Chapter001

# Chapter001 - The Theory of Interest Solutions Manual Chapter 1 1(a Applying formula(1.1 A t = t 2 2t 3 so that and A 0 = 3 a(t = A(t A(t 1 2 = = t

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The Theory of Interest - Solutions Manual 1 Chapter 1 1. ( a ) Applying formula (1.1) ( ) ( ) 2 2 3 and 0 3 At t t A =++ = so that () ( ) ( ) 2 1 23 . 03 at t t kA == = + + ( b ) The three properties are listed on p. 2. (1) 1 1 . 3 a (2) 1 2 2 0 for 0, 3 t t =+ > s o t h a t ( ) is an increasing function. (3) ( ) is a polynomial and thus is continuous. ( c ) Applying formula (1.2) ( ) [ ] ()() 2 2 22 12 31 2 1 3 21 3 . n IA nA n n n n n nn n n =− = + + + + −+− −+ 2. ( a ) Appling formula (1.2) ( ) [ ] ( ) ( ) [ ] ( )( ) [ ] 10 1 0. n II I A A A A A n An A +++= + ++ …" ( b ) The LHS is the increment in the fund over the n periods, which is entirely attributable to the interest earned. The RHS is the sum of the interest earned during each of the n periods. 3. Using ratio and proportion 5000 12,153.96 11,575.20 \$260 11,130 . −= 4. We have ( ) 2 , a t b so that 01 3 9 1.72. ab aa b =+=

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The Theory of Interest - Solutions Manual Chapter 1 2 Solving two equations in two unknowns .08 and 1. ab = = Thus, ( ) ( ) 2 5 5 . 0 8 1 3 a = += ( ) ( ) 2 10 10 .08 1 9 a = . and the answer is ( ) () 10 9 100 100 300. 53 a a == 5. ( a ) From formula (1.4 b ) and ( ) 100 5 A tt = + ( ) 5 5 4 125 120 5 1 . 4 120 120 24 AA i A −− = = ( b ) ( ) ( ) 10 10 9 150 145 5 1 . 9 145 145 29 i A = = 6. ( a ) ( ) ()() 54 5 4 100 1.1 and 5 4 100 1.1 1.1 1.1 1 .1. 4 100 1.1 t At i A = ⎡⎤ ⎣⎦ = = ( b ) () ( ) () () 10 9 10 9 10 9 100 1.1 1.1 1.1 1 .1. 9 100 1.1 i A = = 7. From formula (1.4 b ) ( ) ( ) 1 1 n An i = so that ( ) ( ) ( ) 11 n iAn −= and ( ) ( ) ( ) . n i An = +− 8. We have 567 .05, .06, .07, iii === and using the result derived in Exercise 7 ( )( )( ) 74 1 1 1 1000 1.05 1.06 1.07 \$1190.91. =+ + + 9. ( a ) Applying formula (1.5) ( ) 615 500 1 2.5 500 1250 ii = + so that
The Theory of Interest - Solutions Manual Chapter 1 3 1250 115 and 115/1250 .092, or 9.2%. ii == = ( b ) Similarly, ( ) 630 500 1 .078 500 39 tt =+= + so that 1 3 39 130 and 130/39 10/3 3 years. = = 10. We have ( ) 1110 1000 1 1000 1000 it it =+ = + 1000 110 and .11 it it = = so that () ( ) [] 3 500 1 2 500 1 1.5 4 500 1 1.5 .11 \$582.50. it i t ⎡⎤ ⎛⎞ += + ⎜⎟ ⎢⎥ ⎝⎠ ⎣⎦ = 11. Applying formula (1.6) .04 and .025 11 1 . 0 4 1 n i i in n +− + so that .025 .001 1 .04, .001 .016, and 16 . nn n = = = 12. We have 12 34 5 .01 .02 .03 .04 .05 i ===== and adapting formula (1.5) ( ) ( ) 12345 1000 1 1000 1.15 \$1150. iiiii ++ + + + = = 13. Applying formula (1.8) 2 600 1 600 264 864 i + = which gives 2 1 864/600 1.44, 1 1.2, and .2 i = + = = so that ( ) 33 2000 1 2000 1.2 \$3456. i = 14. We have 1 and 1 1 1 n n n rr j j + +

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The Theory of Interest - Solutions Manual Chapter 1 4 so that ( ) ( ) 11 1 1. 1 ij ii j r j jj +−+ +− =− = = ++ + This type of analysis will be important in Sections 4.7 and 9.4. 15. From the information given: () 12 21 3 1 3 /2 31 15 1 5 61 10 1 5/3 . aa bb cc nn += By inspection 52 1 5. 33 2 =⋅⋅ Since exponents are addictive with multiplication, we have . ncab =−− 16. For one unit invested the amount of interest earned in each quarter is: Quarter: Simple: Compound: 23 2 4 3 1 2 3 4 .03 .03 .03 .03 1.03 1 1.03 1.03 1.03 1.03 1.03 1.03 −− Thus, we have 43 32 4 1.03 1.03 .03 1.523. 3 1.03 1.03 .03 D D ⎡⎤ ⎣⎦ == 17. Applying formula (1.12) 18 19 20 21 : 10,000 1.06 1.06 6808.57 10,000 1.06 1.06 6059.60 Difference \$748.97. A B = 18. We have 2 1 vv + = and multiplying by 2 1 n i + 2 1 = + or ()() 2 1 0 + = which is a quadratic.
The Theory of Interest - Solutions Manual Chapter 1 5 Solving the quadratic () 11 4 15 1 22 n i ±+ + += = rejecting the negative root.

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## This note was uploaded on 11/18/2010 for the course MATH 320 taught by Professor Dr.k during the Spring '10 term at Nevada.

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Chapter001 - The Theory of Interest Solutions Manual Chapter 1 1(a Applying formula(1.1 A t = t 2 2t 3 so that and A 0 = 3 a(t = A(t A(t 1 2 = = t

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