Math 110/003 - Group 4
Members:	Owen Johnston
	Charlene Ng
	Ling Sun
	Dominic Sunga

SUBPAGES Basic Skills Project: http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_04/Basic_Skills_Project

Homework - Due October 20

THOSE THAT POSE NO PROBLEM TO THE GROUP

Basic Functions

Equations Composition of functions

Intersection of functions

Distances of lines Trigonometry and Pythagorean Theorem

Areas and Volume

Mathematical writing

polynomial Long Divison

Construction of graphs

inequalities

properties of functions

THOSE THAT POSE PROBLEMS FOR SOME, BUT NOT ALL

Graphs of functions

Operations of graphs of functions

THOSE THAT NOONE KNOWS HOW TO HANDLE

Homework - Due October 20

1)

We know that the turtle belongs to Suzan because every other pet was given a owner, except for the Turtle.

We are given two names of the animals – Suzan the frog and Jun the Cat. From this, we can deduce the names of the other animals – Jaela the turtle, Bianca the Snake and Tosh the Parrot

With this information, we can create a table:

Owner Gender Pet Pet name Tosh Male Cat Jun Bianca Female Frog Suzan Suzan’s Mom Female Turtle Jaela Jun Male Snake Bianca Jaela Female Parrot Tosh

The question is asking us which pet belongs to Suzan’s mother. The answer is Jaela the Turtle.

3) Use a graph where the column headings are for the names of the people and the row headings are for the different positions. Then use the given information to eliminate and deduce what positions each person will play.

Ed- pitcher

Pascal - left outfielder

Mathieu - right outfielder

Sung - first base

Adam - centre outfielder

Hasan - second base

Bobo - catcher

Jason - third base

Charles - shortstop

4) Understand the problem

The question is asking me who will be playing with Fernanda on the fifth day. The question state that 3 matches are played simultaneously during each day and each player plays each of the others once. 1st Carla vs. Petra 2nd Carla vs. Janet 3rd Janet vs. Li 4th Petra vs. Sandra 5th Fernanda vs. ?

Plan a strategy for solving the problem Since the question already gives us some of the teams during these five days and we already know that each player plays each of the other once, 3 games per day, and played simultaneously which means that one player will only play one game per day. By knowing all these circumstances, we can create a chart to list out all the possible teams that can play on each day by elimination.

Execute the strategy

1st Carla vs. Petra J vs. S F vs. L or J vs. F S vs. L Janet can’t play with Li → they will be playing on the 3rd day, we eliminate the second choice since on the 4th day F vs. J will be playing.

2nd Carla vs. Janet P vs. L S vs. F or P vs. F S vs. L Petra can’t play with Sandra → they will be playing on the 4th day; we eliminate the second choice since on the 3rd day will be P vs. F.

3rd Janet vs. Li C vs. S P vs. F or C vs. F S vs. P Carla can’t play with Petra → they played on the 1st day; we eliminate the second choice since we can tell from the table they will be playing on the 4th day.

4th Petra vs. Sandra C vs. Li F vs. J or C vs. F J vs. Li Carl can’t play with Janet → they played on the 2nd day, and we eliminate the second choice since we can tell from the table Janet vs. Li played on the 3rd day.

→ from the table above we can list out the teams will be playing on the 5th day 5th Fernanda vs. Carla Petra vs. Janet Li vs. Sandra The table above list out all the possible arrangement of each teams, and we eliminate all the impossible guesses. Each of the players only played with each other once during 5 days.

on the 5th day, Fernanda is playing with Carla

Homework - Due October 13

Question 1:

Step 1: The question is asking me why the times are different if the route and speed for travelling to and from the airport is the same

Step 2: To solve this question, we must look at how the time is being measured for the travel to the airport and the travel back

Step 3: It is actually the same time, because 80 mins is equivalent to 1 hour and 20 mins

Step 4: 1 hour = 60 mins, so 80 mins = 1 hour and 20 mins

Question 2: Step 1: The question is asking me why the policeman didn’t stop the lady when she didn’t stop at the stop sign and went three blocks in the wrong direction on a one way street

Step 2: TO solve this question, we must look at the specifics of the words being used and discern what the really mean

Step 3: The lady was not actually driving a car.

Step 4: The question never stated that the woman was driving when she didn’t stop at the stop sign and went three blocks the wrong way.

Question 3: Step 1: The question is asking me how to discern which box has which fruit if we only pick one fruit from one box

Step 2: To solve this question we need to look at the fact that all labels are incorrect and that there are only two different types of fruits

Step 3: If you pick one fruit out of the box that is labeled “apples and oranges,” whatever fruit you picked out, that box is supposed to be labeled that fruit.

