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MBF3C
Foundations for College
Mathematics, Grade 11, College
Preparation
Lesson 6
Simple Interest & Linear Growth
MFB3C – Foundations for College Mathematics
Unit 2 – Lesson 6
Lesson Six Concepts
➢
➢
Solve problems involving the calculation of any variable in the simple-interest
formula (I = Prt), using scientific calculators
Demonstrate an understanding of the relationships between simple interest,
arithmetic sequences, and linear growth
Simple Interest
When you deposit money in a bank account, you lend your money to the bank. The
bank then pays you for the use of the money. The money earned from the bank is
called interest. The formula for Simple Interest is:
I = Prt
Where
I = Interest earned in dollars
P = Principal invested (amount of money you start with)
r = interest rate (in years), expressed as a decimal
t = time (in years)
It is unrealistic to think that we leave our money in the bank for exactly
1, 2, 3, etc years. Sometimes money is only in the account for months
or days. For this reason, it is important to know some conversions for
time.
16
12 months = 1 year
E.g. 16 months ÷ 12 =
years
12
48
52 weeks = 1 year
E.g. 48 weeks ÷ 52 =
years
52
600
365 days = 1 year
E.g. 600 days ÷ 365 =
years
365
Example 1: Sam invested $800 for two years that paid 5.5% per year. How much
interest was earned?
Solution
P = $800
r = 5.5% *** To convert it into a decimal, divide by 100
= 5.5 ÷ 100
= 0.055
t=2
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Page 2 of 42
MFB3C – Foundations for College Mathematics
Now Use the formula I = Prt
I = ($800)(0.055)(2)
I = $22.00
Unit 2 – Lesson 6
$22.00 was earned in interest.
Example 2: A savings account pays interest at 3 3/4% per year. The account has
balance of $2 487.61 on January 1. No deposits or withdrawals happen
during the month. What is the interest that is deposited into the account
on January 31?
Solution
P = $2487.61
3
r = 3 % = 3.75%
4
= 3.75 ÷ 100
= 0.0375
t = 31 days ***time must be in years (there are 365 days in a year)
=
31
***leave t as a fraction so your final answer is more accurate
365
Now Use the formula I = Prt
I = ($2487.61)(0.0375)(
I = $7.92
31
)
365
$7.92 was earned in interest.
Example 3: Paula received $1.25 interest on her savings account in February. She
did not withdraw or deposit any money that month. Her interest rate is
4%. How much was in her account to begin with?
Solution:
Determine what you know and don’t know.
I = $1.25
P=?
r = 4%. As a decimal is 0.04
t = 1 month
=
1
12
Now Use the formula I = Prt
Copyright © 2007, Durham Continuing Education
Page 3 of 42
MFB3C – Foundations for College Mathematics
$1.25 =P(0.04)(
Unit 2 – Lesson 6
1
1
) ***Divide both sides by (0.04)(
)
12
12
$1.25
=P
1
(0.04)( )
12
P = $375
Paula started with $375 in her account.
Example 4: Dan invested $1500 for 18 months. He earned $123.75 interest. What
was the annual interest rate?
Solution
I = $123.75
P = $1500
r=?
18
t=
12
Now Use the formula I = Prt
18
)
12
18
***Divide both sides by ($1500)(
)
12
$123.75
=r
18
($1500)( )
12
r = 0.055 ***Remember to convert it into a percent
r = 0.055 × 100
r = 5.5%
The annual interest rate is 5.5%
$123.75 = ($1500)( r )(
Linear Growth
Linear Growth is represented by a linear relationship and a straight line graph. As in
arithmetic sequences and simple interest, the growth is constant because the common
difference is the same for each interval.
Example 5: Christopher needed some equipment for his landscaping business. He
borrowed $1000 from a bank. Simple interest is charged on the loan at
9%. Christopher plans to pay off the loan in a lump sum at the end of one
year.
a. Create a table of values to show the amount owed after each month for the
first 4 months.
b. Write the sequence of the amount owed at the end of each month. Describe
the sequence.
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Page 4 of 42
MFB3C – Foundations for College Mathematics
Unit 2 – Lesson 6
c. Graph the relationship. Describe the relationship.
Solution
a.
Use the simple interest formula to determine the interest owed each month:
P = $1000
r = 9% = 0.09
t=
1
12
I = Prt
I = ($1000)(0.09)(
I = 7.5
1
)
12
$7.50 interest is owed each month.
