2. Dale, Al and Mario each have a battery-operated car. The cars always travel in a straight line at a constant rate of speed. They
decided to have a race. Each car started at the same time at the beginning of a straight race course.
When Al’s car crossed the finish line, it was ahead of Dale’s car by 24 inches and was ahead of Mario’s car by 32 inches.
When Dale’s car crossed the finish line, it was ahead of Mario’s car by 10 inches. How many inches long is the race course?
AJ
Dale
Mario
8
8
8
8
8
8
8
Dale
Mario
10
If Dale goes 24 inches and Mario goes 22 inches, Mario is 2 inches behind for every 24 inches Dale goes.
First round, Mario is 8 inches behind Dale, Dale must have gone 24 ×
8
= 96 inches
2
Second round Dale goes 24 more inches, so that is 24 + 96 = 120 inches
3. Mrs. Robinson has three kids; Katherine, Benjamin and Elaine. Katherine can drink a quart of milk in two days. Benjamin can
drink a quart of milk in three days. Elaine can also drink a quart of milk in three days.
The kids are expecting five guests for two days next week. How much milk will Mrs. Robinson need for the eight kids?
(Assume the milk preferences for the three children are representative of the five guest’s milk preferences).
7 quarts in 6 days for 3 kids
7
qt / day / 3 kids
6
7
qt / day / kid
18
7 qt
7 qts
18 day
×2 days =
kid
9 kids
7 qts
2
× 8 kids=6 quarts
9 kids
9
4. Three adults and two kids want to cross a river by using a small canoe. The anoe can carry two kids or one adult. How many
times must the canoe cross the river to get everyone to the other side?
Move Adults Kids River Adults Kids
1.
3
0
2.
3
0
3.
2
1
4.
2
1
5.
2
0
6.
2
0
7.
1
1
8.
1
1
9.
1
0
10.
1
0
11.
0
1
12.
0
1
13.
0
0
Result of trip 13
2K

