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If you have not yet registered and received a name tag, please make your way to the front of the Activity Center. Schedule 8:30-9:00 Check In 9:00-9:15 Greeting from Principal 9:15-9:45 Math Medley Exam 9:45-9:50 Break 9:50-10:20 Individual Subject Competition 10:20-10:35 Break 10:35-11:35 Super Quiz Bowl 11:35-12:20 Solutions, Results, and Awards

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This is a group competition in which many problems must be solved through collaboration. Scratch paper is available at your tables, but the final answer for each problem must be written legibly in the box on the provided answer forms. The final solution must be true for all the given clues. Each question is worth 5 points; partial points may be given on some problems.

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Some swimmers are also divers. Some divers are also swimmers. There are a total of 12 swimmers and divers. Two swimmers are also divers. Four divers are not swimmers. 1/3 of all the divers are also swimmers. 1/4 of all the swimmers are also divers. How many swimmers are there? How many divers?

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There are 8 swimmers and 6 divers. 264 Swimmers Divers

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1.The four numbers on a board game spinner are equally likely, but are all different. 2.It’s impossible to get an odd number if you spin this spinner twice, and add the results. 3.The most likely sum of two spins is eight. 4.The smallest sum you can get in the two spins is 2; the largest is 14. 5.If you spin the spinner twice and add the two numbers, you’re just as likely to get 10 as 6. Draw the spinner.

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2. Either ALL numbers are odd or ALL numbers are even. Decipher Clues 1. The spinner is sectioned into fourths. 3. From clue 4, the smallest number must be 1 and the largest number must be 7. 4.This is enough information to finish, since the last two numbers must be odd numbers less than 7. So the remaining numbers are 3 & 5. 5. Check the other clues to make sure they work. 13 57

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1 11 4 Solve for the numeric values of the square, star, and triangle that make ALL equations true.

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7 3 1

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Four men and four women are shipwrecked on an island. Eventually each person falls in love with one other person and is himself loved by one person. 1. John falls in love with a girl who is in love with Jim. 2.Arthur loves a girl who loves the man who loves Ellen. 3.Mary is loved by the man who is loved by the girl who is loved by Bruce. 4.Gloria hates Bruce and is hated by the man whom Hazel loves. Who loves Arthur?

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Boy - Girl - Boy - Girl - Boy - Girl - Boy - Girl Deciphered Clues 1.John - Girl - Jim 2.Arthur - Girl - Boy - Ellen 3.Bruce - Girl - Boy - Mary 4.(NOT Gloria) - Bruce Hazel - Boy - (NOT Gloria) John - Mary - Jim - Gloria - Arthur - Hazel - Bruce - Ellen John - Ellen - Jim - Gloria - Bruce - Hazel - Arthur - Mary 1 st - 5 th - 1 st - - 2 nd - - 3 rd - 4 th 7 th 6 th This is a contradiction of clue 4

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Using EXACTLY (no more, no less) than four 4's, and the symbols +, -, ÷, ×, or ( ), write an expression that is equal to the numbers 2, 3, 5, 6, & 7. (4's can be next to each other like 44 or 444) Example: To get the number 8, you can say : 4 + 4 + 4 - 4 = 8 To get the number 440, you can say : 4 4 4 - 4 = 440 Answers will vary

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2 : (4×4)÷(4+4) 3 : (4+4+4)÷4 or (4×4 - 4)÷4 5 : (4×4+4)÷4 6 : [(4+4)÷4]+4 7 : (4+4) - (4÷4) or (44÷4) - 4

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Assuming A, B, C, D, & E each represent different digits between 0 and 9, find the digits such that : A B C D E × 4 E D C B A Hint : Remember to carry your tens digits.

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Since EDCBA must be a five digit number, A must be either 1 or 2; but it can’t be 1, since all multiples of 4 are even. So A must be 2 2 B C D E × 4 E D C B 2 3 2 B C D 8 × 4 8 D C B 2 A B C D E × 4 E D C B A So E must be 3 or 8 (which make 12 or 32). But E can’t be 3, since no five digit number times 4 is less than 40,000 (product can’t begin with 3). E must be 8 3 2 1 C D 8 × 4 8 D C 1 2 Since EDCBA begins with 8, B × 4 must be either 1 or 0. But when we multiply D × 4 and add 3, we must get an odd number. So B must be 1.

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D must be at least 4, since 1 × 4 = 4. The only remaining multiple of 4 that ends in 1 after adding the carried 3 is 28 (which gives 31). So D must be 7. We must carry a 3, to make 7 in the product. 3 3 3 2 1 C 7 8 × 4 8 7 C 1 2 Finally, the only multiples of 4 that are greater than 30 are 32 and 36 (because of the carried 3). But the digit 8 has already been used. So C must be 9. 3 3 3 2 1 9 7 8 × 4 8 7 9 1 2 Therefore, A = 2 B = 1 C = 9 D = 7 E = 8 3 2 1 C D 8 × 4 8 D C 1 2

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If John gives Paul one apple, they will have the same number of apples. If Paul gives John one apple, John will have twice as many apples as Paul does. How many apples does each have?

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If we let J = the # of apples that John has and we let P = the # of apples that Paul has, then we can rewrite the statements as : J – 1 = P +1 J + 1 = 2(P - 1) J = P +2 (P +2) + 1 = 2(P–1) P + 3 = 2P - 2 P = 5 apples Since J = P + 2, then J = 7 apples.

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Assume Damien High School has 1200 students. Each student at Damien takes 5 classes. Each class at Damien has 30 students and 1 teacher. Every teacher at Damien teaches 4 classes a day (They each have one free period for planning). Every student at Damien has to take a class during each of the five periods. How many teachers are there at Damien High School?

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Since there are 1200 students and 30 students per class, then there are 1200 ÷ 30 = 40 classes per period. Since there are 5 periods per day, then there are a total of 5 × 40 = 200 classes per day. Since each teacher teaches 4 classes per day, then there are a total of 200 ÷ 4 = 50 teachers at Damien High.

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