Course:MATH110/Archive/2010-2011/003/Groups/Group 14/Homework 3
1-5
In each each question I considered analyzing what the topic was, what the question was asking for, key components of the question and of course used logic and abstract thinking to come up with possible scenarios and answers. Drawing diagrams worked to help me as well.
1. We were told that a bus travels from a terminal to an airport at 3o miles per hour and it took 1 hour and 20 minutes. The bus then traveled from the airport back to the terminal at the same speed and it took 80 minutes. Considering the speed is the same and considering the time it took to reach both destinations is the same we can conclude that both trips at 30 miles per hour will take 80 minutes.
2. Here a woman did not stop at a stop sign and went down the wrong way of a one way road, a policeman saw her and did not stop her. Why didn't he stop her? I considered all the points of the question and nowhere under any of the key components did it tell us that indeed the women was driving a car. She could have been walking, riding her bike both of which would not alarm the policeman. If she was driving her car you could mention that the policeman was off duty, maybe the woman was his wife, perhaps he was busy with something else.
3. I have 3 boxes all containing different fruit and each box is labeled incorrectly. Can I determine the right label for each box if I can only look in one box? If I check one box and see which fruit is inside, I can take the corresponding label off the other box and now have a choice between two labels, one of which I know is inncorrect, therefore I will place the label I took off initially and place it on the box that is also incorrect and swap that one with the 3rd box label.
4. This question was simply a riddle, "I am the brother of a blind fiddler, but brothers I have none" the blind fiddler is his sister, and that would explain "but brothers I have none".
5. With the coin rolling around the edge of the first, we are asked how many revolutions the coin has made around the other. Analyzing this question I thought that the question did not provide me with how long the coin was rolling for around the other until it returned to its original position. It could have rolled for an extensive period producing many revolutions or a brief period produing very little
6-10 - Kasia Rasker
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?
When drawing apples, to be sure that you get at least two of a kind, you must draw all of the the different types off apples, plus two extra.
After you select one of all of the types of apples, you are sure that the next two you draw will match two of the apples you already have giving you two of one kind.
This can be written as x2+x3+2. x2 represents one of the types of apples, x3 represents the other type of apples and 2 represents 2 of the extra apples you are selecting.
This way by selecting one of each type of apple and on top of that selecting two more apples you are guaranteed to get at least two of one kind.
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) If You were to draw 2 socks, you have the chance of drawing either two blue socks, one blue one brown, or two brown socks
So, by drawing two you have the possibility of drawing a pair of the same color however, the question asks how many must you draw so that you are SURE that you get a pair of the same color
To be sure that you get a pair of the same color you must draw three socks. By drawing three you are guaranteed either a pair of brown socks or a pair of blue socks
(ii)You must draw 41 socks. This way you will ensure that you will have a pair of one color and a pair of mixed colors. If you draw 40 or less socks you have the chance of choosing socks of all one color, because there is 40 socks of each color. If you choose 41 you will be guaranteed one pair of the same color and one pair of different colors.
8. Reuben says, "Two days ago I was 20 years old. Next year I will be 23 years old." Explain how this is possible.
Reuben's birthday is December 31st
If he celebrates his birthday on December 31st, he will have turned 21. Which means if he says this on January 1st, two days earlier he was 20, and therefore will be 22 that year and the year after that will be 23 years old.
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 rope has ten rungs originally, each a foot apart, and the water rises 5 feet from the bottom of the rungs, then five rungs will be showing after the water rises.
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) No, it does not imply that one-forth of all people are women chocolate eaters because if one-half of all people are chocolate eaters and one-half of all people are women there are endless possibilities of which half of the people are chocolate eaters. For example none of the people could be women chocolate eaters. The question does not tell you how many women are chocolate eaters.
(ii) No, it does not follow that one-half of all men are chocolate eaters because if one-half of all people are chocolate eaters and one-half of all people are women, it doesn't imply that even any of the men are chocolate eaters. It does not specify how many men are chocolate eaters.
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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?
This isn't possible, the information is contradictory. According to the information the best and worst player have the same age so this eliminates the woman and her older brother. So let's say the son is the best player, this making the daughter the worst. So the worst players twin( the son) and the best player( the son) have to be the opposite sex. This can't be true because they're the same person and can't be the opposite sex so the statement is untrue.
