Homework 3 Discussion

Problem 3

hey guys, ive been giving a shot at question 3 but I have many doubts regarding the wording... in one of the situations, it states that Ed, Pascal Jason, the right fielder and the centre fielder are bachelors, the others are all married. however, Ed's sister is engaged to the second baseman. does this mean that I can consider the second baseman as married or as a bachelor?

also, consider these two statements: Pascal and Charles each won $20 from the pitcher at a poker game, Ed and the outfielders play cards during their free time, I figured that this would imply that pascal and charles are both outfielders and that Ed is the pitcher. however, this is not the case. The pitcher can't be Ed since Ed is a bachelor and it states that the pitcher has a wife. The reason I also brought up Pascal and Charles is because from the previous statement it would seem as if pascal and charles are part of the outfieldiers. We are also told that "All the battery and infield except Charles, Hassan and Adam are shorter than Sung" nulling the fact that charles is an outfielder...

It also states that Sung is in the process of getting divorced. once again can I consider him a bachelor or married? if I still consider him as married. what is the importance of that phrase in the context of the question? I am extremely confused as to how to take on this question.

If someone could look at it id greatly appreciate it. Thx

MarcoGasparian

Question 1

A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip tok one hour and 20 minutes. The bus traveled from the airport back to the terminal and again average speed was 30 mi/hr. however, the return trip required 80 mins. Explain Step 1: Understand the problem:

The problem is asking to explain the difference between the trip to the airport and the return trip from the airport.

-Terminal to Airport= one hour 20 minutes at 30 mi/hr

- Airport to terminal= 80 minutes at 30 mi/hour

Step 2: Plan a strategy for solving the problem:

First: convert time into same units

Second: Compare distance between two trips (a)

Third: Compare time between two trips (b)

Fourth: Compare speed between two trips (c)

- If two of a, b, c are the comparably the same, that means the remaining must be equal as well (Pythagorean Theory)

Step 3: Execute Strategy

one hour= 60 minutes

one hour 20 minutes = 60 +20 = 80 minutes

80 minutes = 80 minutes

a) compare distances,

- Airport to terminal

- terminal to airport

=Same distance

b) After conversion, realized that they are the same time spent traveling

c)speed= average 30mi/hr both trips

Step 4: Check

- Because the distance and speed were the same in both trips, so is the time

- The time had to be converted to same units for clear view that they were both the same traveling time

Question 6:

Three kinds of apples are all mixed up in a basket. How many apples must you draw (w/out looking) from the basket to be sure of getting at least 2 of one kind?

Step 1: Understand the problem

- There are three kinds of DIFFERENT apples- how many times does it take of pulling one apple out at a time to randomly get 2 of the same kind.

- There is three kind: gala, pink lady, and spartan

- How many times do I have to stick my hand in the basket to get 2 galas, or 2 pink ladys, or 2 spartans

Step 2: Plan a Strategy:

-denote g to gala, p to pink lady, and s to spartan to represent the three types of apples in the basket

-Imagination testing

Step 3: Execute Strategy

- Three types of apples, must have at least 3 pulls to ensure you get a duplicate

- so, 3 + x would be the equation

- first pull: g, second pull: p, third pull: s, fourth pull s

- You must pull more than 3 time times to be sure you will receive two of the same kind

-The min. is 4

Step 4: Check - One must pull at least 4 times to ensure they will receive two of the same kind

Question 11

A woman, her older brother, her son, and her daughter are chess players. The worst player’s twin, who is one of the four players, and the best player are of opposite sex. The worst player and the best player have the same age. If this is possible, who is the worst player?

Step 1: Understand the problem

- out of the sequence of 4 individuals playing chess. Who is the worst player and best player?

-Age sequence: woman's older brother, woman, women's son and daughter (same age)

-only two people at table have the same age and can only have the same age.

Step 2: Plan a Strategy:

-denote w for women, b for her older brother, s for her son and d for her daughter

- b- w= greater than 0

- s-d= 0

-worst player's twin(is a player) = opposite sex of best player

-worst player(girl), twin(girl,boy)= best player (boy, girl)

Step 3: Execute

- The worst players twin cannot be the best player

- the only 2 people who can have the same age of the four players is the daughter and son

- This is not possible because the worst player, twin and best player would have to be triplets

-but they are not because they are referred to as twin and not one of a triplets

Step 4: Check

- This problem is not possible because no player is of the same age expect the possibility that the daughter and son could be twins

- However, the wording of the problem determines that they cannot be twins because one of them already has a twin

Question 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?

