Hands-on math!

Washington State Math Olympiad
Hints and Solutions
2014 Grade 8 Measurement
Problem
Solution
1) An astronomical unit (AU) is the average distance from the Sun to the Earth. 1 AU is approximately equal to 93 million miles. If Jupiter is about 5.2 AU from the Sun, about how many miles is it from the Sun? Express your answer using scientific notation. 9.3x107 x 5.2 = 48.36x107 = 4.836x108

2) A glass has a height of 12 cm and an inside diameter of 6 cm. Another glass has the same volume but its radius is a third, or about 33.3%, larger. What is the height of the other glass to the nearest centimeter?
  1. Volume of the first glass: v = 12 x 32 = 108 cu. cm.
  2. The radius of the second glass = 3 x 1.333 = 4 cm.
  3. Height of the second glass:
    42 h = 108
  4. h = 108/16 = 6.75 cm. = 7 cm.
3) Nona is intending on making an accurately scaled model of the earth rotating around the sun. The diameter of the sun is about 1.4x106 km and earth has diameter 1.3x104 km. Her earth has diameter 1 cm. What should the diameter of her sun be? Express your answer to the nearest centimeter. Set up the ratio of the earth's diameter to Nona's model's diameter and apply that to the sun's diameter. Let S be the diameter of Nona's sun.
  1. S / 1.4x106 = 1 / 1.3x104 =
  2. S = 1.4x106 / 1.3x104 =
    1.0769x10(6-4) = 107.69 cm = 108 cm

Problem
Solution
4) A cylindrical water tank has a base of 2000 sq. cm. This tank is being filled at the rate of 12 liters per minute. Find the rate at which the height of the water in the tank increases. Express your answer in millimeters per second. 1 cm3 = 1 milliliter of water.
  1. Convert the volume of water flow into cm3/second:
      12 liters x 1000 ml x 1 cm3 x 1 minute = 200 cm3
        minute             liter     ml     60 seconds       second
  2. Translate this volume of water into height in the tank:
      V = 2000 x h
      h = V / 2000 = .1 cm/sec = 1 mm/sec
5) In an old manor house two 3 ft wide hallways meet at right angles as shown in the figure. They both have 7.5 foot high ceilings. To the nearest foot what is the longest ladder you can get through the hallway?
  1. If the ladder remains horizontal, it must fit through the corner as shown by the figure to the right.
  2. The length that will pass horizontally (H) is by pythagorean theorem

      H = √ 6 2 + 6 2  = √72 = 8.48 feet

  3. The hall is 7.5 feet high, so we can fit it diagonally through the corner (see vertical view to the right) , by pythagorean theorem:
      L (the ladder length) =

    √ 8.48 2 + 7.5 2  = √72 + 56.25 = 11.32 feet

  4. The longest ladder that can pass the corner is 11 feet long.