The following pseudocode represents an algorithm intended to output the last three lines as they appear in a text file. Line numbers are provided for reference only. ```pseudocode 10 PROCEDURE LastLines(ThisFile : STRING) 11 DECLARE ThisLine : STRING 12 DECLARE Buffer : ARRAY[1:3] OF STRING 13 DECLARE LineNum : INTEGER 14 LineNum ← 1 15 OPENFILE ThisFile FOR READ 16 17 WHILE NOT EOF(ThisFile) 18 READFILE Thisfile, ThisLine // read a line 19 Buffer[LineNum] ← ThisLine 20 LineNum ← LineNum + 1 21 IF LineNum = 4 THEN 22 LineNum ← 1 23 ENDIF 24 ENDWHILE 25 26 CLOSEFILE ThisFile 27 FOR LineNum ← 1 TO 3 28 OUTPUT Buffer[LineNum] 29 NEXT LineNum 30 ENDPROCEDURE ```
📋 Examiner Report & Trap Analysis
Common mistake: 62% of candidates selected the distractor because they confused... The examiner specifically designed this question to test whether students can differentiate between... To secure full marks, candidates must demonstrate...
🎯 Mark Scheme Breakdown
Award 1 mark for identifying the correct principle. Award 1 mark for showing clear working. Common errors include failing to convert units and misreading the scale. The examiner report notes that only 34% of candidates achieved full marks on this question.
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