Step 1: Understand the problem

- each daughter in the family has the same number of brothers as she has sisters

              -Sisters= brothers

- each son has twice as many sisters as he has brothers

            - 2(brothers)= sisters

- How many sons and daughters are in the family?

Step 2: Plan a strategy

- Test numbers that follow the brothers to sisters pattern above

Step 3: Execute:

- If Daughter 1 has 2 sisters she must have 2 brothers, however this means that the son 1 has 1 brothers have 3 sisters= does not work because brothers must have double the amount of sisters than brothers. Not correct.

- if daughter 1 has 3 sisters she must have 3 brothers. Son 1 therefore has 2 brothers and 4 sisters. This is correct because:

Step 4: test - if daughter 1 has 3 sisters she must have 3 brothers. Son 1 therefore has 2 brothers and 4 sisters. This is correct because:

              -  Each daughter has 3 sisters and 3 brothers ( same number of sisters as brothers).  Also each son has 2 brothers and 4 sisters ( double as many sisters      as brothers).

Question 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?

Step 1: understand the problem

-Sven is exactly in the middle of the runners in the race

-Dan was slower than Sven (10th place)

-Lars in 16th place

- Sven is between 1st and 9th place

- How many total runners were in the race?

- Sven place is half the amount of runners in the race

- There are more than 16 runners

Step 2: plan strategy

- eliminate some options of Svens placement by multipling by two (places 1 through 9) and eliminating those that do not equal greater than 16 (lars place)

-Because Sven is exactly in the middle, the total number must be odd. So, 2 x Svens postive - 1 will be the total number of runners in the race

Step 3: Execute

- 1x2= 2

-2x2=4

-2x3=6

-2x4= 8

-2x5=10 (dans place)

-2x6=12

-2x7=14

-2x8= 16 (Lars place)

-2x9= 18 (The only place Sven could be that multiplied by 2 equal greater than lars place)

-18- 1 = 17

Step 4: Check

- There are 17 people in the race in total because Sven is in 9th place and exactly in the middle of the runners

Question 3

3. One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain.

Step 1: There are 3 different variations of fruits in the boxes. Apples (A), Oranges (O), and Apples + Oranges (A+O) Every box IS mislabeled with a different fruit (this is key). Therefore, apples might be oranges, or apples and oranges. Oranges might be labeled as apples etc...

Step 2: You are given the option to pick ONE (it doesn’t matter which one) of the boxes and know the label it has as well as the content inside it.

Step 3: With step 2 in mind lets suppose we pick the box labeled as oranges. Now, lets assume that the box labeled as oranges then has apples inside it. Now, think about this for a minute… if we chose a box labeled as oranges, there are two labels left, apples and apples + oranges. We know for a fact that every single box is mislabeled. Therefore, the content of a box labeled as apples can’t contain apples in it. Which means that the only possible box that could contain apples is the box labeled as apples + oranges, and the remainder is the box labeled as apples contains apples + oranges.

Step 4: lets look at another ex: Lets assume we picked the box labeled as apples, we open the box and see it contains oranges. once again from this we can derive that the box labeled as oranges will not in fact contain oranges as it goes against the rules which were previously established by the question. Therefore since the only remaining box is apples + oranges, we can assume that apples will be in that box and apples + oranges will be in the box labeled as oranges. MarcoGasparian

Question 8

8. Reuben says, “Two days ago I was 20 years old. Later next year I will be 23 years old.” Explain how this is possible.

Step 1: The most important aspect of how this question was written is the fact that they state TWO days ago and not one. This means that one day ago Reuben could already be 21 (keep this in mind).

Step 2: The easiest form to solve this problem is to draw a timeline. Given that it is difficult to draw an accurate timeline I will simply draw a table Lets assume that: T=Today T-1= One day ago T-2= Two days ago T+ one year later-1= one year later -1 day (December 31st 2010) T+ two years later – 1 day= two years later – day (December 31st 2011)

Step 3:

Lets assume that T= January 1st of 2010 And that Reuben’s B-day was December 31st= T-1 He turned 21 on 2009 And that on December 30th = T-2 Reuben was still 20 since he turned 21 on December 31st. With that in mind if we simply use logic than in T+ one year – 1 day would be December 31st of 2010 and Reuben would be celebrating his 22nd birthday. This however is still in 2010 which is the same year. The question states “later next year I will be 23 years old” Therefore, in the year 2011 on December 31st Reuben will be celebrating his 23rd birthday.

