Course:MATH110/Archive/2010-2011/003/Groups/Group 16/Homework 3
Homework 3 - due Wednesday October 13
Thanks Shamilla,
That sounds like a great idea.
Arabella.
Hey guys, since we did not discuss dividing up the 25 math problems assigned in class, I was thinking everyone could do 5 problems each, going in order by how the group members names have been listed. For instance, my name is first, therefore I would be assigned to complete the first 5 problems. Hopefully you guys agree, if not just let me know. Also I think it would be a great idea if we added our email addresses next to our names above.
NOTE: I plan on adding images... as soon as i figure out how to do so. If someone could explain how to do so, that would be helpful
ShamillaBirring
All you ever wanted to know about adding images can be found here.
--DavidKohler
25 PROBLEMS
QUESTION 1 -
A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain
BUS 1: Terminal Airport to Airport
BUS 2: Airport to Terminal Airport
Bus 1: Speed = 30mi/hr ; Time = 1hour 20mins
BUS 2: Speed = 30mi/hr ; Time = 80mins
By comparing the speed of both buses it is evident that Bus 1 and Bus 2 are travelling at the same speeds, however it is not so clear whether the time taken for each bus to reach their destination is the same. In order to make this more clear we must put both times in the same units.
Bus 1: Time = 1hour 20mins
1hour = 60 mins
Therefore 1hour 20mins = (60mins + 20mins) = 80mins
Hence, Bus 1 took 80mins to reach the airport (Bus 1 time = 80mins)
Now that both the bus times are in the same units we can compare the times
Bus 1: Time = 80mins
Bus 2: Time = 80mins
Therefore it is now evident that both buses took the same amount of time to travel the same distance at the same speed.
Hence, the return trip required 80 mins, because the bus continued to travel at the same speed of 30mi/hr and same distance (between the terminal airport and the airport) as the initial trip which required 80 mins.
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.
Information that can be extracted from the question:
1. Lady does not have her driver's licence with her
2. Lady did not stop at a stop sign
3. Lady went three blocks down a one-way street the wrong way
4. Policeman saw the lady, but did not stop her
After analyzing the question and the conditions, it is clear that it does not state in the problem that the lady was in a vehicle, which if the case she would have to have her driver's licence with her, stop at a stop sign, be prohibited from driving down a one-way street the wrong way and the policeman would be obliged to stop the her. Therefore, it can be understood that the lady is a pedestrian and therefore she does not need her drivers licence to walk, she is not obligated to stop at a stop sign, she can walk down a one-way street the wrong way and under these conditions the policeman has no reason to stop the lady.
Therefore, the lady is a pedestrian.
QUESTION 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.
It is understood from the problem that there are three boxes filled with either/or apples and oranges and the boxes have been incorrectly labeled APPLES, ORANGES, and APPLES AND ORANGES.
Therefore, say the boxes are labelled as described below:
Box 1: APPLES
Box 2: ORANGES
Box 3: APPLES AND ORANGES
Because the boxes have been labelled incorrectly this means that Box 1 does not contain only apples and can therefore only contain either "oranges" or "apples and oranges", this is the same situation for each box.
Hence,
Box 1: Labelled APPLES
Fruit options inside the box: Oranges ; Apples and Oranges
Box 2: Labelled ORANGES
Fruit options inside the box: Apples ; Apples and Oranges
Box 3: Labelled APPLES AND ORANGES
Fruit options inside the box: Apples ; Oranges
Now that we understand the conditions and situation of the questions we must focus on the question; Can you select one fruit from only one box and determine the correct labels?
Consider that you have box 3 which is labelled incorrectly as APPLES AND ORANGES. The fruit options for box 3 are "apples" or "oranges", if an apple were picked out of that box, this would mean that this box contains apples only. Therefore, box 2 which is labelled incorrectly to contain ORANGES and has the fruit options "apples" or "apples and oranges", by the process of elimination one knows that box 2 cannot contain apples since we have already discovered that box 3 contains apples, meaning that the only fruit option left for box 2 is "apples and oranges". Therefore, so far box 3 contains apples and box 2 contains apples and oranges, leaving box 1 with only oranges.
Therefore, with the example above we were able to pick out one fruit from a box and in effect conclude which box contained either apples, oranges, or both apples and oranges.
Hence, yes it is possible to select one fruit from only on box and determine the correct labels.
QUESTION 4 -
I am the brother of the blind fiddler, but brothers I have none. How can this be?
Therefore, this question states that "this person" is a brother to their sibling (whom is not gender defined) and that "this person" does not have any brothers. Hence, because this person has a sibling and has no brothers, one can assume that the sibling is a female. This assumption is accurate because "this person" is a brother to his sister whom is the blind fiddler, and can still have no brothers of his own.
Therefore, this statement "I am the brother of the blind fiddler, but brothers I have none" is true if the blind fiddler is a female and therefore the boy's sister.
QUESTION 5 - Two quarters rest next to each other on a table. One coin is held fixed while the second coin is rolled around the edge of the first coin with no slipping. When the moving coin returns to its original position, how many times has it revolved?
