Question

Hector invests $800 in an account that earns 6.98% annual interest compounded semiannually. Rebecca invests $1000 in an account that earns 5.43% annual interest compounded monthly. Find when the value of Hector's investment equals the value of Rebecca's investment and find the common value of the investments at that time.

Answer 1

A=P (1+r/n)^nt

A= Total amount invested, P=principal amount, r=Interest rate, n=number of time in a year when the interest is earned (for annual, n=1; for semi-annual, n=2, ...), t = time in years

In the current scenario, case 1, n=2; case 2, n=1 and A1=A2, t1=t2
Therefore,
800(1+0.0698/2)^2t = 1000(1+0.0543/1)t
Dividing by 800 on both sides;
(1+0.0349)^2t = 1.25(1+0.02715)^t
(1.0349)^2t = 1.25(1.02715)^t
Taking ln on both sides of the above equation;
2t*ln (1.0349)= ln 1.25 + t*ln (1.02715)
2t*0.0343 = 0.2231+ t*0.0268
0.0686 t = 0.2231+0.0268 t
(0.0686-0.0268)t = 0.2231
0.0418t=0.2231

t=5.337 years

Therefore, after 5.337 years or 5 years and approximately 4 months, their value of investments will be equal.

This value will be,
A=800(1+0.0698/2)^2*5.337 = $1,153.76