MTH 201 Applied Mathematics Sample Final Exam Questions. 1. The augmented matrix of a system of equations (in two variables) is:

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1 MTH 201 Applied Mathematics Sample Final Exam Questions Selected-Solutions 1. The augmented matrix of a system of equations (in two variables) is: [ ] Which of the following is true about the system of equations? (a) The system of equations has a unique solution. (b) The system of equations has infinite solutions. # (c) The system of equations has no solutions. (d) Not enough information to answer the question 2. Which of the following is the augmented matrix of the system of equations? 2x y 7z = 3 x + 3y z = 0 x z = (a) # (b) (c) (d) Let A = Find the matrix after the sequence of row transformations (2)R R 2 R 2 and ( 3)R 3 + R 2 R 2 on A (a) (b) # (c)

2 (d) Solution: 4. Let A = (a) 7 (b) (c) 12 # (d) 9 (2)R 1 + R 2 R 2 ( 3)R 3 + R 2 R 2 and B = What is the (1, 3) entry of A 2B? Solution: The (1, 3) entry of A is 2, and the (1, 3) entry of B is 5. Thus, the (1, 3) entry of A 2B is ( 2) 2(5) = 2 10 = Let A = and B = What is the (2, 3) entry of AB? (a) 6 # (b) 0 (c) 24 (d) (2, 3) entry of AB does not exist (e) None of the above Solution: The second row of A is [ ], and the third column of B is (2, 3) entry of AB is 0( 3) + 2(3) + ( 6)(0) = = Thus, the 6. The reduced form of the augmented matrix of a system of equations in the variables x, y, and z is Which of the following is a solution of the system of equations? (a) (3 r, 2 r, r), where r is any real number. (b) (3 + r, 2r, r), where r is any real number. # (c) ( 3r, 2r, r), where r is any real number. 2

3 (d) The system of equations has no solution. (e) none of these Solution: The system of equations corresponding to this augmented matrix is: x z = 3, y + 2z = 0. Suppose z is the free variable. Let z = r. Then x = 3 + z = 3 + r, and y = 2z = 2r.Hence, the solutions are of the form (3 + r, 2r, r), where r is any real number. 7. The reduced form of the augmented matrix of a system of equations in the variables x, y, and z is Which of the following is true about the system of equations? (a) A solution of this system of equation is of the form (2 + r, 2 r, 4r), where r is any real number (b) A solution of this system of equation is of the form (2 + r, r, 4), where r is any real number # (c) A solution of this system of equation is of the form (2 + r, 2r, 4r), where r is any real number (d) The system of equations has no solution. (e) None of the above Solution: The system of equations corresponding to this augmented matrix is: x y = 2, z = 4. Suppose y is the free variable. Let y = r. Then x = 2 + y = 2 + r. Hence, the solutions are of the form 8. The graph of an inequality is (2 + r, r, 4), where r is any real number. Find the inequality. (a) 3x + y 3 3

4 (b) x 3y 3 (c) x 3y 3 (d) 3x + y 3 # Solution: The line passes through the points (0, 3) and (1, 0). Thus, the slope of the line is: m = = 3. The y intercept of (0, 3). Thus, the equation of the line is y = 3x + 3 or 3x + y = 3. This implies that the inequality is 3x + y 3 or 3x + y 3. Since (0, 0) is not a solution of the inequality, the inequality is 3x + y Which of the following is the feasible region (set) of the system of inequalities? x + 5y 5 2x 3y 6 x 0 (a) Y X 3 (b) Y X 3 4

5 (c) Y X 3 (d) # Y X The corner points of the feasible region (set) of a linear programming problem are given in the following figure. What is the maximum value of the objective function 3x + 2y in the feasible region? Y 3 A( 2,1) 2 1 B(2, 3) C(2, 2) X (a) 8 # (b) 10 (c) 0 (d) 10 Solution: We have Corner Point Objective Function: 3x + 2y A( 2, 1) 3( 2) + 2(1) = = 8 B(2, 3) 3(2) + 2(3) = = 0 C(2, 2) 3(2) + 2( 2) = 6 4 = 10. 5

