ITPEC IP April 2024 Question 6

The final value of x is 5. Start with x = 2 and y = 3, then trace the loop in order.

• •First pass: step (2) changes y from 3 to 2; step (3) sets x = 2 + 2 = 4. Since y is not 1, the procedure repeats.
• •Second pass: step (2) changes y from 2 to 1; step (3) sets x = 4 + 1 = 5. Now y is 1, so the procedure ends.

Thus the accumulated additions are 2 + 2 + 1 = 5.

Answer (b)

Why not others:
- (a) 4 is the intermediate value after only the first pass
- (c) 7 and (d) 8 result from adding values not prescribed by the loop or continuing after the termination condition is met

Key rule: Trace assignments sequentially using the updated variable values, and test the loop termination condition only at the specified step.

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