Question

Situation 1
b) If $100 at Time “0” will be worth $110 a year later and was $90 a year ago, compute the interest rate for the past year and the interest rate next year.
c) Assume that $90 invested a year ago will return $110 a year from now. What is the annual interest rate in this situation?

Answer 1

(b)

• Taking the case from 90$ to 100$ in a year

rate of interest$(R)=(\frac{(A-P)}{P})\times100=(\frac{(100-90)}{90}) \times100 =11.11\%$ per anum (annually),where A=amount and P= principal

• Taking the next case from 100$ to 110$ in a year

rate of interest$(R)=(\frac{(A-P)}{P})\times100=10\%$ per anum (annually)

(c)

• Investment was made a year ago and return was obtained a year from now so it means 2 successive years

.i.e. T=2 years

• Assuming it to be compounded annually from 90$ to 110$

$A=P(1+\frac{R}{100} \times n )^{nT}$ ,where n=no of times it is compounded annually, t= no of years

$110=90(1+\frac{R}{100})2$ , because $n=1$ $(\frac{110}{90})(1/2)=1+\frac{R}{100}$

$R=((\frac{110}{90})1/2-1)\times100=10.55\%$

• Assuming it to be simple interest

$I=P\times R \times T$

$20=\frac{(90\times R \times2)}{100}$

$R=\frac{(20\times100)}{2\times90}=11.11\%$