Step 4: If you picked an apple from the box that is incorrectly labeled “apples and oranges,” the “apples and oranges” box is the box full of apples, the “apples” box is full of oranges, and the “oranges” box is apples and oranges.

If you picked an orange from the box that is incorrectly labeled “apples and oranges,” the “apples and oranges” box is full of oranges, the “oranges” box is full of apples, and the “apples” box is apples and oranges.

Question 4: Step 1: The question is asking me how I can be the brother of a blind fiddler, but have no brothers myself

Step 2: To solve this question we need to determine who the blind fiddler is

Step 3: The blind fiddler is actually my sister

Step 4: To be the brother of a blind fiddler but have no brothers means that the blind fiddler is my sister

Question 5: Step 1: The question is asking me how many revolutions it would take for a quarter to completely roll around the edges of another

Step 2: To solve this equation we need to look at the sizes of the coin, the starting postion, the halfway position and the final position

Step 3: The coin will need to make three revolutions

Step 4: This is because we know that the coins are the same size. When the coin makes one revolution, it means that it is in the halfway point. So to go back to the original position, it has to make one more revolution. Therefore, the coin has to make two revolutions

Question 6: Step 1: The question is asking me how many apples I would have to pick in order to absolutely have two of a kind

Step 2: To solve this problem we need probability, logic and reasoning.

Step 3: You would need to pick 4 apples

Step 4: This is because there is a chance that if you pick 3, you could get three different types. But if you pick four, at least two of the four will definitely be two of a kind

Question 7: Step 1: This question is asking me times I would have to pick a sock to be sure of getting one pair that is the same color and one pair that is different colors

Step 2: To solve this question we need the same things as we did from the previous question

Step 3: You would need to pick a sock 3 three times.

Step 4: This is because when you pick three, two of them of will definitely be the same color, and two will be different

Question 8: Step 1: The question is asking me how Reuben can be 21 two das ago, but 23 later next year.

Step 2: To solve this question, we need to determine his birthday and his present day.

Step 3: His birthday is on December 31st, and his present day is January 1st.

Step 4: This is because if his birthday was on December 31st, 2010, and his present day was January 1st, 2011, then two days ago he would have been 20 at December 30, 2010. So when he says that later next year he will be 23, he means that December 30, 2012, he will turn 23, which is accurate.

Question 9: Step 1: This question is asking me how many rungs will be shown if the tide rises 5 feet.

Step 2: To solve this problem, we have to look at it in two different ways

a) if the rung is at sea level b) if the rung is one foot above sea level

Step 3: If the rung is at sea level, 4 rungs will be shown when the tide rises 5 feet If the rung is one foot above sea leve, then five rungs will be shown when the tide rises 5 feet.

Step 4: This is because when the rung is at sea level, two rungs will be covered when the tide rises one foot, three rungs will be covered when it rises another foot, four rungs will be covered when it rises another foot, five rungs will be covered when it rises another food, and 6 rungs will be covered in total when the tide rises 5 feet, therefore only 4 rungs will be shown.

When the last rung is one foot above sea level, one rung will be covered for every foot the tide rises. So when the tide rises 5 feet, 5 rungs will be covered and 5 rung will be shown.

Question 10: Step 1: The question is asking me if half the people are chocolate eaters, whether or not a quarter of the world are women chocolate eaters and half the men are chocolate eaters.

Step 2: To solve this question, we need to use fractions

Step 3: Yes, it does mean that a quarter of all people are women chocolate eaters and half the men are chocolate eaters, if we follow this very simplified logic

Step 4: This is because if all people are comprised of one half being men and one half being women, then half of each gender would be a chocolate eater of half of all people are chocolate eaters.

Question 11: Step 1: This question is asking me who the worst chess player is

Step 2:

To solve this, need to look closely at the adjectives being used to describe the ages of all players

Step 3: The worst player is either the son or the daughter

Step 4: It cannot be the mother or her older brother, because they are not of the same age (he is older). Therefore, it has to be either the son or the daughter.

Question 12: Step 1: This question is asking me why the Manhattan man’s Bronx Girlfriend left him.

Step 2: We need to look at the train schedules (they both arrive regularly every 10 mins)

Step 3: She left him because the Brooklyn train arrives earlier than the Bronx train. And also because he was cheating on her – not cool.

Step 4:

Question 13: Step 1: This question is asking me how long it would take for a clock to strike 10

Step 2: To solve this question we need to look at the previous information we’re given

Step 3: It would take 10 seconds

Step 4: This is because if it takes 5 seconds for it to strike 5, then it 10 would just be double.

Question 14: Step 1: This question is asking me is how many ways can 2 babies be tagged correctly and 2 babies tagged incorrectly.

