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Problems of the Week – November 5 to November 9

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: Amaryllis takes piano lessons every day. On weekdays, her lessons are 30 minutes long. On Saturdays and Sundays, her lessons are 45 minutes long. How many hours does Amaryllis spend at her piano lessons each week?
Answer: 4 hours
Solution: In a week, there are 5 weekdays. So, Amaryllis spends 30, 60, 90, 120, 150 minutes at her piano lessons on weekdays. On the weekend, she spends 45 + 45 = 90 minutes at her piano lessons. So, Amaryllis spends 150 + 90 = 240 minutes at her piano lessons each week. Since there are and 60 minutes in an hour, 240 minutes is the same as 4 hours.

[6]Upper Elementary:
Question: Tommy and Winthrop are playing a game of pool. If all but 7 of the 15 balls have been pocketed and Tommy has pocketed ¾ of them, then how many balls has Winthrop pocketed?
Answer: 2 balls
Solution: If all but 7 of the balls have been pocketed, then there are 15 – 7 = 8 pocketed balls in total. If Tommy pocketed ¾ of them, then Winthrop pocketed ¼ of them. Since ¼ of 8 is 2, Winthrop pocketed 2 balls.

[7]Middle School:
Question: A marching band includes 76 trombones and 110 cornets. The conductor wants the musicians arranged so that each row has exactly the same number of musicians. He also wants everyone in a row to play the same instrument. What is the greatest number of musicians that can be in each row so that the band is arranged how the conductor wants it?
Answer: 2 musicians
Solution: To find the number of rows there could be so that each row has the same number of people all playing the same instrument, we find the GCF of 76 and 110. The factors of 76 are 1, 2, 4, 19, 38, and 76. The factors of 110 are 1, 2, 5, 10, 11, 22, 55, and 110. The greatest factor they have in common is 2.

[8]Algebra and Up:
Question: A wagon is carrying curtains, a set of dishes, and a double boiler. The double boiler weighs 2 pounds less than 3 times as much as the curtains. The dishes weigh 2 pounds more than twice as much as the double boiler. Altogether, the wagon is carrying 16 pounds. How many times as heavy as the curtains are the dishes?
Answer: 5 times
Solution: First, we find out how much each item weighs by setting up an equation whose sum is the total weight with each individual weight in terms of the weight of the double boiler: x + (x/3 + 2/3) + (2x + 2) = 16. When we solve this, we find that the double boiler weighs 4 pounds. So, the curtains weigh 2 pounds and the dishes weigh 10 pounds. The dishes are 5 times as heavy as the curtains.

3 Comments (Open | Close)

3 Comments To "Problems of the Week – November 5 to November 9"

#1 Comment By BENEDICT ZOE On November 3, 2018 @ 5:05 pm

I think the Middle School problem is confusingly worded. The maximum number of rows consisting of the minimum number of identically arranged players is 2 rows with each row consists of at most 38 trombones and 55 cornets, or a total of 93 musicians.
OR you could have a minimum of 1 row with a maximum of 186 musicians.

#2 Comment By Joshua Edgar On November 4, 2018 @ 1:04 pm

Agreed. Something seems to be off on this one.

#3 Comment By Jane Adams On November 5, 2018 @ 9:01 am

Hey, Benedict and Joshua! I’ve edited the phrasing to make it a bit more clear. I hope this solves your problem.