Workout 7 Solutions. Peter S. Simon. January 19, 2005
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1 Workout 7 Solutions Peter S. Simon January 19, 2005
2 Problem 1 In 1970, the median age of a man marrying for the first time was 23.2 years. By 1998, the median age rose to 26.7 years. Assuming that the same linear trend continues, what will be the median age for a man marrying for the first time in 2005? Express your answer as a decimal to the nearest tenth.
3 Problem 1 In 1970, the median age of a man marrying for the first time was 23.2 years. By 1998, the median age rose to 26.7 years. Assuming that the same linear trend continues, what will be the median age for a man marrying for the first time in 2005? Express your answer as a decimal to the nearest tenth. Med. Age (1970,23.2) (2005,?) (1998,26.7) Year
4 Problem 1, Continued The median age increased by = 3.5 years over a period of = 28 years. This is a rate of change of 3.5/38 = 1/8 year of median age per year of elapsed time. If the trend continues, then in 2005 (after more years), the median age will have increased by 7 1 = years to a 8 new value of = years.
5 Problem 2 James wants to take summer classes but wants to go to class only between 9 a.m. and noon each day. The table below lists the available courses and class times. What is the maximum number of different courses James can take? Course Days Times (am) Course Days Times (am) Archery MWF 9 9:50 Karate MWF 9 10:20 Archery TTh 9 10:20 Karate TTh 9 10:50 Cartooning MWF 11 11:50 Drama MWF 9 9:50 Diving MTWThF 11 11:50 Physics TTh 8:30 9:50 Ceramics MWF 10 10:50 Math Magic TTh 10:30 11:50
6 Problem 2 James wants to take summer classes but wants to go to class only between 9 a.m. and noon each day. The table below lists the available courses and class times. What is the maximum number of different courses James can take? Course Days Times (am) Course Days Times (am) Archery MWF 9 9:50 Karate MWF 9 10:20 Archery TTh 9 10:20 Karate TTh 9 10:50 Cartooning MWF 11 11:50 Drama MWF 9 9:50 Diving MTWThF 11 11:50 Physics TTh 8:30 9:50 Ceramics MWF 10 10:50 Math Magic TTh 10:30 11:50 James can take at most 3 50-minute classes on Monday, Wednesday, and Friday.
7 Problem 2 James wants to take summer classes but wants to go to class only between 9 a.m. and noon each day. The table below lists the available courses and class times. What is the maximum number of different courses James can take? Course Days Times (am) Course Days Times (am) Archery MWF 9 9:50 Karate MWF 9 10:20 Archery TTh 9 10:20 Karate TTh 9 10:50 Cartooning MWF 11 11:50 Drama MWF 9 9:50 Diving MTWThF 11 11:50 Physics TTh 8:30 9:50 Ceramics MWF 10 10:50 Math Magic TTh 10:30 11:50 James can take at most 3 50-minute classes on Monday, Wednesday, and Friday. He can take at most 2 80-minute classes on Tuesdays and Thursdays. So the maximum number of classes he can take is 5.
8 Problem 3 Lisa flies from San Francisco to New York (2500 mi) in 5.5 hours and then flies from New York to Rome (3500 mi) in 7 hours. What is the average speed for her entire trip?
9 Problem 3 Lisa flies from San Francisco to New York (2500 mi) in 5.5 hours and then flies from New York to Rome (3500 mi) in 7 hours. What is the average speed for her entire trip? Avg. Speed = = = Total Distance Total Time 2500 mi mi 5.5hr+ 7hr 6000 mi 12.5hr = 480 mi/hr
10 Problem 4 Munchies Inc. charges $1.57 for a package containing a dozen of Otto s favorite cookies. Crunchies Inc. charges 60 cents for a package containing five of the same cookies. What is the least amount of money Otto could spend to purchase five dozen cookies?
11 Problem 4 Munchies Inc. charges $1.57 for a package containing a dozen of Otto s favorite cookies. Crunchies Inc. charges 60 cents for a package containing five of the same cookies. What is the least amount of money Otto could spend to purchase five dozen cookies? Since five dozen equals 60, then we could obtain the five dozen cookies by buying only Munchies, in groups of a dozen, or by buying only Crunchies, in units of five. So whichever vendor charges the least per cookie should be selected.
12 Problem 4 Munchies Inc. charges $1.57 for a package containing a dozen of Otto s favorite cookies. Crunchies Inc. charges 60 cents for a package containing five of the same cookies. What is the least amount of money Otto could spend to purchase five dozen cookies? Since five dozen equals 60, then we could obtain the five dozen cookies by buying only Munchies, in groups of a dozen, or by buying only Crunchies, in units of five. So whichever vendor charges the least per cookie should be selected. Munchies charges 157/ cents per cookie. Crunchies charges 60/5 = 12 cents per cookie. Therefore, we should purchase the cookies from Crunchies at a cost of 12 cents/cookie 60 cookies = 720 cents = $7.20
13 Problem 5 Corn costs 99 cents per pound, and beans cost 45 cents per pound. If Shea buys 24 total pounds of corn and beans, and it costs $18.09, how many pounds of corn did Shea buy? Express your answer as a decimal to the nearest tenth.
