Hands-on math!

Washington State Math Olympiad
Hints and Solutions
2007 Grade 6 Probability & Statistics
Problem
Solution
1) Onita is pretty sure her semester GPA will be 3.8 for the 6 credits she's taking. Her previous cumulative GPA was 3.5 for the 6 credits she already had from the previous semester. What will be her new cumulative GPA? Round to the nearest hundredth. Take the average of the 2 semester GPAs =
    (3.5 + 3.8)/2 = 3.65

2) Fill in the blanks in the list of numbers such that the only mode for the list will be 9 and the median will be 15.
4, 9, ___, ___, 13, ___, 17, 23, 23, 28
 1. Since there are two 23s in this list, the number of 9s must be 3 to make it the mode.
2. Since there are 10 numbers in this list, the median must be the average of the middle 2 numbers.
3. What number, averaged with 13 is 15?
    This number is (13 + 17)/2 = 15.
4. The completed list is:
    4, 9, 9, 9, 13, 17, 23, 23, 28

3) Joe's dad tells Joe he can have the bag of fruit chews he's holding if he can solve this riddle. There are four flavors: strawberry, orange, lemon, and cherry. There are 5 strawberry chews. The probability of getting orange is 1/7. The probability of getting a lemon is 2/7. What is the fewest number of cherry chews that could be in the bag?		 1. Add the probabilities of orange and lemon chews =
    1/7 + 2/7 = 3/7
2. Subtract this from 1 = 4/7. This is the probability of a strawberry or a cherry chew.
3. Using this, compute the total number of chews =
    Let n = the number of cherry chews and t = total number of chews
    The equation for the total number of chews is:
    5 + n = (4/7) t
    Starting with n = 1, increase the number of cherry chews until you get a whole number for the total number of chews.
    n = 1: 5 + 1 = 4 t / 7; 42 = 4t; t = 11.5
    n = 2: 5 + 2 = 4 t / 7; 49 = 4t; t = 24.5
    n = 3: 5 + 3 = 4 t / 7; 56 = 4t; t = 14

Problem
Solution
4) Mr. Bill arrives at an intersection in the middle of a city and realizes he is totally lost. The streets form a grid with each block a square. They all look the same to him and the fog isn't helping any. At each intersection he has a choice of three directions to go in: left, right or straight ahead. If he chooses a direction completely randomly at each intersection what is the probability that after 4 more moves, he is right back where he started? Consider the figure to the right. Bill is at E and straight ahead is toward B. Left takes him to D and right to F. He can't go to H directly because that is backward. This should help you visualize the moves.
1. So, at each of 4 moves, he has 3 choices, left, right or straight ahead, so the number of possible combinations is
    3 x 3 x 3 x 3 = 81
2. Work out the move combinations that get him back to his start (E in the diagram).
    Using L = left, R = right and S = straight ahead,
    S L L L
    S R R R
    L L L L
    L R R R
    R L L L
    R R R R
    There are 6 of those.
3. The probability of ending up at the start is 6/81 = 2/27
5) A company puts out dial combination locks that open by turning a dial clockwise to get to the first number, counterclockwise to get to the second number and clockwise to get to the last number. If the numbers on the dial are 0 through 5 and neither the first two nor the last two numbers can be the same, how many different combinations are there? 1. There are 6 possible numbers (0,1,2,3,4 and 5).
2. How many ways to get the first number? 6.
    Let's say he got a 1
3. How many ways to get the second number
    if it can't match the first? 5.
    He can get 0,2,3,4 or 5.
    Let's say he got a 2
4. How many ways to get the third number
    if it can match the first number but not the second = 5.
    He can again have a 1 but not a 2.
    He can have a 0,1,3,4 or 5, for a total of 5
5. Multiply these possibilities together =
    6 x 5 x 5 = 150 combinations.