ExpertMedium1 markMultiple Choice
AQA GCSE · Question 02.3 · Fundamentals of algorithms
In the algorithm in Figure 2, what is the largest possible value of the variable counter when the user input is 36?
In the algorithm in Figure 2, what is the largest possible value of the variable counter when the user input is 36?
Answer options:
A.
0
B.
2
C.
4
D.
5
How to approach this question
1. This is another trace question. Start with user input `a = 36`.
2. The `IF a > 0` is true.
3. `counter` is set to 0 (line 5).
4. The inner `WHILE a > 0` (line 6) starts.
- **Pass 1:** `a` is 36. `a ← 36 DIV 3` results in `a = 12`. `counter` becomes 1.
- **Pass 2:** `a` is 12. `a ← 12 DIV 3` results in `a = 4`. `counter` becomes 2.
- **Pass 3:** `a` is 4. `a ← 4 DIV 3` results in `a = 1`. `counter` becomes 3.
- **Pass 4:** `a` is 1. `a ← 1 DIV 3` results in `a = 0`. `counter` becomes 4.
5. The inner `WHILE` loop condition `a > 0` is now false, so it ends.
6. The largest value `counter` reached was 4.
Full Answer
C.4✓ Correct
This question requires tracing the value of the `counter` variable.
- **Input:** `a = 36`
- The condition `a > 0` is true.
- `counter` is initialized to 0.
- The inner `WHILE` loop begins.
- **Iteration 1:**
- `a` is 36. `a > 0` is true.
- `a = 36 DIV 3` = 12.
- `counter` = 0 + 1 = 1.
- **Iteration 2:**
- `a` is 12. `a > 0` is true.
- `a = 12 DIV 3` = 4.
- `counter` = 1 + 1 = 2.
- **Iteration 3:**
- `a` is 4. `a > 0` is true.
- `a = 4 DIV 3` = 1.
- `counter` = 2 + 1 = 3.
- **Iteration 4:**
- `a` is 1. `a > 0` is true.
- `a = 1 DIV 3` = 0.
- `counter` = 3 + 1 = 4.
- **End of loop:**
- `a` is now 0. The condition `a > 0` is false. The inner loop terminates.
The final and largest value of `counter` is 4.
Common mistakes
✗ Stopping the trace too early.\n✗ Making a mistake in the integer division, for example, calculating 4 DIV 3 as 1.33 instead of 1.
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