MATH , Exam 1 Review Spring 2010

Transcription

1 MATH , Exam Review Spring 200 Solutions This review worksheet is longer than the actual exam, and not everything on it will appear on the exam. However, every question on the test will be drawn from this worksheet. This is not a required assignment, and you do not need to turn it in. I. Venn Diagrams and Arguments For problems -5, write each categorical proposition in standard form. Then draw and label a Venn diagram for the given statement. (There is more than one correct answer, but you only need to write down one.). Some philosophers did not live in Greece. Solution: We can restate this in categorical form as Some philosophers are not people who lived in Greece. We can illustrate this statement using the below diagram. A = Philosophers who didn t live in Greece B = People who did live in Greece but were not philosophers C = Philosophers who did live in Greece D = People that are not philosophers and don t live in Greece 2. Herman hates spiders. Solution: We can restate this in categorical form as All spiders are animals Herman hates. We can illustrate this statement using the below diagram. A = Spiders B = Animals that Herman hates C = Animals that Herman doesn t hate

2 Of course, there is more than one correct solution to this problem. Can you think of others?. Cheaters never prosper. Solution: We can restate this in categorical form as All spiders are animals Herman hates. A = Cheaters B = People who prosper C = People who don t prosper but aren t cheaters 4. Some teenagers are responsible drivers. Solution: This statement is already in standard categorical form. A = Teenagers who drive irresponsibly B = Responsible drivers who aren t teens C = Teenagers who drive responsibly D = Irresponsible drivers who aren t teens 5. All widows are women. Solution: This statement is already in standard categorical form. A = Widows B = Women C = People who aren t women For problems 6-0, decide whether an argument is inductive or deductive. If it is inductive, explain why it is strong or why it s not. If the argument is deductive, 2

3 draw the appropriate Venn diagram and decide whether it is valid or sound. 6. Premise: Anne is not in any pain. Premise: Her blood pressure is normal. Conclusion: She is in good health. Solution: Being healthy encompasses more than just not being in pain and having normal blood pressure, so this is an inductive argument because it argues for a general conclusion (Anne s good health) based on the hypothesis that Anne satisfies two specific conditions that are often considered part of being healthy. Since there are a significant number of diseases/health conditions that do not immediately cause pain or negatively affect blood pressure, this argument is relatively weak. 7. Premise: If you swallow food and talk at the same time, you will choke. Premise: Ted is choking. Conclusion: Ted must have swallowed food and talked at the same time. Solution: This problem is essentially the same as example 6 on page 60 of your textbook. The argument is a deductive argument that is not valid and not sound. 8. Premise: All great basketball players must wear expensive shoes. Premise: Shaquille O Neal is a great basketball player. Conclusion: He must wear expensive shoes. Solution: This problem is essentially the same as example 5 on page 60 of your textbook. The argument is a deductive argument that is valid but not sound. 9. Premise: Every winter it has snowed in Utah for all recorded history. Conclusion: It will snow in Utah this coming winter. Solution: This appears to be making an argument for a specific outcome, but notice that without changing the essence of the argument we can expand the conclusion to include the information contained in the premise. So this argument claims that a property (snow) holds for every element in a set (consisting of the coming winter and all recorded winters) based on the hypothesis that the property holds for a subset (consisting of all recorded winters) of that set. Hence, this is an inductive argument. Since the premises support the conclusion fairly convincingly, I would say that this argument is strong. 0. Premise: If you like the book, then you ll love the movie. Premise: You did not like the book.

