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From our friends at DiscoverE, this activity has Mathletes build and test a truss bridge between two chairs. Have Mathletes watch the video first, then get started!
Find instructions for the Truss Bridge Activity on the DiscoverE website here.
Over the summer millions of people around the world visit amusement parks. Previously heralded as the “World’s Largest Amusement Park,” Cedar Point Amusement Park in Sandusky, Ohio earned that title. At one time, Cedar Point had 72 rides, 16 of which were roller coasters. What percent of Cedar Point’s rides were not roller coasters? Express your answer as a decimal to the nearest tenth.
Since 16/72 of the rides at Cedar Point were roller coasters, it follows that 1 − 16/72 » 0.778 = 77.8% of the rides were not.
The GateKeeper is one of the tallest, fastest, longest wing roller coasters in the world. GateKeeper climbs a record 170 feet (tallest inversion of any roller coaster) and reaches a maximum speed of 67 mi/h. The entire 4164-foot ride takes an unbelievable 2 minutes 20 seconds to complete. What is the average rate of speed for GateKeeper, in miles per hour? Express your answer to the nearest whole number.
First, let’s convert 2 minutes 20 seconds to 2 × 60 + 20 = 120 + 20 = 140 seconds. Recall that 1 mile = 5280 feet and 1 hour = 3600 seconds. Therefore, to go 4164 feet in 140 seconds, GateKeeper travels an average of just (4164 ÷ 5280) ÷ (140 ÷ 3600) = (347/440) × (180/7) = 3123/154 » 20 mi/h.
Since standing in line for rides can take up a great deal of time at some amusement parks, serious riders find it necessary to have a good plan of action to make the best use of time. Say Matt is interested in riding only roller coasters, but he has only enough time to ride 12 of the 16 roller coasters at Cedar Point. How many combinations of 12 roller coasters are there, assuming Matt wouldn’t ride any roller coaster multiple times?
There are 16 × 15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 combinations of 12 different rides chosen from the 16 roller coasters at Cedar Point. Each of these combinations can be arranged in 12! different orders. We divide to remove the duplicates and see that there are (16 × 15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5)/(12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) = (16 × 15 × 14 × 13)/(4 × 3 × 2 × 1) = 1820 combinations of 12 roller coasters Matt could ride.
We get the same result using the formula nCr = n!/(r! (n – r)!) to calculate 16C12 = 16!/(12! × 4!) = 1820 combinations.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
Phoenix earns hotel rewards points whenever she stays at a participating hotel. She can then use these points to pay for future hotel stays. For instance, a stay that costs $500 could be paid for using 500 of her accumulated points. Phoenix does not earn rewards points for stays purchased with rewards points.
For a stay of less than 5 nights that costs Phoenix $548, she earns 548 points. If Phoenix’s hotel stay is more than 5 nights, she earns double points. So, for a stay of 7 nights that costs Phoenix $720, she would earn 1440 points.
In June, Phoenix used rewards points to fully pay for a $798 hotel stay. During that same month she stayed at a participating hotel for 6 nights which cost her $844. What was Phoenix’s net gain in rewards points during the month of June?
Phoenix used 798 points, which we can write as −798. For her 6-night stay, she earned double points, so her $844 stay earned her 844 × 2 = 1688 points, which we can write as +1688. That means her net gain in rewards points for June was −798 + 1688 = 890 points.
Ariel participates in a similar rewards program with her favorite airline. She earns rewards points equivalent to the miles she flies that can be used to buy airline tickets. In addition to earning points for flying, Ariel is awarded points for purchases made with her credit card, where each dollar spent earns 1 point. Ariel only uses her credit card to pay bills and has it set up to automatically charge a total of $970 each month.
At the end of June, Ariel had a total of 20,250 rewards points. The next flight Ariel plans to take is for her vacation to Cancun. After how many months will Ariel have enough rewards points to pay for the flight to Cancun, which requires 28,500 points?
Since Ariel doesn’t plan to fly anywhere before her vacation, she will only be awarded rewards points for purchases made with her credit card. At the end of June, Ariel had 20,250 points, which means she needs an additional 28,500 – 20,250 = 8250 points. If she charges $970 each month, she will earn 970 points each month. Since 8250 ÷ 970 ≈ 8.5, we know that after 8 months, she still will be a little short. Therefore, Ariel will have accumulated enough points to purchase the airline ticket for her flight to Cancun after 9 months.
Saxon participates in the same rewards program as Ariel. He is purchasing an airline ticket for his Bermuda vacation, but his accumulated 18,500 rewards points are not enough to purchase his plane ticket, which requires 24,000 rewards points. The airline will allow Saxon to purchase his ticket using both rewards points and money. After applying rewards points, the airline determines what percent of the total points required is left to be paid and charges a dollar amount equivalent to that percent of the actual ticket cost. If the flight to Bermuda actually costs $340, how much money will Saxon need to pay for his plane ticket after applying all of his accumulated points?
