Question

Burt deposits $10,000 into a bank account today. The account earns 4% per annum compounding daily for the first 3 years, then 3.5% per annum compounded quarterly thereafter. No further deposits or withdrawals will be made. For this question, assume all months are of equal length and ignore leap years. (a) Calculate the account balance six months from today. (b) Calculate the account balance 3 years from today. (c) Calculate the account balance 3.5 years from today. (d) Calculate the account balance 10 years from today.

Assume all months are 30 days = 360 days a year.

Answer 1

Burt deposits $10,000 into a bank account today. The account earns 4% per annum compounding daily for the first 3 years, then 3.5% per annum compounded quarterly thereafter. No further deposits or withdrawals will be made. For this question, assume all months are of equal length and ignore leap years.

(a) Calculate the account balance six months from today.

After 6 month, in the account has balance of $ 10000 deposited amount and also 6 months interest of 4% per annum compounded dailly

Calculation of 6 month interest compounded dailly

6 month interest = ( daily interest / r ) * ( ( 1 + r)n - 1 )

Here,

daily interest = 1000 * 4% / 360 = 40 / 360 = .0.1111111

r = interst rate at daily = 4% / 360 = 0.01111%

n = number of period interest compunded = 180

6 month interest = ( 0.1111111 / 0.01111% ) * ( 1 + 0.01111% )180 - 1 )

6 month interest = 1000 * (1.020198 - 1 ) = 1000 * 0.0202 = 20.2

After 6 month, in the account has balance = 1000 + 20.2 = $ 1020.2

(b) Calculate the account balance 3 years from today.

After 3 years, in the account has balance of $ 10000 deposited amount and also 3 years interest of 4% per annum compounded dailly

Calculation of 3 years interest of 4% per annum compounded dailly

6 month interest = ( daily interest / r ) * ( ( 1 + r)n - 1 )

Here,

daily interest = 1000 * 4% / 360 = 40 / 360 = .0.1111111

r = interst rate at daily = 4% / 360 = 0.01111%

n = number of period interest compunded = 1080

3 years interest = ( 0.1111111 / 0.01111% ) * ( 1 + 0.01111% )1080 - 1 )

3 years interest = 1000 * (1.127 - 1 ) = 1000 * 0.127 = 127

After 3 years, in the account has balance = 1000 + 127 = $ 1127

(c) Calculate the account balance 3.5 years from today

After 3 years, in the account has balance = $ 1127 plus 6 month interst @ 3.5% compounded quartely

Calculation of 6 month interst @ 3.5% compounded quartely

6 month interst @ 3.5% compounded quartely = (quarterly interest / r ) * ( ( 1 + r)n - 1 )

quarterly interest on $ 1127 = 1127 * 3.5% / 360 = 39.445 / 4 = 9.86125

r = interst rate at quarterly = 3.5% / 4 = 0.875%

n = number of period interest compunded = 2 ( for 6 months there are 2 quarter)

6 month interst @ 3.5% compounded quartely= ( 9.86125 / 0.875% ) * ( 1 + 0.875% )2 - 1 )

6 month interst @ 3.5% compounded quartely = 1127 * (1.0176 - 1 ) = 1127 * 0.0176 = 19.81

After 3.5 years, in the account has balance = 1127 + 19.81 = $ 1146.81

(d) Calculate the account balance 10 years from today.

After 10 years, in the account has balance = $ 1127 ( for first 3 years) plus 7 years ( remaining years 10 - 3) interst @ 3.5% compounded quartely

Calculation of 7 years ( remaining years 10 - 3) interst @ 3.5% compounded quartely

7 years interst @ 3.5% compounded quartely = (quarterly interest / r ) * ( ( 1 + r)n - 1 )

quarterly interest on $ 1127 = 1127 * 3.5% / 360 = 39.445 / 4 = 9.86125

r = interst rate at quarterly = 3.5% / 4 = 0.875%

n = number of period interest compunded = 28 ( for 7 years * 4 quarter = there are 28 quarter)

7 years interst @ 3.5% compounded quartely= ( 9.86125 / 0.875% ) * ( 1 + 0.875% )28 - 1 )

7 years interest = 1127 * (1.27626 - 1 ) = 1127 * 0.0.27626 = 311.35

After 3.5 years, in the account has balance = 1127 + 1311.35 = $ 1438.35