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Practise 30 exam-style questions on Arithmetic, relational and logical operators for Cambridge IGCSE Computer Science (Paper 2). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.
Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.
Identify the arithmetic operator that gives the remainder when one whole number is divided by another.
MOD, e.g. MOD(17, 5) = 2
MOD returns the remainder after integer division, while DIV returns the whole number of times one number goes into the other. MOD is often used to test whether a number divides exactly, because the remainder is then 0.
MOD
Accept modulus / modulo. Do not accept DIV or /.
This pseudocode should output "Teenager" only for ages 13 to 19:
INPUT Age
IF Age > 12 OR Age < 20
THEN
OUTPUT "Teenager"
ENDIFExplain why it outputs "Teenager" for every age, and write the corrected condition.
OR is TRUE if either comparison is TRUE. Any age above 12 passes the first test and any age below 20 passes the second, so every possible age passes. The condition should be: Age > 12 AND Age < 20
For a value to be inside a range, it must pass both tests, so AND is needed. OR is used for values outside a range, e.g. Age < 13 OR Age > 19 is TRUE for anyone who is not a teenager.
With OR only one comparison needs to be TRUE, and every age is either greater than 12 or less than 20 (or both), so the condition is always TRUE
Age > 12 AND Age < 20
Accept Age >= 13 AND Age <= 19.
Write pseudocode to input two whole numbers, First and Second. Output "Factor" if Second divides exactly into First, otherwise output "Not a factor". If Second is 0, output "Cannot divide by zero" instead.
INPUT First
INPUT Second
IF Second = 0
THEN
OUTPUT "Cannot divide by zero"
ELSE
IF MOD(First, Second) = 0
THEN
OUTPUT "Factor"
ELSE
OUTPUT "Not a factor"
ENDIF
ENDIFA number is a factor when the remainder is 0, which MOD tests. Dividing by zero causes a run-time error, so the zero check must come before MOD is used, not after it.
MP1 inputs both numbers and checks Second = 0 first, outputting "Cannot divide by zero"
MP2 uses MOD(First, Second) = 0 to test whether Second is a factor
MP3 outputs "Factor" or "Not a factor" correctly, only when Second is not 0
Write pseudocode that:
- inputs a whole number, repeating the input until the number is from 100 to 999 inclusive
- uses DIV and MOD to find the three digits of the number
- outputs the sum of the three digits.
For example, 472 outputs 13.
REPEAT
INPUT Number
UNTIL Number >= 100 AND Number <= 999
Hundreds ← DIV(Number, 100)
Tens ← MOD(DIV(Number, 10), 10)
Units ← MOD(Number, 10)
OUTPUT "Sum of digits: ", Hundreds + Tens + UnitsThe validation loop uses relational operators joined by AND to accept only three-digit numbers. DIV by a power of 10 removes digits from the right and MOD by 10 keeps only the last digit, so together they isolate each digit: for 472, 4 + 7 + 2 = 13.
MP1 loop that repeats the input (input inside the loop)
max 5
MP2 correct validation condition: Number >= 100 AND Number <= 999
For a WHILE loop: Number < 100 OR Number > 999.
MP3 hundreds digit found with DIV(Number, 100)
MP4 tens digit found with MOD(DIV(Number, 10), 10)
Accept DIV(MOD(Number, 100), 10).
MP5 units digit found with MOD(Number, 10)
MP6 outputs the sum of the three digits
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