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Problems of the Week – January 7 to January 11

Click here [1] for the Problem Extension Worksheet version of the Problems of the Week.
Click here [2] for an MS Word version of the Problems of the Week.
Click here [3] for the Canadian Problem Extension Worksheet version of the Problems of the Week.
Click here [4] for a Canadian MS Word version of the Problems of the Week.

[5]Lower Elementary:
Question: A sea turtle can swim 5 miles every 15 minutes. How far can the sea turtle swim in 4 hours?
Answer: 80 miles
Solution: Since 15 minutes is a quarter of an hour, there are 4 15-minute periods in each hour. That means the sea turtle swims 5, 10, 15, 20 miles in an hour. So, in 4 hours, the sea turtle swims 20, 40, 60, 80 miles.

[6]Upper Elementary:
Question: An adult blue whale weighs 200 tons. An adult Hector’s dolphin weighs 125 pounds. How many times the weight of the Hector’s dolphin is the blue whale?
Answer: 3,200 times the weight
Solution: A ton is 2,000 pounds, so 200 tons is 200 × 2,000 = 400,000 pounds. If we divide 400,000 by 125, we get 3,200. So, the blue whale is 3,200 times the weight of the Hector’s dolphin.

[7]Middle School:
Question: A giant panda eats 30 pounds of bamboo per day. If 5 giant pandas live in a sanctuary that produces 145 pounds of edible bamboo per day and the sanctuary currently has 60 pounds of edible bamboo available to the pandas, then how long will it take for the pandas to run out of bamboo?
Answer: 12 days
Solution: Each day, the 5 pandas eat 30 × 5 = 150 pounds of bamboo. So, the pandas decrease the amount of bamboo by 150 pounds each day. The bamboo grows 145 pounds each day. So, overall, the bamboo decreases by 5 pounds per day. If the sanctuary currently has 60 pounds of bamboo, that means that the pandas will run out of bamboo in 60 ÷ 5 = 12 days.

[8]Algebra and Up:
Question: When a lemur leaps across the ground, it travels forward 30 feet and reaches a height of 10 feet in the air. Each leap takes 2 seconds from take-off to landing. Model the lemur’s leap with the equation of a parabola in which x = the elapsed time in seconds and y = the height off the ground in feet.
[Hint: A parabola in vertex form is written y = a(x – h)² + k when (h, k) is the vertex.] Answer: y = –10(x – 1)² + 10
Solution: The vertex of a parabola that opens down, like the model of the leap, is the highest point. So, the vertex of our model will be (1, 10). To write our equation, we first plug h and k into our general form: y = a(x – 1)² + 10. Next, we use a known point on the parabola and plug those values in for x and y. Let’s use the origin: 0 = a(0 – 1)² + 10. If we solve for a, we get a = –10. So, our equation that models the leap is y = –10(x – 1)² + 10.

Note: Answers may vary with different domains and forms of the equation.