1. Occasionally a savings account may actually pay interest compounded continuously. For each deposit, find the interest earned if interest is compounded (a) semiannually, (b) quarterly, (c) monthly, (d) daily, and (e) continuously. Use 1 year = 365 days.
Principal $1031
Rate 1.4%
Time 3 years
Answer:
a) $ 44.07
b) $ 44.15
c) $ 44.20
d) $ 44.22
e) $ 44.22
Step-by-step explanation:
The formula to find the total amount earned using compound interest is given as:
A = P(1 + r/n)^nt
Where A = Total amount earned after time t
P = Principal = $1031
r = Interest rate = 1.4%
n = compounding frequency
t = Time in years = 3 years
For each deposit, find the interest earned if interest is compounded
(a) semiannually
This means the interest is compounded 2 times in a year
Hence:
A = P(1 + r/n)^nt
A = 1031(1 + 0.014/2) ^2 × 3
A = 1031 (1 + 0.007)^6
A = $ 1,075.07
A = P + I where
I = A - P
I = $1075.07 - $1031
P (principal) = $ 1,031.00
I (interest) = $ 44.07
(b) quarterly
This means the interest is compounded 4 times in a year
Hence:
A = P(1 + r/n)^nt
A = 1031(1 + 0.014/4) ^4 × 3
A = 1031 (1 + 0.014/4)^12
A = $ 1,075.15
I = A - P
I = $1075.15 - $1031
A = P + I where
P (principal) = $ 1,031.00
I (interest) = $ 44.15
(c) monthly,
This means the interest is compounded 12 times in a year
Hence:
A = P(1 + r/n)^nt
A = 1031(1 + 0.014/12) ^12 × 3
A = 1031 (1 + 0.014/12)^36
A = $ 1,075.20
A = P + I where
I = A - P
I = $1075.20 - $1031
P (principal) = $ 1,031.00
I (interest) = $ 44.20
(d) daily,Use 1 year = 365 days
This means the interest is compounded 365 times in a year
Hence:
A = P(1 + r/n)^nt
A = 1031(1 + 0.014/365) ^2 × 3
A = 1031 (1 + 0.00365)^365 × 3
A = $ 1,075.22
A = P + I where
I = A - P
I = $1075.22 - $1031
P (principal) = $ 1,031.00
I (interest) = $ 44.22
(e) continuously. .
This means the interest is compounded 2 times in a year
Hence:
A = Pe^rt
A = 1031 × e ^0.014 × 3
A = $ 1,075.22
A = P + I where
I = A - P
I = $1075.22 - $1031
P (principal) = $ 1,031.00
I (interest) = $ 44.22
.