Course:MATH110/Archive/2010-2011/003/Groups/Group 02/Homework 3
Homework 3: Solving Problems Using Pólya’s Four-Step Method
Question 1
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.
TO AIRPORT
Going 30 miles an hour: Trip Time: 1 hour, 20 minutes
FROM AIRPORT
Going 30 miles an hour: Trip Time: 80 minutes
As the bus travelled at the same speed (30 miles an hour) for both the trip to the airport and back from airport to the terminal, why is there a difference in the trip times (assuming the route has not changed)?
On a closer look, we see that there is no difference for if we convert 1 hour and 20 minutes from hours to minutes, we obtain 80 minutes.
As 60 minutes is 1 hour. That means, 1 hour and 20 minutes is 80 minutes as 60 minutes(1 hour) + 20 minutes = 80 minutes. Thus: 1 hour and 20 minutes=80 minutes.
Therefore, we can conclude that both the bus that travelled to the airport and the bus that returned from the airport had the exact same trip duration : (80 minutes or 1 hour, 20 minutes).
Question 2
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.
From the question, we know the following information: a.) The lady does not have her driver's license
b.) She failed to stop at a stop sign
c.) She went three blocks down a one-way street...the wrong way.
d.) A policeman fails to apprehend her. (Perhaps, he is lazy but it is more likely that the lady has not broken a law)
In this question, it is crucial not to make assumptions.
a.)Nowhere in the question does it say that the lady was in fact, driving a vehicle. It merely states that the lady did not have her driver's license. We naturally assume that this bit of information about the lady not having her driver's license with her means that she is driving a car.
b.)The lady fails to stop at stop sign meaning she has broken an important driving rule (if she is indeed driving a car.) Also, notice the question states she went down three blocks, not drove.
c.) Travelling three blocks the wrong way on a one-way street is another driving violation. (if she is indeed driving a car.)
d.) A policeman sees her, but does not stop her. This means that the policeman has no reason to stop the lady. This must mean that the lady's actions are within the limits of law, or in other words, perfectly legal.
The logical conclusion to this problem is that the lady was not driving a vehicle for otherwise, the policeman would have been obligated to stop her. Instead, the lady is a pedestrian* for pedestrians do not obey driving regulations.
*Or cyclist, jogger, etc.
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.
BOX ONE: APPLES
BOX TWO: ORANGES
BOX THREE: APPLES AND ORANGES
ALL Boxes have incorrect labels. Therefore:
BOX ONE: Has either ORANGES or APPLES AND ORANGES
BOX TWO: Has either APPLES Or APPLES AND ORANGES
BOX THREE: Has either APPLES or ORANGES
We must select one fruit from only one box and determine the correct labels...
Suppose that box is BOX THREE...
Consider: BOX THREE: APPLES AND ORANGES (incorrect label)
Above, we have figured that BOX THREE has either APPLES or ORANGES. Suppose an orange is picked out of BOX THREE. This would mean that BOX THREE contains ORANGES only.
If BOX THREE has ORANGES, then BOX ONE has APPLES AND ORANGES. This was figured out by the process of elimination for we know now that BOX ONE cannot contain ORANGES for BOX THREE is the box with ORANGES only.
Thus, BOX TWO must contain APPLES only, for that is the only option left by using the process of elimination.
Therefore, we can conclude that by selecting one fruit from one box we are able to determine the correct labels for all three boxes by the process of elimination, as illustrated in the example above.
Question 4
4. I am the brother of the blind fiddler, but brothers I have none. How can this be?
It is crucial in this question not to make assumptions for we may automatically assume that the blind fiddler is male.
However, notice that the gender of the blind fiddler is not defined, therefore the "blind fiddler" may be male or female.
Since the person (I) is the brother of the fiddler, but has no brothers, then the blind fiddler must be his sister.
In the question, he says he has is the sibling of the blind fiddler, but has no brothers (male siblings) thus, his sibling must be female, or in other words, his sister.
Therefore, we can conclude that the statement "I am the brother of the blind fiddler, but brothers I have none" can be true for the blind fiddler is the man's sister.
Question 5
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?
From this question, we can gather three main pieces of information. a.) There are two quarters.
b.) One is fixed (not moving) while the second quarter is rolled around the edge of the first coin. c.) There is no slipping.
Since both coins are quarters, they have the same dimensions. Thus, one rotation will roll the quarter halfway around the fixed quarter, while another rotation will bring the quarter to it's original position (next to the fixed quarter.)
Therefore, the quarter will need to be revolved twice.
Question 6
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?
There are three kinds of apples in the basket. Suppose the three kinds are as follows: red, green and yellow.
Example: 1st draw: RED
2nd draw: GREEN
3rd draw:YELLOW
4th draw: GREEN
You now have two of one kind (GREEN).
To get at least two of one kind, you need to draw from the basket at least 4 times.
Question 7
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?
i.) A pair of the same colour
ii.) To be sure (as in guranteed) of getting a pair of socks with different colours, we must draw 41 times. A person is guranteed of drawing a sock of different colour in the 41st draw, for there are only 40 socks of the same colour.