14 Problem 5 Corn costs 99 cents per pound, and beans cost 45 cents per pound. If Shea buys 24 total pounds of corn and beans, and it costs $18.09, how many pounds of corn did Shea buy? Express your answer as a decimal to the nearest tenth. Let C and B be the number of pounds of corn and beans, respectively. We are told that C + B = 24 99C + 45B = 1809 We want to obtain an equation that involves only C. Since the second equation contains a term 45B, then we multiply the first equation by 45 to obtain 45C 45B = Adding this to the first equation yields (99 45)C+(45 45)B = = 54C = 729 = C = 13.5
15 Problem 6 The product of two positive three-digit palindromes is 436,995. What is their sum?
16 Problem 6 The product of two positive three-digit palindromes is 436,995. What is their sum? A palindrome is a number (or word) that reads the same backwards as forwards. Since the product ends in 5, one of the numbers must end in 5 and the other must end in an odd digit. Since they are palindromes, the number ending in 5 must also begin with 5. Lets try dividing the product by various numbers of the form 5n5:
17 Problem 6, Continued 436, =
18 Problem 6, Continued 436, = , =
19 Problem 6, Continued 436, = , = , =
20 Problem 6, Continued 436, = , = , = , =
21 Problem 6, Continued 436, = , = , = , = , =
22 Problem 6, Continued 436, = , = , = , = , = , =
23 Problem 6, Continued 436, = , = , = , = , = , = , =
24 Problem 6, Continued 436, = , = , = , = , = , = , = , = 747
25 Problem 6, Continued 436, = , = , = , = , = , = , = , = 747 so the sum of the two palindromes is = 1332
26 Problem 7 Take two sheets of 8 1 inch by 11 inch paper. Fold one sheet 2 vertically into fourths to form the sides of a rectangular prism. Fold the other sheet horizontally into fourths to form the sides of a different rectangular prism. How much more volume than the smaller prism does the larger prism have? Express your answer as a decimal to the nearest tenth.
27 Problem 7 Take two sheets of 8 1 inch by 11 inch paper. Fold one sheet 2 vertically into fourths to form the sides of a rectangular prism. Fold the other sheet horizontally into fourths to form the sides of a different rectangular prism. How much more volume than the smaller prism does the larger prism have? Express your answer as a decimal to the nearest tenth. The volume of such a prism or square cylinder is the product of the square base area and the height. So ( ) V 1 = 11 = ( ) 2 11 V 2 = 8.5 = The difference is V 2 V
28 Problem 8 Michael will flip a coin nine times. What is the probability that 2 3 or more of the flips will be heads? Express your answer as a decimal to the nearest thousandth.
29 Problem 8 Michael will flip a coin nine times. What is the probability that 2 3 or more of the flips will be heads? Express your answer as a decimal to the nearest thousandth. The probability of obtaining a head on any one flip of the coin is 1 2, the same as obtaining a tail. So the probability of any particular outcome, say HTHTHTHTT is equal to = }{{ 2 } 9 factors ( ) 9 1 = A successful outcome will result in 6, 7, 8, or 9 heads. We need to count the number of successful outcomes and multiply this by 1/512 to calculate the probability of success.
30 Problem 8, Continued 9 Heads There is only 1 outcome that will result in 9 Heads.
31 Problem 8, Continued 9 Heads There is only 1 outcome that will result in 9 Heads. 8 Heads There are 9 ways to obtain 8 heads, because there are 9 positions that could be occupied by the single tail. Another way to look at this is that we are seeking the number of ways to choose the 8 locations of the tails from the 9 possible slots. This is just the number of combinations of 9 things taken 8 at a time ( 9 choose 8 ) which we write as 9 C 8 or ( 9 8), where ( ) 9 9! 9C 8 = = 8 8! (9 8)! = 9! 8! = 9.