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.
A logical explanation for him always visiting the Brooklyn girlfriend would be that the trains arrive at the same time on the same track but the Brooklyn train is always infront of the Bronx train therefore he always sees the Brooklynn train first and takes it more often. Another possibility could be that the Brooklynn train always arrives 1 minute before then Bronx train, even though they have 10 minute intervals. EX. 1:19 and 1:29 for the Brooklynn Train and 1:20 and 1:30 for the Bronx train.
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)?
It takes 10 second to strike 10:00 because if we double our previous information given that it takes 5 seconds to reach 5:00, it should then take 10 seconds to reach 10:00. Also if you space out the chimes equally 1 second will equal one chime, so 10 chimes equals 10 seconds.
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. Let's say the names of the babies in the correct order are 1234.
So let's look at the different possibilities of naming two babies correctly and two babies incorrectly.
1243
1432
3214
4231
2134
1324
So we have 6 different ways the two babies could be named correctly and two named incorrectly.
ii. There's no way three babies could be tagged correctly and one baby tagged incorrectly because if you have three babies tagged correctly there would only be one nametag left for the only nameless baby remaining therefore that nametag would belong to the remaining baby.
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 shouldn't accept his bet because there are equal number of red cards as there are black cards in the deck, so a half is red and a half is black.A deck of cards has 52 cards so 26 red and 26 black. So let's say there are 10 red cards in the first half of the deck there would be 16 black cards in that first half as well. So in the second half of the deck there would have to be 16 red cards to get the total count of red cards to 26, and there would be 10 black cards to get the total count of red cards to 26. Therefore there are equal numbers of red cards in the first half of the deck as there are black cards in the second hald of the deck.
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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?
We can start by re-stating the conditions given in the problem:
1. Another way of stating the first condition, "each daughter has the same number of brothers as sisters" could be:
Number of Sons = (Number of Daughters) - 1
OR:
S = D - 1
2. Another way of stating the second condition, "each son has twice as many sisters as he has brothers" could be:
Number of Daughters = 2(Number of Sons - 1)
OR:
D = 2(S-1)
We now have two equations and two unknowns, number of sons (S) and number of daughters (D) which we can solve by plugging one variable into the other equation:
D = 2(S-1) D = 2(D-1-1) D = 2D-2-2 D = 2D-4 D = 4
Therefore, there are 4 daughters in this family.
Now, substitute 4 for D in the other equation to determine the number of sons in the family:
S = D - 1
S = 4 - 1
S = 3
Therefore, there are 3 sons in this family.
We can verify our solutions of 4 daughters and 3 sons by comparing them to the conditions stated in the question:
1. Each daughter has the same number of brothers as she has sisters: Yes, each daughter has 3 sisters and 3 brothers. 2. Each son has twice as many sisters as brothers: Yes, each son has 2 brothers and 4 sisters.
Therefore, our solutions satisfy the conditions.
By re-stating the conditions in this problem in terms of two equations, one for each unknown, we were able to solve the problem by solving first for one unknown, then substituting the solution into the other equation to solve for the second unknown. Because the solutions we arrived at using this method satisfied both of the conditions set forth in the question, we can be confident that the solutions are correct.
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.
If we call Dan's true weight A and we call the error on the scale x, then we can come up with an equation that relates Dan's true weight to the error:
A + x = 60 kg
OR:
Dan's true weight A plus the error x equals the displayed weight on the scale, 60 kg.
Next we can define the equation for Sarah in the same way, if we call Sarah's true weight B:
B + x = 50 kg
Next, we know that:
(A + B)+ x = 105 kg
OR:
Dan's true weight plus Sarah's true weight, plus the error equals 105 kg
Now, we have 3 unknowns, Dan's true weight A, Sarah's true weight B and the error x. So we can solve for x and subsitute:
A + x = 60 kg
A = 60 kg - x
Now we do the same for the equation for Sarah:
B + x = 50 kg B = 50 kg - x
Now we substitute these values for A and B into the last equation ((A + B)+ x = 105 kg) and solve for the error x
((60 kg-x)+(50kg-x)) + x = 105 kg 110 kg -x = 105 kg x = 5
Since the true weights A and B plus the error x equals the weight displayed by the scale, and the error x is positive, the scale weighs five pounds too high. This is a reasonable number in relation to the given information.
