Problem 30 · 2023 Math Kangaroo
Stretch
Arithmetic & Operations
Algebra & Patterns
total-then-dividesum-constraint
A rugby team scored 24, 17 and 25 points in their 7th, 8th and 9th game of the previous season. The average number of points per game was higher after 9 games than after their first 6 games. Their average after 10 games was more than 22 points. What is the minimum number of points they could have scored in their 10th game?
Show answer
Answer: C — 24
Show hints
Hint 1 of 2
Turn each 'average' statement into a statement about total points using the number of games.
Still stuck? Show hint 2 →
Hint 2 of 2
Combine the inequality after 9 games with the 'more than 22 after 10 games' condition to bound the 10th score from below.
Show solution
Approach: convert averages to total-point inequalities and minimise
- Let the first-6-game total be T; games 7–9 add 24+17+25 = 66, so the 9-game total is T + 66.
- 'Higher average after 9 than after 6' gives \(\frac{T+66}{9} > \frac{T}{6}\); clearing denominators, \(6(T+66) > 9T\), so \(396 > 3T\) and \(T < 132\).
- 'More than 22 after 10 games' gives \(T + 66 + g > 220\), so \(g > 154 - T\); to make the 10th score \(g\) as small as possible, take \(T\) as large as allowed, \(T = 131\).
- Then \(g > 154 - 131 = 23\), so the smallest whole-number score is 24, option C.
Mark:
· log in to save