|
|
|
|
|
||||||||
|
|
||||||||||||
|
|
UNIT
9 :
MATHEMATICS OF INVESTMENT
LESSON 7:
UNIT 9 TEST SOLUTIONS
1. Compare the following investments. Which is better?
a)
$30 000 invested for 8 years at 7.2%/a, compounded semi-annually.
Solution:
b) $30 000 invested for 8 years at 7.2%/a,
compounded monthly.
Solution:
c) $30 000 invested for 8 years at 7.2%/a,
compounded daily.
Solution:
Hence you
earn more money the more frequently interest is compounded.
The compounded daily option is the best.
2. On the birth of their grandson, Barb and Lee wish to invest for his
education. If the investment pays
6.6%/a, compounded monthly, how much should they invest today in order to
provide $30 000 when he turns 18?
Solution:
Hence
Barb and Lee should invest $9174.73 today.
3. Stuart plans to buy a new trenching machine in
10 years to replace the current one. He
will need $50 000 at this time. How
much should he set aside each month to achieve this goal if interest is 6%/a,
compounded monthly?
Solution:
Let the
yearly payment be $R, with the first payment at the end of the first year.
Interest per. 0 1 2 . . . 118 119
120 Accumulated value
Payment R R
R R R
R
R(1.005)1
R(1.005)2
.
.
.
R(1.005)118
R(1.005)119
4. At the end of each year for 15 years the
Frieda deposits $3000 into an investment account which pays 6.6%/a, compounded
annually. If she then leaves this
amount to accumulate for another 10 years without any further yearly deposits
at the same rate, how much will she have accumulated at this time?
Solution:
Interest period0 1
2 . . . 13 14
15 Accumulated value
Payment 3000 3000 3000 3000
3000
3000
3000(1.066)1
3000(1.066)2
.
.
.
3000(1.066)13
3000(1.066)14
5. The Barton foundation wishes to establish an
academic athletic scholarship to be awarded each year for 15 years. The scholarship will be worth $1200 per
year. How much should be deposited now
in a trust fund that pays 5.4%/a, compounded annually?
Solution:
Interest Period 0 1 2 3 13 14 15
Payment 1200 1200
1200 1200 1200
1500(1.054)-1
1500(1.054)-2
.
.
1500(1.054)-13
1500(1.054)-14
1500(1.054)-15
Hence
$12 125.55 should be deposited now to provide for this scholarship.
6. Evaluate
the following general annuity.
Include a complete time line diagram.
$850 every 6 months for 12 years at 7.2%/a,
compounded quarterly.
Solution:
Here the
payment interval( ½ year ) is different
than the interest period (monthly).
This is a general annuity.
We must
match the interest rate to the payment interval.
Ie. We must
find the semi-annual rate that is equivalent to 7.2%/a, compounded quarterly.
Step 1:
Using the formula A = P(1 + i)n,
find the value of $1 invested at 7.2%/a, compounded quarterly after 1 year.
Step 2:
Let the equivalent ½ year rate be i %. (Note the
equivalent yearly rate would be 2i %.)
Now find the value of $1 invested at i % per ½ year after 1 year.
A = 1(1 + i)2 ** n = 2, the number of times interest is
compounded per year.
Step 3:
These two amounts must be equal.
Hence
Now find
the amount of the annuity using the annuity formula.
Interest per. 0 1 2 . . . 22
23 24 Accumulated value
Payment 850 850 850 850 850
850
850(1.036324)1
850(1.036324)2
.
.
.
850(1.036324)22
850(1.036324)23
Hence
the amount of the annuity is $31 695.31
7.
The Adams family purchased a cottage on Big Straggle Lake for $180
000. They paid 25% down, financing the rest
with a mortgage over 20 years with interest at 6.6%/a, compounded semi-annually
.
a) Determine the monthly payment.
Solution:
Here the
payment interval( monthly ) is different than the interest period (
semi-annual). This is a general annuity.
We must
match the interest rate to the payment interval.
Ie. We
must find the monthly rate that is equivalent to 6.6%/a, compounded
semi-annually.
Step 1:
Using the formula A = P(1 + i)n,
find the value of $1 invested at 6.6%/a, compounded semi-annually after 1 year.
Step 2:
Let the equivalent monthly rate be i %. (Note the equivalent yearly rate would be 12i %.)
Now find the value of $1 invested at i % per month after 1 year.
A = 1(1 + i)12 ** n = 12, the number of times interest is
compounded per year.
Step 3:
These two amounts must be equal.
Hence
Down
Payment = 0.25 x $180 000 = $45 000
Mortgage
amount = $135 000
The money in question is borrowed now – at
point 0 on the time line. Hence this is
a PV general annuity question
Interest Period 0 1 2 3 238 239 240
Payment R
R R R R R
R(1.005425865)-1
R(1.005425865)-2
.
.
R(1.005425865)-238
R(1.005425865)-239
R(1.005425865)-240
This
forms the following geometric series:
R(1.005425865)-240 +
R(1.005425865)-239 + . . . + R(1.005425865)-2 +
R(1.005425865)-1
b) Determine the total interest paid over the
25 year period.
Total amount repaid = 1007.40 x 240
= $241 776
Mortgage amount = $135 000
Interest paid = $241 776 - !35 000 = $106 776
Hence
the total interest paid over 20 years is $106 776.