Course:MATH110/Archive/2010-2011/003/Groups/Group 12/Homework/hw3
Homework #3: Problem Solving
Reading Pólya's method on how to solve problems, our group utilized the four listed steps to solve the following problems. Here are there solutions.
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.
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.
A reason why the policeman did not stop the woman when she failed to stop at a stop sign and went down the wrong side of the street is because she was probably not driving a car. She could have been walking, riding a bike on the sidewalk, wearing rollerblades or even using a skateboard. In this case, the lady not having her driver’s license is irrelevant to the scenario. The policeman would only be obligated to stop her if she was driving a car meaning she would be breaking the law and endangering others.
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.
No you cannot determine the rest of the label. As seen in the diagram 1, if we are given that box 2 is APPLE & ORGANGE then the probability of 50% to be either APPLES or ORANGES. Same applies to the other box which is 50% of being either APPLES or ORANGES therefore it cannot be determine if only one of the box label were given.
Question # 4
I am the brother of the blind fiddler, but brothers I have none. How can this be?
So we can break down this question in 2 parts. The first part is “I am the brother of the blind fiddler” so we can assume that the two related to each other and one of them is related to the blind fiddler. The Second part is “But brothers I have none” so we can assume that the person is not male. So we can conclude that the Blind Fiddler is the brothers 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 moving coin revolves 2 times. A visual presentation will explain the results.
Revolving Coins
Question # 6
Three kinds of apples are all mixed up in a basket. How many apples must you draw (without looking) from the basket to be sure of getting at least two of one kind?
The minimum amount of apples you can take before you get a duplicate of either three would be four apples. With the fourth being the duplicate. Let's say you take three apples out: a, b, and c. No duplicates so far. You take one more out and it will have to be ether of the three since there are only a, b, and c apples in the basket to begin with.
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?
There is a 50-50 chance every time you take a sock out of the drawer one at a time without looking. If you take two socks at a time it will be is a one eighth probability to get a pair of socks of the same color because the probability of getting a pair of blue socks is one fourth and the probability of getting a pair of brown socks is one fourth. When they are combined they make one eighth.
In order to be sure of getting a pair of different colors socks then a person would have to draw all but one of the blue socks (for example) in order to get a blue and brown sock in the same pair. Meaning that it would take 39 socks until getting a pair of different colored socks.
Question # 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.
Let’s assume that Reuben’s birthday is on 31st of December 2000 and he tells you that he was 20 on the 1st of January. As seen in diagram 2, Reuben is 20 years old on December 31st of year 2000, he turns 21 on December 31st 2001. He tells you that he will turn 23 years old later next year, which is on December 31st 2002 but since he made his statement on January 2001, which makes year 2002 the year he turns 23.
Question # 9
A rope ladder hanging over the side of a boat has rungs one foot apart. Ten rungs are showing. If the tide rises five feet, how many rungs will be showing?
Let us assume that the boat is floating on the water. The rope ladder has ten rungs 1 foot apart. The tide rises 5 feet. As the water rises so will the boat. The boast rises 5 feet so the ladder will rise by 5 feet and 10 rungs will still be showing. If we are assuming that the boat is fixed in one spot by a invisible force we will have a different answer. The bottom rung is at feet zero so the second rung is 1 foot high. The ladder is 9 feet tall and as the water rises 5 feet only 4 rungs will be showing. The 10 rungs one of them is touching the water subtract the rising 5 foot tide this will equal 4 feet or 4 rungs.
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.
No, it does not follow that one-fourth of all people are women chocolate eaters and it does not follow that one-half of all men are chocolate eaters. What is the reasoning behind this?
We know two important details regarding this particular population. (1) That half of the population are chocolate eaters and the other half of the population being non-chocolate eaters. (2) That half of the population are women and the other half of the population being men. From the information given, we cannot make these conclusions because being a man or a woman is independent of being a chocolate eater or a non-chocolate eater. Since these two pieces of information are independent of each other, such conclusions given in the question cannot be made.
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?
We will have to assume that the son and daughter are twins since it is stated that the woman has an 'older' brother. Older, as in born at a different time. The worst player is stated to be one of the twins and the best is the opposite gender of the worst twin. The worst and the best player are the same age (see: twin) so we can say that the worst and best player are the son and daughter. But since neither the best or the worst is designated with a gender, the worst can be either one of the twins. In other words, the worst player cannot be identified 100%.
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.
The reason why his Bronx girlfriend left him is because whenever he would go to the train station, the next train was usually going to Brooklyn. If he were to set a certain time to go to the trains stop which would make him visit his girlfriend’s equally then maybe he would not have been dumped.
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)?
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?
First lets name the four girls A, B, C and D
So the first part of this question is In how many different ways could two of the babies be tagged correctly and two of the babies be tagged incorrectly.
