If m and n are positive two-digit integers, what is the value of the

The two numbers m and n lie between 10-99 as per the question.

Analyzing Statement 1 :-
m-n=42
infers m is bigger than n.
now if we assume m = 52; then only lowest value of n = 10 will arrive, since 52-10=42
this puts the difference between 5(tens digit of m) and 1(tens digit of n) = 4;
Now, if we assume m = 62; then n = 20, again difference between ten's digits of m and n is 4.
Hence, a constant difference of 4 is arriving for any value of m and n when Statement 1 stands.
So Statement 1 is sufficient.

Analyzing Statement 2:-
Suppose m=15 and n = 11, hence the difference between their units digits are 5-1 = 4 not a multiple of 3(statement 2 satisfied), hence difference between tens digits is determined to be 1-1 = 0;
Now suppose m 26 and n =11, hence hence the difference between their units digits are 6-1 = 5 not a multiple of 3(statement 2 satisfied), hence difference between tens digits is determined to be 2-1 = 1;
Since the difference between tens digits is not fixed and depends upon values of m and n,
Statement 2 is insufficient.

Thus the answer is A.

Please correct me if i am wrong in my process.

Let m be- 10a+b
and n be- 10c+d

From statement 1, 10a+b - 10c-d= 42
---> 10(a-c) + b-d = 42
Not sufficient

From statement 2, b-d=1,2,4,5,7,8
Not sufficient

Now combining both statements, the only value that fits b-d would 2 such that 10(a-c) + (2) = 42, so 10(a-c) = 40
---> a-c = 4
Therefore C

Can someone tell me what's wrong with this approach? gmatophobia chetan2u KarishmaB Bunuel GMATPill

You have taken cases where b>d, that is 62-20 or 75-33 etc. But what happens when b<d, for example 71-29.
So in those cases your equation will become.
(10a+b)-(10c+d)
Since b<d, we will take one 10 from value of a.
So, 10(a-1)+(10+b)-10c-d=10(a-1-C)+(10+b-d)
=> 10(a-1-c)+2=42
a-1-c = 4 or a-c = 5

Hope it helps