If two 2-digit positive integers have their respective tens

[BTGmoderatorLU]

If two 2-digit positive integers have their respective tens digits exchanged, the difference between the pair of integers changes by 4. What is the greatest possible difference between the original pair of integers?

A. 76
B. 80
C. 82
D. 90
E. 94

The OA is C.

Please, can anyone help me with this PS question? Thanks in advance!

[Scott@TargetTestPrep]

If two 2-digit positive integers have their respective tens digits exchanged, the difference between the pair of integers changes by 4. What is the greatest possible difference between the original pair of integers?

A. 76
B. 80
C. 82
D. 90
E. 94

We can let the first integer be 10a + b and the second be 10c + d originally (where a > c and hence 10a + b > 10c + d). Thus the new pair become 10c + b and 10a + d, respectively (notice that now we have 10a + d > 10c + b. From the information given in the problem, we have:

(10a + b) - (10c + d) = (10a + d) - (10c + b) + 4

10a + b - 10c - d = 10a + d - 10c - b + 4

b - d = d - b + 4

2b - 2d = 4

b - d = 2

We see that the difference between their units digits is 2 regardless of their tens digit. Thus we can let their tens digits be as far apart as possible, i.e. one of them is 1 and the other is 9.

For example, if we let the two numbers be 92 and 10, we see that the original difference is 92 - 10 = 82 and the new difference is 90 - 12 = 78 (notice that 82 is 4 more than 78).

We also could use 99 and 17. We see that the original difference is 99 - 17 = 82 and the new difference is 97 - 19 = 78. Therefore, the greatest possible difference between the original pair is 82.

Answer: C