Course:MATH110/Archive/2010-2011/003/Groups/Group 14/Homework 4
Problem 1
Five persons named their pets after each other. From the following clues, can you decide which pet belongs to Suzan's mother?
- Tosh owns a cat,
- Bianca owns a frog that she loves,
- Jaela owns a parrot which keeps calling her "darling, darling",
- Jun owns a snake, don't mess with him,
- Suzan is the name of the frog,
- The cat is named Jun,
- The name by which they call the turtle is the name of the woman whose pet is Tosh,
- Finally, Suzan's mother's pet is Bianca.
In this problem we are asked to determine which of the five people is Suzan's mother, and what type of pet she owns. We are given many different conditions which the solution must satisfy, but we are also given several clues. We should start by taking a look at what we know. First we will consider the information that is given to us directly, then the additional information we can deduce from what is given:
TOSH owns........CAT (Jun)
BIANCA owns......FROG (Suzan)
JAELA owns.......PARROT (?)
JUN owns.........SNAKE (?)
SUZAN owns....... (?)
We know the above information because it is stated directly in the problem. Now let's take a look at what else we can figure out by reasoning:
1. Suzan (person) owns the turtle, since five different animals and five different people are mentioned in the problem, and Suzan (person) and turtle (pet) are the only person-pet pair not accounted for in the problem.
2. Bianca (person), Jaela (person) and Tosh's (pet) owner are all female. We can reason this because the problem says, "Bianca owns a frog that she loves", "Jaela owns a parrot that keeps calling her "darling, darling", and "the name by which they call the turtle is the name of the woman whose pet is Tosh.
3. Suzan's mother cannot be Suzan.
Given this information we can now determine even more information by further reasoning:
1. Tosh's (pet) owner must be female, so Tosh's owner can be Jaela, Jun, or Suzan. That is, Tosh can be a parrot, snake, or a turtle. Tosh's owner cannot be Tosh or Bianca because we already know they own Jun the cat and Suzan the frog, respectively.
2. Bianca (pet) can only be a parrot or a snake, because she can only be owned by a woman, since the problem states, "Suzan's mother's pet is Bianca." Bianca cannot be a turtle because Suzan owns the turtle andSuzan cannot also be Suzan's mother. Bianca cannot be a cat or a frog because Tosh and Bianca own the cat and the frog, respectively, both of which are already named.
3. Jaela could be the name of the turtle, since "the name by which they call the turtle is the name of the woman whose pet is Tosh" and we have reasoned that it is possible that Jaela is Tosh's owner.
Our known information can now be represented as follows:
TOSH owns........CAT (Jun)
BIANCA owns......FROG (Suzan)
JAELA owns.......PARROT (Tosh, Bianca)
JUN owns.........SNAKE (Tosh, Bianca)
SUZAN owns.......TURTLE (Tosh, Jaela)
Now that we have this additional information, we can determine the names of the parrot, snake, and turtle, and thus the identity of Suzan's mother and her pet Bianca. Remember we stated earlier that Bianca can only be a parrot or a snake. Let's determine which of the two Bianca is by trial and error:
1. Suggestion - Bianca is the parrot, Tosh is the snake, and Jaela is the turtle:
TOSH owns........CAT (Jun)
BIANCA owns......FROG (Suzan)
JAELA owns.......PARROT (Bianca)
JUN owns.........SNAKE (Tosh)
SUZAN owns.......TURTLE (Jun)
This satisfies the condition that "the name by which they call the turtle is the name of the woman whose pet is Tosh" but then the turtle must be named Jun, but: 1) the cat is already named Jun and; 2) this would leave no pet named Jaela.
2. Suggestion - Bianca is the parrot, Jaela is the snake and Tosh is the turtle.
TOSH owns........CAT (Jun)
BIANCA owns......FROG (Suzan)
JAELA owns.......PARROT (Bianca)
JUN owns.........SNAKE (Jaela)
SUZAN owns.......TURTLE (Tosh)
This does not satisfy the condition that "the name by which they call the turtle is the name of the woman whose pet is Tosh" because then the turtle would be named Suzan, but we have called the turtle Tosh here.
We have now shown that Bianca cannot be the parrot because, in this case, the names of the snake and the turtle cannot meet the conditions set forth in the problem.
Earlier we stated that Bianca can only be a parrot or a snake, so Bianca must be a snake and the snake is owned by Suzan's mother, who must be Jun:
TOSH owns.......CAT (Jun)
BIANCA owns......FROG (Suzan)
JAELA owns.......PARROT (Tosh)
JUN owns.........SNAKE (Bianca)
SUZAN owns.......TURTLE (Jaela)
In this case, we can verify that all of the conditions set forth in the problem are met by cross referencing the last table with the list of the stated conditions:
- Tosh owns a cat - YES
- Bianca owns a frog that she loves - YES
- Jaela owns a parrot which keeps calling her "darling, darling" - YES
- Jun owns a snake, don't mess with him - YES
- Suzan is the name of the frog - YES
- The cat is named Jun - YES
- The name by which they call the turtle is the name of the woman whose pet is Tosh - YES
- Finally, Suzan's mother's pet is Bianca - YES
This problem is only solvable when you allow the male snake ("Jun owns a snake, don't mess with him") to take the female name of Bianca. While it is stated in the problem that Bianca is the name of a woman ("Bianca owns a frog that she loves"), it is not stated that a pet cannot be named after a person who is of the opposite gender. However, other things regarding gender are assumed, such as Suzan's mother must be a woman!