6 The maximum value is: A feasible set is described by the following inequalities: x 0, y 0, x y 2, x + 2y 4. Which of the following is a corner point of the feasible region? (a) (1, 1) (b) (0, 2) (c) (2, 1) (d) ( 8 3, ) 2 3 # 12. Suppose that the constraints of a linear programming problem include the inequalities x 0 and y 0. The feasible (solution) set of the linear programming problem is restricted to which of the following quadrants. (a) I quadrant # (b) II quadrant (c) III quadrant (d) IV quadrant 13. A company manufactures two ballpoint pens, silver and gold. The silver requires 1 min in a grinder and 7 min in a bonder. The gold requires 2 min in a grinder and 10 min in a bonder. The grinder can be run no more than 1000 minutes per week and the bonder no more than 5600 minutes per week. The company makes a $4 profit on each silver pen sold and $7 on each gold. How many of each type should be made each week to maximize profits? (a) 0 silver and 500 gold (b) 350 silver and 300 gold (c) 300 silver and 350 gold # (d) 560 silver and 0 gold 14. Suppose that $4, 414 are invested at the simple interest rate of 10%. What is the interest for 5 months? (Round your answer to two decimal places.) (a) $ # (b) $ (c) $ (d) $ Solution: We have P = 4414, r = 0.10 and t = 5/12. The simple interest for 5 months is: I = P rt = 4414(0.1)(5/12) = Find the amount of the monthly payment necessary to amortize the following: Loan amount $35, 000; interest 6% per year compounded monthly; for 8 years. (Round your answer to two decimal places.) (a) $4, (b) $

7 (c) $ # (d) $2, Solution: We have L = 35000, r = 0.06, m = 12, and t = 8. Thus, i = r/m = 0.06/12 = and n = mt = 12(8) = 96. Hence, the monthly payment is Li P = 1 (1 + i) n = 35000(0.005) = ( ) Jason took out a loan, to buy a new plasma TV, for 3 years at the interest rate of 9.6% per year compounded monthly. His monthly payment is $ After making 25 payments, he decided to pay off the remaining loan. What is the amount of his unpaid balance? (Round your answer to two decimal places.) (a) $1, (b) $ # (c) $1, (d) $ Solution: We have R = 57.74, r = 0.096, m = 12, and t = 3. Thus, i = r/m = 0.096/12 = 0.008, n = mt = (12)(3) = 36. Also k = 25, so n k = = 11. The unpaid balance is: [ ] [ ] 1 (1 + i) L (n k) 1 ( ) 11 = R = = i Suppose that $6, 000 are invested at 6.4% interest per year compounded quarterly for 5 years? How much interest is earned at the end of 5 years? (Round your answer to two decimal places.) (a) $8, (b) $2, (c) $2, # (d) $ Solution: We have P = 6000, r = 0.064, m = 4, and t = 5. Thus, i = r/m = 0.064/4 = 0.016, and n = mt = 4(5) = 20. The total accumulated at the end of 5 years is: The interest earned at the end of 5 years is: A = P (1 + i) n = 6000( ) 20 = I = A P = = Justin deposits $50 at the end of every week in an account that pays 7.8% interest per year compounded weekly. How much money is in the account at the end of 5 years? (Round your answer to two decimal places.) (a) $2, (b) $156, (c) $15, (d) $15, # 7

8 Solution: We have R = 50, r = 0.078, m = 52, and t = 5. Thus, i = r/m = 0.078/52, n = mt = (52)(5) = 260. Since the deposit is made at the end of each week, this is ordinary annuity. The total accumulated at the end of 5 years is: [ (1 + i) n ] [( ) ] 1 S = R i = S = = 15, An account pays 9% interest per year compounded semiannually. What is the effective rate? (Round your answer to two decimal places.) (a) 9.38% (b) 9.00% (c) 9.20% # (d) 9.31% Solution: We have r = 0.9 and m = 2. The effective rate is: ( r e = 1 + m) r ( m 1 = re = ) 2 1 = The effective rate as a percentage is: Let p represents a false statement, q represents a true statement, and r represents a false statement. What is the truth value of the statement: (p q) r (p (q r)). (a) True # (b) False (c) Not enough information to evaluate the statement. (d) None of these Solution: (p q) r (p (q r)) = (F T ) F (F (T F )) = T T (F F ) because F T = T, F = T, and T F = F = T (F T ) = T T = T. 21. Use DeMorgan s law to write the negation of the following statement. I did not pay my rent and I went to the football game. (a) I paid my rent and I did not go to the football game. (b) I did not pay my rent and I did not go to the football game. (c) I paid my rent or I did not go to the football game. # (d) I did not pay my rent or I did not go to the football game. 22. What is the contrapositive of the statement: Orange juice contains vitamin C. (a) If it contains vitamin C, then it is not orange juice 8