32 Problem 8, Continued 9 Heads There is only 1 outcome that will result in 9 Heads. 8 Heads There are 9 ways to obtain 8 heads, because there are 9 positions that could be occupied by the single tail. Another way to look at this is that we are seeking the number of ways to choose the 8 locations of the tails from the 9 possible slots. This is just the number of combinations of 9 things taken 8 at a time ( 9 choose 8 ) which we write as 9 C 8 or ( 9 8), where ( ) 9 9! 9C 8 = = 8 8! (9 8)! = 9! 8! = 9. 7 Heads The number of outcomes with 7 heads is 9 C 7 = 9! 7! 2! = 9 8 = 36. 2
33 Problem 8, Continued 9 Heads There is only 1 outcome that will result in 9 Heads. 8 Heads There are 9 ways to obtain 8 heads, because there are 9 positions that could be occupied by the single tail. Another way to look at this is that we are seeking the number of ways to choose the 8 locations of the tails from the 9 possible slots. This is just the number of combinations of 9 things taken 8 at a time ( 9 choose 8 ) which we write as 9 C 8 or ( 9 8), where ( ) 9 9! 9C 8 = = 8 8! (9 8)! = 9! 8! = 9. 7 Heads The number of outcomes with 7 heads is 9 C 7 = 9! 7! 2! = 9 8 = Heads The number of outcomes with 6 heads is 9 C 6 = 9! 6! 3! = =
34 Problem 8, Continued So the number N of ways of obtaining 6 or more heads on 9 tosses of the fair coin is N = 9 C C C C 9 = 9! 6! 3! + 9! 7! 2! + 9! 8! 1! + 9! 9! 0! = ! 9! = = 130
35 Problem 8, Continued So the number N of ways of obtaining 6 or more heads on 9 tosses of the fair coin is N = 9 C C C C 9 = 9! 6! 3! + 9! 7! 2! + 9! 8! 1! + 9! 9! 0! = ! 9! = = 130 and the probability of this happening is N/512 or =
36 Problem 9 If 888x + 889y = 890 and 891x + 892y = 893, what is the value of x y?
37 Problem 9 If 888x + 889y = 890 and 891x + 892y = 893, what is the value of x y? If we subtract the first equation from the second equation we get 3x + 3y = 3 = x + y = 1 and if we multiply this last equation by 888 we get 888x + 888y = 888.
38 Problem 9 If 888x + 889y = 890 and 891x + 892y = 893, what is the value of x y? If we subtract the first equation from the second equation we get 3x + 3y = 3 = x + y = 1 and if we multiply this last equation by 888 we get 888x + 888y = 888. Subtracting this result from the first equation yields (888x + 889y = 890) (888x + 888y = 888) y = 2
39 Problem 9 If 888x + 889y = 890 and 891x + 892y = 893, what is the value of x y? If we subtract the first equation from the second equation we get 3x + 3y = 3 = x + y = 1 and if we multiply this last equation by 888 we get 888x + 888y = 888. Subtracting this result from the first equation yields (888x + 889y = 890) (888x + 888y = 888) y = 2 and, since x + y = 1, then x = 1 y = 1 2 = 1 so that x y = 1 2 = 3
40 Problem 10 In how many ways can Harold, Steve, John, Roslyn, Marian and Connie line up so that no two of the three boys are next to each other?
41 Problem 10 In how many ways can Harold, Steve, John, Roslyn, Marian and Connie line up so that no two of the three boys are next to each other? There are four general arrangents of boy/girl to consider:
42 Problem 10 In how many ways can Harold, Steve, John, Roslyn, Marian and Connie line up so that no two of the three boys are next to each other? There are four general arrangents of boy/girl to consider: BGBGBG,
43 Problem 10 In how many ways can Harold, Steve, John, Roslyn, Marian and Connie line up so that no two of the three boys are next to each other? There are four general arrangents of boy/girl to consider: BGBGBG, GBGBGB,
44 Problem 10 In how many ways can Harold, Steve, John, Roslyn, Marian and Connie line up so that no two of the three boys are next to each other? There are four general arrangents of boy/girl to consider: BGBGBG, GBGBGB, BGGBGB, and
45 Problem 10 In how many ways can Harold, Steve, John, Roslyn, Marian and Connie line up so that no two of the three boys are next to each other? There are four general arrangents of boy/girl to consider: BGBGBG, GBGBGB, BGGBGB, and BGBGGB.
46 Problem 10 In how many ways can Harold, Steve, John, Roslyn, Marian and Connie line up so that no two of the three boys are next to each other? There are four general arrangents of boy/girl to consider: BGBGBG, GBGBGB, BGGBGB, and BGBGGB. For any one of these four basic arrangements, the boys can be rearranged (permuted) among themselves in = 6different ways. Similarly, the three girls can be rearranged among themselves in 6 ways. Since the boys and girls can be rearranged independently, the number of ways to create any one of the four basic arrangements is 6 6 = 36.
47 Problem 10 In how many ways can Harold, Steve, John, Roslyn, Marian and Connie line up so that no two of the three boys are next to each other? There are four general arrangents of boy/girl to consider: BGBGBG, GBGBGB, BGGBGB, and BGBGGB. For any one of these four basic arrangements, the boys can be rearranged (permuted) among themselves in = 6different ways. Similarly, the three girls can be rearranged among themselves in 6 ways. Since the boys and girls can be rearranged independently, the number of ways to create any one of the four basic arrangements is 6 6 = 36. So the total number of ways that 3 boys and 3 girls can line up so that no two of the three boys are next to each other is 4 36 = 144.
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