To calculate the sum of the balances of Account I and Account II at the end of 3 years, we need to calculate the balance in each account separately.
For Account I, the simple interest formula is: A = P(1 + rt), where:
A = the amount in the account after t years,
P = the initial deposit,
r = the annual interest rate,
and t = the number of years.
For Account I:
A = $2,500(1 + 0.04*3)
A = $2,500(1 + 0.12)
A = $2,500(1.12)
A = $2,800
So, the balance in Account I after 3 years is $2,800.
For Account II, the compound interest formula is: A = P(1 + r)^t, where:
A = the amount in the account after t years,
P = the initial deposit,
r = the annual interest rate,
and t = the number of years.
For Account II:
A = $2,500(1 + 0.04)^3
A = $2,500(1.04)^3
A = $2,500(1.124864)
A = $2,812.16
So, the balance in Account II after 3 years is $2,812.16.
The sum of the balances of Account I and Account II at the end of 3 years is:
$2,800 + $2,812.16 = $5,612.16
Therefore, the correct answer is $5,612.16.
Gabriel deposits $2,500 into each of two savings accounts.
Account I earns 4% annual simple interest.
Account II earns 4% interest compounded annually.
Gabriel does not make any additional deposits or withdrawals. What is the sum of the balances of Account I and Account II at the end of 3 years.
$5,624.32
$5,600.00
$5,200.00
$5.612.16
1 answer
1 answer