9 (b) If it does not contain vitamin C, then it is not orange juice. # (c) If it contains vitamin C, then it is orange juice. (d) If it does not contain vitamin C, then it is orange juice 23. Which of the following is the truth table of the statement (p q) p. (a) (b) (c) (d) p q (p q) p T T T T F T F T F F F F p q (p q) p T T F T F T F T F F F F p q (p q) p T T T T F F F T T F F T p q (p q) p T T T T F T F T T F F T (e) None of the above Solution: # p q p q p (p q) p T T T F T T F F F F F T F T T F F F T T 24. Determine whether the following statements are equivalent? (p q) and p q (a) Equivalent (b) Not equivalent # (c) Not enough information to decide (d) None of the above Solution: p q p q (p q) p p q T T T F F F T F F T F F F T T F T T F F T F T F From 4th and the last columns it follows that (p q) and p q are not equivalent. 9

10 25. Determine if the following argument is valid. If Bill is a gambler, then he lives in Las Vegas. If Bill lives in Las Vegas, then Bill has a dog. Bill does not have a dog. Therefore, Bill is not a gambler. (a) The argument is valid. # (b) The argument is not valid. (c) Not enough information to decide. (d) None of these 26. Let X = {math, history, physics, accounting}. Find the number of subsets of X. (a) 8 (b) 4 (c) 15 (d) 16 # Solution: The number of elements in X is 4. Thus, the number of subsets of X is 2 4 = Let A, B, C be sets. Which of the following is the Venn diagram of the set (A B) C. A B A B (a) C (b) C A B A B C C (c) (d) Solution: Note (A B) C = (A B) C. The answer is A. 28. Let A and B be subsets of a universal set U such that n(a) = 50, n(a B) = 9, and n(a B) = 80. What is n(b)? (a) 39 # (b) 37 (c) 41 (d) 30 10

11 Solution: We have n(a B) = n(a) + n(b) n(a B) 80 = 50 + n(b) 9 80 = 41 + n(b) n(b) = = A survey of 150 students was done to find out the language classes they were taking. Let A be the set of students taking Spanish, B be the set of students taking French, and C be the set of students taking Latin. The survey revealed the following information: n(a) = 45; n(b) = 55; n(c) = 40; n(a B) = 12; n(a C) = 15; n(b C) = 23; n(a B C) = 2. How many students were not taking any of these languages? (a) 60 (b) 58 # (c) 68 (d) A die is rolled twice. Find the event that the sum of the numbers rolled is either 4 or 5. (a) {(2, 2), (2, 3)} (b) {(1, 3), (3, 1), (2, 2), (1, 4), (4, 1), (2, 3), (3, 2)} # (c) {(1, 3), (1, 4), (2, 2), (2, 3)} (d) {(2, 2), (3, 1), (3, 2), (4, 1)} Solution: {(1, 3), (3, 1), (2, 2), (1, 4), (4, 1), (2, 3), (3, 2)} 31. Each letter of the word MISSISSIP P I is written on a different card and the cards are placed in a bag. One card is drawn at random. What is the probability that the drawn card shows an I. (a) 1 (b) 4 11 # (c) 1 11 (d) 3 11 Solution: The number of letters in the word MISSISSIP P I is 11. Since there are two I, the probability that the drawn card shows an I is Pr[I] = A bag contains 5 red, 2 yellow, and 4 blue marbles. One marble is drawn at random. Find the odds in favor of drawing a blue marble. (a) 1 to 4 11