K

1A

K

2K

K

1A

K

2K

K

1A

K

2K

0
0
0
1
0
1
1
0
1
0
1
1
1
1
2
0
2
0
2
1
2
1
3
0
3
0
3
2
4 trips to get an adult
across and boat back
8 trips, 2 adults
12 trips, 3 adults
5. Two people are playing a game. Each factor of the number 72 is written on one of a series of cards and placed on the game
board. The players alternate turns, with each player choosing one of the factors of 72. When a player chooses a number, he
gets to take that number and all the factors of that number off the board.
For example, if Player One started the game by choosing 18, he would get to take 18, 9, 6, 3, 2 and 1 off the board, since
those are the factors of 18. It would then be Player Two’s turn. Player Two would choose one of the remaining markers, and
take that number and all of the factors of that number off the board.
The players continue to alternate turns until only the number 72 is left. The player who has to take 72 loses the game. It is
possible, by careful play, for Player One to win the game. What number should Player One choose first and how should Player
One then play the rest of the game?
1)
Factors
Player
1
Player
1
1
Factors
Player
1
Player
2
Player
1
1
1
Factors
Player
1
Player
2
Player
1
1
1
2
3
4
6
8
9
12
18
24
36
72
72
12
12
18
18
24
36
36
72
36
72
Loses
2)
2
2
3
3
4
4
6
6
8
9
9
8
Loses
24
3)
2
2
3
3
4
4
6
6
8
8
9
12
12
9
Loses
18
24
24
18
36
A) Partial strategy from tables 2 & 3: if opponent takes 24 you take 36; if opponent takes 36 you take 24
4)
Factors
Player
1
Player
2
Player
1
1
1
2
2
3
3
4
4
6
6
8
9
9
8
Loses
12
12
18
18
24
36
36
72
5)
Factors
Player
1
Player
2
Player
1
Player
1
1
1
2
2
3
3
4
4
6
6
8
9
12
12
18
24
36
72
9
If player 1 loosing strategy: take 24 or 36 (see partial strategy (A) above)
If player 1 loosing strategy: take 18 or 8. Then player 2 takes 8 or 18. Player 1
looses by partial strategy (A)
Loses
B) Partial strategy from table 5: If remaining factors are 8, 18, 24, 36, 72, then if opponent takes 18 you take 8 and if opponent
takes 8 you take 18. Then follow strategy (A)
**Notice that if we do not include 72, then there are 2 groups with an equal number of related factors:
Group 1: 8 & 24 (8 is a factor of 24)
Group 2: 18 & 36 (18 is a factor of 36)
6)
Factors
1
2
3
4
6
8
9
12
18
24
36
Player
1
1
3
9
Player
2
2
4
6
12
Player
1
Player 1 will eventually lose: See partial strategy (B) above
Player
1
72
Loses
7)
Factors
Player
Player
Player
Player
1
2
1
2
1
1
2
2
1
1
2
2
3
3
4
4
6
8
9
12
8
Chooses either 24 or 36
Uses partial strategy A
Loses
18
24
36
72
3
4
6
8
9
12
18
24
36
3
4
8
Chooses either 12 or 18
Chooses either 18 or 12 – opposite of player 2’s choice
(Similar to partial strategy B)
Loses
72
8)
Factors
Player
1
Player
2
Player
1
Player
2
9)
Factors
Player
1
Player
2
Player
1
Player
2
1
1
2
2
3
4
6
8
9
12
18
24
3
4
8
Chooses either 6 or 9
Chooses either 9 or 6 – opposite of player 2’s choice
(Similar to partial strategy B)
Loses
36
72
Overall notice how when player 1 chooses 8, then there are 2 groups each with an equal number of related factors (excluding 72):
Group 1: 6, 12, 24 (6 & 12 are factors of 24)
Group 2: 9, 18, 36 (9 & 18 are factors of 36)
So, whatever player 2 chooses any factor, then player 1 chooses the corresponding factor to make both groups have
an equal number of factors.
Example: Player 1 chooses 8, Player 2 chooses 9, Player 1 chooses 6 (now both groups have 2 factors)
5. There was once a farmer who, as part of his route to town, used a rickety old boat to cross a wide river. One day he took his
dog and went to town just to buy corn. However, in addition to buying corn, he bought a goose that he intended to take home
and use to start raising geese. (The farmer was no fool; he already had another goose.) But he also knew that his boat was not
reliable, it could handle only himself and one of the other three things he had with him. He feared that if left alone, the dog
would eat the goose or the goose would eat the corn. How could he get himself and everything else across the river safely?
They just did this for homework
F
D
F
D
C
D
G
C
1. F – G
F
G
G
G
C
C
C
C
C
C
C
F
2. F
3. F - D
F
4. F - G
5. F - C
F
6. F
F
G
7. F - G
F
D
D
D
D
D
D
D
D
D
G
G
G
G
G
G
C
C
C
C
C
6. There are 48 people at a party. There are seven times as many adults as children. There are twice as many females as males.
If there are 5 girls (female children) at the party, ho many men (male adults) are there?
MA = # of Male Adults
FA = # of Female Adults
MC = # of Male Children
MA + FA + MC + FC = 48
(MA + FA) = 7(MC + FC)
One possible solution:
FC = 5
MA + FA + MC + 5 = 48
(MA + FA) = 7(MC + 5)
MA + FA + MC = 43
(MA + FA) = 7(MC + 5)
