Hands-on math!

Washington State Math Olympiad
Hints and Solutions
2013 Grade 6 Measurement
Problem
Solution
1) Megan is wrapping a birthday present for her friend with a very expensive shiny neon pink wrapping paper. The width of the box is 6 inches. The height is 2/3 of the width and the length is double the width. If Megan plans on covering the entire outside of the box without wasting any paper, how many square inches of paper will Megan need? 1. Compute the length and height of the box given it's width =
    4 in. height and 12 inches in length.
2. Compute the area of the 6 sides of the box and add =
    2 x L x W = 2 x 12 X 6 = 144 sq. in.
    2 x L x H = 2 x 12 X 4 = 96 sq. in.
    2 x W x H = 2 x 6 X 4 = 48 sq. in.
    Total = 288 sq. in.

2) Molly drives an average of 52 miles per hour. At this rate, how long would it take Molly to drive 429 miles? Express your answer in hours and minutes to the nearest minute.		 1. Divide 429 miles by 52 miles per hour = 8.25 hours
2. Convert the fraction of an hour to minutes and round = 15 min
3. Total time = 8 hours and 15 minutes.

3) In a sudden downpour, rain falls at a rate of 1 inch per hour for 20 minutes on Mr. Duggan's roof. One piece of his roof is a rectangle that is 40 feet by 30 feet and all the rain rolls off into a gutter that feeds into a down spout that flows into a rain barrel with no backing up. How fast is the water flowing into the rain barrel during the middle of this downpour? Give your answer to the nearest gallon per minute.
(1 gallon = 0.134 cubic feet)		 1. Compute the area of the roof = 1200 sq. ft.
2. Convert the rain rate into feet/minute =
    1 in/hr / (12 in./ft.) / (60 min/hr) = 1/720 ft. min.
3. Multiply these two to get the number of cu. ft./min =
    = 1.6667 cu. ft./min
4. Divide by the cubic feet per gallon to get the gallons per minute = 1.667 cu. ft./min / 0.134 gal/cu. ft.= 12.438 gal/min
5. Round to the nearest gallon = 12 gal/min

Problem
Solution
4) A hollow cement brick measures 12 inches tall and 8 inches wide and 5 inches deep on the outside, see figure. The brick is 1 inch thick. Cement weighs 0.077 pounds per cubic inch. Find the weight of the brick to the nearest tenth of a pound. Method 1: Use volume subtraction:
The idea is to compute the volume of the brick, ignoring the hollow center, and then subtract the volume of the hollow center.
1. Compute the volume of the brick, ignoring the hollow space =
    5 x 12 x 8 = 480 cu. in.
2. Reduce the height and the width (but not the depth) by 1 inch and compute the volume of the hollow interior =
    5 x 10 x 6 = 300 cu. in.
3. Subtract the two = 480 - 300 = 180 cu. in.
4. Multiply by the weight per cubic inch of cement and round to the nearest tenth of a pound =
    180 x 0.077 = 13.9 pounds

Method 2: Add up the volume of the 4 sides:
1. Top: 8 X 5 x 1 = 40
2. Bottom: 8 X 5 X 1 = 40
3. Left side: (12 - 2) x 5 x 1 = 50 (subtract 2 thicknesses)
4. Right side: (12 - 2) x 5 x 1 = 50 (subtract 2 thicknesses)
5. Total = 40 + 40 + 50 + 50 = 180 cu. in.
6. Multiply by the weight per cubic inch of cement and round to the nearest tenth of a pound:
    180 x 0.077 = 13.9 pounds

5) A farm has a rectangular field. The field is 32 yards wide. The area of the farm is 2,048 square yards. How much fencing is needed to surround the whole field?		 1. Using the area and the width of the field, compute its length =
    2048/32 = 64 yards.
2. Compute the perimeter of the field = 2 x 64 + 2 x 32 =
    192 yards.