Question

Q1) Suppose you invest $20,958 today in an account that earns 02.00% interest annually. How much money will be in your account 19 years from today?

Q2) What is the value today, of single payment of $6,189 made 12 years from today, if the value is discounted at a rate of 14.00%?

Q3) How many years would it take an investment of $264 to grow to $2,845 at an annual rate of return of 08.00%?

Q4) How much money would you need to deposit today at 05.00% annual interest compounded monthly to have $13,601 in the account after 5 years?

Q5) If you deposit $562 into an account paying 06.00% annual interest compounded quarterly, how many years until there is $57,321 in the account?

Answer 1

1.We use the formula:
A=P(1+r/100)^n
where
A=future value
P=present value
r=rate of interest
n=time period.

A=$20958*(1.02)^19

=$20958*1.456811173

=$30531.85(Approx).

2.

We use the formula:
A=P(1+r/100)^n
where
A=future value
P=present value
r=rate of interest
n=time period

6189=P*(1.14)^12

P=6189/1.14^12

=6189*0.207559102

=$1284.58(Approx).

.3.

We use the formula:
A=P(1+r/100)^n
where
A=future value
P=present value
r=rate of interest
n=time period.

2845=264*(1.08)^n

(2845/264)=(1.08)^n

Taking log on both sides;

log (2845/264)=n*log (1.08)

n=log(2845/264)log(1.08)

=30.89 years(Approx).

4.

We use the formula:
A=P(1+r/12)^12n
where
A=future value
P=present value
r=rate of interest
n=time period.

13601=P*(1+0.05/12)^(12*5)

P=13601/(1+0.05/12)^(12*5)

=$13601*0.77920539

=$10597.97(Approx).

5.

We use the formula:
A=P(1+r/4)^4n
where
A=future value
P=present value
r=rate of interest
n=time period.

57321=562*(1+0.06/4)^4n

(57321/562)=(1.015)^4n

Taking log on both sides;

log(57321/562)=4n*log(1.015)

n=1/4[log(57321/562)/log(1.015)]

=77.66 years(Approx).