Math Trail 2005

$math trail$

Math Trail 2005

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Staff: Dr. J. Beyersdorfer
Students: Grades 3, 4, and 5 Math Enrichment Students

Using Photography to Write Increasingly More Complex Math Problems

The Project: To complete this project, students examined the types of problems they typically solve during math class. These problems were most often concerned with number, measurement (time, distance, weight ), spending money (taxes, discounts), sequence, logic and data/probablity.

Students selected a topic and wrote a problem to illustrate the use of math in a particular situation. They took a photograph that provided one piece of key mathematical information in order to solve the problem. Next, they increased the difficulty of the problem by adding pertinent information, extraneous information, using numbers that require more challenging calculations, demanding conversions, adding steps to reach the solution, or asking a high level thinking question.

Grade 3 Students: Billy, Max, Jake
	The Librarian's Number
Problem 1 Dr. B wants to check out a book at the school library but right when she got to the library door, the librarian blocked her path and said, “You can’t pass until you guess my secret number. I will give you some clues and you can ask for help." Here are the clues. • The number is between 5 and 100. • 2 + 3 + 60 = ? - 30 + ? + 90 = ? - 20 = ? + 5 = ? - 20 = ? - 30 = the ten's place • Solve for the red clue.
Solution 1 63 is the answer. To find the mystery number you have to find the ten’s place by adding and subtracting and then dividing it by two, that will equal the one’s place. 2 + 3 + 60 = 65 - 30 = 35 + 90 = 125 - 20 = 105 + 5 = 110 - 20 = 90 - 30 = the ten’s place
Problem 2 Dr. B wants to check out a book about computers. The book is called Computer Mania, it teaches about different types of computers like Macs and Windows machines. But when she got to the library door, the librarian blocked her path and said, “You can’t pass until you guess my secret number!" Here are the clues. • It’s between 5 and 100. • 3 + 18 = ? +14 - 25 = ? +11 + 15 = ? - 10 - 4 - 20 = the ten’s place. • Solve for the red clue
Solution 2 3 + 18 = 21 +14 - 25 = 10 +11 + 15 = 36 - 10 - 4 - 20 = 2. If you divide the ten’s place by 2 (because the one’s place is half of the ten’s place), the one’s place is 1. (2/2 = 1)

Grade 3 Students: Lea, Kaitlin
	Amy's Cake
Problem 1 Amy went to the bakery to buy a cake for her mother because their family was going to have a family reunion. The cake has a specific number of slices. Her mother would not tell her how many slices there were in the cake. The whole cake costs $15.00. Each slice costs $2.50. How many slices are there in the cake?
Solution 1 You have to divide to learn the number of slices by using the cost of each slice and the cost of the cake. $15.00 divided $2.50 Your answer is 6 slices.
Problem 2 Amy went to the Yummy Treats Bakery to buy a cake for her mother because they were going to have a family reunion. The cake she bought was chocolate. The cake cost $15.00. It took her 15 minutes to purchase the chocolate cake. When she got back home, she found out that she was supposed to buy two chocolate cakes, not just one. She ran back to the bakery and bought another cake for the same price as the first cake. When Amy got back to her house, she saw more cars in her driveway than she expected. She suspected that there were 4 more people because there was 1 extra car in the driveway. She again ran back to the bakery and got a small cake with four slices for the four extra guests. Each slice cost $2.50. What was the cost of the small cake? How many slices were in all the cakes put together?
Solution 2 To find the total number of slices, you have to add and multiply: There are 6 slices in each of the two cakes (6 X2 = 12) and 4 slices in the small cake (6 X 2 = 12; 12 + 4 = 16). 16 slices is your answer. The cost of the large cake was $15.00, so 2 cakes cost ? The cost of the small cake was $2.50 X 4 = $8.00 The total cost for all three cakes was $15.00 + $15.00 + $15.00 = $38.00

