Concept:Subtraction with borrowing and place value.
Explanation:The given expression is 1957 – a9 = 18b8.
Start from the units: 7 – 9 is not possible, so borrow 1 from the tens (5 becomes 4, 7 becomes 17). Then 17 – 9 = 8. The units digit of the result is 8, which matches.
In the tens place, we now have 4 – a. Since the result’s tens digit is b, and 4 is less than a (because we needed to borrow again), borrow 1 from the hundreds (9 becomes 8, 4 becomes 14). Then 14 – a = b.
In the hundreds place, after borrowing we have 8 – 8 = 0, which matches the result’s hundreds digit 8? Actually careful: The result’s hundreds digit is 8 (from 18b8, the hundreds place is 8). The subtraction in hundreds: 8 (borrowed from 9) minus nothing (since a9 has no hundreds digit) gives 8, which is correct.
From the tens equation: 14 – a = b, so a + b = 14.
Answer:The sum of digits a and b is 14 (Option B).