Step 2: We need to use probability

Step 3: There are 6 ways two babies can be tagged correctly and two babies tagged incorrectly. It is impossible to incorrectly tag a baby wrong if three were tagged correctly

Step 4: If three babies are tagged, it is impossible to tag the last baby incorrectly

Question 15: Step 1: This question is asking me if I should accept this extremely risky bet

Step 2: We need to use logic and common sense

Step 3: I should take the bet

Step 4: This is because there are 26 red cards and 26 black cards. If there are 15 red cards in the first half and 12 black cards in the first half, there will be 11 red cards in the second half and 14 cards. Unless the guy is a magician or something.

16. Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family has twice as many sisters as he has brothers. How many sons and daughters are in the family?

Let x represent the number of daughters Let y represent the number of brothers

From the information given in the question, we can deduce the following equations. 1) y = x-1

2) x = 2(y-1)

  = 2y - 2

Now let us substitute equation 1 into equation 2 to solve for x. x = 2y – 2 x = 2(x-1) – 2 x = 2x – 2 – 2 x = 2x – 4 x = 4

Substitute x = 4 into equation 1. y = x – 1 y = 4 – 1 y = 3

Therefore, the family has 4 girls and 3 boys

17. The zero point on a bathroom scale is set incorrectly, but otherwise the scale is accurate. It shows 60 kg when Dan stands on the scale, 50 kg when Sarah stands on the scale, but 105 kg when Dan and Sarah both stand on the scale. Does the scale read too high or too low? Explain.

Let x represent Dan’s actual weight. Let y represent Sarah’s actual weight. Let z represent the error on the scale.

Then using the information given in the question we can come up with the following 3 equations.

1) 60kg = x + z 2) 50kg = y + z 3) 105kg = (x + y) + z Now we will reorganize the variables in equations 1 and 2.

1) 60kg = x + z

           x = 60 – z

2) 50kg = y + z

     y = 50kg – z

Now substitute x = 60 – z and y = 50kg – z into equation 3 105kg = (x + y) + z 105kg = [(60-z) + (50kg – z )] + z 105kg = (60 – z + 50kg – z) + z 105kg = 110kg – z z = 5kg

Therefore, the scale reads too high by 5kg.

18. Alice takes one-third of the pennies from a large jar. Then Bret takes one-third of the remaining pennies from the jar. Finally, Carla takes one-third of the remaining pennies from the jar, leaving 40 pennies in the jar. How many pennies were in the jar at the start?

Let x represent the total number of pennies in the jar. Let y represent the number of pennies Alice took from the jar. Let z represent the number of pennies Bret took from the jar. Let w represent the number of pennies Carla took from the jar.

From the information given in the question we can deduce the following 4 equations. 1) y = x/3 2) z = (1/3)(x - x/3) 3) w = (1/3)[x – (1/3)(x - x/3) – x/3] 4) x = y + z + w + 40

Substitute the values of y, z and w into equation 4. x = (x/3) + (1/3)(x - x/3) + (1/3)[x – (1/3)(x - x/3) – x/3] + 40 x = (x/3) + (2/9)x + (4/27)x + 40 x = (19/27)x + 40

       (8/27)x = 40	      
     	      x = 135

Therefore, there are 135 pennies in total at the start.

19. One morning each member of Angela's family drank an eight-ounce cup of coffee and milk, with the (nonzero) amounts of coffee and milk varying from cup to cup. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. What is the least number of people in the family?

Let b represent the total amount of milk. Let a represent the total amount of coffee. Let n represent the number of people in Angela’s family.

From the information given in the question we can deduce the following equation.

(a/6 + b/4)n = a + b 2a(6-n) = 3b(n-4) Since both a and b are positive values, it follows that 6-n and n-4 are also positive which is only possible when n=5.

Therefore, there are at least 5 people in the family.

20. Of two clocks next to each other, one runs 5 min per hour fast and the other runs 5 min per hour slow. At midnight the clocks show the same time. At what time are they are one hour apart?

If one clock is 5 minutes faster per hour, and the other is 5 minutes slower per hour, we can deduce that after 6 hours have passed, they will be 1 hour apart. The first clock would be at 6:30, while the other at 5:30. This is because in 6 hours, one clock would have gone ahead by half an hour while the other slowed down by half an hour

21. Sven placed exactly in the middle among all runners in a race. Dan was slower than Sven, in 10th place, and Lars was in 16th place. How many runners were in the race?

Given that: Sven is the median of the sequence, Dan is the 10th and Lars is the 16th. We know that there must be at least 16 runners. But as 16 is an even number, an exact median cannot be found. Therefore, there should be 17 placements for the answer to be reasonable where Sven is the 9th.

22. During a vacation, it rained on 13 days, but when it rained in the morning, the afternoon was sunny, and every rainy afternoon was preceded by a sunny morning. There were 11 sunny mornings and 12 sunny afternoons. How long was the vacation?

