Question

 

a) How many years would you have to put $500 per year into an account that earns 4% annually in order to accumulate $6,000?

b) Your friend withdrew $2,078.9 from an account into which she had invested $1,000. If the account paid interest at 5% per year, she kept her money in the account for how many years?

c) A machine will need to be replaced 10 years from today for $10,000. How much must be deposited now into an account that earns 5% per year to cover the replacement cost?

d) An effective rate of 10% per quarter is closest to what effective annual rate?

e) A savings account’s value today is $150, and it earns interest at 1% per month. You want to know how much will be in the account one year from today. Which of the following is correct to solve for the unknown value?

Answer 1

A.

Deposits per year = $500

R = 4%

Future value = $6000

Let, number of years = n

Then,

6000 = 500*(1.04^n - 1)/.04

6000*.04/500 + 1 = 1.04^n

1.04^n = 1.48

n = log 1.48 / log 1.04

n = 10 years

So, time required is 10 years.

======

B.

Let, number of years = n

Then,

2078.9 = 1000*(1+5%)^n

1.05^n = 2078.9/1000

1.05^n = 2.0789

n = log 2.0789 / log 1.05

n = 15 years

So, time required is 15 years.

===

C.

Let, amount to be deposited now = P

Time = 10 years

Future value = $10000

R = 5%

Then,

10000 = P*(1+5%)^10

P = 10000/(1+5%)^10

P = $6139.13 or $6139

So, $6139.13 or $6139 should be deposited now.

===

D.

Effective annual interest rate = (1+10%/4)^4 - 1

Effective annual interest rate = 10.38%

===

E.

Alternatives are not given as mentioned.

Let, amount to be one year from today = F

Time = 12 months ( 1 year)

Monthly rate = 1%

Then,

F = 150*(1+1%)^12

F = $169.02 or $169

So,

amount to be one year from today is $169.02 or $169.