Question

Regina deposits $3500 in a savings account that pays 1.05% interest, compounded semi-annually.
b) What is the balance at the end of the first 6 months?
c) How much interest does the account earn in the second six months?
d) What is the balance at the end of the year?
e) How much interest does the account earn the first year?
f)How much interest would $3500 earn in one year at 1.05% interest, compounded annually?
g) How much more interest does Regina earn at an interest rate of 1.05% compounded semi-annually than compounded annually?

Answer 1

Using the compound formula A = P(1+r/n)^tn

We get:

b)

A = 3500(1 + 0.0105/2)^2*0.5

A = 3500 x 1.00525^1

A = 3518.38

Interest = 3518.38 - 3500 = $18.38 for the first 6 months.

c) 3518.38 x 1.00525^1

Interest = $18.47 in the second 6 months.

d) 3518.38 + 18.47 = $3,536.85 at the end of the first year.

e) 3536.85 - 3500 = $36.85 interest in first year.

f) 3500 x 1.0105 = 3536.75

3536.75 - 3500 = $36.75 interest in one year.

g)36.85 - 36.75 = $0.10 more.

Answer 2

The result of the compound interest operation on a principal of $3500 earning at a rate of 1.05% are :

• Balance after 6 months = $3518.375
• Interest after second six month = $18.471
• Year end balance (semiannual) = $3536.846
• Interest at year end (semiannual) = $36.846
• Year end interest (annual compounding) = $36.75
• Difference in interest = $0.096

The compound interest formula :

$A = P(1 + \frac{r}{n} )^{nt}$

Where ;

• A = Final amount
• P = Principal
• r = Rate
• n = number of compounding times per period
• t = time

A.)

Balance at the end of 6 months

t = 0.5 year ; n = semiannually(2)

Hence,

$3500(1 + \frac{0.0105}{2})^{1} = 3518.375$

B.)

Interest earned in the second 6 month :

Final amount after 1 year :

t = 1

$3500(1 + \frac{0.0105}{2})^{2} = 3536.846$

Interest in 2nd 6 month = (3536.846 - 3518.375) = $18.471

C.)

Balance at the end of the year = $3536.846

D.)

Interest earned in first year :

Balance at year end - Principal

$(3536.846 - 3500) = $36.846

E.)

If interest is compounded annually, ; n = 1

Then, Balance at year end will be ::

$3500(1 + \frac{0.0105}{1})^{1} = 3536.75$

Interest earned = $(3536.75 - 3500) = $36.75

F.)

Difference in interest earned at semiannual and annual compounding interest :

Year end Interest(semiannual) - Year end Interest (annual)

($36.846 - $36.750) = $0.096

Therefore, year end balance for semiannual compounding period is marginally higher than that of annual compounding.

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