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Concepts in Time Value of Money

Please refer to the attachment/s for missing diagrams.
Answers are provided for the problems listed below. You may not find the solutions of all problems listed in the attachment/s.

1. Tom borrows $ 6000 for 5 five at 8% interest compounded annually. He repays the loan in a single payment at the end of the 5 year period. How much is his payment?
$7680
$8400
$8820
$9140

2. Tom borrows $ 6000 for 5 years at 8 % per year simple interest. He repays the loan with a single payment at the end of the 5 years. The value of the payment is most nearly
$8250
$8400
$8820
$9250

3. The interest rate is 10 % compounded annually. Ten thousand dollars 3 years from today is equivalent to a value today that is most nearly
$7513
$8140
$8477
$13310

4. Gloria borrows $ 3,000 for three years at 6 % simple interest per year. She pays the principle and interest in a single payment at the end of three years. The amount of interest ($) Gloria pays is most nearly
$500
$540
$565
$590

5. Gloria borrows $ 3,000 for three years at 6 % interest per year, compounded annually. She pays the principle and interest in a single payment at the end of three years. The amount of interest ($) Gloria pays is most nearly
$565
$575
$585
$595

6. Harry borrows $ 10,000 for ten years at 12 % interest. He pays the principle and interest at the end of the ten years. His total payment ($) at the end of ten years is most nearly
$29,000
$31,000
$32,000
$33,000

7. Harry borrows $ 10,000 for ten years at 12 % interest, compounded every six months. He pays the principle and interest at the end of the ten years. His total payment ($) at the end of ten years is most nearly

$29,000
$31,000
$32,000
$33,000

8. Harry borrows $ 10,000 for ten years at 12 % interest, compounded every three months (quarterly). He pays the principle and interest at the end of the ten years. His total payment ($) at the end of ten years is most nearly
$29,000
$31,000
$32,000
$33,000

9.
The value of (F/P, 6%, 6) is most nearly
0.6663
0.7050
1.419
1.501

10.
The value of (P/F, 4%, 9) is most nearly
0.6756
0.7026
0.7084
0.7307

11.
The value of (F/P,11%,10) is most nearly
0.3310
0.3574
2.845
2.940

12. The interest rate is 7%.
$ 4,000

P
The value of P ($) is most nearly
$2,850
$3,050
$3,550
$ 4,110

13. Interest rate = 9 %.
$ 2,000

Q
The value of Q ($) is most nearly
$ 1,500
$ 2,145
$ 2,825
$ 2,930

14. The interest rate is 10 %. I deposit $ 100 today. One year from today I deposit $ 200. How much will I have ($) four years from today?
$ 413
$ 419
$ 425
$ 439

15. At the end of two years, what will be the balance in the local bank?
$ 3,150
$ 3.300
$ 3,306
$ 3,312

16. At the end of two years, what will be the balance in the out of town bank?
$3,150.
$3,300
$3,306.
$3,312

17. How much additional interest will be earned by using the out of town bank?
$ 3
$ 5
$ 6
$12

18. How much is the lump sum payment that Marty owes under the original contract?
$371.
$385.
$394.
$423.

19. How much is the alternate payment proposed?
$371.
$385.
$394.
$423.

20. Which alternative should be selected?
Original Contract
Alternate Payment

21. Consider the following Cash Flow Table
Year Cash Flow
0 -1,000
1 0
2 0
3 - W
4 0
5 0
6 + $ 4,000
Interest rate is 8%. The value ($) of W is most nearly
$ 1,522
$ 1,780
$ 1,915
$ 2,110

22. P = $ 4,000 i = 10 % n = 5 .The value of F ($) is most nearly
$ 5,450.
$ 5,900
$ 6,250
$ 6,450.

23. P = $ 2,000 F = $ 2,800 n = 6 i (%) is most nearly
5.22%
5.76%
5.89%
6.24%

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Solution Preview

1
PV=$6000, i=8%, n=5
FV=PV*(1+i)^n=6000*(1+8%)^5=$8816
Correct option is C. $8820

2
Simple interest=PV*i*n=6000*8%*5=$2400
Total payment at the end of 5 years=6000+2400=$8400
Option B i.e. $8400 is correct

3.
FV=$10000, i=10%, n=3
PV=FV/(1+i)^3=10000/(1+10%)^3=$7513
Option A i.e. $7513 is correct

4.
PV=$3000, i=6%, n=3
S.I.=PV*i*n=3000*6%*3=$540
Option B i.e. $540 is correct

5.
PV=$3000, i=6%, n=3
FV=PV*(1+i)^n=3000*(1+6%)^3=$3573
Interest=FV-PV=3573-3000=$573
Option B i.e. $575 is ...

Solution Summary

There are 23 basic problems in finance. Solution to each problem depicts the methodology to find out the value of desired parameter.

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