Question

6. Brian makes a deposit into an account which credits interest at a nominal interest rate of 10% per annum, convertible semiannually. At the same time, John deposits $1,000 into a different account, which is credited with simple interest. At the end of 5 years, the forces of interest on the two accounts are equal, and John's account has accumulated to $X. Determine X. A. 1953 B. 1943 C. 1933 D. 1923 E. 1913

Answer 1

Force of interest: As the number of compounding periods reaches infinity in continuous compounding, the continuous compound interest rate is referred to as the force of interest.

The force of interest can be calculated by differentiating the natural logarithm of accumulation function (i.e. the formula for amount) with respect to time.

So for the first case of Brian (compound interest), the accumulation function can be written as

a(t) = (1+nominal rate per year / compounding periods per year)^(time in years*compounding periods per year)

=(1+10% / 2)^2*t

= (1.05)2t

So the force of interest

= $\frac{\mathrm{d} (ln(1.05)^{2t})}{\mathrm{d} t}$

=2ln(1.05)              ---Eqn 1

Similarly, the amount at time t in case of simple interest can be written as : P + P*r*t

So accumulation function a(t) = (1+ rt)

and force of interest = $\frac{\mathrm{d (ln(1+rt))} }{\mathrm{d} t}$

= r / (1+rt)            ---Eqn 2

Equating both 1 & 2 and putting t=5 , we get

2ln(1.05)   = r / (1+r*5)

r = 2 * ln(1.05) / (1-10 ln(1.05))

r= 0.19055

So, the rate of interest offered by John's account is 19.055% p.a. simple interest

So the amount after 5 years = 1000 ( 1+ 19.055% *5)

= 1952.75

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