Hands-on math!

Washington State Math Olympiad
Hints and Solutions
2010 Grade 7 Number Sense
Problem
Solution
1) What is the smallest exponent n for which the following expression is true?
    (0.5)3 > (0.6)n 		Use your calculator that does exponents for this problem.
  1. (0.5)3 = 0.125 = 1⁄8
  2. For n = 2:  (0.6)2 = 0.36
    For n = 3:   (0.6)3 = 0.216
    For n = 4:   (0.6)4 = 0.1296
    For n = 5:   (0.6)5 = 0.07776
  3. n = 5
2) In Mr. Hemple's orchestra there are 4/7 as many boys as girls. 1/4 of the girls wear glasses, and 2/7 of the girls who wear glasses also have braces. If 2 girls have glasses and braces, how many students are in Mr. Hemple's orchestra? This is a classic 'work backwards' problem.
  1. 2⁄7 of the girls who wear glasses also have braces and that fraction is 2 girls, therefore the number of girls who wear glasses are:
        2⁄7 X = 2; X = 7 girls who wear glasses
  2. 7 girls who wear glasses are 1⁄4 of the girls, therefore there are 4x7 = 28 girls total.
  3. There are 4⁄7 as many boys, so
    4⁄7 x 28 = 16 boys.
  4. The total number of students is
    28 + 16 = 44 students
3) On a number line, what fraction is halfway between
    1 - 3 and   -3 + 5 ?
    3   4           8     6
  1. First term:   1⁄3 - 3⁄4 = -5⁄12
  2. Second term:   -3⁄8 + 5⁄6 = +11⁄24
  3. Halfway is (-5⁄12 + +11⁄24 ) x 1⁄2 =
    (-10⁄24 + +11⁄24 ) x 1⁄2 = 1⁄48

Problem
Solution
4) Liz was playing with a pile of pennies. She noticed when she stacked them in piles of three there was 1 left over. When she stacked them in piles of five there were 2 left over. When she stacked them in piles of seven there were 3 left over. What is the smallest number of pennies she could have? Start with stacks of 7 and compute the remainders of stacking in 5s and 3s:
# stacks
of 7
(N)		 Total
pennies

(7N + 3)		 # stacks of 5 + remainder # stacks of 3 + remainder
2173+25+2
3244+48+0
4316+17+3
5387+312+2
6459+015+0
75210+217+1
Liz has 52 pennies.
5) It is the end of market day and Grandma Wilcox wants to sell all her remaining eggs. One at a time to the last two customers come to her stand. To the first customer she sells half of all of the eggs she has right then plus half an egg without breaking any eggs. To the last customer she sells half of all her remaining eggs plus half an egg without breaking any eggs. She's now out of eggs. What is the smallest number of eggs she could have had left at the time the last two customers came? Very confusing explanation.
What the problem means is that when you take half the remaining eggs (which can involve a fraction of an egg) and add half an egg you end up with a whole number of eggs with each customer transaction. Therefore the number of eggs she started with must be odd to have half an egg remaining after dividing by 2.
  1. So, the number of eggs must be one of: 3,5, 7 ...
  2. With 3 eggs/2 = 11⁄2 eggs + 1⁄2 egg = 2 eggs.
  3. She now has 1 egg left, so that, divided by 2 is 1⁄2 egg + 1⁄2 egg = 1 egg the last customer gets.
  4. The least number of eggs she could have had before these 2 customers arrived was 3