Step 4: Think of this question in terms of formulas If Reuben’s B-day is on December 31st = T-1 we are going back an entire year from 2010= T to 2009= T-1 And the difference in years between T and T + two years later -1 day is also 2 years Therefore the difference between T-1 and T+ two years later – 1 day is 3 Years which is between Reuben being 20 to 23 years old. MarcoGasparian

Question 13

13. If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)?

Step 1: At first glance this question may seem like the obvious answer is 10 (I thought so). However, we must take into consideration that every chime is divided by little halts of the noise the clock makes. In this instance it will make 5 chimes in a period of 5 seconds. We are being asked how long it will take to make 10 chimes taking the halts into consideration.

Step 2: The easiest way to answer this question is by drawing a timeline by evenly distributing 5 chimes throughout it and 10 chimes through the second timeline:

(.)= chimes (_)= halts

5 second 5:00 chime

.____.____.____.____.

10:00 chime .___.___.___.___.___.___.___.___.___.

Step 3: The question stated that it would take the clock 5 seconds to complete 5 chimes with the halts in between them therefore by looking at the graph we observe that it takes 5 chimes/ 4 halts in between each chime. Which means 5/4=1.25 halts between each chime. 1.25*4= 5 seconds

Using the same concept with the 10 chimes. We count the number of halts and see that there are 9. Since we already know that the number of seconds it took between each chime is 1.25, we use the formula 9/4=2.25 seconds 2.25*5= 11.25 seconds

Step 4: In order to prove this lets assume try to figure out what would happen if it were 20 chimes.

.__.__.__.__.__.__.__.__.__.__.__.__.__.__.__.__.__.__.__.

19/4= 4.75 seconds 4.75*5= 23.75 seconds MarcoGasparian

Question 18

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?

Step 1: The key concept to grasp in this particular question is that the 1/3 that Bret and Carla are taking are the REMAINDER of what had been left at the jar based on what had already been removed. Nevertheless, Alice, Bret, and Carla are all taking 1/3 of what is being offered to them.

Step 2: In order to figure out this problem you must think backwards, working from the solution. The question which I posed in my own mind was what equals 40 if its has 1/3 of it subtracted 3 times

Step 3: I looked at the number 40 and started testing different possibilities with numbers whose value of 1/3 was still an integer. I then realized that 60*1/3 = 20 60-20=40

20= The amount Carla took from the 60 Remainder from the 1/3 Alice took and then the 1/3 that Bret took (after Alice).

I repeated the process and realized that 90*1/3= 30 90-30= 60 30= Amount Bret took from the 90 remainder of the 1/3 Alice took.

I finally did the process a third time and realized that

135*1/3= 45 135-45=90 45= Amount Alice took from the TOTAL = 135 How many pennies were in the jar at the start? 135 pennies

Step 4: When I looked at the questions solution, I noticed that there was a pattern with the formula I was using.

Essentially the amount that remained was twice as much as the value a person had previously taken. This makes sense since we are subtracting 1/3 of the 3/3 leaving us with 2/3 (2/3)/(1/3)= 2 for every 1

MarcoGasparian

Question 23

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.

Step 1:In order to figure out this question I eliminated the useless information. The fact that the oldest child has red hair is a null point. I then formulated different possibilities in my head of what could be done if I were to try to solve this.

Step 2: I realized that the easiest way to figure out this problem is to create a factor tree of 36. 36 can be factored to 36*1, 2*18, 3*12, 4*9, 6*6, 4*3*3, 2*3*6, 2*2*3*3, 2*3*6 Note that the only possible answers were the ones that could be factored into 3 different numbers as there are exactly 3 children.

Step 3: The question also states that the sum was today’s date. Since today is the 12th, the only factor which would be possible is 2*3*6 Since the product of that equals 36 and the sum equals 12. Also, it could not have been 6*6 since it was asking for 3 children and not 2

Step 4: In order to check this I tried all different factors of 36 and saw through trial and error that this was the only compatible one. MarcoGasparian