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Programming Constructs and Operators: Question 1

Syllabus 8.1

Multiple choice 1 mark

A conference organiser uses this pseudocode to work out how many minibuses are needed to transport delegates to a venue. Each minibus seats exactly 9 people.

DECLARE Delegates : INTEGER
DECLARE FullBuses : INTEGER
DECLARE Remaining : INTEGER
Delegates ← 47
FullBuses ← Delegates DIV 9
Remaining ← Delegates MOD 9
OUTPUT FullBuses, " full minibuses and ", Remaining, " delegates left over"

What is output when this algorithm is run?

Choose an answer to check it, then compare with the worked solution below.
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Worked solution

Step 1: Recall what DIV and MOD do

In Cambridge pseudocode, for two integers A and B:

  • A DIV B gives the whole number of times B divides into A, with any remainder discarded (integer division, always rounded down towards zero for positive integers).
  • A MOD B gives the remainder left over after A DIV B whole groups of B have been removed from A.

Both operators use the same two operands, Delegates (47) and 9, so they must be worked out from the same pair of numbers.

Step 2: Work out FullBuses

FullBuses ← Delegates DIV 9

Find the largest whole number of 9s that fit into 47:

  • 9 × 5 = 45 (fits, since 45 is less than or equal to 47)
  • 9 × 6 = 54 (too big, since 54 is greater than 47)

So 47 DIV 9 = 5. There are 5 complete minibuses.

Step 3: Work out Remaining

Remaining ← Delegates MOD 9

This is what is left over after removing those 5 complete groups of 9:

47 - (5 × 9) = 47 - 45 = 2

So 47 MOD 9 = 2. There are 2 delegates left over after filling the 5 full minibuses.

Step 4: Confirm the output

The OUTPUT statement prints FullBuses, the fixed text " full minibuses and ", Remaining, then " delegates left over". Substituting the traced values gives:

5 full minibuses and 2 delegates left over

Step 5: Why the other options are wrong

  • Option B (6 full minibuses and 2 delegates left over) comes from doing ordinary division, 47 ÷ 9 = 5.22, and rounding this up to 6, but DIV truncates rather than rounds, so it must give exactly 5, not 6.
  • Option C (2 full minibuses and 5 delegates left over) comes from swapping the two operators: computing FullBuses with MOD (giving 2) and Remaining with DIV (giving 5). The pseudocode clearly assigns DIV to FullBuses and MOD to Remaining.
  • Option D (5 full minibuses and 47 delegates left over) comes from misunderstanding MOD as simply returning the original value of Delegates unchanged, instead of the true remainder of 2 after 5 whole groups of 9 have been removed.

Final answer

  • The algorithm outputs “5 full minibuses and 2 delegates left over”, option A.