Engineering Economics Board Exam Reviewer 1 PDF

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Engineering Economics Exam/ Board Exam Reviewer...

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1. A man wishes to accumulate 3722P after 5 years, 8 months and 28 days. How much should be deposited by the man in a bank if the ordinary simple interest is 15% per annum? Solution: 𝐹 = 𝑃 ( 1 + 𝑖𝑛 ) 𝑃=

𝐹 (1+𝑖𝑛)

Where: F = 3,722P i = 15% = 0.15 n = ??? P = ??? Solving for n: 𝑛 = 5 𝑦𝑒𝑎𝑟𝑠 + (8 𝑚𝑜𝑛𝑡ℎ𝑠)

1 𝑦𝑒𝑎𝑟 1 𝑦𝑒𝑎𝑟 + (28 𝑑𝑎𝑦𝑠) 360 𝑑𝑎𝑦𝑠 12 𝑚𝑜𝑛𝑡ℎ𝑠

𝑛 = 5.74 𝑦𝑒𝑎𝑟𝑠 Solving for P: 𝐹 𝑃 = (1+𝑖𝑛) = 𝑷 = 𝟐, 𝟎𝟎𝟎𝑷

3722 1+(0.15)(5.74)

= 2000

2. A man deposited 2000P in a bank at the rate of 15% per annum from March 21,1996 to October 25,1997. Find the exact simple interest. Solution: 𝐼 = 𝑃𝑖𝑛 Where: P = 2000P I = 15% per year = 0.15 n = ??? Solving for n: March 21-31 April 1996 May 1996 June 1996 July 1996 August 1996 September 1996 October 1996 November 1996 December 1996

10 30 31 30 31 31 30 31 31

January 1997 February 1997 March 1997 April 1997 May 1997 June 1997 July 1997 August 1997 September 1997 October 1-25

30

Total: 583 days

𝑛= Solving for I: 𝐼 = 2000 (0.15)(1.6) 𝑰 = 𝟒𝟖𝟎 𝑷

583 = 1.597 ≈ 1.6 𝑦𝑒𝑎𝑟 365

31 28 31 30 31 30 31 31 30 25

3. A bank charges 1.5% per month on a loan. Find the equivalent nominal rate of interest. 𝑗

Solution: 𝑖 = 𝑛 Where: i = 1.5% n = per month, 12 Solving for i: 𝑗 1.5% = 12 𝒋 = 𝟏𝟖% 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅𝒆𝒅 𝒎𝒐𝒏𝒕𝒉𝒍𝒚

4. A financing company charges 1.5% per month on a loan. Find the equivalent effective rate of interest.

5. A nominal rate of 12% compounded monthly is equal to an effective rate of ____. Solution: 𝑖𝑒 = (1 + 𝑖)𝑛1 − 1 Where: 𝑗 = 12% 𝑝𝑒𝑟 𝑚𝑜𝑛𝑡ℎ 𝑛𝑖 = 12 𝑖 =? Solving for i: 𝑗

𝑖 =𝑛 𝑖=

.12 12

𝑖 = 0.01 Solving for ie: 𝑖𝑒 = (1 + 0.01)12 − 1 𝑖 = 0.1268 𝒊𝒆 = 𝟏𝟐. 𝟔𝟖% 𝒑𝒆𝒓 𝒚𝒆𝒂𝒓

6. Convert 16% compounded semi-annually to equivalent nominal rate which is compounded daily.

7. Find the accumulated amount of 1000P after 5 years when deposited in a bank at a rate of 16% compounded monthly. Solution: 𝑖 =

𝑗 𝑛1

𝐹 = 𝑃 ( 1 + 𝑖 )𝑛 Where: j = 16% n1 = 12 P = 1,000P n = 5 years = 60 months i = ??? F = ??? Solving for i: 𝑗

𝑖=𝑛 = 1

16% 12

= 1.33% = 0.0133

Solving for F: 𝐹 = 𝑃 ( 1 + 𝑖 )𝑛 𝐹 = 1000 ( 1 + 0.0133 )60 𝑭 = 𝟐, 𝟐𝟎𝟗. 𝟒𝟒𝑷

8. How long in years will a certain sum of money to triple its amount when deposited at a rate of 12% compounded annually? Solution: F = P (1 + i)n Where: F = 3P i = 12% = 0.12 (compounded annually) Solving for n: 3P = P (1 + 0. 12)n P (1 + 0. 12)n 3P = P P 3 = (1 + 0.12)𝑛 ln (3) 𝑛= ln (1 + 0.12) 𝒏 = 𝟗. 𝟕 𝒚𝒆𝒂𝒓𝒔

9. How much should be deposited in a bank at a rate of 12% compounded continuously for 5 yrs if its accumulated amount is 9110.60P? Solution: F = Pejn Where: j= 12%, 0.12 F= 9110.60P n=5 Solving for P: F = Pejn 9110.60 𝑃𝑒 (0.12)(5) = (0.12)(5) 𝑒 𝑒 (0.12)(5) 𝑷 = 𝟓𝟎𝟎𝟎𝑷

10. An effective rate of interest, which is 12.75%, is equivalent to what percent if compounded continuously.

11. How much is expected to be received by a man that makes a loan of 851.06P, which is payable at once, if the bank gave him a discount of 6%. Solution: 𝐷 = 𝐹𝑑 P = F-D Where: 𝑑 = 0.06 𝐹 = 851.06𝑃 Solving for D: D = 851.06P x 0.06 D = 51.06 Solving for P: P = 851.06P - 51.06P P = 800.00P

12. Find the cash price of a generator which was bought in installment basis that requires a down payment of 50,000P and payment of 30,000P after 1 year, 40,000P after 2 years and a final payment of 76,374.34P after 4 years at a rate of 15% per annum.