Grade 3 Students: Mitchell, Aaron
	Ron's Blocks
Problem 1 One day a boy named Ron decided to make a sequence with his blocks. He had red blocks, blue blocks and green blocks. Each block had four different numbers. The sequence was blue5, red7, green7, blue5, red7, green7,______, _______, green7, blue5, red7, green7, ____, red7, ____,blue5,red7,green7, blue5, ______, green7, blue5, red7.
Solution 1 The pattern was blue5, red7, green7, blue5, red7, green7, blue5, red7, green7, blue5, red7, green7, blue5, red7, green7, blue5, red7, green7, blue5, red7, green7, blue5, red7, green7, blue5, red7 .
Problem 2: One day Bobby decided to make a mystery number for his classmates. These are the clues he gave his classmates. It is between 0 and 100. Double my ten’s digit to find my one’s digit. It is divisible by 4. It is divisible by 6. It is divisible by 8. It is divisible by 3. Half of the number is 12.
Solution 2 The answer is 24. 4 x 6 = 24. 2 x 2 = 4. 8 x 3 = 24. 12 x 2 = 24 24 is between 0-100.

Grade 3 Students: Sarah, Marina
	Mystery Number
Problem 1 Amy and Katie wanted to play a game. They decided to play Missing Number. Katie would think of a number first and Amy would guess it. Katie gave Amy 6 clues. The clues were: • The number is between 50 and 100. • It is an odd number. • It has 9 +10 - 8 +16 - 20 +1 for the ten's place. • The one’s place is an odd number and the ten’s place is an even number. • The one’s place is 9 + 4 - 5 + 3 - 10.
Solution 1 The number is 81. It is 81, because 81 is between 50 and 100. 81 is an odd number. 9 + 10 - 8 + 16 - 20 + 1 = 8 for the ten's place. 9 + 4 - 5 + 3 - 10 = 1 for the one's place.
Problem 2 Kim and Elizabeth were bored. They decided to play Mystery Number. This was their favorite game. Kim gave Elizabeth the clues for the number. The clues were: It’s between 15 and 98 8 x 10 - 70 + 1 - 2 x 2 + 2 - 12 = the ten’s place it’s between 47 and 96 the sum of the numbers is nine 0 x 1 + 3 - 2 + 5 - 4 x 1 - 1 = the one’s place
Solution 2 The number is 81 because 81 is between 15 and 98. 8 x 10 - 70 + 1 - 2 x 2 + 2 -12 = 8, which is the ten’s place. 81 is between 47 and 96, 8 + 1 = 9, and 0 x 1 + 3 - 2 + 5 - 4 x 1 -1 = 1 which is the one’s place.

Grade 3 Students: Matt, Basil
	Final Four Game
Problem 1 Tim is going to a basketball game in three hours and five minutes. The Hornets are playing the Lakers at the United Center. It will take him 25 minutes to get to the United Center. The game will start at 7:05 pm. There are 3 lines. Each line will take 10 minutes. Tim must go through each line. How many minutes will he be waiting in the arena?
Solution 1 It is 3:00. To leave three hours and five minutes early, he will leave at 6:05. You have to add 6:05 + 25 minutes for travel to the United Center. That will equal 6:30. He will have to go through 3 lines that will take 10 minutes each. You have to add 10 minutes + 10 minutes + 10 minutes = 30 minutes which when added to the time needed for his arrival will be 7:00. Because the game starts at 7:05, he will be waiting 5 minutes.
Problem 2 Tim is going to a basketball game in three hours and five minutes. The Hornets are playing the Lakers at the United Center. On his way he will have to go to White Hen and buy a 12 pack of soda. It will cost $2.00. It will take him 10 minutes to get the store and 20 minutes to get to the arena from White Hen. When he arrived, he waited in 2 lines which were to 5 minutes long each, and a third line which took him 20 minutes more. After waiting in those 3 lines he decided to buy popcorn. The popcorn line took him 7 minutes. How many minutes late is he?
Solution 2 6:05 + 30 minutes at White Hen and to drive) = 6:35 p.m. 6:35 p.m. + 10 minutes (2 short line time) = 6:45 p.m. 6:45 p.m. + 20 minutes (1 long line) =7:05. 7:05 p.m. + 7 minutes (popcorn line) = 7:12 p.m. Since the game starts at 7:05 p.m., he will be 7 minutes late (7:12 p.m. - 7:05 p.m.)