By first reviewing the information given directly in the problem, then considering the information that could be deduced from the given information, it was possible to solve this problem by trial and error. However, given that there are five people and five pets, as well as several conditions which the solution must satisfy, simply solving the problem with trial and error could be very time consuming and ineffective. Therefore it was both necessary and helpful to first determine some missing information through reasoning to shorten and simplify the trial and error process.
Problem 2
Question: Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?
Lets start by writing out all the given information in a list:
*2 are left handed
*3 are right handed
*2 are over 2m tall
*3 are under 2m
*Bohao and Dylan are of the same handedness
*Tim and Chan use different hands
*Stewart and Chan have the same height range
*Dylan and Tim have different height ranges
Find the player who is over 2m tall and is left handed.
Use the clues to put some of the information together:
*2 are left handed--- either Tim or Chan
*3 are right handed--- either Tim or Chan
*2 are over 2m tall--either both Stewart and Chan or neither..... either Dylan or Tim
*3 are under 2m--- either both Stewart and Chan or neither..... either Dylan or Tim
From the clues you can put together that both Stewart and Chan are under 2m because only 2 people can be over 2m and one of those 2 has to be either Dylan or Tim. The clues say that two players are left handed, one of which has to be Tim or Chan.
Thus the only two players that have both the possibility of being left handed (Tim and Chan) and Tim is the only player out of the two of them that over 2m tall and left handed.
Tim is the player who is over 2m tall and is left handed.
Problem 3
Adam, Bobo, Charles, Ed, Hassan, Jason, Mathieu, Pascal, and Sung have formed a basketball team. The following facts are true:
- Adam does not like the cather,
- Ed's sister is engaged to the second baseman,
- The center fielder is talelr than the right fielder,
- Hassan and the third baseman live in the same building,
- Pascal and Charleds each won $20 from the pitcher at a poke game,
- Ed and the outfielders play cards during their free time,
- The pitcher's wife is the third baseman's sister,
- All the battery and infield except Charles, Hassan and Adam are shorter than Sung,
- Pascal, Adam and the shortstop all $100 each at the race track,
- The second baseman beat Pascal, Hassan, Bobo and the catcher at billiards,
- Sung is in the process of getting a divorce,
- The catcher and the third baseman each have two legitimate children,
- Ed, Pascal, Jason, the right fielder and the center fielder are bachelors, the others are all married,
- The shortstop, the third baseman and Bobo all attended the fight,
- Mathieu is the shortest player of the team,
Determine the positions of each player on the baseball team.
Note: on a baseball team there are three outfielders(right, center and left), four infielders(first baseman, second baseman, third baseman and shortstop), and the battery(pitcher and catcher).
From this information we can make the deductions:
Adam: Is in the infield or battery, he's not the catcher, he's not the shortstop, and he's not second base
Bobo: Is not 2nd or 3rd base
Charles: Is in the infield or battery, he's not the pitcher,and he's not 2nd base
Ed: Is in the infield or battery, he's not second base, he's not the cather, he's not the pitcher
Hassan: Is in the infield or battery, he's not second or third base
Jason: Isn't right or center field
Mathieu
Pascal: Is not pitcher, he's not shortstop, not second base, and not right or center field
Sung: Isn't right or center field
From this information we can group the players into the battery and infield, and the outfield.
The infield and battery are: Adam, Charles, Ed, Hassan, Jason, and Pascal
The outfield are: Sung, Mathieu, and Bobo.
From here we can now start applying our previous knowledge and cancelling players out and slotting them into their positions
So once we do this the players and their positions are as follows:
Adam: third basemsan
Charles: catcher
Ed: shortstop
Hassan: pitcher
Jason: second baseman
Pacal: first baseman
Sung: left field
Mathieu: right field
Bobo: center field
Problem 4
Six players - Petra, Carla, Janet, Sandra, Li and Fernanda - are competing in a chess tournament over a period of five days. Each player plays each of the others once. Three matches are played simultaneously during each of the five days. The first day, Carla beats Petra after 36 moves. The second day, Carla was again victorious when Janet failed to complete 40 moves within the required time limit. The third day had the most exciting match of all when Janet declared that she would checkmate Li in 8 moves and succeeded in doing so. On the fourth day, Petra defeated Sandra. Who played against Fernanda on the fifth day?