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.
In order for Reuben's statement to make sense, he must have been born on the 31st of December. Suppose Reuben was born Decemeber. 31st, 1980
That means on December 31st, 2000, he turns 20. And on December. 31st, 2001, he turns 21. On January. 1st. 2001, Reuben states "Two days ago I was 20 years old" - (December.30th)
"Later next year I will be 23 years old"...
On December 31st, 2001, Reuben turns 22. (The same year as his statement --2001) On January 1st, 2002- The Year Reuben turns 23..."later this year..." -Reuben's 23rd Birthday: December 31st, 2002.
Question 9
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?
If the tide were to rise five feet, the boat would also rise with it. Therefore, the tide has no relationship to how many rungs of the rope ladder would be showing.
Thus, we can conclude that there will be still 10 rungs showing.
Question 10
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.
i) The above fact may happen (it makes sense), but it is not necessarily true all the time. This is because half of the population can eat chocolate but the other half can be woman. Or, the half that eats chocolate can contain only a small percentage of women in it.
pink: women
Brown:Chocolate eaters
Red: combination of women and chocloate eaters
ii) All men can be chocolate eaters, but also can’t in different situations. For example, the half of the chocolate eaters that are being referred to can be men, women or both.
Use the diagram above but instead of pink symbolizing women it repesents men. The same concept applies
Question 11
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?
There is not enough information to say which one of the four players is the worst one. It is impossible for the woman to have the same age as her older brother or her son and daughter. The only two people who have the same age are the son and the daughter. At the end, the worst player is debated between the daughter and the son. So, lets say that the daughter is the best player which means that the son is the worst. OK, the worst player's twin, which is the daughter, and the best player, the daugher, have to be the opposite sex. They are the same person, so this can't be true. This doesnt work either way.
Question 12
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 he might chose the Brooklyn train more often might be because he always sees the Brooklyn train first. He might arrive at the train station at the point where the Brooklyn train passes by, a couple of seconds or 1 min before the Bronxs train.
BROOKLYN TRAIN: 2:59, 3:09, 3:19, 3:29, 3:39, 3:49, 3:59
BRONX TRAIN: 3:00, 3:10, 3:20, 3:30, 3:40, 3:50, 4:00
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)?
To find x you must first divide 5 (which are the total seconds it takes for the clock to chime) by 4 (which are the spaces between chimes) to find the actual time between each chime. 5/4=1.25
When we have found the time for which each chime takes to play, we multiply that by 9 (which are the spaces between the chimes at 10 chimes) so we can get the total seconds in which the 10 chimes played in. 1.25X9=11.25
It takes a clock 11.25 seconds to chime 10 chimes.
Question 14
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?
These are all the combinations possible that the tags can have on 4 babies:
ABCD √
ABDC √
ACDB
ADCB √
BACD √
BCAD
BADC
BDAC
BDCA
BCDA
CDBA
CBDA
CDAB
CADB
CABD
CBAD √
DCBA
DBCA √
DACB
DCAB
DABC
DBAC
i) We have to count how many combinations include 2 tags in the right order for the 4 babies, considering that the right order is ABCD (look at the √)= there are 6 ways that 2 of the four babies can be correctly tagged
iii) Using the same template we can count how many combinations include 3 tags in the right order for the 4 babies, considering that the right order is ABCD= there are no ways in which three of the four babies can be correctly tagged
Question 15
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?
TOTAL CARDS IN DECK: 52
RED CARDS: 26
BLACK CARDS: 26
Suppose Alex splits the deck and this is what happens: First Deck: 12 BLACK CARDS, 14 RED CARDS. Second Deck: 14 BLACK CARDS, 12 RED CARDS.
Since there are 26 red cards and 26 black cards, everytime Alex splits the deck 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, because every colour adds up to 26.
Thus, we can come to the conclusion that everytime that Alex splits the deck, the red cards will always equal the quantity of the black cards, thus it would be a terrible mistake to accept Alex`s bet.
Question 16
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 girls be X
Let boys be Y
Y= X-1 X= 2(Y-2)
Solve the equations, we get X=4, Y=3.
The family has 4 daughters and 3 sons.
Question 17
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.
Since Dan`s weight is 60kg, and Sarah`s weight is 50kg, then there combineded weight should be 110 kilograms.
60kg + 50kg =110kg
Since the scale reads 105 kg when both Dan and Sarah are on it, that means the scale reads 5 kilograms too high.
This means Dan`s actual weight is 55kg while Sarah`s actual weight is 45kg. Their actual combined weight is 100kg; however, the scale reads it as being 105kg: 5 kilograms higher than it actually is.
Thus, we can conclude that the scale reads too high, by five kiograms to be precise.
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?
In order to work out this problem, we must work backwards.
When Carla takes her third, she leaves 40 pennies in the jar. When someone removes one third of the pennies, there is always two thirds of the previous amount left in the jar.
