Course:MATH110/Archive/2010-2011/003/Homework/3/Solutions

Here are solutions for this problem set. We've chosen the answers that seemed to us the most complete and helpful.

Problem 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.

Solution by Group 1

There is no time difference! It is how the time has been written! One hour consists of 60 minutes. and When we add 20 minutes to this it adds up to 80 minutes bus drive. And this is the same exact amount that the driver needed for returning to the terminal

Problem 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.

Solution by Group 2

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.

Solution by Group 15

Since the policeman never had the opportunity to check the driver’s license, he wouldn’t have known that she did not have it on her. Therefore, that mistake was irrelevant to the question why the policeman didn’t stop her at her antics. There can be a few possibilities why the policeman didn’t stop the driver despite seeing her apparently breaking road etiquettes (failing to stop at a stop sign and going down a one-way street the wrong way). The policeman could have been preoccupied with another road incident and therefore didn’t manage to stop the female driver in time. Or it could be that execution of road policies was not usually strictly carried out in that vicinity (or city, or country). Or it could be that the woman was not driving at all- there isn’t any indication in the question that the woman was in fact conducting a vehicle when she did all those things.

Reading the question “superficially”, as demonstrated by the solutions we have seen above, it is very easy to automatically assume that the woman broke the road codes while she was driving. We expect to be given the context in the beginning of any paragraph of texts. In this example, the first sentence seems to set the scene for the readers, “The lady driver did not have her driver’s license with her…”, to be immediately followed by a series of driving behaviours that we all know are breaking the road code. We link the lady and her driver’s license to the assumption that “she must be behind the wheel”. However, as one should be able to see, that piece of information was there to mislead instead of inform. I found that to understand the problem (Polya’s Step 1) really useful in thinking up solutions for this problem. Specifically it helped me to re-read the question carefully and especially not taking anything in it for granted. As we all know, a pedestrian who fails to stop at a stop sign (presumably those signs designed for vehicles), and “going” (walking) down a one way street the wrong way (wrong only for vehicles) does not get stopped by the police!

Problem 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.

Solution by Group 2

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.

Problem 4

I am the brother of the blind fiddler, but brothers I have none. How can this be?

Solution by Group 2

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.

Problem 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?

Solution by Group 2

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.

Solution by Group 13

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?

Just follow the diagram and you will see that the coin is revolved twice when it returns to its original position.

Problem 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?

Solution by Group 12

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.

Problem 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?

Solution by Group 2

i) In order to be sure you have a pair of the same colour you must take 3 socks from the drawer. If you only pick two socks then it is possible that they are different. However if you pick 3, at least 2 of them must be different. ii.) To be sure (as in guaranteed) 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.

Problem 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.

Solution by Group 2

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.

Problem 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?

Solution by Group 12

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.

Solution by Group 10

Alternatively: if the ladder is hung horizontally then all the rungs would show as well

Solution by Group 15

It is easy to be distracted by the presence of actual figures in the problem. It would have been the easiest route to solve the problem – one has the figures, one simply needs to do some mathematical calculations. However, not only does that approach miss the point of the question totally, it also lacks creativity on the part of the problem-solver. Step 2 of the Polya method emphasises that creativity is one of the most important element (even before organisation and experience) one should adopt while planning for a strategy to tackle a problem.

Problem 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.

Solution by Group 6

Understand: All people suggest a population of people. This can be interpreted as any group. Suddenly, half of all people are women. This forces the population to be half women and half ~women (not women). Half of all people are chocolate eaters. Half of the population (composed of half women, half ~women) are chocolate eaters. I picture this by imagining two different coloured circles (red and blue), one superimposed over the other. Each circle is bisected into 2 equal parts. The red circle represents women (shaded) and ~women (not shaded). The blue circle represents chocolate eaters (shaded) and ~chocolate eaters (not shaded). Notice that the circles, when one is rotated, the colour mix moves with it - the circles are not locked together; they do not necessarily spin together. 1. Does it follow that 1/4 of all women are chocolate eaters? No. It doesn't. This problem suggests no linkage between (chocolate eaters, ~chocolate eaters) and (women, ~women). 2. Does it follow that 1/2 of all men are chocolate eaters? No. Same as above.

Problem 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?

Solution by Group 2

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.

Problem 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.

Solution by Group 2

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

Problem 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)?

Solution by Group 1

Although it seems that it would just take 10 seconds for the clock to strike ten, simply double, this cannot be right because between 5 chimes, there is only 4 intervals of time so, letting C=chimes and I=intervals, it can be said that 5C+4I=5 and then the formula for the second one would be 10C+9I=x. So, each interval for the 5 chimes is equal to 5/4. Since there is 9 intervals when the clock strikes 10, you would have 5/4*9 which equals 11.25 therefore, it takes 11.25 seconds for the clock to strike ten.