The first match is played between Carla and Petra and because 3 other matches are played simultaneously it is possible for Janet vs Sandra, Li vs Sandra,Fernanda vs Sandra, Janet vs Fernanda, and Li vs Fernanda. Let's say Li and Janet are playing and Sandra and Fernanda are playing. The second match consists of Carla vs Janet and potential matchups between Petra vs Fernanda,Li vs Petra,Fernanda vs Sandra, and Li vs Sandra. We assume that Petra vs Fernanda and Li vs Sandra play at the same time as Carla vs Janet. The third round will be Janet vs Li while the possible matches are Petra vs Sandra or Petra vs Fernanda. The 4th round match is Petra vs Li with the possibility of matches being are Carla vs Janet or Janet vs Fernanda. By the end of all the matches played, the only two possible players who have not played are Carla and Fernanda, assuming all possible scenarios during the tournament are correct.
Problem 5
Homer finally had a week off from his job at the nuclear power plant and intended to spend all nine days of his vacation (Saturday through the following Sunday) sleeping late. But his plans were foiled by some of the people who work in his neighborhood.
On Saturday, his first morning off, Homer was wakened by the doorbell; it was a salesman of magazine subscriptions.
On Sunday, the barking of the neighbor's dog abruptly ended Homer's sleep.
On Monday, he was again wakened by the persistent salesman but was able to fall asleep again, only to be disturbed by the construction workers next door.
In fact, the salesman, the neighbor's dog and the construction workers combined to wake Homer at least once each day of his vacation, with only one exception.
The salesman woke him again on Wednesday; the construction workers on the second Saturday; the dog on Wednesday and on the final Sunday.
No one of the three noisemakers was quiet for three consecutive days; but yet, no pair of them made noise on more than one day during Homer's vacation. On which day of his holiday was Homer actually able to sleep late?
To solve this problem, we should first consider the information we are given in the problem, then consider the clues provided:
Saturday - Salesman
Sunday - Dog
Monday - Salesman/Construction
Tuesday -
Wednesday - Salesman/Dog
Thursday -
Friday -
Saturday - Construction
Sunday - Dog
by displaying the information given to us in the problem as above, we are able to eliminate 6 of the 9 days as possibilities for when Homer could sleep late. So, we are left with Tuesday, Thursday, or Friday. Let's look at the clues given in the problem to see if we can eliminate any of these choices:
Clue 1: No one of the three noisemakers was quiet for three consecutive days
Clue 2: No pair of them made noise on more than one day during Homer's vacation
Let's first consider Clue 2 and the possible pairs of noisemakers: Dog/Salesman, Dog/Construction, Salesman/Construction. Since no pair of them made noise on more than one day, we can eliminate Salesman/Construction and Salesman/Dog as noisemaker pairs on Tuesday, Thursday, and Friday. This means the only noisemaker pair possible for these three days is Dog/Construction, keeping in mind that the pair can only be noisemakers on Tuesday, Thursday, or Friday:
Saturday - Salesman
Sunday - Dog
Monday - Salesman/Construction
Tuesday - Dog/Construction? Salesman?
Wednesday - Salesman/Dog
Thursday - Dog/Construction? Salesman?
Friday - Dog/Construction? Salesman?
Saturday - Construction
Sunday - Dog
Now that we have these possibilities given to us by Clue 2, let's consider Clue 1: No one of the three noisemakers was quiet for three consecutive days:
The three noisemakers are Salesman, Dog, Construction. Starting with the Salesman, he does not have to be the noisemaker on Tuesday, since he is the noisemaker on Monday and Wednesday. But, he must be the noisemaker on Friday, since he is not the noisemaker on Saturday or Sunday and according to Clue 1, no one is quiet for three consecutive days.
We have just ruled out Friday as 1 of 3 possible days Homer can sleep late.
Moving on to the Dog, he also does not have to make noise on Tuesday, since he is the noisemaker on Sunday and Wednesday. He must, however, be a noisemaker on Thursday, because he does not make noise on Saturday and again, according to Clue 1, no one is quiet for three consecutive days. He cannot be a noisemaker on Friday because we have shown that Salesman is the noisemaker on Friday and according to Clue 2 no pair of them made noise on more than one day.
Summarizing to this point, we have that Salesman must be noisemaking on Friday and Dog must be noisemaking on Thursday. Salesman and Dog do not have to be noisemakers on Tuesday. This leaves Tuesday as the only day Homer can sleep late.
Now let's look at Construction to verify this answer. Construction does not have to be a noisemaker on Tuesday either, since Construction is the noisemaker on Monday. But, Construction must be a noisemaker on Thursday or Friday because Clue 1 states that no one is quiet for three consecutive days. Since Clue 2 states that no pair of them made noise on more than one day, Construction cannot pair with Salesman again on Friday because they are a noisemaking pair on Monday, so Construction is a noisemaker on Thursday:
Saturday - Salesman
Sunday - Dog
Monday - Salesman/Construction
Tuesday - No noisemakers
Wednesday - Salesman/Dog
Thursday - Dog/Construction
Friday - Salesman
Saturday - Construction
Sunday - Dog
We have now confirmed that on Tuesday, no noisemaker needs to be "noise making" in order to satisfy the information given in the problem. Therefore, if Homer gets to sleep late during one of his nine days off, he could only have slept late on Tuesday.