A two-digit number exceeds the number formed by interchanging its digits by 36. If the ratio between its digits is 1 : 2, what is the difference between the sum and the difference of its digits?

Problem restatement
Digits (tens = a, units = b) satisfy 9(a − b) = 36 ⇒ a − b = 4. The digits are in the ratio 1 : 2. Find (a + b) − (a − b).

Given data

• a − b = 4
• Digits in ratio 1 : 2 ⇒ the two digits are proportional to 1 and 2 (smaller : larger).

Concept/Approach
Since a − b > 0, the tens digit a is larger. Thus the pair must be (8, 4) up to the 1:2 ratio (4 : 8).

Step-by-step calculation
Take digits as 8 and 4 so that a − b = 4 and smaller : larger = 1 : 2 Sum = a + b = 12; Difference = a − b = 4 Required = (a + b) − (a − b) = 12 − 4 = 8

Verification/Alternative
Check the interchange condition: 9(a − b) = 9×4 = 36 (satisfied). The ratio condition holds with digits {4, 8}.

Common pitfalls

• Interpreting the ratio as a : b = 1 : 2 directly, which would contradict a − b = 4 (since that would make a < b). The ratio refers to values of the digits (smaller : larger).

Final Answer
8