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Practise 24 exam-style questions on Library routines: MOD, DIV, ROUND and RANDOM 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.
State the value of each of these expressions:
DIV(29, 4)
MOD(29, 4)
DIV(29, 4) = 7
MOD(29, 4) = 1
4 goes into 29 seven whole times (4 × 7 = 28), so DIV gives 7. The amount left over is 29 - 28 = 1, so MOD gives 1. A useful check: DIV(29, 4) × 4 + MOD(29, 4) gets you back to 29.
DIV(29, 4) = 7
MOD(29, 4) = 1
Look at this pseudocode:
Count ← 0
FOR Number ← 10 TO 30
IF MOD(Number, 4) = 0 OR MOD(Number, 6) = 0
THEN
Count ← Count + 1
ENDIF
NEXT Number
OUTPUT CountState the values of Number that cause Count to increase, and state the output.
Count increases for 12, 16, 18, 20, 24, 28 and 30.
The output is 7.
MOD(Number, 4) = 0 picks out multiples of 4 (12, 16, 20, 24, 28) and MOD(Number, 6) = 0 picks out multiples of 6 (12, 18, 24, 30). With OR, a number counts once even if it is a multiple of both, so 12 and 24 are not counted twice.
12, 16, 18, 20, 24, 28, 30
All seven values and no others.
Output: 7
Follow through from a list given.
This pseudocode should simulate rolling a six-sided dice 50 times and output the percentage of rolls that were a 6, rounded to 1 decimal place.
01Sixes ← 0
02FOR Roll ← 0 TO 50
03 Dice ← ROUND(RANDOM() * 6, 0)
04 IF Dice = 6
05 THEN
06 Sixes ← Sixes + 1
07 ENDIF
08NEXT Roll
09OUTPUT ROUND(Sixes / 50 * 100)There are three errors. Identify the line number of each error and write the corrected line.
Line 2: FOR Roll ← 1 TO 50
Line 3: Dice ← ROUND(RANDOM() * 5, 0) + 1
Line 9: OUTPUT ROUND(Sixes / 50 * 100, 1)
A loop from 0 to 50 runs 51 times, not 50. ROUND(RANDOM() * 6, 0) gives 0 to 6, so 0 is possible and 6 is less likely than it should be; multiplying by 5 and adding 1 gives 1 to 6. ROUND always needs a second parameter, the number of decimal places, which here is 1.
Line 2: FOR Roll ← 1 TO 50
The line number and the correction are both needed for each mark. Accept FOR Roll ← 0 TO 49.
Line 3: Dice ← ROUND(RANDOM() * 5, 0) + 1
Accept any correct statement giving whole numbers 1 to 6 only.
Line 9: OUTPUT ROUND(Sixes / 50 * 100, 1)
Write an algorithm in pseudocode that inputs 20 whole numbers, one at a time, and then outputs how many of the numbers were even and how many were odd.
Evens ← 0
Odds ← 0
FOR Count ← 1 TO 20
INPUT Number
IF MOD(Number, 2) = 0
THEN
Evens ← Evens + 1
ELSE
Odds ← Odds + 1
ENDIF
NEXT Count
OUTPUT "Even numbers: ", Evens
OUTPUT "Odd numbers: ", OddsMOD(Number, 2) is 0 for every even number and 1 for every odd number, so one IF ... ELSE decides which counter to increase. The counters start at 0 before the loop and are output only once all 20 numbers have been processed.
MP1 two counters (even and odd) initialised to 0 before the loop
max 5; accept one counter with the other worked out as 20 minus it
MP2 loop that repeats 20 times
MP3 inputs a number inside the loop
MP4 uses MOD(Number, 2) = 0 (or equivalent) to test for an even number
MP5 adds 1 to the correct counter inside the loop
MP6 outputs both counts, with messages, after the loop
Do not accept output inside the loop.
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