12 (b) 4 to 11 (c) 7 to 11 (d) 4 to 7 # Solution: The probability of drawing a blue marble is: Pr[B] = This implies that Pr[B ] = = Thus, the odds in favor of drawing a blue marble is: that is 4 to 7. Pr[B] Pr[B ] = 4/11 7/11 = 4 7, 33. Let E and F be events in a sample space such that Pr[E] = 0.40, Pr[F ] = 0.36, and Pr[E F ] = Find Pr[E F ]. (a) 1.00 (b) 0.76 (c) 0.52 # (d) 0.39 (e) None of the above. Solution: Pr[E F ] = Pr[E] + Pr[F ] Pr[E F ] = = Find the number of ways to select five cards from a deck of 52 cards such that 2 are aces and 3 are face cards. (Only picture cards are face cards.) (a) C(4, 2) C(12, 3) # (b) C(4, 2) + C(12, 3) (c) C(16, 5) (d) C(16, 3) Solution: The number of ways to select 2 aces from 4 aces, and 3 picture cards from 12 picture cards is: C(4, 2) C(12, 3). 35. Find the number of distinct permutations of the letters of the word ENGINEERING. (a) 39, 916, 800 (b) 277, 200 # (c) 138, 600 (d) 69,

13 Solution: In the word ENGINEERING, E occurs 3 times, N occurs 3 times, I occurs 2 times, G occurs 2 times and R occurs 1 time. Hence, the number of distinct permutations are 11! = 277, ! 3! 2! 2! 1! 36. A debate club contains 6 students from Arts and Sciences, 3 students from Business Administration, and 2 from the Law school. To participate in the next debate competition, 3 students are selected at random. Find the probability that 1 student is from Arts and Sciences and 2 are from the Law school. (Round your answer to four decimal places.) (a) (b) # (c) (d) Solution: The number of ways to choose 3 students such that 1 is from Arts and Sciences and 2 are from Law school is C(6, 1) C(2, 2) = 6 1 = 6. The number of ways to select 3 students from 11 students is C(11, 3) = 165. Thus, the probability that 1 student is from Arts and Sciences and 2 are from the Law school is C(6, 1) C(2, 2) C(11, 3) 6 = 165 = At a Humane Society, 30% of the dogs are considered large, 45% are medium, and 25% are small. Some of the dogs have been raised as outside dogs, those that are not allowed inside the owner s house. The rest are allowed to come inside. The percents of animals in each category are summarized on the following tree. Large 0.50 outside in or out 0.45 Medium outside in or out outside Small 0.80 in or out A dog is selected at random. What is the probability that the dog is a small outside dog? 13

14 (a) 0.25 (b) 0.20 (c) 0.05 # (d) 0.25 Solution: Let O-outside dog, and S-small sized dog. We want to find Pr[O S]. Thus, Pr[O S] = Pr[O S] Pr[S] = (0.20)(0.25) = Refer to the tree diagram of Exercise 37. A dog is selected at random. If the dog selected is medium sized, what is the probability that it is an in or out dog? (a) 0.45 (b) 0.60 # (c) 0.27 (d) 0.75 Solution: Let I-in or out dog and M-medium sized dog. We want to find Pr[I M]. Thus, Pr[I M] = Of groundhogs living in a Midwestern state, 70% hibernate all winter. Forty percent are afraid of their own shadows, and 28% both hibernate and fear their own shadows. A groundhog is chosen at random. If the groundhog fears its shadow, what is the probability that it hibernates? (a) 0.40 (b) 0.70 # (c) 0.28 (d) 0.42 Solution: Let H-hibernate all winter and S-afraid of their own shadow. Then We want to find Pr[H S]. Thus, Pr[H] = 0.70, Pr[S] = 0.40, and Pr[H S] = Pr[H S] = Pr[H S] Pr[S] = = Of groundhogs living in a Midwestern state, 70% hibernate all winter. Forty percent are afraid of their own shadows, and 28% both hibernate and fear their own shadows. Which of the following statements is true regarding the events the groundhog hibernates all winter and the groundhog does not fear its shadow.? (a) The events are mutually exclusive. (b) The events are independent. # (c) The events are not independent. (d) The events are pairwise discreet. statements is true. 14