4 Conclusion: You will not love the movie. Solution: See example 7 on page 6 of your textbook. The argument is a deductive argument that is not valid and not sound. II. Fundamentals of Geometry. If the radius of a tennis ball is one third the radius of a bowling ball and if the bowling ball has a volume of ft, what is the surface area and volume of the tennis ball in square inches and cubic inches, respectively? Solution: Recall that the volume of a sphere with radius r is given by the equation 4π r. Since we know what the volume of the bowling ball is, we can use this formula to calculate the radius of the bowling ball. 4π r = ft r = π ft Divide by 4π. (r ) = ( r = ( π ( 0.9 = π 4π ft ) ) ft. ) ft Raise both sides to the power. Since the radius of the tennis ball is one third the radius of the bowling ball, the radius of the tennis ball is ( ) 0.9 ft = ( ) ( ) in. 0.9 ft = 4 in. π π ft π The surface area of a sphere is equal to 4πr 2, so the surface area of the tennis ball is ( ( ) ) π 4 in in. 2 π Using the equation for volume from before, we have that the volume of the tennis ball is ( ( ) ) 4π in. =.56 in. π 4

5 2. A running track is shaped like a giant doughnut. The distance from a point on the track s outer edge to the center of the region enclosed by the track is mile. The track is comprised of 5 lanes, each 6 feet wide. Jerry runs along the outside of the track, and Gary runs along the inside of the track. They each run a single lap. How many more feet has Jerry run than Gary? Solution: The length that Jerry runs is equal to the perimeter of a circle of radius mile (or 5,280 ft), so Jerry runs 2π(5, 280) ft, ft. On the other hand, the length Gary runs is equal to the perimeter of a circle of radius mile (5)6 ft = 5, 250 ft, so Gary runs 2π(5, 250) ft 2, ft. This means that Jerry runs about, ft 2, ft = ft more than Gary.. For the track described in the previous problem, how much more area does the outside lane have than the inside lane? Solution: The area of the outside lane is and the area of the inside lane is π(5, 280) 2 ft 2 π(5, 274) 2 ft 2 98, 98.2 ft 2, π(5, 256) 2 π(5, 250) 2 98, 0.45 ft 2. Hence, the area of the outside lane is about 98, 98.2 ft 2 98, 0.45 ft 2 = ft 2 more than the that of the inside lane. 4. Find the area and perimeter of a parallelogram with sides of length 5 feet and 27 feet, where the distance between the 27-foot sides is 8 feet. Solution: The perimeter of the parallelogram is 2(5) ft + 2(27) ft = 84 ft. The area of the parallelogram is 27(8) ft 2 = 26 ft Suppose that I hold a can of soup in each hand. The can in my right hand has a height of 4 inches and a radius of.8 inches. The can in my left hand has a height of.6 inches and a radius of. inches. (Assume that the cans are perfect right circular cylinders.) Justify your answers to each of the following questions by doing the appropriate calculations. 5

6 (a) Which can can hold more soup? (b) Which can will need a larger label? (c) Which can needed more metal to make? Solution: (a) To answer this question, we need to calculate the volume of each can. The volume of the right can is and the volume of the left can is π(.8) 2 (4) in in., π(.) 2 (.6) in. 9. in. So the can in my left hand can hold more soup. (b) The area of the label is the surface area of the can minus the area of the top and the bottom. So the right can needs a label with an area of 2π(.8)(4) in in. 2, while the left can needs a label with an area of 2π(.)(.6) in in. 2 So the can in my left hand needs a larger label. (c) We ll assume that the metal used in both cans has the same thickness. Then this question becomes a question of which can has more surface area. The the surface area of the right can is The the surface area of the left can is 2π(.8)(4) + 2π(.8) 2 in in. 2 2π(.)(.6) + 2π(.) 2 in in. 2 So the can in my left hand also needs more metal to make. 6. Suppose you build an architectual model of a new concert hall using a scale factor of 0. (a) How will the height of the actual concert hall compare to the height of the scale model? Explain. 6