Create a table of values:
Month Interest Owed $ (I = Prt)
1 4500
7.50 x +1000
f( x) = 10000.09
2
15.00 12
3 4000
22.50
4
30.00
Amount Owed $ (A = P + I)
1007.50
1015.00
1022.50
1030.00
( )
b.
3500
$1007.50, $1015.00, $1022.50, $1030.00
3000 owed increases by $7.50 each month. This forms an
The amount
arithmetic sequence.
c.
2500
Amount
Owed ($)
2000
1500
1000
500
5
10
15
20
Time in months
The relationship between time in months and the amount owed is linear because
the rate of change is constant and the line on the graph is a straight line.
Copyright © 2007, Durham Continuing Education
Page 5 of 42
25
MFB3C – Foundations for College Mathematics
Unit 2 – Lesson 6
Support Questions
1.
Calculate the missing item in the chart.
a)
b)
c)
d)
Principal
$485
$895
$925
Rate
2.75%
4 ¾%
Time
1.5 years
90 days
8months
4%
Interest
$16.98
$23.84
$18.25
2.
Troy had an outstanding balance of $1236.90 on his credit card for 80 days. The
annual interest rate is 18.2%. How much interest did Troy pay?
3.
A principal of $500 was invested for 3 years. The interest earned was $26.25.
What was the annual interest rate?
4.
An investment earns 8.75% per year. What principal will earn interest of $75.75
in 14 months?
Key Question #6
1.
Calculate the missing item in the chart.
a)
b)
c)
d)
Principal
$500
$1387
$1100
Rate
4%
5%
3.4%
Time
6 years
3 months
720 days
Interest
$18
$360
$10.40
2.
Dave has a savings account that pays interest at 3 ½% per year. His opening
balance for May was $1374.67. He did not deposit or withdraw money during the
month. The interest is calculated daily. How much interest did the account earn
in May?
3.
Lori has $500 in a savings account. She earned $1.54 interest in 25 days. What
annual rate of interest does her account pay?
4.
Nadine has a term deposit of $5250 at 4.8% per year. She receives the interest
from the deposit each month.
a. Write a sequence to show the accumulated interest Nadine will receive for
the first 5 months.
b. Draw a graph to show the terms of the sequence.
c. What type of growth does the graph display? Explain.
Copyright © 2007, Durham Continuing Education
Page 6 of 42
MFB3C – Foundations for College Mathematics
Unit 2 – Lesson 6
Key Question #6 (con’t)
5.
Mohammed invested $875 at 5.2% per year. When the investment matured,
Mohammed received $22.44. Determine the term of Mohammed’s investment.
6.
Suppose the term of an investment at simple interest is doubled. Does the
interest received double? Explain.
Copyright © 2007, Durham Continuing Education
Page 7 of 42
MBF3C
Foundations for College
Mathematics, Grade 11, College
Preparation
Lesson 7
Compound Interest & Exponential Growth
MFB3C – Foundations for College Mathematics
Unit 2 – Lesson 7
Lesson Seven Concepts
➢
Solve problems involving the calculation of the amount (A) in the compoundinterest formula A = P(1 + i)n, using scientific calculators
Demonstrate an understanding of the relationships between compound interest,
geometric sequences, and exponential growth
➢
Compound Interest
When interest is earned on interest, we say the interest compounds; thus the term
compound interest. The formula for Compound Interest is:
A = P(1 + i)n
Where
A = the amount the investment “grows”
P = principal invested (amount of money you start with)
i = interest rate, as a decimal, per compounding period OR
= interest rate as a decimal
# of interest periods in 1 year
n = number of compounding periods OR
= # of compounding periods per year X # of years
A compounding period means the time frame in which compound interest is
calculated.
Many times, the compounding periods are less than 1 year. (For example, interest on
mortgages is usually compounded semi-annually or interest on some savings accounts
is compounded monthly). The following is a chart of commonly used compounding
periods:
Compounding
Number of
frequency
compounding periods
per year
Annually
1
Semi-annually
2
Quarterly
4
Monthly
12
Weekly
52
Daily
365
Example 1: Determine the interest (i) and the number of compounding periods (n) for
each scenario. Do not solve.
