Solution

BANK 1002 PERSONAL FINANCIAL MANAGEMENT

TUTORIAL SOLUTIONS

TOPIC 2

INTRODUCTION TO FINANCIAL MATHEMATICS

Problems

1. Determine the simple interest on a 90-day loan of $8,000 at 8.5%.

We have P = $8,000, r = 0.085, numerator of t = 90 days.

Simple interest, I = Prt = $8,000 x 0.085 x (90/365) = $167.67

2. A loan of $100 is to be repaid with $120 at the end of 2 months. What was the annual simple interest rate?

We have P = $100, I = $20, t = 2/12, and

r = I/Pt = $20 / [$100 x (2/12)] = 120%

3. At what rate of simple interest will $500 accumulate $10 interest in 85 days?

We have P = $500, I = $10, t = 85/365

r = I/Pt = $10 / [$500 x (85/365)] = 0.0859 = 8.59%

4. How long will it take $3,000 to earn $60 interest at 6% simple interest?

We have P = $3,000, I = $60, r = 0.06, and

t = I / Pr = $60 / ($3,000 x 0.06) = 1/3 = 4 months = 120 days

BANK 1002 PERSONAL FINANCIAL MANAGEMENT

Tutorial Solutions Topic 2 Introduction to Financial Mathematics 2

5. How many days will it take:

(a) $1,000 to accumulate to at least $1,200 at 5.5% simple interest,

(b) $1,600 to earn at least $30 of interest at 3.5% simple interest, and

(c) $5,000 to accumulate to at least $5,100 at 9% simple interest?

t = I / Pr

(a) We have P = $1,000, I = $200, r = 0.055

t = $200 / ($1,000 x 0.055) = 3.6364 = 1,327 days

(b) We have P = $1,600, I = $30, r = 0.035

t = $30 / ($1,600 x 0.035) = 0.5357 = 196 days

(c) We have P = $5,000, I = $100, r = 0.09

t = $100 / ($5,000 x 0.09) = 0.2222 = 81 days

6. Determine the maturity value of a $10,000 loan for 64 days at 7% simple interest.

S = P (1 + rt) = $10,000 {1 + [0.07 x (64/365)]} = $10,122.74

7. A student lends his friend $50 for 1 month. At the end of the month he asks for repayment of the $50 plus purchase of a chocolate bar worth $2.50. What simple interest rate is implied?

We have P = $50, I = $2.50, t = 1/12, and

r = I/Pt = $2.5 / [$50 x (1/12)] = 0.6 = 60%

8. A sum of $2,000 is invested from 18 May 2010, to 8 April...