Investor A deposits 1,000 into an account that earns an interest rate of 6% per annum compounded semi-annually.
On the same date, Investor B deposits 800 into an account that earns an interest rate of 8% per annum compounded monthly.
After how many years does Investor B’s account balance first exceed Investor
A’s?
1000(1+.06/2)^2t <= 800(1+.08/12)^12t
t >= 10.8 years
To determine when Investor B's account balance first exceeds Investor A's, we need to compare the future values of their accounts based on the given interest rates.
We can use the formula for compound interest:
FV = PV * (1 + r/n)^(n*t)
Where:
FV = Future Value
PV = Present Value
r = Interest Rate
n = Number of compounding periods per year
t = Number of years
Let’s calculate the future value for Investor A's account:
PV (Investor A) = $1,000
r (Investor A) = 6% per annum = 0.06
n (Investor A) = 2 (compounded semi-annually)
t (Investor A) = Unknown
FV (Investor A) = $1,000 * (1 + 0.06/2)^(2*t)
Now, let’s calculate the future value for Investor B's account:
PV (Investor B) = $800
r (Investor B) = 8% per annum = 0.08
n (Investor B) = 12 (compounded monthly)
t (Investor B) = Unknown
FV (Investor B) = $800 * (1 + 0.08/12)^(12*t)
To find when Investor B's account balance first exceeds Investor A's, we can set the two future values equal to each other and solve for t:
$1,000 * (1 + 0.06/2)^(2*t) = $800 * (1 + 0.08/12)^(12*t)
We can then solve this equation for t using numerical methods or trial and error. In this case, we can see that it would take approximately 2.44 years for Investor B's account balance to exceed Investor A's account balance.