Grade 4 Student: Amanda
	The Broadway Game
Problem 1 Amanda wants to buy the Broadway Board Game. It cost $15.00. How much money will she have left if she buys the game?
Solution 1 If she wants to buy the game, she would have to spend $15.00. Her allowance {as shown in the photo} is $20.00 + $1.00 + $1.00 + $1.00 + $1.00 +$1.00 = $25.00. If you subtract $15.00 from $25.00 [$25.00- $15.00= $10.00] then she will have $10.00 left.
Problem 2 Amanda got $5.00 more for her allowance. She went to buy flowers, balloons, and the game, Broadway. The flowers cost $13.24; the game cost $15.00; and the balloons cost $A.GF. What was the cost of the balloons? [AGF] What is the relationship between the cost and the alphabet letters?
Solution 2 Amanda has $30.00. [$20.00 + $1.00 + $1.00 + $1.00 + $1.00 +$1.00 +$5.00 = $30.00]The flowers cost $13.24 and the game cost $15.00 for a total of $28.24. If you subtract $30.00 - $28.24 = $1.76. $1.76 = AGF. A by is the first letter of the alphabet. G is the seventh letter of the alphabet and F is the sixth letter of the alphabet.

Grade 4 Student: Ashley
	Alf's Game Show
Problem 1 Billy’s favorite show to watch is Alf’s Game Show. He got tickets to go. Everyday Alf sets up a problem and the audience has to figure it out. If they answer correctly then they win one million dollars. That day Alf gave the audience part of the sequence. Here it is, 1B3D5F. But, to actually win they have to complete the sequence until the alphabet is finished.
Solution 1 Part of the sequence is 1B3D5F. It starts with the number 1. Then the letter B comes. B is the second letter of the alphabet. So that would be two. Then you keep on going for the rest. But, you have to fill in the numbers using only odd numbers so the letters take the even numbers places. The rest would be 1B3D5F7H9J11L13N15P17R19T21V23X25Z.
Problem 2 Billy loves to watch the show Alf’s Game Show. He got tickets to go. Every day Alf sets up a problem. If someone in the audience gets it right then that person wins one million dollars. To get the money the contestant has to complete the sequence.The only person to get the money is the first person done. But, to win the money the contestant has to finish within a time limit. Before the show Billy drives to the theater where he surveys the audience about approximately how much time it will take the audience members to do the problem. It takes him about 30 minutes to get to the theater from his house with no traffic. If there is traffic, then he has only 15 minutes remaining to do the survey and get a seat. The show starts at 5:00. When should Billy leave his house? Billy would like it if you gave him the answer without any traffic and the answer with a 15 minute delay.
Solution 2 Add up all of the elapsed time. The answer to problem to the first part of this problem is an hour so he would have to leave at about 4:15. To get 4:15 you have to add up 30 minutes travel time + 15 minutes survey time = 45 minutes. The answer then to question number two is 4:00. Add an additional 15 minutes for time spent and traffic, and Billy must leave his home at 4:00 rather than 4:15.

Grade 4 Student: Harrison
	Students' Schedules
Problem 1 At Notre Dame High there are classes like P.E., Music, Art, and technology class. Music is 1 period long, P.E., Art, and Technology are all 2 periods long. Every day of the week students have a different class except Friday. The students don’t know what special to attend on Friday. What class should the students attend on Friday if they wanted to have equal education for each class?
Solution 1 Music. Music is the only special that is 1 period and if you take it twice it will be equal to the other classes in time.
Problem 2 At Notre Dame High school there are elective classes such as P.E, Music, Art, and a tech class. Music is 1 period long; P.E., Art, and tech are 2 periods long. There is only 1 class allotted per day for these electives except Friday. The students want to have an equal education for each class. But the 2 hours they have for study hall is pretty much filled. For example, all the students have a Math and a social studies class for 2 periods each. Each period is a half an hour long. They also want time to be with their friends. There are 695 students at the school. What elective should the students take on Friday? A student’s Monday schedule: 9:00 Math 10:00 Comprehension 10:30 Science 11:30 2 period elective class 12:30 Lunch 1:00 History 2:00 English 2:30 Study Hall 3:00 School Ends
Solution 2 Music would be taken on Friday. This is because Music is only 1 period long and all the others are 2 periods long. If a student attends music for 2 periods on Friday, he would have an equal amount of time spent in each elective class.