By determining equations for the true weights of Dan and Sarah that take into account the unknown error, x, and by determining an equation for their total weight that also takes into account the error x, we were able to determine the error, x.
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?
Another way to state the information given in the problem is, "Alice takes 1/3 of the full jar, Bret takes 1/3 of what remains, and Carla takes 1/3 of what remains of that, leaving 40 pennies in the jar at the end." This means that each person takes fewer pennies than the last.
We can show how much of the total jar each person, Alice, Bret, and Carla, takes in reference to the original amount and to the amount of the original that is left over by the previous person who took from the jar:
Total Original Amount of Pennies = ((1/3 of original amount taken by Alice)(Total Original Amount of Pennies) + (1/3 of what is left taken by Brett)(2/3 because 2/3 of the original amount is left over after Alice had taken 1/3 of the original amount)(Total Original Amount of Pennies) + (1/3 taken by Carla)(2/3 X 2/3 because 2/3 of 2/3 of the original amount is what is left over after Alice takes 1/3 of the original amount and Brett has taken 1/3 of the remaining 2/3)(Total Original amount of pennies)) + 40 because 40 pennies remain after all three people have taken their pennies from the jar.
Since we have determined how much of the penny jar each person has taken, we can write this verbal description as an equation to simplify things:
T = The Total Original Amount of Pennies, Where the Total Original Amount of Pennies = The portion taken by Alice ((1/3)(T)) + The portion taken by Brett ((1/3)(2/3)(T)) + The portion taken by Carla ((1/3)(2/3)(2/3)(T)) + 40 pennies that remain at the end:
T = ((1/3)(T) + (1/3)(2/3)(T) + (1/3)(2/3)(2/3)(T)) + 40
Next, we need to multiply the fractions in the brackets so that we can isolate T and solve for T:
T = (1/3T) + (2/6T) + (4/27T) + 40
Since each of the denominators is a multiple of 27, 27 can be the common denominator:
T = (9/27)T + (6/27)T + (4/27)T + 40
T = (19/27)T + 40
We subtract (19/27)T from both sides:
T-(19/27T) = 40
(T/1) - (19/27)T = 40
We can use the common denominator of 27 to simplify the left side of the equation:
(27/27)T - (19/27)T = 40
(8/27)T = 40
Now we divide both sides by (8/27) to isolate T:
T = 40/(8/27)
To divide 40 by (8/27) we simply multiply 40 by the inverse of (8/27)
T = (40/1)(27/8) = 320/27 T = 135
Therefore, there were 135 pennies in the jar to start with.
As a double-check, we can use this number, 135, to determine how many pennies each of the individuals took, by substituting 135 back into our original equation:
135 = (1/3)(135) + (1/3)(2/3)(135) + (1/3)(2/3)(2/3)(135) + 40
135 = 45 + 30 + 20 + 40
135 = 135
Therefore, from the jar containing 135 pennies, Alice took 45 pennies, Bret took 30 pennies, and Carla took 20 pennies, leaving 40 pennies remaining in the jar at the end.
By re-stating this problem, then defining an equation in words, we were able to determine the total number of pennies the jar started with, given only the number of pennies remaining at the end, and the proportion of pennies each individual took. By using a numerical equation to describe this problem, we were able to double check the result.
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?
First, we can describe Angela's cup numerically:
Angela: 8 = (1/4)M X (1/6)C
OR:
1/4 of Angela's 8 oz. cup is milk and 1/6 of Angela's 8 oz. cup is coffee.
This leaves us with (1-1/4 Milk = 3/4 Milk) and (1-1/6 Coffee = 5/6 Coffee) for the remaining family members (n-1 to take Angela into account), each of which have an 8 oz. cup:
(3/4)M + (5/6)C = (n-1)(8 oz.)
Thirdly, we can define an equation for the number of people there must be in the family:
n = (M + C)/8
OR:
The number of people there must be in the family is equal to the total milk multiplied by the total coffee divided by 8 oz.
