Hands-on math!

Washington State Math Olympiad
Hints and Solutions
2008 Grade 6 Probability & Statistics

Problem
Solution
1) A dart hits the dartboard shown at random. The board is a square and the horizontal lines are equally spaced. Find the probability of the dart landing in the shaded region. You need to find the total area of the two shaded areas. Use 1 for the square side length
  1. The large lower triangle and the shaded triangle that is inside of it are similar because they have parallel bases.
  2. The height of the large triangle is .5 and the height of the included triangle is 12 x 13 = 16
  3. Therefore the ratio of their side lengths is 12 ÷ 16 = 3
  4. The base of the small triangle is, therefore, 13
  5. The area of the 2 small shaded triangles and the probability of the dart hitting them) is
        13 x 16 = 118
2) Jesse makes 50% of his free throws and 10% of his half court shots in basketball. What are the chances that Jesse can make three free throws and a half court shot one right after the other? Multiply the probabilities of 3 free throws and one half court shot together (4 numbers):
    1/2 x 1/2 x 1/2 x 1/10 = 1/8 x 1/10 = 1/80
3) In a certain carnival game, a player gets to spin each of these spinners once. What is the probability (express as a fraction in lowest terms) of getting two numbers that have a sum of 7?
Fill out this table of sums and find the ones that have a sum of 7, then divide by the total number of sums:
+   1     2    3    4    5    6 
1234567
2345678
3456789
45678910
There are 4 of them in the above table.
The probability of getting a sum of 7 is 4/24 = 1/6

Problem
Solution
4) After three tests Max has a test average of 83. He is nervous for the fourth test and wants to keep an average of at least 80. What is the lowest score he can get on the fourth test if each test is worth 100 points? This is actually an algebra problem!
Use N3 for the sum of the first 3 tests and N for the next test score.
1. So, if the 3 scores average 83, then N3 must be 249.
2. Then (N + N3)/4 = 80.
3. (N + 249)/4 = 80
    N + 249 = 320
    N = 320 - 249 = 71
5) The bar graph below shows the heights for a 4th grade class that has 11 boys in it measured to the nearest inch. What is the difference between the mean and the median? 1. Write out all the heights:
  47 49 49 51 51 51 53 53 54 54 60
2. Compute the mean = 572/11 = 52.
3. Compute the median (6th boy's height) = 51.
4. Their difference is 52 - 51 = 1.