Weekly Dose 10 - Maths Olympiad Practice
- 1. In a class of students, 80% participated in basketball, 85% participated in football, 74% participated in baseball, 68% participated in volleyball. What is the minimum percent of the students who participated in all the four sports events? Solution: Let say only two sports: basketball and football. To get the minimum percent of students who participated in both: 80% play basketball 20% not playing 85% play football 15% not playing Scenario 1: if the 15% not playing football also not playing basketball, then 5% not playing basketball but play football. We get this: 100% − 20% = 80% playing both. Scenario 2: if the 15% not playing football and 20% not playing basketball are not the same students, we get the below venn diagram: 100% − 20% − 15% = 65% playing both. So, need to use scenario 2 to get the minimum percent of students participated in both. Meaning, we must assume each student misses at most one sport.
- 2. In a class of students, 80% participated in basketball, 85% participated in football, 74% participated in baseball, 68% participated in volleyball. What is the minimum percent of the students who participated in all the four sports events? Solution (continue): For our question, to minimize the number of students playing in all four sports, we make each students misses at most one sport. The number of students not playing basketball, football, baseball and volleyball are 20% + 15% + 26% + 32% = 93%. The percent of students who participated in all the four sports events are 100% − 93% = ____% Answer: ____%
- 3. Ten whole numbers (not necessarily all different) have the property that if all but one of them are added, the possible sums (depending on which one is omitted) are: 82, 83, 84, 85, 87, 89, 90, 91, 92. The 10th sum is a repetition of one of thesse. What is the sum of the ten whole numbers? Solution: Assuming the 10 numbers are 𝑎, 𝑏, 𝑐, 𝑑, 𝑒, 𝑓, 𝑔, ℎ, 𝑖, 𝑗. Given: 𝑎 + 𝑏 + 𝑐 + 𝑑 + 𝑒 + 𝑓 + 𝑔 + ℎ + 𝑖 = 82 𝑎 + 𝑏 + 𝑐 + 𝑑 + 𝑒 + 𝑓 + 𝑔 + ℎ + 𝑗 = 83 𝑎 + 𝑏 + 𝑐 + 𝑑 + 𝑒 + 𝑓 + 𝑔 + 𝑖 + 𝑗 = 84 𝑎 + 𝑏 + 𝑐 + 𝑑 + 𝑒 + 𝑓 + ℎ + 𝑖 + 𝑗 = 85 𝑎 + 𝑏 + 𝑐 + 𝑑 + 𝑒 + 𝑔 + ℎ + 𝑖 + 𝑗 = 87 𝑎 + 𝑏 + 𝑐 + 𝑑 + 𝑓 + 𝑔 + ℎ + 𝑖 + 𝑗 = 89 𝑎 + 𝑏 + 𝑐 + 𝑒 + 𝑓 + 𝑔 + ℎ + 𝑖 + 𝑗 = 90 𝑎 + 𝑏 + 𝑑 + 𝑒 + 𝑓 + 𝑔 + ℎ + 𝑖 + 𝑗 = 91 𝑎 + 𝑐 + 𝑑 + 𝑒 + 𝑓 + 𝑔 + ℎ + 𝑖 + 𝑗 = 92 𝑏 + 𝑐 + 𝑑 + 𝑒 + 𝑓 + 𝑔 + ℎ + 𝑖 + 𝑗 = 𝑥 (𝑡ℎ𝑒 10𝑡ℎ 𝑠𝑢𝑚) 9𝑎 + 9𝑏 + 9𝑐 + 9𝑑 + 9𝑒 + 9𝑓 + 9𝑔 + 9ℎ + 9𝑖 + 9𝑗 = 82 + 83 + 84 + 85 + 87 + 89 + 90 + 91 + 92 + (10𝑡ℎ 𝑠𝑢𝑚) 9(𝑎 + 𝑏 + 𝑐 + 𝑑 + 𝑒 + 𝑓 + 𝑔 + ℎ + 𝑖 + 𝑗) = 783 + (10𝑡ℎ 𝑠𝑢𝑚) 9(𝑎 + 𝑏 + 𝑐 + 𝑑 + 𝑒 + 𝑓 + 𝑔 + ℎ + 𝑖 + 𝑗) = 9(87) + (10𝑡ℎ 𝑠𝑢𝑚) Since 783 is also a multiple of 9, the 10𝑡ℎ 𝑠𝑢𝑚 must be a multiple of 9 also. Of the nine known sums, only 90 is a multiple of 9, therefore, the 10𝑡ℎ 𝑠𝑢𝑚 is 90. 9(𝑎 + 𝑏 + 𝑐 + 𝑑 + 𝑒 + 𝑓 + 𝑔 + ℎ + 𝑖 + 𝑗) = 9(87) + 90 𝑎 + 𝑏 + 𝑐 + 𝑑 + 𝑒 + 𝑓 + 𝑔 + ℎ + 𝑖 + 𝑗 = 87 + 10 Answer: 97
- 4. Solution: The number of tiles in 2004th square = 20042 The number of tiles in 2005th square = 20052 The difference = 20052 − 20042 = (2005 − 2004)(2005 + 2004) = __________ Answer: ________ A sequence of square is made of identical square tiles. The edge of each square is one tile length longer than the edge of the previous square. The first three squares are shown. How many more tiles does the 2005th square have than the 2004th ?
- 5. Solution: Assume the money Obet received from each of them is 𝐴. 1 5 𝑀 = 𝐴 → 𝑀 = 5𝐴 1 4 𝐿 = 𝐴 → 𝐿 = 4𝐴 1 3 𝑁 = 𝐴 → 𝑁 = 3𝐴 Total received by Obet = 3𝐴 Total money = 𝑀 + 𝐿 + 𝑁 = 12𝐴 Fraction = 3𝐴 12𝐴 = 1 4 Answer: 1 4 Lucky, Michael, Nelson and Obet were good friends. Obet had no money. Michael gave one-fifth of his money to Obet. Lucky gave one-fourth of his money to Obet. Finally, Nelson gave one-third of his money to Obet. Obet received the same amount of money from each of them. What fraction of the group’s total money did Obet have at the end?