Hands-on math!

Washington State Math Olympiad
Hints and Solutions
2016 Grade 7 Measurement
Problem
Solution

1) The container used for gardening measures 14 in across and 3 ft wide, capped with a half circle at each end as shown below. You want to cover it with a layer of fine topsoil. What is the area covered by the topsoil?
(Express your answer to the nearest tenth of a square inch)
 The area of the container is the sum of a 14x36 inch rectangle and a circle of radius 7 inches. (each of the 2 half circles makes a whole one)
  1. Area of the rectangle: 14x36 = 504 sq. in.
  2. Area of the circle = 72 = 49 = 153.86 sq. in.
  3. The total = 504 + 153.86 = 657.86 = 657.9 sq. in. (rounded)

2) Marcus is a marathon runner. After running 10 miles at 12 miles per hour (mph), he walks at 1/6th that rate for 30 minutes. If he repeats this cycle, how long will it take Marcus to complete a 27 mile marathon??
  1. The time it takes Marcus to run 10 miles is:
        10 miles/(12 miles/hour) = 5⁄6 hour = 50 minutes
  2. The distance Marcus runs in the 30-minute walk is:
        2 mi/hr x .5 hr = 1 mile
  3. In each run-walk cycle Marcus completes 11 miles in 80 minutes
  4. The marathon has 27/11 = 2 cycles with 5 miles remaining, so the last 5 miles are while he is running at 12 miles/hour = :
        5 mi/12 mi/hr = 5/12 hour = 25 minutes
  5. Total time = 2(80) + 25 =
        185 minutes = 3 hours 5 minutes

Problem
Solution

3) Kerry uses 80 meters of fencing to enclose two separate squares each with an integer side length. The total area of the fenced spaces is 250 square meters. How many times larger is a side of the big square compared to the small square? 		Call the side length of the first square s1 and the second s2.
  1. The equation for the total length of fencing is:
        4(s1 + s2) = 80
  2. The equation for the total areas is:
        s12 + s22 = 250
  3. This is 2 equations with 2 unknowns, so we have:
    Equation 1:     4(s1 + s2) = 80
    Equation 2:     s12 + s22 = 250
  4. Rewriting equation 1 for s1:
    s1 = 20 - s2
  5. Substituting this is the second equation:
    (20 - s2)2 + s22 = 250
  6. Using the distributive property:
    400 -40 s2 + s22 = 250
  7. So, simplifying:
    s22 -40 s2 + 150 = 0
  8. Using the quadratic equation formula:
        s2 = +40 ± √ (-40)2 - (4) (1) (150)      So:
                      2
        s2 = 40 ± √1600 -1200     =
                      2
        s2 = 40 ± 20     = 30 or 10
                      2
  9. Substitute back in the first equation:
    4(s1 + 10) = 80
    4s1 + 40 = 80
    4s1 = 40
    s1 = 10
  10. The ratio of the sides of the 2 squares is s2/s1 =
    30/10 = 3

Problem
Solution
4) Andre drops a ball from a height of 64 feet. The ratio between any height of drop and subsequent rise is constant. He records the third bounce reaching a height of 27 feet. What fraction of a drop distance does the ball reach with each bounce? Use f for the fraction of the previous height that the current height reaches.
  1. The first bounce goes to 64 f feet
  2. The second bouce goes to 64 f 2 feet
  3. The third bounce goes to 64 f 3 feet = 27 feet
  4. So f 3 = 27⁄64
  5. f = 3⁄4
5) Brynn is using a ruler to determine the measurements of her tablet computer's screen. She measures the tablet to be 8.1 in wide and 10.8 in long. What is the measurement of the diagonal of her screen? Let d = the length of the diagonal
Using the pythagorean theorem:

    d = √ (8.1 2 + 10.8 2) 

    d = √ (65.61 + 116.64)     = √ 182.25 = 13.5 inches