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To satisfy condition 2, we have to eliminate set 1
and hence difference between the tens digit is always 2.
Am I missing something OR anything wrong in my analysis

C
assuming they are integers (positive or negative)
-10,17 , -10,-37 and 10,37 are all possibilites. hence tens digit difference will be 2 as -1-1 or -1-(-3) or 1-3

the 2 statements require that the units digits of the 2 no.s be a combination of 0,7 or 1,8 or 2,9 only. these limts ensure that the tens digits are different by only 2 as the 7 in 27 can never round up into the tens digit or the second no.
for eg 33,60 wont do as 3-0 = 3 but 32,59 will as 9-2 = 7 > 3

x - y = 10(a-c) + (b - d)
1) x - y = 27 Insuff
2) b - d > 3 Insuff

Combined we get 27 = 10(a - b) + (b - d)
or 27 - ( b - d) = 10 (a - b)
or a - b = (27 - (b -d))/10
This equation must be satisfied and a - b must be an integer

From above we can safely assume b - d = 7 , hence a - b = 2

I would say that the question should have specified that the two numbers are positive integers, otherwise the answer would be E. With positive integers we would get C. (Basically the only thing we need to know is if we have borrowed 10 from the tenth digit when we calculate the unit digit). _________________

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