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Page 9 of 42
MFB3C – Foundations for College Mathematics
Unit 2 – Lesson 7
a. A principal of $400 is invested at 5% compounded semi-annually for 6 years.
b. A principal of $625 is invested at 8.3% compounded weekly for 10 years.
c. A $185 GIC pays 6 3/4% compounded quarterly. How much interest will the GIC
earn in 7.5 years?
Solution
a. The annual interest rate is 5% = 0.05
1
The semi-annual interest rate is the annual rate.
2
0.05
i=
= 0.025
2
Interest is compounded 2 times a year for 6 years.
n = 2 x 6 = 12
Helpful Hint: If i has more than 5
numbers after the decimal, keep i as a
0.083
b. i =
fraction and enter it into your calculator as
52
it is!
n = 52 x 10 = 520
0.0675
4
n = 4 x 7.5 = 30
c. i =
3
% compounded annually for 8 years.
4
a. Determine the amount when the investment matures.
Example 2: Jose invested $1250 at 5
b. How much interest does the investment earn?
Solution
P = $1250
3
i = 5 % = 5.75 ÷ 100 = 0.0575
4
n=1×8=8
a. Now use the formula A = P(1 + i)n
A = 1250(1 + 0.0575)8
A = $1955.03
The investment is worth $1955.03 at maturity.
b. Use Interest = Amount – Principal
= 1955.03 – 1250
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MFB3C – Foundations for College Mathematics
Unit 2 – Lesson 7
= $705.03
The investment earned $705.03 in interest.
Example 3: Claire invested $500 at 4.5%compounded monthly for 3 years. What is
the amount of the investment at maturity?
Solution
P = $500
0.045
i=
12
n = 12 x 3 = 36
Now use the formula A = P(1 + i)n
0.045 36
)
12
A = $572.12
The investment is worth $1955.03 at maturity.
A = 500(1 +
Exponential Growth
Exponential Growth is represented by an equation with an exponent and the graph will
form an upward exponential curve. As in geometric sequences and compound interest,
the growth is not constant because there is a common ratio between consecutive terms.
Example 4: A principal of $100 is invested at 8% compounded annually for 6 years.
a. Create a table of values to show the amount of the investment at the end of
each year.
b. Graph the relationship.
c. Is the growth of the investment linear? Explain.
Solution
a. Make a table of values. Use the compound interest formula to determine the
amount for each year:
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MFB3C – Foundations for College Mathematics
Year
0
1
2
3
4
5
6
Unit 2 – Lesson 7
Amount ($)
100
100(1 + 0.08)1 = 108
100(1 + 0.08)2 ≈ 116.64
100(1 + 0.08)3 ≈ 125.97
100(1 + 0.08)4 ≈ 136.05
100(1 + 0.08)5 ≈ 146.93
250
100(1 + 0.08)6 ≈ 158.69
b.
200
y( x) = 100( 1+0.08) x
150
Amount ($)
100
50
5
10
Time in years
-50
c. Using the table of values from part a, calculate the differences in the amounts.
-100
Amount ($)
100.00
-150
108.00
116.64
125.97
136.05
146.93
158.69
Time in Years
Difference
($)
8.00
8.64
9.33
10.08
10.88
11.76
Since the differences are not constant, the growth is not linear. Also, since the
points on the graph do not lie on a straight line, the growth is not linear.
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MFB3C – Foundations for College Mathematics
Unit 2 – Lesson 7
Support Questions
1.
Determine each amount:
a. $375 at 3.5% compounded monthly for 4 years.
b. $100 000 at 5 1/4% compounded semi-annually for 6 years.
c. $235 at 7.68% compounded daily for 20 years.
2.
Karen purchased a $2500 compound interest CSB (Canadian Savings Bond) with
an annual rate of 4 ¼% and a 7-year term.
a. What is the amount of the investment at maturity?
b. How much interest was earned?
3.
A principle of $350 is invested at 3 3/4%compounded annually for 5 years.
a. Draw a graph to show the amount of the investment at the end of each year.
b. Is the growth of the investment linear? Explain.
Key Question #7
1.
Phil invested $600 at 4% compounded monthly for 6.5 years. How much interest
did the investment earn?
2.
Julia invested $875 at 2.8% compounded quarterly for 10 years. What is the
amount of the investment at maturity?
3.
Mark plans to invest $500 in a GIC for 2 years. He has a choice of 2 plans:
Plan A: 6.75% compounded annually
Plan B: 6.60% compounded quarterly
In which plan should Mark invest? Explain.