13. A man made a loan of 100,000P at a rate of 15% per annum and promise to pay it according to the following manner, 30,000P at the end of 1st yr, unknown payment at the end of 2nd yr and a final payment of 76,374.38P at the end of 4th yr. Find the unknown payment made by the man. 100,000P

i = 15% per year

76,374.38P

Unknown = P

30,000P

Solving for P: 100,000 = 76,374.38 (1.15)−4 + P (1.15)−2 + 30,000 (1.15)−1 P (1.15)−2 100,000 − 76,374.38 (1.15)−4 − 30,000 (1.15)−1 = (1.15)−2 (1.15)−2 −4 100,000 − 76,374.38 (1.15) − 30,000 (1.15)−1 𝑃= (1.15)−2 𝑃 = 39,999.99622P 𝑷 = 𝟒𝟎, 𝟎𝟎𝟎𝑷

14. Find the present worth of the following payments, 5000P after 1 year, 4000P after 2 years, 8000P after 4 years at a rate of 12% per annum.

15. Find the amount of the following payments at the end of 5th yr, 3000P at the end of 1st yr, 4500P at the end of 2nd yr and 6000P at the end of 4th yr if money worth 12% per annum. Solution: F = P(1+i)n Where: 5th yr=? 3000P=1st yr 4500P= 2nd yr 6000P=4th yr Solving for P: 5th yr= 3000(1+0.12)4 + 4500(1+0.12)3 + 6000(1+0.12)1 5th yr = 17762.73 5th yr = 17763P

16. How many yrs will it take for a certain sum of money to triple its amount when deposited at a rate of 12% compounded continuously?

17. A bank is advertising 9.5% accounts that yield 9.84% annually. How often is the interest compounded? Solution: 𝑖𝑒 = (1 +

𝑗 𝑛1

)𝑛1 − 1

Where: 𝑗 = 9.5% 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑𝑒𝑑 𝑛 =? 𝑖𝑒 = 9.84% By ES, Shift solving for ie: 0.0984 = (1 +

0.095 𝑛 ) 1 𝑛1

−1

𝑛1 = 3.88 ≈ 4 Therefore: 𝟗. 𝟓% 𝒊𝒔 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅𝒆𝒅 𝒒𝒖𝒂𝒓𝒕𝒆𝒓𝒍𝒚

18. A man borrows 2000P for 6 years at 8%. At the end of 6 years, he renews the loan for the amount due plus 2000P more for two years at 8%. What is the lump sum due?

19. You deposit 1000P into a 9% account today. At the end of two years, you will deposit another 3000P. In five years, you plan a 4000P purchase. How much is left in the account one year after the purchase? Solution: 𝐹 = 𝑃 ( 1 + 𝑖 )𝑛 Where: P1 = 1,000P i = 9% = 0.09 n1 = 2 years P2 = 3,000P n2 = 5 years P3 = 4,000P n3 = 1 year F = ??? Solving for F: 𝐹 = 𝑃 ( 1 + 𝑖 )𝑛 𝐹 = 1000 ( 1 + 0.09)1 𝐹 = 1188.1 𝑃 = 1188.1 + 𝑃2 𝑃 = 1188.1 + 3000 𝑃 = 4188.1 𝐹 = 4188.1 ( 1 + 0.09)5 𝐹 = 6443.91 𝑃 = 6443.91 − 𝑃3 𝑃 = 6443.91 − 4000 𝑃 = 2443.91 𝐹 = 2443.91 ( 1 + 0.09)1 𝑭 = 𝟐, 𝟔𝟔𝟑. 𝟖𝟔𝑷

20. Consider a business which involves the investment of 100,000P now and 100,000P at the end of one year. Revenue of 150,000P will be generated at the end of years 1 and 2. What is the net present value of this business if the effective annual interest rate is 10%? Solution: 𝑁𝑃𝑉 =

𝐹1 1+𝑖

+

𝐹2 (1+𝑖)2

−𝑃

Where: F1 = 150,000P F2 = 150,000P i = 10% = 0.10 P = 200,000P (100,000P + 100,000P) Solving for NPV: 𝑁𝑃𝑉 = 𝑁𝑃𝑉 =

𝐹1

+

𝐹2

−𝑃

(1+𝑖)2 1+𝑖 150000 150000 + 1+0.10 (1+0.10)2

𝑵𝑷𝑽 = 𝟔𝟎, 𝟑𝟑𝟎. 𝟓𝟖𝑷

− 200000...