7 (b) How will the surface area of the actual concert hall compare to the surface area of the scale model? Explain. (c) How will the volume of the actual concert hall compare to the volume of the scale model? Explain. Solution: Remember that lengths always scale with the scale factor, areas always scale with the square of the scale factor, and volumes always scale with the cube of the scale factor. So the answers are (a) 0, (b) 0 2 = 900, and (c) 0 = A water canal has a rectangular cross section meters wide and 2 meters deep. How much water is contained in a 0-meter length of the canal? How much water does it hold after 60% of the water has evapolated? Solution: The volume of a -meter by 2-meter by 0-meter rectangular prism is (2)(0) m = 80 m. Because 60% of the water is = 08 m, it holds = 72 m after 60% of the water has evapolated. III. Unit Conversions. Convert 2 decimeters into decameters. Solution: Remember that decimeter = 0 meters and decameter = 0 meters. 2 decimeters 0 decimeters decameters 0 = meters So 2 decimeters is.2 decameters decameters =.2 decameters 2. How many joules are in one kilowatt-hour ( watt = joule/second)? Solution: We first notice that joule is equal to watt-second; just multiply top and bottom of the hint by second. Now, convert kilowatt-hours into kilowattseconds kilowatt- hour 60 minutes seconds hour 60 = 60(60) kilowatt-seconds = 600 kilowatt-seconds minutes Next, convert kilowatt-seconds into watt-seconds 600 kilowatt-seconds watts 000 kilowatt = 600(000) watt-seconds =.6 06 watt-seconds Since watt-seconds is equivalent to joules, there are joules in kilowatthour. 7

8 . Convert 25 miles per hour into meters per second ( mile =.6 kilometers). Solution: First, convert miles per hour into meters per hour (remember that meter =,000 kilometers). 25 miles hour.6 kilometer miles 000 meters kilometer = 25(.6)000meters hour Second, convert meters per hour into meters per second. 40, 000 meters hour hour 60 minutes minute 60 seconds So 25 miles per hour is about meters per second. 4. Convert 8 grams per milliliter into grams per 00 milliters. Solution: We have grams 8 millimeter = 8 grams millimeter = 40, 000 meters hour 40, 000 meters = 60(60) second.meters second = 800 grams 00 milligrams In other words, 8 grams per millimeter is 800 grams per 00 millimeters. 5. Suppose you bought an imported Italian car but are concerned about its gas mileage. When you fill up its 5.8 gallon gas tank, the odometer reads 500 kilometers, and when you need to refuel, the odometer reads 2,0.26 kilometers. What is your car s gas mileage, in miles per gallon? Remember that there are miles in one kilometer. Solution: Your car can drive 2, =60.26 kilometers before refueling. Converting this to miles gives kilometers miles kilometer = miles So your car can travel miles per 5.8 gallons of gas. That is, miles 5.8 gallons = miles gallons. This means that your car gets about 24 miles per gallon. 6. Scientists come through for once and discover that it takes 47 licks to get to the center of a Tootsie roll pop. If you lick at the rate of 5 licks per minute and your 8

9 friend licks at the rate of 79 licks per minute, how many more seconds will it take you to finish your lollipop than your friend? Solution: First, let s calculate how many minutes it takes you to get to the center of a Tootsie roll pop. 47 licks minute 5 licks = 47 5 minutes = minutes Second, let s calculate how many minutes it takes your friend to get to the center of a Tootsie roll pop. 47 licks minute 79 licks = minutes = minutes This means that it takes you = more minutes than your friend. In seconds, this is minutes 60 seconds minutes = (60) seconds = seconds So it takes you more seconds to get to the center of a Tootsie roll pop. 7. A new coal-burning power plant can generate.5 gigawatts (billion watts) of power. Burning kilogram of coal yields about 450 kilowatt-hours of energy. (a) How much energy, in kilowatt-hours, can the plant generate each month? (Assume that there are 0 days in a month.) Solution: In kilowatts,.5 gigawatts (i.e.,.5 billion= watts) is watts In hours, a 0-day month is kilowatt 0 watts =.5(09 ) kilowatts = kilowatts 0 0 days 24 hours day = 0(24) hours = 720 hours So the energy the plant generate each month is kilowatts 720 hours = kilowatt-hours (that is, 080 million kilowatt-hours). (b) How much coal, in kilograms, is needed by this power plant each month to generate this much energy? 9

