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ENGR 345- Summer 2002 The Quiz Book

ENGR 345- Summer 2002 The Quiz Book. Prepared by: Eng. Ahmed Taha. T. F. I believe that my role is a learning facilitator. T. F. Exam dates may be changed if enough students support it. T. F. I assume that my students are only interested in earning getting a good grade. T. F.

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ENGR 345- Summer 2002 The Quiz Book

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  1. ENGR 345- Summer 2002 The Quiz Book Prepared by: Eng. Ahmed Taha

  2. T F I believe that my role is a learning facilitator. T F Exam dates may be changed if enough students support it. T F I assume that my students are only interested in earning getting a good grade. T F Grade changes will only be considered if requested within two classes. T F AUC policy allows for using mobile phones as calculators. T F You are determined to cheat on some aspect of this course. T F You may not leave the classroom unless you get my permission. T F Attendance is optional. T F It is OK to eat or drink as long as I get a piece of the act. T F You are responsible for announcements made in the class only if you attend. ENGR 345- Summer 2002QUIZ 1 • If you deposit $100 today in an account that pays 10% in profit, how much would the account have after: a. One year? b. Two years? • F= P(1+i)n F= 100(1.1)1 = $110 • F= P(1+i)n F= 100(1.1)2 = $121 • How much should you deposit today to get a total of $600 a year later if the interest rate is 20%? • P = F/ (1+i)n P = 600/1.2 = $500 • Please indicate whether each of the following statements is True of False by circling T or F, respectively.

  3. ENGR 345- Summer 2002QUIZ 2 • Construct a cash flow diagram of the following activities on the same bank account: deposit $3000 now, withdraw $1000 2 years from now. 1000 1 2 • Construct a cash flow diagram of the following activities on the same bank account: deposit $3000 now, withdraw $1000 2 years from now. 3000 • In the scenario above, how much do you expect to have in the account 5 years from now if the interest rate is 10%? • F = P (F/P,10%,5) – 1000 (F/P,10% , 3) • F = 3000 (1.6105) – 1000 (1.3310) = $3500.5 • At what interest rate would $100 today become $150.3657 after 5 years? • 150.3657 = 100 (F/P, i , 5)  (F/P, i , 5) = 1.503657 • Using interpolation I F/P • @ n = 5 8%  1.4693 • i  1.503657 • 9%  1.5386 •  i% = 8.53 %

  4. ENGR 345- Summer 2002QUIZ 3 • Show that: • F= A (1+i)n-1 +A (1+i)n-2 +…..+A taking A as a common factor • = A [(1+i)n-1 + (1+i)n-2 + …. + 1] • = A [(1+i)n-1 (1+i) -1/ ((1+i)-1)  • Show that: • How much should you invest each year to get $30,000 after 5 years if the expected annual profit rate is 15%? • F = A (F/A, 15% , 5) • A = 30000/ (F/A, 15% , 5) = 4449.46 • As you know, I went yesterday to renew my car registration. I opted for an option to pay the registration fee for 3 years in one lump sum for the convenience of not having to visit the traffic department every year. If the registration fee is L.E. 200 per year and does not change over time, how much did I pay for this convenience in present terms at a 10% rate? • F1= 600 (F/P,10%, 2) = $726 • F2= 200 (F/A,10%, 2) + 200 (F/P,2,10%) = $662 • The difference using F = $64 • The difference using P = $52.9 0 0 1 2 600 200 200 200 F1 F2

  5. ENGR 345- Summer 2002QUIZ 4 • How long does it take to pay off a loan of $3,000 (received today) when making equal payments of $1000 annually: • starting a year from today? • starting 3 years from today? • (Assume i = 10%, you may use factor tables from the textbook) • 3000 = 1000 (P/A, 10%, n) • (P/A, 10%, n) = 3  n  4 years. • 3000 = 1000 (P/A, 10%, n-2) (P/F, 10%, 2) •  (P/A, 10%, n-2) = 3.63  n-2  5 years  n = 7. 3000 1 n 0 1000 3000 1 2 3 n 0 1000 • Draw a cash flow diagram for an arithmetic gradient with a base equal to the gradient at $200 for n = 5 0 1 2 3 4 5 200 400 600 800

  6. F8 F9 1 4 2000 0 2 8 ENGR 345- Summer 2002QUIZ 5 • A deposit of $2,000 will be made every 2 years, starting after 2 years, in an account that pays an annual interest rate of 12%. How much would this account have: • After 8 years? • After 9 years? (you may use factor tables from the textbook) • Assuming a unit time of 2 years • ia =(1+.012)2 – 1 = 25.44% • F8 = 2000 (F/A,25.44,4) • = 2000 ((1.2544)4 – 1)/0.2544) = $11603.48 • F9 = F8 (F/P,12%,1) = $12995.9 Actual time • What is r, if ia = 0.26247696 and compounding is quarterly? • ia = (1+r/m)m • ln (ia+1) = m ln (1+r/m) • (ln (ia+1))/4 =ln (1+r/4) • Let (ln (ia+1))/4 = X • ex = 1 + r/4 •  r = 24 %

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