15 Solution: Let H-hibernate all winter and S-afraid of their own shadow. Then Now Pr[H] = 0.70, Pr[S] = 0.40, and Pr[H S] = Pr[H] Pr[S] = (0.70)(0.40) = 0.28 = Pr[H S]. This implies that the events H and S are independent. 41. Of the outstanding bills at a local dentist s offi ce, only 22% of those over 90 days overdue are expected to be paid. Six bills are chosen at random. What is the probability that 2 of these bills will be paid? (a) (b) (c) (d) 0.37 # Solution: This is binomial probability. Here n = 6, p = 0.22, and r = 2. Thus, the probability that 2 of bills out of 6 will be paid is: C(6, 2)(0.22) 2 (0.78) 4 = Find the mean. Round your answer to the nearest tenth. Value Frequency (a) 20.1 # (b) 40.2 (c) 9.8 (d) 10.8 Solution: The mean is Value (x) Frequency(f) xf Σf = 10 Σxf = 201 x = Σxf Σf = = Find the median of the following data: 5, 3, 27, 13, 50, 47, 35 (a) 35 (b) 13 (c) 25 (d) 27 # 15

16 Solution: The numbers in the increasing order are: 3, 5, 13, 27, 35, 47, 50. Since there are 7 number, the median is the fourth element. Hence, the median is Find the total area under the standard normal curve between the z-scores z = 0.60 and z = (a) 2.58 (b) 1.38 (c) # (d) Solution: Pr[0.60 z 1.98] = Pr[z 1.98] Pr[z 60] = = Find the z-score satisfying the following condition: 25.78% area is to the right of z. (a) 0.65 # (b) 0.65 (c) 1.65 (d) 1.65 Solution: Finding the z-score a such that the area under the standard normal curve to the right of a is 25.78% is the same as finding the z-score a such that Now Pr[z a] = Pr[z a] = 1 Pr[z a] = = In the table, is in row labeled 0.6 and column labeled Hence, a = Suppose that a fair coin is tossed 500 times. Use a normal approximation to a binomial distribution to find the following probability: Less than or equal to 230 tosses are heads. (Round your answer to three decimal places.) (a) (b) (c) # (d) Solution: Note that the probability of tossing a head is p = 0.5. Now n = 500. So np = 500 (0.5) = 250 > 5 and n(1 p) = 500 (0.5) = 250 > 5. So a normal approximation to binomial distribution is valid. Now µ = np = 250 and σ = np(1 p) = 500 (0.5) (0.5) =

17 Thus, we want to find Pr[x 230]. Now [ ] Pr[x 230] = Pr z ( ) = Pr [z 1.74] = Suppose that two percents of the MP3 players manufactured are defective. Find the probability that in a shipment of 10, 000 MP3 players exactly 225 are defective. (a) # (b) (c) (d) Solution: Note that the probability of a defective MP3 player is p = Now n = So np = (0.02) = 200 > 5 and n(1 p) = (0.98) = 9800 > 5. So a normal approximation to binomial distribution is valid. Now µ = np = 200 and Thus, we want to find Pr[x = 225]. Now σ = np(1 p) = (0.02) (98) = 196 = 14. Pr[x = 225] = Pr [ ] ( ) z ( ) = Pr [1.75 z 1.82] = Pr [z 1.82] Pr [z 1.75] = =

18 Answers: Form A 1. B 2. A 3. B 4. C 5. A 6. B 7. B 8. D 9. D 10. A 11. D 12. A 13. C 14. A 15. C 16. B 17. C 18. D 19. C 20. A 21. C 22. B 23. C 24. B 25. A 26. D 27. A 28. A 29. B 30. B 31. B 32. D 33. C 18

19 34. A 35. B 36. B 37. C 38. B 39. B 40. B 41. E 42. A 43. D 44. C 45. A 46. C 47. A 19