ITPEC FE Subject B April 2025 Question 4

Binary to Decimal Conversion — fill in blanks for a function that interprets a decimal-digit string of 0s and 1s as a binary number.

The function peels off one digit at a time from the right using n mod 10. Each digit is either 0 or 1 — it is the current bit. To get its decimal contribution, multiply it by the positional weight (place), which starts at 1 and doubles each iteration (powers of two: 1, 2, 4, 8, …).

Trace with number = 1100:

Result: decimal = 12 ✓

• •A = `remainder × place` — current bit × positional weight
• •B = `2 × place` — double the weight for the next binary position

Why not others:
- (a) 2 × place for A would ignore the actual bit value; n × place for B makes no sense as a weight update
- (b) 2 × place for A is wrong — you need the bit, not a constant multiplier
- (c) 10 × place for A treats it as decimal positional notation, not binary
- (d) 10 × place for A is decimal, not binary
- (e) n × place for A uses the entire remaining number, not just the current digit
- (f) n × place for A — same issue as (e)
- (h) 10 × place for B would give decimal positional weights (1, 10, 100…), not binary

Key rule: Binary positional weights are powers of 2 — the weight doubles each step (place ← 2 × place), and each bit's contribution is bit × place.

AI-generated — may contain errors