Question

A passbook saving account has a rate of 6%. find the effective annual yield, rounded to the nearest tenth of a percent, if the interest is compounded
a. semianually

b. quarterly

c. monthly

d. daily( assume 360 days in a year)

e. 1000 times a year

f. 100000 times per year

Answer 1

Answer:

a,b,c,d,e,f) The effective annual yield is 1.1P-P = 0.1P = 10%P.

Step-by-step explanation:

This is a compound interest problem

Compound interest formula:

The compound interest formula is given by:

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

A: Amount of money(Balance)

P: Principal(Initial deposit)

r: interest rate(as a decimal value)

n: number of times that interest is compounded per unit t

t: time the money is invested or borrowed for.

a)

r = 0.06

n: 2(semianually means that the interest is compounded twice a year).

t = 1.

$A = P(1 + \frac{0.06}{2})^{2*1}$

$A = 1.1P$

The acount started the year with P, and it ended with 1.1P, so the effective annual yield is 1.1P-P = 0.1P = 10%P.

b)

Now we have n = 3, since if the interest is compounded quarterly, is is compounded three times a year(a year has 3 quarters). So:

$A = P(1 + \frac{0.06}{3})^{3*1}$

$A = 1.1P$

The effective annual yield is 1.1P-P = 0.1P = 10%P.

c) Now we have n = 12, since the interest is compounded monthly, and there are 12 months a year. So:

$A = P(1 + \frac{0.06}{12})^{12*1}$

$A = 1.1P$

The effective annual yield is 1.1P-P = 0.1P = 10%P.

d) Since the interest is compounded daily, and we assume 360 days in a year, n = 360. So:

$A = P(1 + \frac{0.06}{360})^{360*1}$

$A = 1.1P$

The effective annual yield is 1.1P-P = 0.1P = 10%P.

e) The interest is compounded 1000 times a year, so n = 1000

$A = P(1 + \frac{0.06}{1000})^{1000*1}$

$A = 1.1P$

The effective annual yield is 1.1P-P = 0.1P = 10%P.

f) The interest is compounded 100000 times a year, so n = 100000

$A = P(1 + \frac{0.06}{100000})^{100000*1}$

$A = 1.1P$

The effective annual yield is 1.1P-P = 0.1P = 10%P.