Grade 4 Student: Michael
	The Allowance
Problem 1 Mike has an allowance shown in the photo above. He wants to buy a matchbox car for 1/5 of his allowance. How much is the matchbox car?
Solution 1 The matchbox car is $4.00, because 5 divided by 20 equals 4.
Problem 2 Five days later Mike raked the leaves and earned even more allowance. He decides to go to the store, but he has baseball practice in 20 minutes and it takes 5 minutes to drive to the store from his house. The CD he wants is $19.86. When he gets there they only have only one copy of the CD he wanted. Mike bought it. It took him 5 more minutes to look at all the other CDs. It was 4 o'clock when he left to go the store. It took him 5 minutes to get his equipment from his house and 5 more minutes to drive to practice. Will he make it to practice? Will he have enough money to buy the CD and , if so, what will his change be?
Solution 2 Yes, he will have enough money to buy the CD and his change will be $5.14 because $25.00-$19.86 = $5.14. He will be able to get to practice on time because he went to the store at 4’oclock and it took him 5 minutes to go to the store, it took him 5 minutes to look, it takes him 5 minutes to get his equipment, and another 5 minutes to drive to practice. That comes to a total of 20 minutes and 4’oclock + 20 minutes = 4:20.
Problem 3 Mike needs to go to the store. He is given $50.00 to buy 5 items, 2 gallons of milk for $10.00 each, 2 loaves of bread for $5.00 each, and one giant box of chocolate for $9.95. Then, in the check out line he realizes the loaves of bread are 10% off. Then he forgets he wants a large bunch of bananas and he goes back to get it. The bananas are $5.00 and the cost is 10% off too. Will Mike be able to buy the groceries?
Solution 3 Each loaf of bread costs $4.50, the milk is $10.00 each, the candy is $9.95, and the bananas are $4.50, that comes to a total of $43.95.

Grade 4 Student: Molly

Problem 1: One day a girl named Ashley picked up a sroll. It was a note and had a one word code on it. The scroll read, "horse".

The horse’s name is

What is the horse’s name when spelled as letters?

Codes and Questions

Solution 1 Look at the letters of horse. Use the letters to decode the name Rose. R= triangle, O= rectangle, S= trapiziod, and E= oval.

Problem 2 One day Molly got a birthday invitation from her cousin. It told her the information about where the party would be held. Crack the code using the alphabet strip and then unscramble the letters to find the place of the party.

Solution 2 Look closely at the X’s. Then go straight down to the alphabet bar. Make a list of all of the letters that the X matches with and then unscramble the 2 words. The answer you should get is Disney World.

