Course:MATH110/Archive/2010-2011/003/Groups/Group 17/Problem Solving 1
Problem Solving
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 2
A lady did not have her driver's license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.
Step 1: Why would a police officer not stop a woman who doesn't have a license, travelling the wrong way down a street?
Step 2: In what manner would this be legal?
Step 3: If the woman was walking, then there is no reason the officer should be concerned.
Step 4: This does, indeed, make sense.
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.
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 7
Suppose you have 40 blue socks and 40 brown socks in a drawer. How many socks must you take from the drawer (without looking) to be sure of getting (i) a pair of the same color, and (ii) a pair with different colors?
Step 1: 40 blue socks and 40 brown socks
i) how many do you need to pick to get two of one colour?
ii) how many do you need to pick to get one of each colour?
Step 2: What are the possible pulls each time?
You can get either a brown sock or a blue sock from each pull.
Step 3:
i) If you pull a brown sock, then your next will either be brown or blue. Given blue, then your next will, by extension, complete one of the two pairs. Therefore, the maximum number of socks you will have is 3.
ii) In the most extreme example, you could, with some strange luck, manage to pull every sock of one colour before winding up with one of the other colour. If this were the case, then the 41st pick will, by necessity, be the opposite colour, as there are no socks of the original colour remaining. So it will take a maximum of 41 socks to get a non-matching pair.
Step 4: Though I am not about to try this with my own socks until it works, it does make sense.
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.
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 12
A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at the train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man's Bronx girlfriend left him because he rarely visited. Give a (logical) explanation.
Step 1: Taking the first train to either A or B, after arriving at a random time, why would a man be more likely to wind up in A than B, given that the trains arrive every 10 minutes.
Step 2: If the trains have the same frequency, why would one train occur more frequently?
Step 3: If the trains arrived on a random schedule, then it's likely that he would arrive at both places with equal frequency. However, given that the schedule is fixed, it is likely that the time between the pair of trains is not equal.
Step 4: If the Brooklyn train arrives at 5:10 and every 10 minutes thereafter, and the Bronx train arrives at 5:12 and every 10 minutes thereafter, then the man would have to arrive after the Brooklyn train left, but before the Bronx train arrived in order to arrive at the Bronx. This gives him a span of about 2 minutes. Any time from 5:12 (after the Bronx train leaves) to 5:20 will result in the Brooklyn train being the next. This means that 1/5 times he will go to the Bronx, 4/5 time he will go to Brooklyn, and his Bronx girlfriend will likely be rather surprised when he shows up.
Of course, another logical explanation is that the Bronx girlfriend found out about the Brooklyn girlfriend, and was none too impressed.
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
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 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.
Step 1: The scale is calibrated incorrectly. Two separate measurements total to 5kg more than a combined measurement of the same weights.
Step 2:
Step 3:
Step 4: If the scale were calibrated to read too high by 5 kg, then Dan actually weights 55 kg and Sarah weighs 45kg. Their combined weight of 100 kg would show as 5 kg higher, 105 kg, when they both were on the scale simultaneously. The scale reads too high by 5kg.
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
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 (Dan's 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 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?
Step 1: Rainy morning -> Sunny afternoon. Sunny morning <- rainy afternoon. It rained 13 days, there were 11 sunny mornings, and 12 sunny afternoons.
Step 2:
rm = rainy mornings
ra = rainy afternoons
rd = rainy days
sm = sunny mornings
sa = sunny afternoons
sd = sunny days
Step 3:
Since it can't have rained for a full day (given that if it rained in the morning it was sunny in the afternoon, then
Failed to parse (syntax error): {\displaystyle rm = 13 – ra}
The number of sunny afternoons can be given by sa = sd + rm, since both sunny days and rainy mornings will result in sunny afternoons.
Therefore: Failed to parse (syntax error): {\displaystyle sd = sa – rm = 12 - (13 – ra)}
Of course, the number of sunny mornings must be equal to the number of sunny days plus the number of rainy afternoons, since every rainy afternoon was preceded by a sunny morning.
So: Failed to parse (syntax error): {\displaystyle sd = sm – ra = 11 - ra}
Failed to parse (syntax error): {\displaystyle sd = 12 – (13 – ra) = 11 - ra}
Failed to parse (syntax error): {\displaystyle 12 – (13 - 6) = 5 = 11 - 6}
Therefore,
Since sFailed to parse (syntax error): {\displaystyle d = sm – ra = 11 – 6 = 5,} there we 5 sunny days on the trip.
Since the entire vacation would be , this gives us , or 18 days.
Step 4: Checking by mornings, we know that there are 11 sunny mornings. This means that there must have been 7 rainy mornings. Since we know that there were 6 rainy afternoons and 13 total rainy days, this makes sense.
Shame about their vacation, though...
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