Maths olympiad question · P6★★★★ Expert
Averages: olympiad question 9
An expert question on averages for P6, with a step-by-step solution, diagrams and the answer. Averages appear in the Hua Xia Cup's P4 and P5 first rounds, qualifying rounds and finals.
The question
A school speech-festival event has 5 judges, each giving a whole mark out of 100; a pupil gets five different marks.
Without the highest, the other four average 6 less than all five; without the lowest, the other four average 3 more than all five; without both, the middle three average 59.
What is the highest mark?
Show the hints (3)
- Hint 1: Without the highest, the other four average 6 lower. How far is the highest above the five-mark average? (Who was pulling the average up?)
- Hint 2: Likewise, how far is the lowest below it? Call the average A: highest = A + 4 × 6, lowest = A − 4 × 3.
- Hint 3: The five add to 5 × A; without the two, 5 × A − (A + 24) − (A − 12) is left, which equals 3 × 59 = 177.
Worked solution
Have a go first, then read the solution step by step.
- 1.Let the five average A, so they total 5A. Without the highest, the four average A − 6: highest = 5A − 4(A − 6) = A + 24.
- 2.Likewise, lowest = 5A − 4(A + 3) = A − 12.
- 3.The middle three = 5A − (A + 24) − (A − 12) = 3A − 12, which is also 3 × 59 = 177.
- 4.So 3A = 177 + 12 = 189, A = 63 (total 315).
- 5.Highest = 63 + 24 = 87 (lowest = 63 − 12 = 51).
- 6.Check: say the middle three are 52, 61, 64. All five add to 315, average 63; without 87: 228 ÷ 4 = 57, 6 lower ✓; without 51: 264 ÷ 4 = 66, 3 higher ✓; middle three 177 ÷ 3 = 59 ✓
- ⚠️Common mistake: thinking the highest is only 6 above the average. Taking it away drops the other four's average by 6, so it is 4 × 6 = 24 above.
Answer: 87 marks
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