Click here for the Problem Extension Worksheet version of the Problems of the Week.
Click here for an MS Word version of the Problems of the Week.
Click here for the Canadian Problem Extension Worksheet version of the Problems of the Week.
Click here for a Canadian MS Word version of the Problems of the Week.

small tree

Lower Elementary:
Question: William has 3 fruit trees in his garden. He picks 8 apples from the apple tree, 6 pears from the pear tree, and 7 limes from the lime tree. How many pieces of fruit did William pick?
Answer: 21 pieces of fruit
Solution: To find the total, add the apples and the pears, and then add the limes: 8 + 6 = 14. Then, 14 + 7 = 21.

small watering can

Upper Elementary:
Question: Harrison has a full 3–gallon watering can. He uses 5/8 of the water on the gardenias in his garden. How much water did the gardenias get?
Answer: 15 pints
Solution: First we converted the amount of water from gallons into pints because a pint is 1/8 of a gallon. Harrison had 24 pints of water. Then we found 5/8 of 24 and determined Harrison used 15 pints of water.

small cards

Middle School:
Question: Lily is making labels for the plants in her garden. Each label is made of a white 3″ × 5″ laminated against a colorful 5″ × 8″ card. What is the area of the front of each label that isn’t covered by the white card?
Answer: 25 square inches
Solution: First, we need to find the area of the larger card. That’s 5 × 8 = 40 square inches. Next, we subtract the part that’s covered by the white card, which is 3 × 5 = 15 square inches. Then we subtract the area of the small card from the larger card, which gives us 40 – 15 = 25 square inches.

small bouquet

Algebra and Up:
Question: Cameron is selling flowers from his garden. By selling large bouquets for $21 and small bouquets for $15, he makes $540. If he sells a total of 30 bouquets, how many large bouquets did Cameron sell?
Answer: 15 large bouquets
Solution: We can use the equations 21L + 15S = 540 and L + S = 30 to solve for both variables. First, we solve for S in the second equation. That gives us that S = 30 – L. We can plug that value of S into the original equation, so 21L + 15(30 – L) = 540. If we solve for L, we get 15.