Trace tables and errors — GCSE Computer Science practice questions

Question 1

A café loyalty card holds stamps. When a card has 9 or more stamps, the customer gets a free drink and 9 stamps are taken off the card.

This pseudocode runs when a customer with 7 stamps buys 5 drinks:

```pseudocode
Stamps ← 7
Bought ← 5
Stamps ← Stamps + Bought
IF Stamps >= 9
   THEN
      Stamps ← Stamps - 9
      OUTPUT "Free drink!"
ENDIF
```

What is the value of Stamps after the pseudocode has run?

Accepted Answer

B: 3. The third line changes Stamps to 7 + 5 = 12. The condition 12 >= 9 is true, so the THEN branch takes 9 stamps off, leaving 3 (and outputs the free-drink message).

Question 2

A football academy times 20 players over 100 m. This algorithm should input each player's time, in seconds, and output the fastest (lowest) time.

```pseudocode numbered
DECLARE Fastest : REAL
DECLARE Time : REAL
DECLARE Runner : INTEGER
Fastest ← 0
FOR Runner ← 1 TO 20
   INPUT Time
   IF Time < Fastest
      THEN
         Fastest ← Time
   ENDIF
NEXT Runner
OUTPUT "Fastest time: ", Fastest
```

When it is tested with real times, it always outputs 0. Which line contains the error?

Accepted Answer

A: Line 04. Fastest starts at 0. Every real sprint time is greater than 0, so Time < Fastest is never true and Fastest is never replaced: 0 is output. Line 04 must give Fastest a starting value larger than any possible time (for example Fastest ← 999), or the first time input.

Question 3

A bus-pass top-up machine accepts only coins worth 1, 2 or 5. A top-up costs 10.

```pseudocode
Paid ← 0
Rejected ← 0
REPEAT
   INPUT Coin
   IF Coin = 1 OR Coin = 2 OR Coin = 5
      THEN
         Paid ← Paid + Coin
      ELSE
         Rejected ← Rejected + 1
   ENDIF
UNTIL Paid >= 10
```

The coins entered are, in this order: 2, 3, 5, 20, 1, 2, 2, 5

What are the values of Paid and Rejected when the loop ends?

Accepted Answer

B: Paid = 10, Rejected = 2. 2 is accepted (Paid 2), 3 is rejected (Rejected 1), 5 is accepted (Paid 7), 20 is rejected (Rejected 2), 1 is accepted (Paid 8) and 2 is accepted (Paid 10). Now Paid >= 10 is true, so the loop ends and the last two coins are never read: Paid = 10 and Rejected = 2.