We come up with all the different combinations for the first part which is:
ABDC ADCB ACBD DBCA CBAD BACD
Assuming that A, B, C and D are the correct order the letters are in the order tagged wrong. So for example in ABDC, D and C are switched and thus wrong. The number of possible combinations for the first part is 6
All the possible combinations for the second part is:
ACBD ACDB ADBC CBDA DBAC CBDA DACB BDCA ADCB CABD BCAD CBAD
So assuming again that A,B, C and D are the correct order. The number of possible combinations for the second part is 12.
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?
No, you should not accept the bet with Alex because no matter how Alex shuffles the 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. How is this possible?
As an illustration, let's imagine this particular scenario. Let us assume that Alex shuffles the deck of cards and cuts the deck of cards in exactly two halves, leaving us with 26 cards in the first pile and the remaining 26 cards in the other pile. Examining the first pile, we see that there are 20 Red cards and 6 Black cards. Without even looking at the other pile, we know there has to be 20 Black cards and 6 Red cards because the definition of a deck of cards consist of 52 cards in total with 26 Red cards and 26 Black cards. From this particular example, we can see that Alex is indeed correct, there are as many red cards in the first half of the deck as there are black cards in the second half.
Generalizing from this particular situation, we can apply the same logic to any derivation of cards. No matter how many red cards are in the first half, there has to be the exact same of black cards in the second half as there are red cards in the first half. A half of a deck totals to 26 cards and since there are two colours, red and black, the number of red and black cards will be mirrored oppositely in the second half by definition of what a deck of cards is.
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?
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.
The information given cannot let me determine whether the scale is set too high or too low. The question states that the scale is set incorrectly but still accurate, however, it also claims that the weight of Dan and Sarah together is 5 kilos off than what it would be. Meaning that this information is inaccurate and cannot be applied.
Question # 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?
The question stated that Alice takes 1/3 of the pennies from the original jar (X), and Bret takes 1/3 from whatever is left from Alice, also Carla takes another 1/3 from whatever is left in the jar from Alice and Bret. There will be a multiple of (2/3) to Bret’s 1/3 because 2/3 is what is left from after Alice has taken 1/3 of pennies from the jar, and two 2/3 is multiplied to Carla’s 1/3 because it is whatever is left from both Alice and Bret. Therefore we can put this into an equation:
X = (1/3X) + ((1/3)(2/3)X) + ((1/3)(2/3)(2/3)X) + 40
X = (1/3X) + (2/9X) + (4/27X) + 40
X = (9/27X) + (6/27X) + (4/27X) + 40
X = (19/27X) + 40
(8/27)X = 40
X = 135
The process and the checking can be found on diagram 4
Therefore we can conclude that the total pennies are 135 at the start.
Question # 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?
With eight ounce cup of coffee and eight ounce cup of milk. Since everyone in her family drank 8 ounces of coffee and its 1.6 of the total amount of coffee there must be at least 6 people. If Angela drinks only 1/6 of the whole coffee we can conclude that the rest of the 5 people are her family members. There are 5 people in Angela’s family not counting Angela.
Question # 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?
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?
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?
SA = Sunny Afternoons
RM = Rainy Mornings
SM = Sunny Mornings
RA = Rainy Afternoons
SA RM* SM RA*
SA RM* SM RA*
SA RM* SM RA*
SA RM* SM RA*
SA RM* SM RA*
SA RM* SM RA*
SA RM* SM RA
SA RM SM RA
SA RM SM RA
SA RM SM RA
SA RM SM RA
SA RM SM RA
SA RM
* marks the day that it actually rained
Since there is the same amount of sunny afternoons (6) left as sunny mornings (6) then it can be assumed that there were 6 completely sunny days.
13 rainy days + 6 sunny days = 19 days
Making the vacation, 19 days long.
Question # 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.
Since the question stated that the product of the three children is 36, we can write down a set of product of 3 numbers that will give the result of 36 after multiplication (as seen in diagram 3a.). We came up with 8 different groups of product of 36, and now we are being told that the sum of three children is the same as today’s date that means the sum of the three children must be less or equal to 31 (as seen in diagram 3b), we can now eliminate the first option which is 1+1+36 (because the result is 38 and the maximum day in a month is 31). Now we are left with 7 options, now looking back into the question when Paul said “that is not enough information” we can assume that Paul worked out the math but were stuck since there were sets of options that gives him the same result which is 13 (2+2+9 and 1+6+6). Therefore we can again eliminate other options and we are now left with [2+2+9 and 1+6+6], looking again back into the question Paula stated, “the OLDEST child also has red hair.” The OLDEST gives us the clue that one of the three children must have been older than the other two which means we are now left with one equation, which is 2+2+9 the oldest being 9 years old and the other two are 2 years old.
Question # 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?
So first we can we notice that one candle burns twice as fast as the other. The first step to solve this problem is to make the 6hr candle equal to 2 of the 3hr candle. So the equation we come up with is 1-1/6x that rate of which one candle burns is 1/6. 1/6x is how much it burs after x amount of time. So we can conclude that since the 3hr candle is equivalent to 2 of the 6hr candle we make 1/6x = 2 (1/3x). So after solving the equation x will equal 4 hours.
Question # 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.