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?