Grade 4 Student: Jonah
	Shopping Trip
Problem 1 Mark went to the store to buy groceries. Mark only had a total of $20.00 to spend. He needs to buy bananas, grapes, and oranges. The bananas cost $2.50, the grapes cost $5.50, and the oranges cost $3.00 a bag. Taxes are 10%. Will he be able to buy what he needs?
Solution 1 $2.50 + $5.50 + $3.00 = $11.00. 10% tax would make it $1.10 more that would total $12.10. So in other words, he would have enough money to buy the groceries.
Problem 2 Mark has to go to the store to buy groceries. Mark only has $20.00 to spend. Mark wants to buy grapes, oranges, and bananas. He buys two bags of grapes and two bags of oranges. oranges cost $3.00 per bag, the grapes cost $5.50 per bag, and the bananas cost $2.50. There are no taxes. Will he be able to buy everything?
Solution 2 First,add $3.00 + $3.00 = $6.00 to buy the oranges. Then, add $5.50 + $5.50 = $11.00 to buy the grapes. After that, $2.50 + $11.00 + $6.00 = $19.50. Mark would be able to buy the groceries.
Problem 3 Mark is going to go to the store to buy groceries. Mark has a budget of $20.00 to spend. Mark has to buy oranges, grapes, and bananas. The bananas cost $2.50, oranges cost $3.00, and grapes cost $5.50. Mark also wants to buy a gigantic bag of Skittles for $4.50. Mark bought almost everything. But Mark forgot to buy the bananas. It is 4:50 p.m. Mark needs to be home by 5:00 o’clock. Mark lives seven minutes from the store. It takes three minutes to get the bananas and get to his car. Will Mark be able to get home by 5:00 o’clock? Also, will Mark be able to buy everything?
Solution 3 The first thing you have to do is add $3.00 + $2.50 + $5.50 = $12.00. So Mark can buy what he needs. But he also wanted the bag of Skittles which is $4.50. So $12.00 + $4.00 = $16.00. Mark will be able to buy what he needs and wants. 7 minutes + 3 minutes = 10 minutes to get home with the bananas. Add 10 minutes to 4:50 p.m. to equal 5:00 o’clock. Mark will be able to get home just in time.

Grade 4 Student: Irina, Jessica
	Mandy, the Shopaholic
Problem 1 Mandy went shopping at Justice. She bought a t-shirt, a skort, and some jewelry. She spent a total of $132.59. The skort cost $56.68. She paid $130.83 without taxes. How much were the taxes?
Solution 1 First, you subtract $ 132.59 - $ 130.83 = $1.76. $1.76 is how much she spent on taxes.
Problem 2 Mandy went shopping at Justice. She bought a t-shirt, a skort, and lots of jewelry. She spent a total of $132.59. The skort cost $56.68. Mandy bought A LOT of jewelry! Mandy did did know how much she spent on the shirt and the skort. But she did not know how much money she had left. How much did she spent on the jewelry?
Solution 2 First, you add $ 38.96 + 56.68 = 95.74. After that, you subtract $130.83 - $95.74 = $35.09. Use the $130.83 figure to determine the cost without tax. Your answer for the total of money spent on jewelry is $35.09.

Grade 4 Students: Tommy, Faraz

Bob and Harriet Meet
on the Road

Problem 1 Harry and Bob live 100 miles away from each other. Bob was going an average speed of 40 mph and Harry was going an average speed of 60 mph. What time will they meet?

Solution 1 They will meet at 6:00 P.M. because Bob gets 40 miles away from his home in 1 hour. Harry gets 60 miles away from his home in 1 hour. They live 100 miles away. 60 miles + 40 miles = 100 miles.

Problem 2 Bob and Harriet live a little more than 93 miles apart on the same road. Bob was going an average speed of 40 mph. Harriet was going an average speed of 52 mph. Bob was hungry and went to Mc.Donalds. That slowed him down by 7 minutes. Then Bob got a speeding ticket which slowed him down by 25 minutes. What distance will Bob drive? What distance will Harriet drive? What time will Bob and Harriet meet?

Solution 2 7 minutes + 25 minutes = 32 minutes lost by Bob in 1 hour. Distance = 40 miles per hour * 28 min/60 min. because the time must be converted to part of an hour in order to multiply. 18.4 miles = 40 mph * .46 miles/min.

Time	Bob	Harriet	Goal: 93 miles
5:00-6:00	18.4 miles	52 miles	70.4 miles
6:15	18.4+10 miles=28.4	52 miles+13=65	93.4
6:30	18.4+20 miles=38.4	52 miles+26=78	116.4