4.
Elizabeth has $937.21 in her savings account. The account pays 4.5%
compounded monthly. Elizabeth does not make any deposits or withdrawals
over the next 6 months. How much interest does the account earn?
Copyright © 2007, Durham Continuing Education
Page 13 of 42
MFB3C – Foundations for College Mathematics
Unit 2 – Lesson 7
Key Question #7 (con’t)
5.
A principle of $500 is invested at 7.5% compounded monthly for 7 years.
a. Calculate the accumulated interest at the each of each year.
b. Draw a graph to show the accumulated interest.
c. Use the graph to determine when the accumulated interest is $250.
Copyright © 2007, Durham Continuing Education
Page 14 of 42
MBF3C
Foundations for College
Mathematics, Grade 11, College
Preparation
Lesson 8
Compound Interest Formula
MFB3C – Foundations for College Mathematics
Unit 2 – Lesson 8
Lesson Eight Concepts
➢
➢
Solve problems involving the calculation of the principal (P) in the compoundinterest formula A = P(1 + i)n, using scientific calculators
Solve problems involving the calculation of the interest rate per period (i) and
the number of periods (n) in the compound-interest formula A = P(1 + i)n,
using a spreadsheet**
Please note: In the interest of this lesson, spreadsheets will not be used. Instead, all
problems will be solved using a guess-and-check method. Because this is a variation of
the Ministry Expectations, only a couple of Key Questions will be given.
Finding the Principal (P)
The principal (P), is the money invested today so that you have a certain amount in the
future (A).
Example 1: Mrs. Kim has some money to invest. She would like to give her grandson,
Carl, $10 000 on this 16th birthday. Carl is celebrating his 10th birthday
today. How much must Mrs. Kim invest today at 6% compound monthly?
Solution
This is a compound-interest problem so write down what you know:
0.06
A = $10 000
P=?
i=
12
Carl will get the money in: 16 – 10 = 6 years so
n = 6 x 12 = 72
Now use the formula
A = P(1 + i)n:
0.06 72
$10 000 = P(1 +
)
**Rearrange the formula to solve for P
12
10000
P=
.06 72
1 + 012
P = $6 983.02
( ( ))
Mrs. Kim must invest $6 983.02 today in order for Carl to have
$10 000 in 6 years (by his 16th birthday).
Finding the Interest Rate (i)
We will use guess and check to calculate i.
Example 2: A principal of $625 amounts to $1028.37 in 4 years. The interest was
compounded monthly. What is the annual interest rate?
Copyright © 2007, Durham Continuing Education
Page 16 of 42
MFB3C – Foundations for College Mathematics
Unit 2 – Lesson 8
Solution
Use the formula A = P(1 + i)n.
A = $1028.37
P = $625
n = 4 x 12 = 48
i=?
A = P(1 + i)n.
1028.37 = 625(1 + i)48
1028.37
= (1 + i)48
625
1.645392 ≈ (1 + i)48 **Remember i will be divided by 12.
st
1 guess:
i = 0.005, then 1.00548 ≈ 1.28 (this is too small!)
2nd guess:
i = 0.010, then 1.01048 ≈ 1.61 (this is too small)
3rd guess:
I = 0.0102, then 1.010248 ≈ 1.6276 (this is getting closer)
4th guess:
I = 0.0104, then 1.010448 ≈ 1.64316 (this is just under!)
5th guess:
I = 0.01043, then 1.0104348 ≈ 1.645505 (good enough!)
The monthly rate is 0.01043 as a decimal.
0.01043 x 100% = 1.043%
The annual rate is 1.043% x 12 = 12.516%
The annual interest rate is approximately 12.516%
Helpful Hint: When using the guess and check method, it is important to be
patient and try to get as close to the expected target as possible!
Finding the Number of Compounding Periods (n) We will use guess and
check to calculate n.
Example 3: How long does it take money to triple at 5% compounded semiannually?
Solution
For money to triple, $1 amounts to $3.
Use the formula A = P(1 + i)n.
i=
0.05
= 0.025
2
A = $3
Copyright © 2007, Durham Continuing Education
P = $1
n=?
Page 17 of 42
MFB3C – Foundations for College Mathematics
Unit 2 – Lesson 8
A = P(1 + i)n
3 = 1(1 + 0.025)n
3 = 1.025n **Remember n is the number of semi-annuals of a year
1st guess:
n = 10, then 1.02510 ≈ 1.28 (this is too small!)
