Question

A couple wants to accumulate $10 000 by December 31, 2023. They make 10 annual deposits starting January 1, 2014. If interest is at j1 = 2%, what annual deposits are needed?

Deposits of $350 are made every 3 months from June 1, 2014, to June 1, 2017. How much has been accumulated on September 1, 2017, if the deposits earn j4 = 6%?

Melissa takes her inheritance of $25 000 and invests it at j12 = 6%. How many monthly payments of $250 can she expect to receive and what will be the size of the concluding payment?

A loan of $10 000 is to be repaid by monthly payments of $400, the first payment due in one year's time. If j12 = 12%, determine the number of regular monthly payments needed and the size of the final smaller
payment.

Answer 1

Let the annual deposit required be $A

then A*1.02^10+A*1.02^9+.....+A*1.02 = 10000

=> A/0.02*(1.02^10-1)*1.02 = 10000

=> A*11.1687 = 10000

A =$895.36

Annual Deposits required is $895.36

1st deposit of $350 is made on June 1, 2014,and last on June 1, 2017 (total 13 deposits)

interest rate per 3 months = 6%/4 =1.5% or 0.015

So, Amount accumulated by September 1, 2017

= 350*1.015^13+350*1.015^12+.....+350*1.015

=350/0.015*(1.015^13-1)*1.015

=$5057.63

Monthly interest rate = 6%/12 =0.005

Let the no of payments be n

then present value of n payments = 25000

=> 250/0.005*(1-1/1.005^n) = 25000

1/1.005^n = 0.5

1.005^n = 2

n = ln(2) /ln(1.005) = 138.98

So, no of payments of $250 Meissa can expect to receive is 138

and Final 139th payment of X is given by

250/0.005*(1-1/1.005^138)+X/1.005^139= 25000

X = $243.95

size of the concluding payment is $243.95

Monthly interest rate =12%/12 =0.01

No of payments n is given by

400/1.01^12+400/1.01^13+...+400/1.01^(n+11) = 10000

=>400/1.01^12 *(1-1/1.01^n)/(1-1/1.01) = 10000

=> 35852.9487* (1-1/1.01^n) = 10000

=>  1/1.01^n =0.72108

=> 1.01^n =1.3868

n = ln(1.3868)/ln(1.01) = 32.86

So, No of regular monthly payments of $400 needed is 32

Final Smaller payment X is given by

400/1.01^12 *(1-1/1.01^32)/(1-1/1.01) + X/1.01^44 =10000

=> 9776.95+ X/1.01^44 = 10000

=> X = $345.57

Size of the final Smaller payment is $345.57