10 Solution: Based on our solution to part (a), we can can rephrase the question as follows: How much coal in kilograms needs to be burned to give us kilowatt-hours? kilowatt-hours kilograms of coal = kilograms of coal 450 kilowatt-hours 450 = kilograms of coal So 2.4 million kilograms of coal are needed. (c) If a typical home uses 000 kilowatt-hours per month, how many homes can the power plant supply? Solution: We have kilowatt-hours home = 000 kilowatt-hours 000 So the power plant can supply,080,000 homes each month. homes = homes 8. China has a population of,2,85,888 people (July 2007 estimate). If the birth rate is.45 births per,000 population per year and the death rate is 7 deaths per,000 population per year, what will China s population be next year? Solution: First, let s calculate how many people will die this year., 2, 85, 888 people 7 deaths 000 people =, 2, 85, 888(7) 000 deaths 9, 252, 96 deaths Second, let s calculate how many people will be born this year., 2, 85, 888 people births, 2, 85, 888(.45).45 = births 7, 778, 908 births 000 people 000 So China will have about, 2, 85, 888 9, 252, 96+7, 778, 908 =, 0, 77, 8 people next year. 9. Suppose you own an aquarium 6 in. long, 5 in. wide, and 8 in. high. How many gallons of water can it hold? (Hint: There are 7.48 gallons in one cubic foot.) Solution: The volume of the aquarium is 6 in.(5 in.)(8 in.) = 9, 720 in. in. in. = 9, 720 in.. In gallons, this is 9, 720 in. ft in gallons = ft 0 9, 720(7.48) 2 gallons 42 gallons

11 So the aquarium can hold about 42 gallons of water. 0. You re a paramedic behind the wheel of an ambulance and need to drive to the scene of an accident 4 blocks away. If you need to get there within 5 minutes, how fast do you need to drive, in miles per hour? (Assume that there are 2 blocks in one mile.) Solution: First, convert 4 blocks into miles. IV. Percentages 4 blocks mile 2 blocks = 4 2 Second, convert 5 minutes into hours. 5 minutes hour 60 minutes = 5 60 miles = 2.8 miles hours =.25 hours Finally, divide the number of blocks in miles by the number of minutes in hours. 2.8 miles.25 hours =. miles hour So the we need to drive faster than about miles per hour. (This seems fairly manageable.) Suppose a typical McDonald s meal consists of a Big Mac burger, medium fries, and a medium chocolate shake. Use the table provided below for problems -. Food Calories Fat (in grams) Big Mac 56 Fries Shake 77 2

12 . A typical McDonald s meal contributes what percentage of my recommended daily Calorie intake. Assume that my recommended daily Calorie intake is 200 Calories. (This figure is not unusual.) Solution: The total amount of Calories in a McDonald s meal is =, 78 Cal. So a McDonald s meal contributes of my recommended Calorie intake., 78 00% = % 75% 2, Suppose I order a typical McDonald s meal except that I ask for the Big Mac sauce to be left off my Big Mac. In this case, my Big Mac (without its signature sauce) contributes about 6% of the fat in my meal. So how much fat does a Big Mac with the sauce left off have? Solution: Let x be the amount of fat in a Big Mac (without the sauce). Then we know that x = 6% of the fat in my meal =.6(the amount of fat in my fries, shake, and Big Mac) =.6( x) =.6(4) +.6x Subtracting.6x from both sides gives x.6x =.6(4).64x =.6(4) Divide both sides by.64: x =.6(4) 2.64 So a Big Mac without the sauce contains about 2 grams of fat.. If a Big Mac contains 28.7% less Calories than a Whopper (with cheese) from Burger King, how many Calories does a Whopper contain? Solution: Let x be the Calories in a Whopper. Then we have that Calories in a Big Mac = 56 = x 28.7 % of x = x.287x =.7x Dividing by.7 gives us that x = 56/.7 = Cal are in a Whopper (with cheese). 2