Using this logic, we figure out the following:
forty is two thirds of sixty, and sixty is two thirds of ninety which is two thirds of 135.
X = 135
There were 135 pennies in the jar at the start.
Question 19
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 c be the total amount of coffee, m of milk, and n the number of people in the family.
Then each person drinks the same total amount of coffee and milk (8 ounces), so we get
(c/6+m/4)n= c + m
Then we get: 2c(6-n)=3m(n-4), both 6-n and n-4 are only made positive when n=5.
Therefore, we can conclude that Angelas family has at least 5 members.
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?
CLOCK A CLOCK B
FIVE MINUTES PER HOUR FAST VERSUS 5 MINUTES PER HOUR SLOW
With every passing hour, Clock A gains five minutes per hour (1 hour + 5 minutes) while Clock B loses five minutes (1 hour - 5 minutes). This means the time differences between Clock A and Clock B continually increase by 10 minutes each passing hour. Thus, the two clocks will be 1 hour apart when the 10 minute difference is multiplied by six.
10minutes x 6Hours = 60 minutes (time difference)
60minutes=1 hour
Start of Clock A and B: 12:00AM (midnight)
End (when there is a 1 hour difference): 6:00AM
The Time on the Clocks will be different: Clock A= 6:30AM Clock B= 5:30AM
In conclusion, we can say that after 6 hours or at 6:00 AM the two clocks will be one hour apart in time.
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?
Since Sven is said to be exactly in the middle among all the runners, we know that the amount of runners in the race are an odd number, for we cannot have someone exactly in the middle if it were an even amount of runners in the race.
From the question, we have also gathered that Sven is higher than 10th place, for Dan is slower than him.
Lars is in 16th place so we know that there at least 16 runners in the race. Lars cannot be in last place because sixteen is an even number and Sven is exactly in the middle among all the runners.
However, if you add one more runner, making the total to 17 (an even number), we can see that Sven is exactly in the middle (9th place), and that he is faster than Dan (10th place). If we increase the amount of players, then Sven will fall lower among the rankings then Dan who is supposed to be slower than him, so thus, we can say that there are a grand total of 17 runners in the race.
By using the process of guess and check, we can conclude that there are 17 runners 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?
Let y= Number of rainy mornings (and sunny afternoons)
Let x= Number of rainy afternoons (and sunny mornings)
Let z= Number of non-rainy days
Since every rainy afternoon was preceded by a sunny morning (but not necessarily vice-versa) and there were 11 sunny mornings, we can say that x+7=11
Also: y +z=13 ( rainy mornings leading to sunny afternoons - but not vice-versa)
Also: x + y = 13 (number of rainy afternoons + rainy mornings= rainy days)
Solve:
1.) y + x = 13
2.) x = 13 − y Sub 2.) into 3.)
3.) x + z = 11
= 13 − y + z = 11 and y + z = 12
=12 − 11 = y + z − (13 − y + z)
=1 = y + z − 13 + y − z
=1 = 2y − 13 therefore, y = 7
SUB y=7 into 1.) y + x = 13
7 + x = 13 x = 6
y + z = 12 7 + z = 12
z = 5
Solutions: x=6, y=7 and z=5. 6 + 5 + 7 = 18
The length of the vacation was 18 days.
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.
We can represent the ages of all three children with variables a, b, c So, the first clue that is given is that the product of the age of all three children is 36. We look at all the numbers that are multiplied with two other numbers gives us 36.
1x1x36=36
2x18x1=36
3x12x1=36
9x4x1=36
6x6x1=36
3x2x6=36
3x3x4=36
9x2x2=36
The second clue that is given is that they sum up to today’s date. This means that the sum needs to be less that 31 because there is no month that ultra pass that. Automatically 1x1x36 is discarded.
Paul then says that there is not enough information given, meaning that there is something wrong. Two sets that are both equal to 13 are used and the other sets discarded: 6x6x1 and 9x2x2 since their sum is 13.
Later on Paula mentions that the oldest one has red hair. This might be insignificant at first, but we must consider the fact that there actually is an oldest child. Since there is only one older child in the sequence 9,2,2 we suppose that this is the age of the children.
A=9 b=2 c=2
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?
we can clearly see that the first candle is burning twice as fast as the second candle. This means that we can equal the 6hr candle to 2 of the 3hr candle. This leaves us with the rate of 1/6. Since the 3 hour candle equals 2 of the 6hr candle, we have the eqaution (1/6)x = 2, and the same will apply for (1/3)x we can solve the eqatuons.
The candles will be the same hight after 4 hrs.
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.
We have two candles. I first looked at how long both candles lasted for.
I figured out that the 1st candle (L + 1) lasted for 6 hours. The other candle lasted 4 hours.
Henceforth, the first candle lasted 3/2 longer than the second candle.
So, (L + 1) =6 and L=4 we can cross multiply that so we can get the 6L = 4L + 4.
The equation then becomes 2L=4 (subtracting 4L from both sides)
L is then equal to 2