Peter made a deposit of R6000 into his savings account. One year later he made a second deposit of R1000. How much is in his account 3 years after the first deposit, if the account earns at a rate of 10% compounded annually?
To solve the problem, we need to use the formula:
A = P(1 + r/n)^(nt)
Where:
A = Final amount
P = Initial amount (the sum of the two deposits)
r = interest rate (10%)
n = number of times compounded per year (annually)
t = time elapsed (3 years)
Let's start by adding the two deposits:
P = R6000 + R1000 = R7000
Now we can plug in the numbers:
A = R7000(1 + 0.1/1)^(1*3)
A = R7000(1.1)^3
A = R7000(1.331)
A = R9317
Therefore, there is R9317 in Peter's account 3 years after the first deposit.