Now that we have our three equations, in order to solve the problem, we must find a value for n that satisfies the equations.
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 one hour apart?
The clocks show the same time at midnight, so we can use midnight (0000) as a starting point. Given the above information, let’s determine what times clocks A and B will be displaying at one hour intervals following midnight:
At Actual Time 0000: A: 0000 B: 0000
At Actual Time 0100: A: 0105 B: 0055
At Actual Time 0200: A: 0210 B: 0150
At Actual Time 0300: A: 0315 B: 245
… … …
The table above shows a pattern. At each one hour interval, Clock A displays a time five minutes faster than Actual Time and Clock B displays a time five minutes slower than Actual Time. We want to determine at what time these clocks are one hour apart. Since one hour equals sixty minutes, the question can be re-stated as, “At what time is clock A thirty minutes (sixty minutes divided by two = thirty minutes) fast and clock B thirty minutes slow?”
If we allow x to represent the number of hours past midnight, at how many hours past midnight will clocks A and B be thirty (30) minutes fast and thirty minutes slow, respectively? This formula can be used for both clocks A and B, because they are both five (5) minutes per hour off the actual time. We can answer this question as follows:
5(x) = 30
Solving for x, we get:
5(x)/5 = 30/5 x = 6
Remember, x represents the number of hours past midnight at which clocks A and B will be thirty minutes fast and thirty minutes slow, respectively. Therefore, X = 6 means that at six hours past midnight, or 0600, clocks A and B will be one hour (or sixty minutes) apart.
We can complete the table from above to verify our answer:
At Actual Time 0000: A: 0000 B: 0000
At Actual Time 0100: A: 0105 B: 0055
At Actual Time 0200: A: 0210 B: 0150
At Actual Time 0300: A: 0315 B: 0245
At Actual Time 0400: A: 0420 B: 0340
At Actual Time 0500: A: 0525 B: 0435
At Actual Time 0600: A: 0630 B: 0530
At Actual Time 0600:
(Time Displayed by Clock A) – (Time Displayed by Clock B) = 60 minutes:
(0630) - (0530) = 60 minutes
Therefore, at 0600, or six hours past midnight, Clocks A and B will be 1 hour, or 60 minutes, apart.
By recording the given information in the problem and setting up a table, we are able to come up with an equation to solve a seemingly difficult problem. Using both a table and an equation also allows us to look at the problem in two different ways in order to verify that the solution we arrived at is correct.
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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?
There were 17 runners in the race. We find this by examining the information given to us. Sven placed exactly in the middle of all the racers so therefore the number must be odd. There must be more than 16 people in the race because Lars was in 16th place and 16 isn't and even number. Finally Sven must have placed lower that 10th place because that's what Dan placed. So after examining this information we can start seeing which place between 1-10 that will equal greater than 16 when doubled and it has to be odd. The only number to do this is 9 but final number must be odd so we give 2x9=18-1 which equals 17. So there were 17 runners in the race and Sven placed 9th, which means there were 8 runners before him and 8 runners after him.
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 x be the # of rainy mornings, you can find out this # by taking the # of sunny afternoons, because the question informs you that "when it rained in the morning, the afternoon was sunny".
Let y be the # of rainy afternoons, you can find this number by the # it gives you for sunny days because in the question it informs you that "every rainy afternoon was preceded by a sunny morning".
Let z be the # of days it didn't rain at all.
Now you can come up with equations to represent this information:
One equation: x+z=13, since every rainy morning led to a sunny afternoon, but every sunny afternoon does not entail a rainy morning before it.
Next equation: y+7=11, because in the question it states every rainy afternoon was followed by a sunny morning,however not every sunny morning led to a rainy afternoon.
Last equation: y+x= 13, because the number of days it rained equals the number of rainy mornings and rainy afternoons.
With these equations you can solve for the number of days in the vacation:
y + x = 13 x = 13 − y x + z = 11
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?
After 3 hours candle 1 will be twice as large as candle 2. This is true because after 3 hours candle 2 is burnt out, and candle 1 is at exactly half its length, so therefore candle 1 is twice as large as candle 2.