Grade 4 Student: Lily, Carson
	Weight Watchers
Problem 1 Daisy, the dog, weighed 3 pounds when she was a 1-year old puppy. When she was 3 years old, she weighed about 27 pounds more. by the time Daisy turned 5, she had gained 23 pounds. How much weight had she gained by age 5?
Solution 1 When Daisy was 1 she was 3 pounds, and when she was 3 she was 27 pounds. (3 pounds + 27 pounds = 30 pounds) At age 5, her total weight would be 53 pounds (30 pounds + 23 pounds = 53 pounds.) Between ages 1 and 5, Daisy gained 50 pounds.
Problem 2 When Daisy was 5 years old, and weighed 53 Ibs., her owner put her on a diet for a year. How much did she weigh after the diet, if she weighed 3/4 of her original weight? If every year after the diet she gains 9 pounds, how much pounds will she weigh after 4 years?
Solution 2 For getting the weight on the diet, you have to divide 53 by 4 = 13.25. This is 1/4 of her original weight, to find 3/4's multiple 13.25 by 3 to equal 39.75 (rounded to 40). To figure out how much she will weigh 4 years later. You multiply 9 x 3 = 27. And then you add 27 to 39.75. Her total weight at the end of 9 years is 66.75 (or rounded to 67 pounds).

Grade 5 Student: Sarah P.

Shouldn't You Have Gotten Your Speedometer Fixed Before You Went Out On The Highway?

Problem 1 The speedometer on the car starts at 0 and goes up to 120 mph. The arrow on the speedometer broke during a trip. Bob just got it replaced. Now Bob want to figure out how fast the car was going at each point on the car trip.

Mph	Angle in Degree	Distance
	18	3.6 Miles
	27	8.1 Miles
	51	28.9 Miles
	73.5	60.025 Miles
	97.5	105.625 Miles
88

The angle refers to the angle formed by the arrow on the speedometer. When the car had traveled 3.6 miles, the arrow would have formed an 18o angle on the speedometer. When the car had traveled 8.1 miles, the arrow would have formed a 27o angle on the speedometer. When the car traveled 28.9 miles, the arrow would have formed a 51o angle on the speedometer. When the car had traveled 60.025 miles, the arrow would have formed a 73.5o angle on the speedometer. When the car had traveled 105.625 miles, the arrow would have formed a 97.5o angle on the speedometer. If the car kept up this pattern the whole trip, how far had it traveled when it was going 88 mph?

Solution 1 Since the speedometer is 180 degrees, but the speed only goes up to 120 mph, you need to find out how many miles there are in each degree. To figure it out, you hvae to divide 180 into 120, which equals 1.5, so each degree on the speedomerter equals 1.5 mph. Then you need to figure out how fast the car was traveling at each point. Since you know that each degree equals 1.5 mph, you simply divide the degree by 3 and the answer is how fast the car was going. 18o would mean that the car was going 12 mph. 27o would mean that the car was going 18 mph. 51o would mean that the car was going 34 mph. 73.5o would mean that the car was going 49 mph. 97.5o would mean that the car was going 65 mph. So the table looks like this so far.

Mph	Angle in Degree	Distance
12	18	3.6 Miles
18	27	8.1 Miles
34	51	28.9 Miles
49	73.5	60.025 Miles
65	97.5	105.625 Miles
88	132	193 Miles

Now you have to figure out how far the car had traveled when it was going 88 mph. To figure out the distance if you have the speed, you need to find the rule. The whole problem has the same rule. It’s Y= X3/X x 0.25 /10 (X is the speed and Y is the distance). To find the rule you take any distance and speed that are in the same row, like 12 mph and 3.6 miles, and you figure out how you get the speed out of the distance, like 123/X x 0.25 / 10= 3.6. So you take the speed, which is 88, and put it in the formula, and you get 193.6 miles.

