Basics Of Engineering Economy
2nd Edition
ISBN: 9780073376356
Author: Leland Blank, Anthony Tarquin
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Question
Chapter 1, Problem 42P
a.
To determine
The loan amount with more interest rate.
b.
To determine
The difference in interest paid between the two loans.
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Q1: A sum of $2000 borrowed at 16% compounded daily for 2 years,
state
(a) The nominal rate of interest (G);
(b) The number of compounding periods per year (m);
(c) The periodic rate of interest (i);
Find the effective bimonthly interest rate equivalent to:
(a) nominal annual interest of 9%, compounded 6 times per year;
(b) nominal annual discount of 6%, compounded quarterly;
(c) 1/2 nominal annual interest of 8%, compounded continuously.
The only time you change the original cash flow diagram in problems involving uniform series cash flows is when the: (a) payment period is longer than the compounding period (b) payment period is equal to the compounding period (c) payment period is shorter than the compounding period (d) stated interest rate is a nominal interest rate
Chapter 1 Solutions
Basics Of Engineering Economy
Ch. 1 - Prob. 1PCh. 1 - Which of the following would be considered...Ch. 1 - Prob. 3PCh. 1 - Prob. 4PCh. 1 - Prob. 5PCh. 1 - Prob. 6PCh. 1 - Prob. 7PCh. 1 - Prob. 8PCh. 1 - Prob. 9PCh. 1 - What is meant by the term time value of money?
Ch. 1 - Prob. 11PCh. 1 - Of the fundamental dimensions length, mass, time,...Ch. 1 - Prob. 13PCh. 1 - When an interest rate statement does not include a...Ch. 1 - Prob. 15PCh. 1 - Prob. 16PCh. 1 - Prob. 17PCh. 1 - Prob. 18PCh. 1 - Prob. 19PCh. 1 - Prob. 20PCh. 1 - Prob. 21PCh. 1 - Prob. 22PCh. 1 - Prob. 23PCh. 1 - Prob. 24PCh. 1 - Prob. 25PCh. 1 - Prob. 26PCh. 1 - Prob. 27PCh. 1 - Prob. 28PCh. 1 - How many years does it take for an investment of...Ch. 1 - Prob. 30PCh. 1 - Prob. 31PCh. 1 - Prob. 32PCh. 1 - Prob. 33PCh. 1 - Prob. 34PCh. 1 - Prob. 35PCh. 1 - Prob. 36PCh. 1 - Prob. 37PCh. 1 - Prob. 38PCh. 1 - Prob. 39PCh. 1 - Prob. 40PCh. 1 - Prob. 41PCh. 1 - Prob. 42PCh. 1 - Prob. 43PCh. 1 - Prob. 44PCh. 1 - At 9% per year simple interest, 1000 is equivalent...Ch. 1 - Prob. 46PCh. 1 - Prob. 47PCh. 1 - Prob. 48PCh. 1 - Prob. 49PCh. 1 - Prob. 50PCh. 1 - Prob. 51PCh. 1 - Prob. 52PCh. 1 - Prob. 53PCh. 1 - Prob. 54PCh. 1 - Prob. 55PCh. 1 - Prob. 56PCh. 1 - Prob. 57PCh. 1 - Prob. 58PCh. 1 - Prob. 59PCh. 1 - Prob. 60PCh. 1 - Prob. 61PCh. 1 - Prob. 62PCh. 1 - Prob. 63PCh. 1 - Prob. 64PCh. 1 - Prob. 65PCh. 1 - Prob. 66PCh. 1 - Prob. 67PCh. 1 - Prob. 69APQCh. 1 - Prob. 70APQCh. 1 - Prob. 71APQCh. 1 - Prob. 72APQCh. 1 - An investment of 8,000 nine years ago has...Ch. 1 - Prob. 74APQCh. 1 - Prob. 75APQCh. 1 - Prob. 76APQCh. 1 - Prob. 77APQ
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- A problem on the midterm exam asked you to calculate the monthly payment on a $15,000 car loan at an interest rate of 9% compounded monthly to be repaid over 60 months, where 9% is the nominal annual interest rate. Some students incorrectly used 9% as the monthly interest rate. A monthly interest rate of 9% is equivalent to what effective annual interest rate? (a) Effective annual interest rate = 9% (b) Effective annual interest rate = 9% per month x 12 months per year = 108% (c) Effective annual interest rate = 9% per month ÷ 12 months per year 0.75% (d) Effective annual interest rate = (1+ effective monthly rate)12 – 1 = (1.09)12 – 1 = 181%arrow_forwardUsing linear interpolation, determine the value of (a) (F/A, 11%, 15) and (b) (F/P, 16%, 25) from the compound interest tables. Compute this same value using the equation or a 5-BUTTON SOLUTION. Why do the values differ?arrow_forwardEXAMPLE 1.4 / HP borrowed money to do rapid prototyping for a new ruggedized computer that targets desert oilfield conditions. The loan is $1 million for 3 years at 5% per year simple interest. How much money will HP repay at the end of 3 years? Tabulate the results in $1000 units. EXAMPLE 1.5 / If HP borrows $1,000,000 from a different source at 5% per year compound interest, compute the total amount due after 3 years. Compare the results of this and the previous example. Comment : The difference between simple and compound interest grows signif- icantly each year. If the computations are continued for more years, for exam- ple, 10 years, the difference is $128.894; after 20 years compound interest is $653.298 more than simple interest.arrow_forward
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