Transcript Welcome to the 1st Annual Damien High School Math Competition

WELCOME TO THE 1ST ANNUAL
DAMIEN HIGH SCHOOL
MATH COMPETITION
Schedule
8:30-9:00
9:00-9:15
9:15-9:45
9:45-9:50
9:50-10:20
10:20-10:35
10:35-11:35
11:35-12:20
Check In
Greeting from Principal
Math Medley Exam
Break
Individual Subject Competition
Break
Super Quiz Bowl
Solutions, Results, and Awards
If you have not yet registered and received a name tag, please
make your way to the front of the Activity Center.
SUPER QUIZ BOWL
•
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.
PROBLEM #1
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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?
PROBLEM #1 : SOLUTION
There are 8 swimmers and 6 divers.
6
Swimmers
2
4
Divers
PROBLEM #2
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.
PROBLEM #2 : SOLUTION
Decipher Clues
1. The spinner is sectioned into fourths.
2. Either ALL numbers are odd or
ALL numbers are even.
3. From clue 4, the smallest
number must be 1 and the
largest number must be 7.
1
3
5
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.
PROBLEM #3
1
11
4
Solve for the numeric values of the square, star,
and triangle that make ALL equations true.
PROBLEM #3 : SOLUTION
1
3
7
PROBLEM #4
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?
PROBLEM #4 : SOLUTION
Boy - Girl - Boy - Girl - Boy - Girl - Boy - Girl
John - Ellen - Jim - Gloria - Bruce - Hazel - Arthur - Mary
1st - 5th - 1st -
7th
-
2nd - 6th -
3rd
- 4th
John - Mary - JimThis
- Gloria
is a contradiction
- Arthur - Hazel
of clue
- Bruce
4
- Ellen
Deciphered Clues
1. John - Girl - Jim
2. Arthur - Girl - Boy - Ellen
3. Bruce - Girl - Boy - Mary
4. (NOT Gloria) - Bruce
Hazel - Boy - (NOT Gloria)
PROBLEM #5
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
PROBLEM #5 : SOLUTION
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
PROBLEM #6
Assuming A, B, C, D, & E each represent
different digits between 0 and 9, find the
digits such that :
ABCDE
×4
EDCBA
Hint : Remember to carry your tens digits.
PROBLEM #6 : SOLUTION
• 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
• 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
• 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.
3
ABCDE
×4
EDCBA
2BCDE
×4
EDCB2
2BCD8
×4
8DCB2
3
21CD8
×4
8DC12
PROBLEM #6 : SOLUTION
• 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.
• 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
21CD8
×4
8DC12
3
3
3
21C78
×4
87C12
3
3
3
21978
×4
87912
Therefore,
A=2
B=1
C=9
D=7
E=8
TIEBREAKER
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?
TIEBREAKER #1 : SOLUTION
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 = P +2
J – 1 = P +1
J + 1 = 2(P - 1)
(P +2) + 1 = 2(P–1)
P + 3 = 2P - 2
P = 5 apples
Since J = P + 2 , then J = 7 apples .
TIEBREAKER #2
• 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?
TIEBREAKER #2 : SOLUTION
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.