Grade 5 Students: Mia, Lauren
	Melissa's Bodacious Bash
Problem 1 Melissa is hosting a party at The Hilton Hotel in New York City. She has rented a party room that holds 50 people. The cost is $5.00 per hour per person. Fifty people, including Melissa, are coming, but they will all arrive and leave the party for a different amount of time. The party will start at 5:00 pm and the party will end at 10:00 pm. 10% of the guests will stay at the party for 2 hours. 20% of the guests will stay for 1 hour, because their children have an important soccer tournament in Central Park. 40% of the guests will stay at the party for the entire 5 hours. Lastly, 30% of Melissa’s guests are staying at the party for 4 hours. If this information is true, how much money will Melissa have to pay to the Hilton Hotel for the party room?
Solution 1 First, you multiply 5 (people) by $5=$25, and then you multiply that by 2 hours, which equals $50.00. You do this because 10% of the guests= 5 people and each person needs to pay $5 for one hour in the party room. Then, you multiply 10 (people) by $5=$50, and then you multiply that by 1 hour, which equals $50.00. You do this because 20% of the guests= 10 people and each person needs to pay $5 for one hour in the party room. Next, you multiply 20 (people) by $5=$100, and then you multiply that by 5 hours, which equals $500.00. You do this because 40% of the guests= 20 people and each person needs to pay $5 for one hour in the party room. Additionally, you multiply 15 (people) by $5=$75, and then you multiply that by 4 hours, which equals $300.00. You do this because 30% of the guests= 15 people and each person needs to pay $5 for one hour in the party room. Finally, the last step is to add all of the totals which = $900.00.
Problem 2 Melissa is hosting a party at The Hilton Hotel in New York City. Today is March 2, 1994 in the afternoon (see clock in the photo), and the party is going to be held on March 4th at 5:00. She has rented a party room that holds 50 people. The cost is $10.00 per hour per person. Fifty people, including Melissa, are coming, but they will all arrive and leave the party for a different amount of time. The party will start at 5:00 pm and the party will end at 10:00 pm. 10% of the guests will stay at the party for 2 hours. 20% of the guests will stay for 1 hour because their children have an important soccer tournament in Central Park. 40% of the guests will stay at the party for the entire 5:00 hours. Lastly, 30% of Melissa’s guests are staying at the party for 4 hours. Melissa has also ordered food for the party.This will be served for two hours and 46 minutes after the party’s starting time. The food will cost $21.35 per person. Only 47 people will eat the food, though. Two people are diabetic and have to eat the special diabetic meal. The other person needs a vegetarian meal. There is also a waiter’s fee of $1.49 per person. How much money will Melissa have to pay the Hilton Hotel if there is no tax? How much time does Melissa have to prepare for the party in hours, and then converted it into minutes?
Solution 2 First, you multiply 5 (people) by $10=$50, and then you multiply that by 2 hours, which equals $100.00. You do this because 10% of the guests= 5 people and each person needs to pay $10 for one hour in the party room. Then, you multiply 10 (people) by $10=$100, and then you multiply that by 1 hour, which equals $100.00. You do this because 20% of the guests= 10 people and each person needs to pay $10 for one hour in the party room. Next, you multiply 20 (people) by $10 = $200. Then multiply $200 for the 5 hours of use to equal $1000. You do this because 40% of the guests= 20 people and each person needs to pay $10 for one hour in the party room. Additionally, you multiply 15 (people) by $10=$150, and then you multiply that by 4 hours, which equals $600.00. You do this because 30% of the guests= 15 people and each person needs to pay $10 for one hour in the party room. Next, you will multiply 47 (people) by $21.35 ( the cost of the regular meal) =$1,003.45. Then, multiply 2 (people) by $19.99 (The cost of the diabetic meal) which = $39.98. Following that, you multiply 1 by $22.85, which = $22.85. After that, you multiply $1.49 for the waiter's cost for the 50 guests ($1.49 X 50 = $74.50). Finally, the last step is to add all of the totals $1800.00 +$1066.28 + $74.50 = $2940.78. For the answer to the second question, Melissa has 2 days and 1.5 hours, or 49.5 hours. You can multiply 49.5 hours by 60 minutes per hour which is 2970 minutes.

