Hands-on math!

Washington State Math Olympiad
Hints and Solutions
2016 Grade 6 Algebra
Problem
Solution
1) Charlie's class has 28 students in it (including himself). Of the 28 students, 15 have dogs, 12 have cats, and 6 have both. How many students have neither a cat nor a dog? Method 1: Use a Venn diagram
1. Draw a Venn diagram of the two classes of students and fill in the numbers. The number that don't have a cat or a dog are members of the Universal set that don't fall into either circle.
Your Venn diagram should look like this:
1. Compute the sum of the 3 parts =
    9 + 6 + 6 = 21.
2. 28 - 21 = 7 students.

Method 2: Use logic:
1. If 6 have both cats and dogs and there are 12 students that have cats then there are 6 students that have cats only. (12 - 6 = 6)
2. If 6 have both cats and dogs and there are 15 students that have dogs, then there are 9 students that have dogs only. (15 - 6 = 9)
3. Compute the sum of
    both dogs and cats 6,
    dogs only 9 and
    cats only = 6
    sum = (9 + 6 + 6) = 21
4. Subtract this from 28 = 28 - 21 = 7 students
2) Evaluate the following equation:
    If x = 4.7, y = 6.2, z = 8.3;
(Express to the nearest hundredth.)
3xy - 4y + 2 (z - 1.9) = ?
 Substitute the values for x, y, and z into the equation and solve.
    3 (4.7) (6.2) - 4(6.2) + 2(8.3 - 1.9) =
    87.42 - 24.8 + 2(6.4) =
    62.62 + 12.8 = 75.42
3) Mira's friend is having a birthday this month, but Mira always forgets the exact birthday. Mira's friend reminds her that 1/3 of the birth date plus 52 equals two times as much as the birth date. What is her friend's birthday? Let D be the birth date.
Solve the equation (1/3)D + 52 = 2 X D.
    D/3 + 25 = 2D
    2D - D/3 = 25
    5D/3 = 25
    5D = 75;   D = 15

Problem
Solution
4) 3x2 + 7 = 199. Evaluate for x.

 Method 1 (easier way):
1. Rewrite the equation with only x2 on the left side of the equation:
    3x2 = 199 - 7
    x2 = 192/3 = 64.
2. Take the square root of the right side.
    x = 8.

Method 2: Use the quadratic formula
1. Plug in the following values for the constants in the quadratic formula and solve:
    a = 3;   b = 0 (there is no x term) and c = -192
The quadratic formula is:
    x = -b ± √ b 2 - 4 a c 
                  2a
Putting in our values:
    x = ± √ - 4 (3)(-192) 
                  2(3)

    x = ± √ +2304 
                6
    x = ± 48 
              6
    x = +8 or -8 (Note: allow a solution of - 8)
5) Jill likes insects and wants to be an entomologist when she grows up. For now, she keeps a collection of ants and spiders. Ants have 6 legs and spiders have 8, and together there are 380 legs in her collection. If she has 59 total ants and spiders, how many of each, ants and spiders, does she have?

Editor's note: I know, I know, spiders are not insects. Let it go! 		Method 1: Use guess-and-check:

Guess ## Ants# SpidersTotal
heads		Total legs
146135946x6 + 13x8 = 380

Method 2: Use 2 equations:
Use A for the number of ants and
    S for the number of spiders.
1. The equation for heads is A + S = 59
2. The equation for legs is 6A + 8S = 380
3. Make the first equation an equation for A.
    A = 59 - S
4. Substitute this for A in the second equation and solve for S
    6(59 - S) + 8S = 380
    354 - 6S + 8S = 380
    2S = 26
    S = 13 spiders
5. Subtract this from 59 to get the number of ants = 46 ants