Problem Solving

Answer - 9.

Yes as per the trial and error method, I got the answer as D. But just be careful in one of the cases...

19 + 92 = 111 cannot be the case that can be suited into our given equation...

We need A B C and D as distinct positive integers...

All others are genuine cases present...

A has to be 1 as sum of 9+9 = 18 < 20.
Hence B+D = 11
and C+A+1 = 11 => C = 9

Would this be 700-800 level question?

The only three digit possibility among 111,222,333 etc is 111

That can be obtained by

18+93=111
17+94=111
16+95=111
15+96=111
14+97=111
13+98=111

Hence 9

Answer is D

AB+CD=AAA
A(10)+B+C(10)+D=A(100)+A(10)+A=111A
C10+B+D=101A

At Maximum, C10+B+D must equal less than 90+8+7=105 since all integers are different. Therefore A must equal 1 because it is the only value that will make it less than 105.

If A=1.. Then C must equal 9. If C equaled 8, then 8(10)+9+7=96 does not put it at 101. Nor does anything less than 8. And because C is an single digit integer, then it can't be greater than 10. Therefore, 8<C<10... C=9.

Answer is D.

. AB
+ CD
AAA

if we separately evaluate the left half, A+C must give AA
Logically, take the highest single digit number for ex: 9 =>9+9 = 18 that means, when two sngle numbers are added, the sum can be 18 or less than 18.

so, out of the many possibilities of AA i.e., 11, 22,33 etc, AA can only take the value "11" which is less than 18

in the options, check which value can produce AA = 11.
Ans: 9

(D) 9

A=1

A+C+1=11

A+C=10

C=10-1
C=9

B+D=11
B/D=6/5,7/4,8/3

Good Question, 9 it is.

Hi,

I think you are committing one mistake while picking values like 12 & 99 and 19 & 92, here you need to remember that ABCD are distinct integers, In the first case C and D are same and in second B & C are same.

Thanks
Deepak

albatross86 wrote:The maximum value of AB + CD is = 98 + 76 = 174

Thus A must be 1, as 111 is the only 3 digit number that could have all 3 digits equal

So, 1B + CD = 111

Now consider B = 1 => The number AB is 11, which makes the number CD = 100 which is impossible

Thus B must be atleast 2, yielding AB as 12 and CD as 99
The maximum value of B is 9, yielding AB as 19 and CD as 92

In all possible values of AB, the value of CD is in the nineties. Thus C must be equal to 9

Pick D

Simple way Adding any 2 two digit numbers will be more than 198 so A is 1 always

so AAA is 111 always,

13 + 98 or 14 + 97 or 15 + 96 or 16 + 96 or 17 + 94 or 18 + 93

C is always 9

Bingo.

Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.


==> AB
+CD
-----
in order to be AAA, a two digit number can never be greater than 199 thus A = 1. If A = 1, then C = 9 and B+D must be 11. Therefore C = 9, and the answer is D.


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