Since Saxon’s ticket requires 24,000 points, after he applies all of his accumulated points, he will still be short 24,000 – 18,500 = 5500 points. The amount Saxon will need to pay for his plane ticket is 340(5500/24,000) ≈ 77.916 = $77.92.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
The nation’s top 12 middle school Mathletes go head-to-head in this culminating event of the 2023 Raytheon Technologies MATHCOUNTS National Competition. Students race against the clock and each other to answer challenging questions and determine the 2023 National Champion.
After polling each student in her homeroom about their summer plans, Mrs. Baker discovered that each student planned to participate in one or more of the following activities:
- Attend summer school
- Work part-time
- Attend sports camp
Mrs. Baker learned that 17 of the students in her homeroom plan to participate in at least two of these activities, and exactly two students plan to participate in all three activities this summer. If the number of students participating in exactly two of these activities is equally divided among the three pairs of activities, how many students plan to both attend sports camp and work part-time this summer, but not attend summer school?
We let x, y and z represent the number of students planning to participate in sports camp and work part-time, work part-time and attend summer school, and attend summer school and sports camp, respectively. We are told there are 2 students who plan to participate in all three activities. Since there are 17 students planning to participate in two or more activities, we have x + y + z + 2 = 17. We are also told that x = y = z. Therefore, 3x + 2 = 17 → 3x = 15 → x = 5 students plan to attend sports camp and work part-time this summer.
If 1/4 of Mrs. Baker’s homeroom students who plan to attend summer school do not plan to participate in either of the two other activities, how many students plan only to attend summer school?
Let S represent the number of students who plan only to attend summer school. From the previous problem, we learned that 5 + 5 + 2 = 12 students plan on participating in summer school and at least one other activity, so S = ¼(S + 12). Solving for S yields 4S = S + 12 → 3S = 12 → S = 4. So, 4 students plan only to attend summer school.
According to this poll, 16 of Mrs. Baker’s homeroom students plan to participate in only one of these activities. If the number of students who plan only to attend sports camp is three times the number of students who plan only to work part-time, what is the total number of students in Mrs. Baker’s homeroom?
We know that 17 students plan to participate in two or more activities, and 16 plan to participate in only one activity. So, there must be 17 + 16 = 33 students in Mrs. Baker’s homeroom.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
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From our friends at DiscoverE, this activity has Mathletes build and test a truss bridge between two chairs. Have Mathletes watch the video first, then get started!
Find instructions for the Truss Bridge Activity on the DiscoverE website here.
Over the summer millions of people around the world visit amusement parks. Previously heralded as the “World’s Largest Amusement Park,” Cedar Point Amusement Park in Sandusky, Ohio earned that title. At one time, Cedar Point had 72 rides, 16 of which were roller coasters. What percent of Cedar Point’s rides were not roller coasters? Express your answer as a decimal to the nearest tenth.
Since 16/72 of the rides at Cedar Point were roller coasters, it follows that 1 − 16/72 » 0.778 = 77.8% of the rides were not.
The GateKeeper is one of the tallest, fastest, longest wing roller coasters in the world. GateKeeper climbs a record 170 feet (tallest inversion of any roller coaster) and reaches a maximum speed of 67 mi/h. The entire 4164-foot ride takes an unbelievable 2 minutes 20 seconds to complete. What is the average rate of speed for GateKeeper, in miles per hour? Express your answer to the nearest whole number.
First, let’s convert 2 minutes 20 seconds to 2 × 60 + 20 = 120 + 20 = 140 seconds. Recall that 1 mile = 5280 feet and 1 hour = 3600 seconds. Therefore, to go 4164 feet in 140 seconds, GateKeeper travels an average of just (4164 ÷ 5280) ÷ (140 ÷ 3600) = (347/440) × (180/7) = 3123/154 » 20 mi/h.
Since standing in line for rides can take up a great deal of time at some amusement parks, serious riders find it necessary to have a good plan of action to make the best use of time. Say Matt is interested in riding only roller coasters, but he has only enough time to ride 12 of the 16 roller coasters at Cedar Point. How many combinations of 12 roller coasters are there, assuming Matt wouldn’t ride any roller coaster multiple times?
There are 16 × 15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 combinations of 12 different rides chosen from the 16 roller coasters at Cedar Point. Each of these combinations can be arranged in 12! different orders. We divide to remove the duplicates and see that there are (16 × 15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5)/(12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) = (16 × 15 × 14 × 13)/(4 × 3 × 2 × 1) = 1820 combinations of 12 roller coasters Matt could ride.