Let R represent a rainy day. Let S represent a sunny day. It is given that it rained on 13 days. There were 11 sunny morning and 12 sunny afternoons. I will now list out the combinations of rainy/sunny mornings and afternoons. Morning - R S S S S R R S S S S S R R S S R R Noon - S S R R S S S R R S R S S S R S S S There are 11 sunny mornings, 12 sunny afternoons and it rained on 13 days. Therefore, there were 18 total days of vacation.

23. Suppose you overhear the following conversation: Paul: How old are your three children? Paula: The product of their ages is 36 and the sum of their ages is the same as today's date. Paul: That is not enough information. Paula: The oldest child also has red hair. If you were Paul could you determine the ages of Paula's children? Explain.

Let x represent the oldest child. Let y represent the second child. Let z represent the third child.

Given that the produce of their ages is 36. We can deduce equation 1. 1) 36 = xyz

Given that the sum of their ages is the same as today’s date. I will now make a list listing out every possible value for x, y and z which will result in a product of 36.

Ages Product Sum 1 x 1 x 36 36 38 2 x 3 x 6 36 11 1 x 3 x 12 36 16 2 x 2 x 9 36 13 1 x 4 x 9 36 14 3 x 3 x 4 36 10 1 x 2 x 18 36 21 1 x 6 x 6 36 13

The only possible ages that meet the conditions given in the question are the ages of 1, 6, 6 and 2, 2, 9. But if we assume that the oldest child and the second child are not the same age, the only possible ages of the children are 2, 2, and 9.

24. Two candles of equal length were lit at the same time. One candle took 6 hr to burn out and the other candle took 3 hr to burn out. After how much time was one candle exactly twice as long as the other candle?

Let A and B represent the 2 candles. Let A represent the candle that takes 6 hours to burn out. Let B represent the candle that takes 3 hours to burn out.

Candle A takes twice as long to burn out than Candle B. We can form equation 1. 1) A = 2B

Assume that time starts at 0 and that the burning of the candle is constant. Then we can say that Candle A burns 1/6 per hour and Candle B burns 1/3 per hour. Let x represent the time in hours. We can form equation 2 and 3. 2) A = (6-x)/6 3) B = (3-x)/6

Substitute equation 2 and 3 into equation 1. A = 2B (6-x)/6 = 2[(3-x)/6] x = 2

Therefore, it takes 2 hours for one candle to be half of the other.

25. Two candles of length L and L + 1 were lit at 6:00 and 4:30, respectively. At 8:30 they had the same length. The longer candle died at 10:30 and the shorter candle died at 10:00. Find L.

Given: Candle X has length L and was lit at 6 hour. Candle Y has length L + 1 and was lit at 4.5 hour. At 8.5 hour, the length of Candle A is equal to the length of Candle B. Candle A burns out at 10 hours. Candle B burns out at 10.5 hours.

Then the total burning time of Candle A is: 10 – 6 = 4 hours Then the total burning time of Candle B is: 10.5 – 4.6 = 6 hours

Burning time till 8.5 hours for Candle A is: 8.5 – 6 = 2.5 hours. Burning time till 8.5h for Candle B is: 8.5 - 4.5 = 4 hours.

Let x represent the length of Candle A Let y represent the length of Candle B

y = x + 1

Let B1 represent the burning constant for x Let B2 represent the burning constant for y We know that x=y at 8.5 hours.

(x-4) B1 = x + 1 – 2.5B2

y = 6 B2 at end time x = 4B1 at end time 6B2 = 4B1 + 1 B1 = (6B1 -1)/4

Using the above equations and substituting back into the original. 4B2 = 1 – 2.5B1 4B2 = 1 – 2.5((6B1 -1)/4) B2 = 1.5

Substitute B2 into y = 6B2 4((4B1 + 1)/6) = 1 – 2.5B1 B1 = 2 Substitute B1 into x = 4B1

y = 6B2 y = 6(1.5) = 9

x = 4B1 x = 4(2) = 8

Therefore, the length of Candle B is 9 units longer than the length of Candle A which is 8 units.

Basic Skills Project

Our group plans to contribute to his project by doing the following:

Providing examples of how to solve particular problems, which includes a step by step procedure of how to go about it as well as descriptions at each step.

When it comes to theory, providing real world examples so as to remember the theories more clearly and be able to apply them The real world examples will also come with example problems and methods.

Creating a program of some sort to create a study group when people want them. Basically, if people are having a hard time with a section of the course, they can go to the project, request a study group and then they can meet up and work out the problems together. This way, students learn from each other, times are flexible because everyone is involved and it will be very dynamic as it won't be the same exact study group every single time.

Find resources on the internet that give concise instructions as to how to solve problems that students can refer to at their leisure.