13 Use the table provided below for problems 4-6. Gender Elementary and middle school teachers Secondary teachers Men 562, 499,098 Women 2,80, , What percentage of all secondary teachers are female? Solution: There are 499, , 902 =, 58, 000 total secondary teachers. So the percentage of these teachers who are female is given by 658, % = 56.9%., 58, If 56% of secondary teachers were female the previous year, by what percent did the number of female secondary teachers change (assuming that the total number of secondary teachers remained the same both years)? Solution: Let x be the number of secondary teachers (for each year). It is important to notice that although we know the number of female secondary teachers, we don t need that information to solve this problem. It is also important to notice that we are not solving for x here; we re just using it as a temporary placeholder. Then our reference value is 56% of x (.56x), and our new value is, by the previous problem, 56.9% of x (.569x). So the absolute difference is.569x.56x =.009x, and the relative difference is.009x.56x 00% = % So the number of female secondary teachers increased by about 2%. 6. How would you answer the previous problem, if you were told that the total number of secondary teachers actually increased by 5.5% from the previous year to this year? Solution: Again let x be the number of secondary teachers (for the year before). Then our reference value is 56% of x (.56x), and our new value is 56.9% of 05.5% of x (.569(.055)x= x). (Note that a 5.5% increase means that the new number is 05.5% of the original value.) So the absolute difference is x.56x = x,

14 and the relative difference is x.56x 00% = % So in this case, the number of female secondary teachers increased by about 7%. 7. Last year, oil paintings comprised 7% of a certain art gallery s collection. This year, oil paintings comprise 70% of the gallery s collection, but the total number of the gallery s art pieces is % more than the number last year. By what percent has the number of oil paintings changed since last year? Solution: Let T be the total number of pieces in the art gallery last year, so the total number of oil paintings last year was.7t. Since the total number of the gallery s art pieces is % more than the number last year, the total number of art pieces this year is.0t. So the number of oil paintings this year is 70% of the art pieces this year =.70(.0(T )) This means that the percent by which the number of oil paintings has changed since last year is new number of oil paintings - old number of oil paintings old number of oil paintings That is,.70(.0(t )).7T 00%.7T We can cancel out the T s and simplify to get our answer:.70(.0) %.2% 00% 8. If a 4-ounce can of olives is 8% more money than a 2-once can of olives, which is a better deal? Explain. Solution: The larger can of olives has 4 2 = 22 ounces more than the smaller can of olives, so the 4-ounce can contains 22 00% 8.% more ounces 2 of olives than the 2-ounce can; this means that per ounce, the 4-ounce can is about the same price as the 2-ounce can. So neither can is really a better deal than the other in terms of price per ounce. 4

15 9. If you have a 20% off coupon on an item that is already 40% off, my friend thinks this means the item is 60% off. Is he correct? Explain. Solution:No. Since the item is 40% off, its price is 00% 40% = 60% of its old value. Careful consideration gives new price = 60% of old price - 20% of 60% of old price =.6(old price).2(.6)(old price) = (.6.2.6)(old price) = 0.48(old price) = 48% of the old price So the item s new price is 48% of its old price. In other words, the item is now 00% 48% = 52% off. If this is difficult to see, try calculating the amount off with a specific price (like $00). 0. I have an informant working in a major department store. She tells me that about a month before the store s annual 25% off sale the store increases prices by 0%. Should I buy now or during the sale? What percentage will I save by choosing whichever is the cheaper option? Solution: Let p be the original price of an item. If the store increases prices by 0%, then the new price is 0% of the original price. This means that the new price is.p. If the store then decreases this new price by 25%, the resulting price is (.25).p =.75(.)p = 0.975p. Because we re multipying p by a number less than, this means that the item during the sale is cheaper than it is right now. The problem is asking us to find how much less the cheaper option is than the other option, so in this case, we use the formula for relative difference to get.0975p p p 00% = 2.5%. This means that if I buy during the sale, I will save 2.5%. (In other words, the price of the item on sale is 2.5% less than the original price of the item.). Profits at the Sleeping Inn have fluctuated greatly over the last few years. Three years ago profits increased by 6%, two years ago profits decreased by %, and last year profits increased by 9%. By what percent did profits increase over the last three years combined? Solution: This example was done during the lecture. 5