D)
also, the sum of any two distinctive digits can't be greater than 17 and C is also the carry for the second column. that means C must be 1 .
Secondly, this gives the sum of A and B must not be greater than 11 as both C should be the same, which gives D = 2
in this case, A and B can be as follows :
9 and 2 (vice versa)
8 and 3 (vice versa)
5 and 6 (vice versa)
7 and 4 (vice versa)

Although any combination of A+B+C+D will be 14 only provided C=1 and D =2.

So The Answer is 14