Lists: Practice Questions (Computer Science Principles 3.10)

These are original Marlbridge practice questions, not past exam questions. All scenarios are fictional. The code is Marlbridge’s rendering of the exam reference notation (← assignment, = equality test, lists start at index 1, DISPLAY shows a value). Mark points are a suggested Marlbridge rubric, not an official scoring guideline. No calculator is needed or allowed. Try each question before opening its answer.

Question 1 (multiple choice · core)

What is displayed?

nums ← [5, 6, 7]
INSERT(nums, 1, 4)
REMOVE(nums, 3)
APPEND(nums, 9)
DISPLAY(nums[3])
DISPLAY(LENGTH(nums))

(A) 9 4
(B) 7 3
(C) 7 4
(D) 6 4

Answer and explanation

Answer: (C). The list changes: [5, 6, 7] → insert 4 at index 1 gives [4, 5, 6, 7] → remove index 3 (the 6) gives [4, 5, 7] → append 9 gives [4, 5, 7, 9]. So nums[3] is 7 and the length is 4.

(A) reads the last element.
(B) forgets the final APPEND for the length.
(D) removes the wrong element or reads the wrong index.

Question 2 (multiple choice · core)

What is displayed?

vals ← [12, 5, 18, 10, 11]
count ← 0
FOR EACH v IN vals
{
  IF(v > 10)
  {
    count ← count + 1
  }
}
DISPLAY(count)

(A) 5
(B) 0
(C) 2
(D) 3

Answer and explanation

Answer: (D). The loop counts values greater than 10: 12, 18 and 11. The values 5 and 10 are not greater than 10.

(A) counts every element.
(B) would result if the count were never updated.
(C) would result from using > 11.

Question 3 (multiple choice · core)

A programmer tries to swap the first two elements of aList, which holds [3, 8], with the two lines aList[1] ← aList[2] then aList[2] ← aList[1]. What does aList hold afterwards?

(A) [3, 3]
(B) [8, 8]
(C) [3, 8]
(D) [8, 3]

Answer and explanation

Answer: (B). After the first line the list is [8, 8]; the original 3 is lost. The second line copies 8 into index 2, which is already 8.

(A) would need the second element to be overwritten first.
(C) is the original list.
(D) is the intended result; it needs a temporary variable.

Question 4 (multiple choice · core)

Which segment correctly swaps aList[1] and aList[2]?

(A) aList[1] ← temp, then temp ← aList[2], then aList[2] ← aList[1]
(B) temp ← aList[1], then aList[2] ← temp, then aList[1] ← aList[2]
(C) temp ← aList[1], then aList[1] ← aList[2], then aList[2] ← temp
(D) aList[1] ← aList[2], then aList[2] ← aList[1]

Answer and explanation

Answer: (C). The first value is saved in temp before it is overwritten.

(A) uses temp before it has been given a value.
(B) overwrites aList[2] with the first value before copying it back, so both hold the first value.
(D) loses the first value (see Question 3).

Question 5 (multiple choice · core)

What happens when this code runs?

aList ← [4, 5, 6]
REMOVE(aList, 0)
DISPLAY(aList)

(A) The list becomes [5, 6].
(B) The program stops with an error at the REMOVE statement.
(C) The list becomes [4, 6].
(D) The list becomes [4, 5].

Answer and explanation

Answer: (B). The list [4, 5, 6] has indices 1 to 3. Index 0 does not exist, so an error is produced and the program terminates.

(A) assumes index 0 is the first element and removes the 4.
(C) and (D) assume a different element is removed.

Question 6 (multiple choice · core)

How many elements does the search examine, and what does it display for found?

scores ← [14, 9, 21, 17, 9]
target ← 17
found ← false
i ← 1
REPEAT UNTIL(found OR i > LENGTH(scores))
{
  IF(scores[i] = target)
  {
    found ← true
  }
  i ← i + 1
}
DISPLAY(found)
DISPLAY(i - 1)

(A) 4 elements, true
(B) 5 elements, false
(C) 5 elements, true
(D) 3 elements, true

Answer and explanation

Answer: (A). The loop examines 14, 9, 21 and 17, then found becomes true and the loop ends. The last line shows i - 1, which is 4.

(B) would be the result only if 17 were not in the list.
(C) assumes the loop continues to the end of the list.
(D) stops one element too early.

Question 7 (multiple choice · stretch)

What is displayed?

nums ← [1, 2, 3]
FOR EACH item IN nums
{
  item ← item * 2
}
DISPLAY(nums[1])

(A) 2 4 6
(B) 0
(C) 2
(D) 1

Answer and explanation

Answer: (D). item is a copy of each element. Assigning to item does not change nums, so nums[1] is still 1.

(A) and (C) assume the list was doubled.
(B) assumes the list was cleared.

Question 8 (multiple select, choose two · core)

A list aList currently holds [10, 20, 30]. Which two statements are true?

(A) After REMOVE(aList, 2), LENGTH(aList) is 2.
(B) After REMOVE(aList, 3), aList[3] is 30.
(C) aList[0] is 10.
(D) After APPEND(aList, 40), aList[1] is 40.
(E) After INSERT(aList, 1, 5), aList[2] is 10.

Answer and explanation

Answers: (A) and (E). Removing one element leaves 2. Inserting 5 at index 1 shifts 10 to index 2.

(B) After removing index 3 the list has only two elements, so index 3 is out of range.
(C) Index 0 is an error.
(D) APPEND adds at the end; aList[1] is still 10.

Question 9 (constructed response · core)

Write code that displays the average of the list marks using FOR EACH and LENGTH. Trace it for [8, 6, 10, 4].

Answer and explanation

One correct answer:

total ← 0
FOR EACH mark IN marks
{
  total ← total + mark
}
DISPLAY(total / LENGTH(marks))

Trace: total goes 0 → 8 → 14 → 24 → 28. LENGTH(marks) is 4, so the average is 28 / 4 = 7. 7 is displayed.

Suggested Marlbridge rubric (5 marks)

Mark point	Marks
Initialises `total` to 0	1
Traverses every element with FOR EACH (or an index loop to LENGTH)	1
Accumulates `total ← total + mark`	1
Divides by `LENGTH(marks)` after the loop	1
Trace values 8, 14, 24, 28 and result 7	1

Question 10 (constructed response · stretch)

The list temps holds [18, 25, 31, 22, 31, 19]. Write code that finds the highest temperature and counts how many times it occurs, then displays both. State what is displayed.

Answer and explanation

One correct answer:

highest ← temps[1]
FOR EACH t IN temps
{
  IF(t > highest)
  {
    highest ← t
  }
}
count ← 0
FOR EACH t IN temps
{
  IF(t = highest)
  {
    count ← count + 1
  }
}
DISPLAY(highest)
DISPLAY(count)

The highest is 31, and it occurs twice (indices 3 and 5), so 31 2 is displayed. Two passes are needed because the maximum must be known before it can be counted.

Suggested Marlbridge rubric (5 marks)

Mark point	Marks
Starts `highest` at the first element, not 0	1
First loop updates `highest` using `>`	1
Second loop counts elements equal to `highest`	1
Displays both values	1
Result 31 and 2	1

How did you do?

Count the questions you got right first time, then read the explanation for every one you missed, including the wrong options you rejected correctly, because the distractors show you the typical slips. Anything you missed twice goes onto your revision list for the revision notes; re-read the matching section of the study guide before trying again.