S6 ICT Databases and SQL, Paper 2: filtering, sorting and aggregates
Paper 2 uses a five-row Book table priced 120, 85, 150, 60 and 95 to count 5 rows, three rows above 90, an average of 102 and a total of 510, then a range of 90, D first when sorted ascending and three rows between 80 and 120.
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 25 |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: S6 ICT: Databases and SQL |
| Level | S6 |
| # | Answer | One-line note |
|---|---|---|
| 1 | 5 | COUNT(*) counts rows, so five rows return 5 regardless of the column contents. |
| 2 | 3 rows | The prices 120, 150 and 95 are above 90, while 85 and 60 fail, so three rows remain. |
| 3 | 102 | The five prices add to 510 and dividing by 5 gives 102. |
| 4 | 510 | SUM adds the five prices, and 120 plus 85 plus 150 plus 60 plus 95 is 510. |
| 5 | 90 | The highest is 150 and the lowest 60, so the difference, the range, is 90. |
| 6 | D (priced 60) | ASC sorts from smallest to largest, so D at 60 comes first. |
| 7 | 3 rows | The prices inside the range are 120, 85 and 95, while 150 and 60 fall outside. |
| 8 | DELETE removes matching rows and keeps the table; DROP removes the whole table | DELETE removes rows and keeps the table, while DROP removes the table itself. |
| 9 | UPDATE with SET | UPDATE with SET assigns new values, and WHERE limits the rows, otherwise the whole table changes. |
| 10 | NULL cannot be compared with equals; use IS NULL | NULL means unknown, so any equals test against it is never true; use IS NULL or IS NOT NULL. |
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