ExpertMedium1 markMultiple Choice
AQA GCSE · Question 02.2 · Fundamentals of algorithms
In the algorithm in Figure 2, what will be output when the user input is 10?
In the algorithm in Figure 2, what will be output when the user input is 10?
Answer options:
A.
0
B.
1
C.
2
D.
4
How to approach this question
1. This is a trace question. Start with the user input `a = 10`.
2. The outer `WHILE` loop (line 2) starts.
3. The `IF a > 0` (line 4) is true.
4. `counter` is set to 0 (line 5).
5. The inner `WHILE a > 0` (line 6) starts.
- **Pass 1:** `a` is 10. `a ← 10 DIV 3` results in `a = 3`. `counter` becomes 1.
- **Pass 2:** `a` is 3. `a ← 3 DIV 3` results in `a = 1`. `counter` becomes 2.
- **Pass 3:** `a` is 1. `a ← 1 DIV 3` results in `a = 0`. `counter` becomes 3.
6. The inner `WHILE` loop condition `a > 0` is now false, so it ends.
7. The algorithm proceeds to line 13, `OUTPUT a`.
8. The current value of `a` is 0. So, 0 is output.
Full Answer
A.0✓ Correct
The correct answer is A (0). The algorithm repeatedly divides the input `a` by 3 until `a` is no longer greater than 0. The final value of `a` is then output. For an input of 10, the sequence for `a` is 10 -> 3 -> 1 -> 0. The final value is 0.
A dry run or trace is needed to solve this.
- **Input:** `a = 10`
- The condition `a > 0` is true.
- The inner `WHILE` loop begins.
- **Iteration 1:**
- `a` is 10. `a > 0` is true.
- `a = 10 DIV 3` (integer division) which is 3.
- `counter` becomes 1.
- **Iteration 2:**
- `a` is 3. `a > 0` is true.
- `a = 3 DIV 3` which is 1.
- `counter` becomes 2.
- **Iteration 3:**
- `a` is 1. `a > 0` is true.
- `a = 1 DIV 3` which is 0.
- `counter` becomes 3.
- **End of loop:**
- `a` is now 0. The condition `a > 0` is false. The inner loop terminates.
- The code then reaches line 13: `OUTPUT a`.
- The final value of `a` is 0.
Common mistakes
✗ Confusing the output variable. The code outputs `a` (line 13), not `counter`.\n✗ Making an error in the integer division (DIV) calculation.
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