The conditions of the question:
1. two quarters
2. one quarter is fixed while the second coin is rolled around the edge of the first coin
3. there is no slipping
It is important to recognize that it is two quarters being used, meaning that they are the same size (have the same diameter, perimeter and area). Hence, for the second quarter to go around the first quarter it will only revolve once. A more simpler manner to picture this question is to imagine two identical squares (as opposed to the quarters) under the same conditions (have same width and length and therefore the same perimeter and area). Each sides of the second square will get to touch a side of the first square, while only revolving once.
Therefore, the second quarter will only revolve once in order to return to it's original position.
"QUESTION 10" - Suppose one-half of all people are chocolate eaters and one-half of all people are women. (i) Does it follow that one-fourth of all people are women chocolate eaters? (ii) Does it follow that one-half of all men are chocolate eaters? Explain.
Being a man or a woman is independent of preference for chocolate.
i) No, it does not follow. ii) No, it does not follow.
"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?
The situation is not possible. The woman's daughter and son must be either the best or worst players. They are of opposite sex and they have the same age, since the woman is older than both of them (she is their mother) . The woman's older brother is the oldest player.
"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.
For example, the Bronx train arrives at 12.00, 12.10. 12.20...and the Brooklyn train arrives at 12.09, 12.19, 12.29...
"QUESTION 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)?
Considering the time between chimes: it takes 45/4 = 11.25 seconds to strike 10.00.
"QUESTION 14" - One day in the maternity ward, the name tags for four girl babies became mixed up. (i) In how many different ways could two of the babies be tagged correctly and two of the babies be tagged incorrectly? (ii) In how many different ways could three of the babies be tagged correctly and one baby be tagged incorrectly?
i) There are six ways that two of the four babies can be correctly tagged. ii) There are no ways in which three of the four babies correctly tagged.
"QUESTION 15" - Alex says to you, “I'll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet?
Alex is right, I shouldn't accept his bet. The deck of cards is made up of an equal number of black and red cards (respectively 1/2 of the total number of cards).
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?
Looking at the question, we are the given positions of the runners as followed:
Sven: Middle place of all runners
Dan: 10th place
Lars: 16th place
From the details of the questions, we can first of all try to determine a range for which place Sven came in as we only know he is in the middle amongst all runners. First we know that Dan, who was slower than Sven came in 10th place thus we can assume that Sven came in places ranging from 1st to 9th. Taking into account that Sven is in the middle we can say that the maximum number of runners is 18 as the Sven can only be placed up to the 9th at max. Next, as we are told that Sven is placed exactly in the middle therefore we can deduce that the number of runners is an odd number, as an even number cannot produce a runner exactly in the middle. Lastly we take into account the placement of Lars who came in 16th, as mentioned earlier we deduced that the maximum number of runners is 18 therefore shortening the range to 16-18.
In conclusion, the only possible number that fits with all the criteria is 17; therefore there were 17 runners in the race.
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?
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. We are told that Paula has three children and that the product of their ages is 36, therefore our first step would be to find the factors of 36 in sets of 3. This gives us the following sets:
• 1,1,36
• 1,2,18
• 1,3,12
• 1,4,9
• 1,6,6
• 2,2,9
• 2,3,6
• 3, 3,4
Next we are told that the sum of their ages is the same as a date, therefore we know the sum cannot exceed 31 as that is the maximum number of days in a month. By calculating the sums of each set of factors we are given the following:
• 1+ 1+36 = 38
• 1+2+18= 21
• 1+3+12 = 16
• 1+4+9 = 14
• 1+6+6 = 13
• 2+2+9= 13
• 2+3+4 = 11
• 3+3+4= 10
From this we see that we can eliminate the set 1,1,36 as its sum is greater than 31. Next we are told that the information is still not enough, the only reason for this may be that there are sums of the same number thus making Paul unable to determine which the correct set is. Looking at the sums we can see that the number 13 is duplicated making 1,6,6 and 2,2,9 possible answers.
Lastly, we are told that the oldest child has red hair, the clue here is that there is a eldest sibling, ruling out the set 6,6,1 as there is not one single elder sibling. Therefore, this we can determine the ages of Paula’s children to be 9,2 and 2.
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? From the details of the question we are given the following conditions:
1.) Two candles are of equal length
2.) One candle took 6hr to burn out (we can call this Candle A)
3.) One candle took 3hr to burn out (we can call this Canlde B)
Firstly to make this question more clearly, we can give the candles an actual length giving that both candles are of equal length. Let’s assume that the candle length is 60cm.Taking into account the time it takes each candle to burn out we can come up rate of how fast each candle burns by dividing the length by the time, thus giving us the following information.
Candle A: 60cm/6hr = 10cm/hr
Candle B: 60cm/3hr = 20cm/hr
Second, we can shorten our time to under 3hrs as we know that by 3hrs, Candle B would’ve already burned out leaving only one candle left. Now we can use a trial and error method, by trying the times of each hour. After the first hour, the length of Candle A would be 50cm while the length of Candle B would be 40cm; one is not exactly twice as long as the other. Next after the second hour, the length of Candle A is 40cm while the length of Candle B is 20cm; one candle is exactly twice as long as the other.
Therefore after 2 hours, one candle will be exactly twice as long as the other candle.
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.