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Business Mathematics

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Week 7 Lecture Quiz

1. Wayne Rooney received the amount of money Z at the end of 4 years when X dollar is invested at compound rate r . Z = X (1+r)t, where t is investment period. What range of interest rate (in percent) will provide the expected return, if: a. A firm is investing 80,000 for 4-year period and expects to get a return of no less than $ 200,000 and less than 500,000. (5) b. A firm is investing 100,000 for 4-year period and expects to get a return of no less than $ 200,000 and no more than 500,000. (5)

2. Raisa Andriana purchased a sofa for her company for $160,000. Her production company depreciated the sofa at the end of each year at the rate of 15% (annual rate) of its current value. For question (a-c) follow above information: a. What is the value of the sofa at the end of 3rd year. (2.5) b. What is the value of the sofa at the end of 5th year. (2.5) c. What is the value of total sofa depreciation at the end of 4th year. (2.5) d. What is the value of the sofa at the end of 3rd year, if the sofa is depreciated at the end of each quarter at the rate of 20% of its current value. (2.5)

3. In 1950, Frank Sinatra purchased a sedan from Elvis Presley for $50. Suppose the $50 was invested by Elvis Presley in 1960 at 8% rate. How much money would the (i) investment be worth & (ii) total interest accumulated in 2005 if the interest was: a. Simple Interest (2) b. Compounded annually (2) c. Compounded monthly (2) d. Compounded quarterly (2) e. Compounded continuously (2)

Answer Quiz A

(1. A.) Z = X (1+r)t Z = 80,000 (1+r)4 Z is between 200,000 and 500,000 So, 200,000 ≤ 80,000 (1+r)4 < 500,000 Divide into two groups of calculation: 1st group:


2nd group:

200,000 ≤ 80,000 (1+r) 4(divide by 40,000) 5 ≤ 2 (1+r)4 2.5 ≤ (1+r)4 (1+r)4 ≥ 2.5 4 r ≥ ± 2.5 -1 4 r ≥ ± 2.5 -1 r ≥ 1.25 – 1 or r ≥ -1.25 – 1 r ≥ 0.25 or r ≥ -2.25 ( not possible)

80,000 (1+r)4 < 500,000 (divide by 40,000) 2 (1+r)4 < 12.5 (1+r)4< 6.25 4 r <± 6.25 -1 r < 1.58 – 1 or r <-1.58 – 1 r < 0.58 or r <-2.58 ( not possible)

The solution set is 0.25 ≤ r < 0.58. The interest rate must range from 25% to less than 58%

(1.B.) Z = X (1+r)t Z = 100,000 (1+r)4 Z is between 200,000 and 500,000 So, 200,000 ≤ 100,000 (1+r)4 ≤500,000 Divide into two groups of calculation: 1st group:

2nd group:

200,000 ≤ 100,000 (1+r) 4(divide by 100,000) 2 ≤ 1 (1+r)4 2 ≤ (1+r)4 (1+r)4 ≥ 2 4 r ≥ ± 2 -1 4 r ≥ ± 2 -1 r ≥ 0.19 – 1 or r ≥ -1.19 – 1 r ≥ 0.19 or r ≥ -1.19 ( not possible)

100,000 (1+r)4≤ 500,000 (divide by 100,000) 1 (1+r)4 ≤ 5 (1+r)4≤ 5 4 r ≤± 5 -1 r ≤ 1.49 – 1 or r ≤-1.49 – 1 r ≤ 0.49 or r ≤-2.49 ( not possible)

The solution set is 0.19 ≤ r ≤ 0.49. The interest rate must range from 19% to 49%

Value after depreciation formula A = Initial Value 1 − Similar to Compound interest formula (page 420), but the sign is negative r = interest rate (expressed in decimal number) n = number of times interest is compounded each year, t = number of years
������ ������(������) ������

a. A = 160,000 1 − b. A = 160,000 1 − c. A = 160,000 1 − d. A = 160,000 1

Total Depreciation for 4 years = Initial Value – A (Value after depreciation) = $160000 – $83521 = $76479

0.15 1(3) =$98260 1 0.15 1(5) =$70992 1 0.15 1(4) =$83521 1

0.20 4(3) − = 4

160,000 (0.54) =$86400

Simple Interest formula I = Prt A(t) = P + Prt A = amount after t years P = principal r = interest rate (expressed in decimal number) n = number of times interest is compounded each year, t = number of years Compound Interest formula A=P 1
������ ������(������) − ������

A = amount after t years P = principal r = interest rate (expressed in decimal number) n = number of times interest is compounded each year, t = number of years

Continuous Compound Interest formula

A = Pert
A = amount after t years P = principal r = interest rate (expressed in decimal number)

3. a) SIMPLE INTEREST I = 50 (0.08) 45 = $180 The total interest from 1995 to 2005 = $180 A = P + Prt = P + I = 50+180 = 230 The total investment in 2005 =$230
b) ANNUAL COMPOUNDING A=P 1+ 50 1 + = $50 * 31.92 = $1596 The total investment in 2005 =1596 I = A – P = $1596 – $50 = $1546 The total interest from1950 to 2005 = $1546 c) QUARTER COMPOUNDING A =P 1+ 50 1 + = 50 *35.32 =1766 The total investment in 2005 =1766 I = A – P = $1766 – $50 = $1716 The total interest from 1950 to 2005 = $1716
������ ������(������) = ������ 0.08 4(45) 4 ������ ������(������) = ������ 0.08 1(45) 1

d) MONTHLY COMPOUNDING A=P 1+ 50 1 + = 50*36.16 = 1808 The total investment in 2005 =1808 I = A – P = $1808 – $50 = $1758 The total interest from1950 to 2005 = $1758
������ ������(������) = ������

0.08 12(45) 12

e) CONTINUOUS COMPOUNDING A = Pert = $50 * 2.718^0.08*45 = $50 * 36.58 = $1829 The total investment in 2005 =$1829 I = A – P = $1829 – $50 = $1779 The total interest from1950 to 2005 = $1779

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