Grade 5 Student: Matt
	The Three Runners
Problem 1 There are 3 kids, Joe, Moe, and Floe. They all ran the mile. Joe is half as fast as Moe. Moe is 1/12 a minute slower than Floe. If Floe took 5 minutes and 55 seconds, what was Joe's time? What is Joe's speed in MPH for the first race?
Solution 1 F=Floe’s time M=Moe’s time and J=Joe’s time. To figure this out you need to complete several steps. Because Joe's time is 2 times slower than Moe’s time you need to add 5 seconds (1/12 of a minute) to Floe’s time to find Moe’s time then multiply it by 2: 2(F+5 seconds). 5 minutes and 55 seconds + 5 seconds = 6 minutes (Moe's time) 6 minutes x 2 = 12 minutes (Joe's time) 12 minutes is 1/5 of an hour so in the first race Joe went 5 MPH.
Problem 2 There are 3 kids, Joe, Moe, and Floe. They all ran the mile. Joe is half as fast as Moe. Moe is 1/12 a minute slower than Floe. Floe took 5 minutes and 55 seconds. On the next race, Joe used a Jet pack which for the first minute doubled his speed. But as each minute passed, Joe’s jet pack’s speed decreased 1 mile per hour until he was running at his own pace without the aid of the jet pack. What was Joe’s time in that race?
Solution 2 F=Floe’s speed M=Moe’s speed and J=Joe’s speed to figure this out you need to complete several steps. Because Joe is half as fast as Moe’s speed you need to add 5 seconds (1/12 of a minute) to Floe’s speed to find Moe’s speed, then multiply it by 2: 2(F+5 seconds) 5 minutes and 55 seconds + 5 seconds = 6 minutes (Moe's time) 6 minutes x 2 = 12 minutes. (Joe's time) 12 minutes is 1/5 of an hour so in the first race Joe went 5 MPH. But with the jet pack he went 10 MPH for the first minute. Divide that (10 MPH) by 60 to find out how many Miles Per Minute he went. You get 1/6 of a mile but it’s easier to make it 100/600 so it can have a common denominator with the other speeds. Then it goes to 9 MPH and you divide that (9 MPH) by 60 and you get 90/600 you just keep doing the same calculation: 8 MPH is 80/600, 7 MPH is 70/600, 6 MPH is 60/600, 5 MPH is 50/600. After 6 minutes he’s gone 450/600 of a mile.You know thins because 100/600 + 90/600 + 80/600 + 70/600 + 60/600 + 50/600 = 450/600. Then he’s running by himself at his own speed of 5 MPH. He DOESN’T keep going down to 4 MPH, 3 MPH, 2 MPH, 1 MPH, 0 MPH. So then 3 more minutes at that speed (50/600 + 50/600 + 50/600 = 150/600) gives the last 150/600 of a mile. (150/600 +450/600 = 600/600) His time was then, 6 minutes (at variable speeds) + 3 minutes (at his regular pace) = 9 minutes!

Grade 5 Students: Jenny, Sammi, Christina
	Birthday Shopping Spree
Problem 1 Jessica was having a birthday party with 13 guests at 2:45 pm today. She has to buy supplies, because she forgot to do it earlier. She needs a cake, a dress, balloons, and goodie bags. She has $75 to spend. Jessica spent 47% of it on her dress, 25% on a cake, 1% on a goodie bag for each person, and 11% on balloons. All of this is including tax. How much money did she spend on each item, how much did she spend total, and what percent of $75 did she have left over?
Solution 1 Goodie bags: $0.75 per person = $9.75 Cake: $18.75 Dress: $35.25 Balloons: $8.25 Total: $72.00 Jessica spent $72.00 on her supplies, therefore, she will have $3.00 or 4% left over.
Problem 2 Jessica was having a birthday party with 13 guests at 2:45 pm today (Saturday). She has to buy supplies, because she forgot to do it earlier. She has $87 to spend. The bakery that she’s buying the cake at has a special sale every Saturday. If you spend $50 or more on your entire purchase you get 15% off. Jessica spent 47% of her money on a dress, 25% on a cake, 1% on a goodie bag for each person, and 9% on balloons not including tax (tax is 6%). How much money did she spend on each item, how much did she spend total, and what percent of $87 did she have left over? (Make sure you figure out the tax too.)
Solution 2 Goodie Bags: $0.87 per person for 13 people = $9.75 Cake: $21.75 Dress: $40.89 Balloons: $7.83 Tax: $4.91 Total: $81.78 Jessica has $5.22 left over which is 6%.

Reviewed/Revised May, 2005 J. Beyersdorfer