(MA + FA) = 43 – MC
(MA + FA) = 7MC + 35
43 – MC = 7MC + 35
8 = 8MC →MC = 1
→
MA + (2MA – 3) = 43 – 1
3MA = 45 → MA = 15
FC = # Female Children
(FA + FC) = 2(MA + MC)
→
(FA + 5) = 2(MA + MC)
(FA + 5) = 2(MA + MC)
FA + 5 = 2MA + 2MC
FA = 2MA + 2MC – 5
FA = 2MA + 2(1) – 5 → FA = 2MA – 3
Find MA = ?
7. Al, George, Kathleen and Chad counted ballots in Florida. They were each given a box of ballots. Each box contained the
same number of ballots. Each person counted ballots at a constant rate. However, the rate at which each person counted
ballots was different. When Al finished counting all of his ballots, George had 54 ballots left to count, Kathleen had 90 ballots
left to count and Chad had 106 ballots left to count. When George finished counting all his ballots, Kathleen had 45 ballots left
to count and Chad had 65 ballots left to count. When Kathleen finished counting all of her ballots, how many ballots did Chad
have left to count? How many ballots were in each box to start with?
Think of this group of people counting ballots in 4 rounds:
End of round 1(R1) – A finishes, G has 54 left, K has 90 left, C has 106 left
End of round 2(R2) – G finishes, K has 45 left, C has 65 left
End of round 3(R3) – K finishes, C has ?? left
R1 to R2: K counts 45 ballots (90 – 45 = 45), C counts 41 ballots (106 – 65 = 41)
R2 to R3: K counts 45 ballots so then C must have counted 41 ballots. So then C has 65 – 41 = 24 Ballots
1st answer: C has 24 ballots left to count
End of R1 G has 54 and K has 90, so G has counted 36 more than K at end of R1.
R1 to R2 we know that G counted 54 while K counted 45 for a difference of 9. So for each group of 45 that K counts, G will count
9 more.
Work backwards: At the end of R2 the difference between G and K was 45, so how many rounds did it take for K to fall behind G
by 45? Remember G does 9 more than K each round, so 45 ÷ 9 = 5. So it must have taken 5 rounds for G to finish and K to still
have 45.
Thus 5 rounds of G counting at 54 ballots per round is 5 x 54 = 270 total ballots.
2nd answer: 270 total ballots for each person.
8. Abbie, Bridget, Cynthia and Dena are women whose professions are water quality engineer, soil contamination scientist, air
pollution consultant and biological diversity advocate. Match each woman to her expertise, using the follow clues.
1. Bridget and the biological diversity advocate are both from Oregon.
2. Abbie, the soil contamination scientist and the air pollution consultant all love to garden.
3. The air pollution expert, the water quality engineer and Dena all met one another at a global warming conference.
4. Bridget has never met the person who works on air pollution.
Abbie is H2O Engineer
Bridget is Soil scientist
Cynthia is Air consultant
Dena is Bio advocate
Abbie
Bridget
Cynthia
Dena
H2O Engineer
Yes
X3&4
X
X3
Soil scientist
X2
Yes
X
X
Air consultant
X2
X4
Yes
X3
Bio advocate
X
X1
X
Yes
9. Human blood is classified as O, A, B or AB, depending on whether the blood contains no antigen, an A antigen, a B antigen or
both the A and B antigens. A third antigen, called the Rh antigen, is important to human reproduction. Blood is said to be Rhpositive if it contains this Rh antigen, otherwise, the blood is Rh-negative. For instance, blood is type A+ if it contains the A
antigen and the Rh antigen, blood is type AB– if it contains the A and B antigens, but does not contain the Rh antigen and
blood is type O– if it contains no antigens.
In a hospital, the following data were recorded:
• 22 people were either type A or type AB, 16 of whom had the Rh antigen.
• 27 were either type B or type AB, 18 of whom had the Rh antigen.
• 8 were type AB, 2 of whom were AB-.
• 35 were type O, 5 of whom were O-.
How many patients are listed here? How many have type B+ blood? How many have type A– blood? How many have exactly
two antigens?
Use Venn Diagram
1. 22 people were either type A or type AB, 16 of whom had the Rh antigen.
2. 27 were either type B or type AB, 18 of whom had the Rh antigen.
3. 8 were type AB, 2 of whom were AB–.
4. 35 were type O, 5 of whom were O–.
A = 22
Has Rh = 6+10+12+30 = 58
1. A ∩ B ⊂ A so Number in A = 22
22–2–6–10 = 4
1. Number in A ∩ Rh = 16
2. A ∩ B ⊂ B so Number in B = 27
16–6 =10
2. Number in B ∩ Rh = 18
O = 35
35–5 =30
2
6
5
3. Number in A ∩ B = 8 (This is all of AB)
3. Number in Only A ∩ B = 2, (This is AB–)
18–6 =12
so then number in A ∩ B ∩ Rh = 8 – 2 = 6 (This is AB+)
4. Number in O = 35
27–2–6–12 = 7
4. Number in Only O = 5 (this is O–)
B = 27
# patients: 4+10+6+2+7+12+30+5 = 76
# w/ B+: 12
# w/ A–: 4
# w/ Exactly 2 antigens: [# w/(A ∩ B)– ] + [# w/A+ only] +[# w/B+ only] = 2 + 10 + 12 = 24