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

2. ENGR 345- Summer 2002 QUIZ 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? a)F= P(1+i) n  F= 100(1.1) 1 = $110 b)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. TFI believe that my role is a learning facilitator. TFExam dates may be changed if enough students support it. TFI assume that my students are only interested in earning getting a good grade. TFGrade changes will only be considered if requested within two classes. TFAUC policy allows for using mobile phones as calculators. TFYou are determined to cheat on some aspect of this course. TFYou may not leave the classroom unless you get my permission. TFAttendance is optional. TFIt is OK to eat or drink as long as I get a piece of the act. TFYou are responsible for announcements made in the class only if you attend.

3. ENGR 345- Summer 2002 QUIZ 2 Construct a cash flow diagram of the following activities on the same bank account: deposit $3000 now, withdraw $1000 2 years from now. 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 interpolationI F/P @ n = 58%  1.4693 i  1.503657 9%  1.5386  i% = 8.53 % Construct a cash flow diagram of the following activities on the same bank account: deposit $3000 now, withdraw $1000 2 years from now. 12 3000 1000

4. ENGR 345- Summer 2002 QUIZ 3 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? F 1 = 600 (F/P,10%, 2) = $726 F 2 = 200 (F/A,10%, 2) + 200 (F/P,2,10%) = $662 The difference using F = $64 The difference using P = $52.9 Show that: F= A (1+i) n-1 +A (1+i) n-2 +…..+Ataking 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)  200 600 F1F1 F2F2 0120

5. ENGR 345- Summer 2002 QUIZ 4 Draw a cash flow diagram for an arithmetic gradient with a base equal to the gradient at $200 for n = 5 How long does it take to pay off a loan of $3,000 (received today) when making equal payments of $1000 annually: a.starting a year from today? b.starting 3 years from today? (Assume i = 10%, you may use factor tables from the textbook) a)3000 = 1000 (P/A, 10%, n) (P/A, 10%, n) = 3  n  4 years. b)3000 = 1000 (P/A, 10%, n-2) (P/F, 10%, 2)  (P/A, 10%, n-2) = 3.63  n-2  5 years  n = 7. 0 1n 1000 3000 0 1 n 1000 3000 23 200 400 600 800 012345

6. What is r, if i a = 0.26247696 and compounding is quarterly? i a = (1+r/m) m ln (i a +1) = m ln (1+r/m) (ln (i a +1))/4 =ln (1+r/4) Let (ln (i a +1))/4 = X e x = 1 + r/4  r = 24 % 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 i a =(1+.012) 2 – 1 = 25.44% F 8 = 2000 (F/A,25.44,4) = 2000 ((1.2544) 4 – 1)/0.2544) = $11603.48 F9 = F 8 (F/P,12%,1) = $12995.9 ENGR 345- Summer 2002 QUIZ 5 2000 0 14 28 F8F8 F9F9 Actual time