We get the same result using the formula nCr = n!/(r! (n – r)!) to calculate 16C12 = 16!/(12! × 4!) = 1820 combinations.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
Phoenix earns hotel rewards points whenever she stays at a participating hotel. She can then use these points to pay for future hotel stays. For instance, a stay that costs $500 could be paid for using 500 of her accumulated points. Phoenix does not earn rewards points for stays purchased with rewards points.
For a stay of less than 5 nights that costs Phoenix $548, she earns 548 points. If Phoenix’s hotel stay is more than 5 nights, she earns double points. So, for a stay of 7 nights that costs Phoenix $720, she would earn 1440 points.
In June, Phoenix used rewards points to fully pay for a $798 hotel stay. During that same month she stayed at a participating hotel for 6 nights which cost her $844. What was Phoenix’s net gain in rewards points during the month of June?
Phoenix used 798 points, which we can write as −798. For her 6-night stay, she earned double points, so her $844 stay earned her 844 × 2 = 1688 points, which we can write as +1688. That means her net gain in rewards points for June was −798 + 1688 = 890 points.
Ariel participates in a similar rewards program with her favorite airline. She earns rewards points equivalent to the miles she flies that can be used to buy airline tickets. In addition to earning points for flying, Ariel is awarded points for purchases made with her credit card, where each dollar spent earns 1 point. Ariel only uses her credit card to pay bills and has it set up to automatically charge a total of $970 each month.
At the end of June, Ariel had a total of 20,250 rewards points. The next flight Ariel plans to take is for her vacation to Cancun. After how many months will Ariel have enough rewards points to pay for the flight to Cancun, which requires 28,500 points?
Since Ariel doesn’t plan to fly anywhere before her vacation, she will only be awarded rewards points for purchases made with her credit card. At the end of June, Ariel had 20,250 points, which means she needs an additional 28,500 – 20,250 = 8250 points. If she charges $970 each month, she will earn 970 points each month. Since 8250 ÷ 970 ≈ 8.5, we know that after 8 months, she still will be a little short. Therefore, Ariel will have accumulated enough points to purchase the airline ticket for her flight to Cancun after 9 months.
Saxon participates in the same rewards program as Ariel. He is purchasing an airline ticket for his Bermuda vacation, but his accumulated 18,500 rewards points are not enough to purchase his plane ticket, which requires 24,000 rewards points. The airline will allow Saxon to purchase his ticket using both rewards points and money. After applying rewards points, the airline determines what percent of the total points required is left to be paid and charges a dollar amount equivalent to that percent of the actual ticket cost. If the flight to Bermuda actually costs $340, how much money will Saxon need to pay for his plane ticket after applying all of his accumulated points?
Since Saxon’s ticket requires 24,000 points, after he applies all of his accumulated points, he will still be short 24,000 – 18,500 = 5500 points. The amount Saxon will need to pay for his plane ticket is 340(5500/24,000) ≈ 77.916 = $77.92.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
The nation’s top 12 middle school Mathletes go head-to-head in this culminating event of the 2023 Raytheon Technologies MATHCOUNTS National Competition. Students race against the clock and each other to answer challenging questions and determine the 2023 National Champion.
After polling each student in her homeroom about their summer plans, Mrs. Baker discovered that each student planned to participate in one or more of the following activities:
- Attend summer school
- Work part-time
- Attend sports camp
Mrs. Baker learned that 17 of the students in her homeroom plan to participate in at least two of these activities, and exactly two students plan to participate in all three activities this summer. If the number of students participating in exactly two of these activities is equally divided among the three pairs of activities, how many students plan to both attend sports camp and work part-time this summer, but not attend summer school?
We let x, y and z represent the number of students planning to participate in sports camp and work part-time, work part-time and attend summer school, and attend summer school and sports camp, respectively. We are told there are 2 students who plan to participate in all three activities. Since there are 17 students planning to participate in two or more activities, we have x + y + z + 2 = 17. We are also told that x = y = z. Therefore, 3x + 2 = 17 → 3x = 15 → x = 5 students plan to attend sports camp and work part-time this summer.
If 1/4 of Mrs. Baker’s homeroom students who plan to attend summer school do not plan to participate in either of the two other activities, how many students plan only to attend summer school?
Let S represent the number of students who plan only to attend summer school. From the previous problem, we learned that 5 + 5 + 2 = 12 students plan on participating in summer school and at least one other activity, so S = ¼(S + 12). Solving for S yields 4S = S + 12 → 3S = 12 → S = 4. So, 4 students plan only to attend summer school.
According to this poll, 16 of Mrs. Baker’s homeroom students plan to participate in only one of these activities. If the number of students who plan only to attend sports camp is three times the number of students who plan only to work part-time, what is the total number of students in Mrs. Baker’s homeroom?
We know that 17 students plan to participate in two or more activities, and 16 plan to participate in only one activity. So, there must be 17 + 16 = 33 students in Mrs. Baker’s homeroom.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
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