Problem 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 6 ways. How do you know? It's a simple combinatorics problem: we do not care about the order, so we use combinatorics. We have $4 \choose 2$ $={\frac {4!}{2!2!}}=6$

Solution by Group 1

ii)There is no way that three of the four babies can be directly tagged, since if three are correctly tagged, the last must also be correctly tagged

Problem 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?

Solution by Group 1

No, you should not accept his bet. 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 colors, red and black, the number of red and black cards will be mirrored oppositely. Example: if Alex splits the deck of cards, and we count what we have in the first half, say 20 black cards and 6 red cards, we know without looking that there are going to be 20 red and 6 black in the other half, simply because there are only 2 colors and 26+26=52

Problem 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?

Solution by Group 1

First we must translate this information into an equation for the daughters and sons. Let S= sisters and let B= brothers then our equation for the daughters is: S-1=B, and for the sons is: S=2(B-1) Next we solve for B by substituting the information we have that S=B+1: B+1=2(B-1), 1=2B-2-B, 3=2B-B, B=3 therefore by substituting B=3 into S-1=B we get: S-1=3 so S=4. We can then see that there are 4 sisters and 3 brothers.

Problem 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.

Solution by Group 1

Let D= Dan's weight, let S= Sarah's weight and let x= the amount the scale is off by. Then our equations will be D+x=60, S+x=50, and D+S+x=105. Then we can do some simple algebra and substitution to get D=60-x, S=50-x, and (60-x)+(50-x)+x=105. Finally, we can solve for x: -2x+x=105-60-50, -x=-5, x=5. So, the scale is adding 5 kilograms.

Problem 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?

Solution by Group 10

In this situation, Alice leaves two thirds of a whole, Bret leaves two thirds of two thirds (4/9), and Carla leaves two thirds of four ninths (16/81). Adding the fractions of the total amount left in the jar, we find that it equals 8/27. Now we need to determine what value of x satisfies the equation that (8/27)x = 40 This can be done by dividing 40 by 8/27. The value for this is 135. Therefore, the jar started with 135 pennies originally.

Problem 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?

Solution by Group 2

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.

Problem 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?

Solution by Group 1

Let the gap between them is increased by 10 minutes each hour. 60(minutes in an hour)/10(minutes clocks move apart)=6 so, after 6 hours the clocks will be an hour apart. Therefore, the clocks will be 6 hours apart at 6 am.

Problem 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?

Solution by Group 1

Sven is the median of the sequence. Dan is the 10th and Lars is the 16th, so there must be at least 16 runners in order to have a 16th placement. Since 16 is an even number the isn't an exact median in the sequence. So 17, the next number would be reasonable. The median would be 9. Sven is placed exactly the 9th, which is the middle among all 17 runners, faster than Dan and Lars.

Problem 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?

Solution by Group 2

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.

Problem 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.

Solution by Group 2

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. Hence a=9, b=2, c=2.

Problem 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?

Solution by Group 13

Both candles were at equal length when they were lit, given by L. With one burning out after 6 hours and the other after 3 hours. From this we can see a progression of the first candle having the equation:

$L=((6-t)/6)$ with t being the number of hours elapsed since the candle was lit

The second candle's equation is as follows:

$L=((3-t)/3)$ with t again being the number of hours after the candle was lit

Following the values given by substituting t with increasing multiples of 1 we achieve the criteria in the question after 2 hours:

$((6-1)/6)=5/6$ <----> $((3-1)/6)=2/3$

$((6-2)/6)=4/6$ <----> $((3-2)/6)=1/3$

$4/6=2/3$ $2/3=2(1/3)$ with 1/3 being the length of the second candle.

Thus giving the answer as after 2 hours.

Problem 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 $L$.

TA's Solution

Let candle A be the one with length L+1, Let candle B be the one with length L. We have for the rates $r_{a}={\frac {L+1}{6}},r_{b}={\frac {L}{4}}$

At 8:30, candle A has been burning for 4 hours, candle B has been burning for 2.5 hours. We know that:

rate*time passed=length burned, and total length-length burned = length left.

So if they have the same length left at 8:30, we get $L+1-{\frac {L+1}{6}}*4=L-{\frac {L}{4}}*2.5$.

Subtract L from both sides and change the decimals into fractions, we get

$1-{\frac {2L+2}{3}}={\frac {1-2L}{3}}=-{\frac {5L}{8}}$.

Add ${\frac {2L}{3}}$ to both sides, we get

${\frac {1}{3}}={\frac {2L}{3}}-{\frac {5L}{9}}={\frac {L}{24}}$,

multiply 24 to both sides, we get L = 8.