B2 Programming in Python (data types, control, functions, files, recursion) – IB Computer Science exam questions with mark scheme

1Core5 marks

State the value produced by each Python expression.

(a) 17 // 5
(b) 17 % 5
(c) 2 ** 3
(d) "hello"[1:4]
(e) 7 / 2

[5]

(a) 3(1)
(b) 2(1)
(c) 8(1)
(d) 'ell' (accept ell or "ell", with or without quotes)(1)
(e) 3.5 (reject 3: / is true division in Python 3)(1)
Answer: 3, 2, 8, 'ell', 3.5
Examiner tip: The slice stops before index 4; students often include the final character or start at index 0.

2Standard4 marks

Consider the following Python program.

total = 0
for i in range(1, 6):
    if i % 2 == 0:
        total = total + i * i
    else:
        total = total - i
print(total)

(a) Complete a trace table showing the value of total at the end of each iteration, and state the output. [3]

(b) State the output if range(1, 6) is changed to range(1, 5). [1]

i = 1 → total = −1; i = 2 → total = 3(1)
i = 3 → total = 0; i = 4 → total = 16(1)
i = 5 → total = 11, so the output is 11(1)
(b) 16(1)
Answer: (a) −1, 3, 0, 16, 11; output 11 (b) 16
Examiner tip: Odd values are subtracted, even values have their square added; check the sign on every row.

3Standard4 marks

Write a Python function count_vowels(text) that returns the number of vowels (a, e, i, o, u) in the string text. Upper-case and lower-case vowels must both be counted. For example, count_vowels("Education") returns 5.

[4]

Function header with a parameter, e.g. def count_vowels(text):(1)
Initialises a counter to 0 and loops through every character of the string (allow an equivalent built-in approach, e.g. adding text.lower().count(v) for each vowel v)(1)
Converts case (or tests both cases) and tests whether the character is a vowel, incrementing the counter if so (allow ch in "aeiouAEIOU" or a chain of ==/or tests)(1)
Returns the counter (after the loop) rather than printing it(1)

Examiner tip: Return the count after the loop ends; a return inside the loop exits on the first character.

4Hard6 marks

A text file scores.txt stores one integer test score on each line, for example:

60
75
90

Write a Python function average_score(filename) that:

returns the mean of all the scores in the file;
returns 0 if the file contains no scores;
returns None if the file does not exist.

[6]

Opens the file for reading (using open, preferably with with so it is closed)(1)
Reads the file line by line (or all lines) with a loop (allow a list comprehension over the file or over f.read().split())(1)
Converts each line to an integer using int() (allow with or without .strip(), since int() ignores surrounding whitespace and the newline)(1)
Accumulates the total and the number of scores (allow building a list of the scores and using sum() and len())(1)
Avoids division by zero: returns 0 when there are no scores, otherwise returns total ÷ count(1)
Uses try/except FileNotFoundError (or equivalent check) to return None for a missing file(1)

Answer:

def average_score(filename):
    try:
        with open(filename, "r") as f:
            total = 0
            count = 0
            for line in f:
                line = line.strip()
                if line != "":
                    total += int(line)
                    count += 1
    except FileNotFoundError:
        return None
    if count == 0:
        return 0
    return total / count

Examiner tip: Check for an empty file before dividing, otherwise total / count raises ZeroDivisionError; int() copes with the trailing newline on its own.

5HardHL only5 marks

Consider the recursive function below.

def digit_sum(n):
    if n < 10:
        return n
    return n % 10 + digit_sum(n // 10)

(a) State the base case. [1]

(b) Show the sequence of calls made by digit_sum(4729). [1]

(d) State what would happen if the base case were removed. [1]

(a) n < 10 (a single digit is returned unchanged)(1)
(b) digit_sum(4729) → digit_sum(472) → digit_sum(47) → digit_sum(4)(1)
(c) digit_sum(4) returns 4; digit_sum(47) returns 7 + 4 = 11; digit_sum(472) returns 2 + 11 = 13(1)
digit_sum(4729) returns 9 + 13 = 22 (final result 22)(1)
(d) Infinite recursion: the function keeps calling itself (n // 10 reaches 0 and digit_sum(0) calls digit_sum(0) for ever), so it never terminates / the call stack overflows and a RecursionError is raised (accept any one of: infinite recursion, never stops, stack overflow, RecursionError)(1)
Answer: (a) n < 10 (b) 4729, 472, 47, 4 (c) 4, 11, 13, 22 (d) RecursionError (stack overflow)
Examiner tip: Show the value returned at each level as the recursion unwinds, not only the final 22; in (d) say the recursion never stops (stack overflow), not just 'it breaks'.

6StandardHL only4 marks

Consider the Python program below.

words = "the cat and the dog and the bird".split()
freq = {}
for w in words:
    freq[w] = freq.get(w, 0) + 1
print(freq)
print(max(freq, key=freq.get))

(a) State the output of the program. [3]

(b) Outline the purpose of freq.get(w, 0). [1]

First line of output is the dictionary with the, cat, and, dog, bird in order of first appearance(1)
Counts are correct: the 3, cat 1, and 2, dog 1, bird 1 → {'the': 3, 'cat': 1, 'and': 2, 'dog': 1, 'bird': 1} (ignore quote style and spacing)(1)
Second line of output is the (the key with the highest value)(1)
(b) Returns the current count for w, or 0 if w is not yet a key, so a new word starts at 0 and no KeyError is raised (accept either idea)(1)
Answer: {'the': 3, 'cat': 1, 'and': 2, 'dog': 1, 'bird': 1} then the
Examiner tip: Dictionaries keep insertion order in Python 3.7+, so the printed order follows first appearance.

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