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S5 ICT: Programming Arrays and Functions – Practice Paper 1 (Foundation, 10 Questions)
S5 ICT: Programming Arrays and Functions – Practice Paper 1 (Foundation, 10 Questions)
Paper one builds the basics: zero-based indexing, length and last index, totals and means, function arguments and return values, and linear search counts – ten items with answers and one-line notes.
Quick answer Most array marks are lost on zero-based indexing; separate index from length first, then handle search and functions.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | 25 minutes |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: S5 ICT | Programming: Arrays and Functions |
| Level | Foundation |
02 Questions (10 items)
- FoundationArray A = [12, 7, 25, 3, 18]. What is the value of A[2]?
- FoundationWhat is the length of A, and what is the index of its last element?
- IntermediateWhat are the total and the mean of all elements in A?
- FoundationA function double(x) returns x multiplied by 2. What does double(7) return?
- FoundationA function add(a, b) returns a plus b. What does add(3, 4) return?
- IntermediateA linear search starts at A[0] and looks for 25. How many comparisons are needed, and at which index is it found?
- IntermediateAfter swapping A[1] with A[3], what is the value of A[1]?
- ChallengeWhat is the value of the expression 25 % 4?
- ChallengeA program reads A[5] on an array of length 5. What happens?
- Error redoA total loop starts sum at 1 instead of 0. Summing the five elements of A, how far off is the result?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 25 | Indexing starts at 0, so A[0]=12, A[1]=7 and A[2]=25. |
| 2 | Length 5; last index 4. | An array of length 5 has indices 0 to 4, so the last index is the length minus 1. |
| 3 | Total 65; mean 13.0. | 12+7+25+3+18 = 65, and 65 divided by 5 gives 13.0. |
| 4 | 14 | The argument 7 replaces x, so the function returns 7 times 2, which is 14. |
| 5 | 7 | The arguments go in by position, a=3 and b=4, so the result is 7. |
| 6 | Three comparisons; index 2. | It compares A[0]=12, A[1]=7 then A[2]=25, so the third comparison succeeds. |
| 7 | 3 | A[1] held 7 and A[3] held 3, so after the swap A[1] holds 3. |
| 8 | 1 | 25 divided by 4 gives quotient 6 and remainder 1, and the % operator returns the remainder. |
| 9 | The index is out of range and the program raises an error. | Valid indices are 0 to 4, so A[5] does not exist and most languages raise an index error. |
| 10 | One too many, giving 66. | Starting at 1 adds a spurious 1, so the total becomes 66 instead of 65. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
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