2nd guess:
n = 50, then 1.02550 ≈ 3.44 (this is too big, but close)
3rd guess:
n = 40, then 1.02540 ≈ 2.69 (this is too small)
4th guess:
n = 44, then 1.02544 ≈ 2.96 (this is just under!)
5th guess:
n = 45, then 1.02545 ≈ 3.04 (this is just over!)
It is 0.04 under 3 for 44 semi-annuals of a year and it is 0.04 over 3
for 45 semi-annuals of a year for money to triple. Because the
number is the same, use common sense – in this case, semiannual means 2, so use the even number 44 to get a round
number.
This is
44
years = 22 years
2
At 5% compounded semi-annually, money triples in 22 years.
Support Questions
1.
Tricia has $1500 in her bank account. She wants to buy a CD player and invest
the remainder at 6 1/2% compounded quarterly for 4 years. At the end of the 4
years, Tricia wants $1500 in her bank account. Approximately how much can
Tricia spend on the CD player?
2.
Pat borrowed some money from a bank and will repay the loan in 3 years. The
interest rate is 12.5% compounded monthly. Pat must repay $1426.73 in 3
years. How much did Pat borrow?
3.
How long will it take for money to double at 6.5% compounded weekly?
4.
At what rate compounded semi-annually will $200 grow to $349.82 in 8 years?
Copyright © 2007, Durham Continuing Education
Page 18 of 42
MFB3C – Foundations for College Mathematics
Unit 2 – Lesson 8
Key Question #8
1.
What principal invested today at 4.28% compounded semi-annually will amount
to $3,500 in 9 years?
2.
Haley won $25,000 in a lottery. She will spend some of her winnings now and
save the rest. The money Haley saves must amount to $45,000 in 25 years.
Haley can invest the money at 6.35% compounded monthly. About how much
could Haley spend now?
3.
How long will it take for money to double at 4 1/2% compounded quarterly?
4.
A principal of $750 amounts to $915.14 after 5 years. The interest rate is
compounded quarterly. What is the annual interest rate?
5.
Which is the better investment: 5% compounded monthly or 5.25% compounded
annually? Explain your answer using examples.
Copyright © 2007, Durham Continuing Education
Page 19 of 42
MBF3C
Foundations for College
Mathematics, Grade 11, College
Preparation
Lesson 9
Loans & RRSP’s
MFB3C – Foundations for College Mathematics
Unit 2 – Lesson 9
Lesson Nine Concepts
➢
➢
➢
Calculate the cost of borrowing to purchase a costly item (e.g., a car, a stereo)
Demonstrate, through calculation, the advantages of early deposits to long-term
savings plans (e.g., compare the results of making an annual deposit of $1000 to
an RRSP, beginning at age 20, with the results of making an annual deposit of
$3000, beginning at age 50)
Explain the process used in making a decision and justify the conclusions
reached
Loans
Most people borrow money at some time to finance the purchase of items we want to
have now, but cannot afford to pay in full. Loans are usually repaid by making equal
monthly payments for a certain length of time and the interest rates tend to be very high.
When all the payments have been made, you not only pay the original amount
borrowed, but also the accumulated interest that accrued over the time period.
For loan payments, we use the Present Value of an Annuity Formula:
PV =
Where
R 1 − (1 + i)-n
i
or
R=
PVi
1 − (1 + i)-n
PV = the amount in dollars that must be invested now
R = the regular payment (withdrawal)
i = interest rate, as a decimal, per compounding period OR
= interest rate as a decimal
# of interest periods in 1 year
n = number of withdrawals made OR
= # of withdrawals per year X # of years
Example 1: A DVD player can be purchased for no money down and 24 equal monthly
payments of $18. The interest charged is 11.5% compounded monthly.
Determine the equivalent cash price of the DVD player.
Solution
R = $18
0.115
i=
12
n = 24
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Page 21 of 42
MFB3C – Foundations for College Mathematics
Now use the annuity formula
PV =
PV =
Unit 2 – Lesson 9
R 1 − (1 + i)
i
-n
(
)
0.115 -24
12
0.115
12
18 1 − 1 +
PV ≈ $384.28
The equivalent cash price of the DVD player is about $384.28.
Example 2: Pamela wants to buy a car for $35 000. She can finance the car through
the dealership at 8.7% compounded monthly for 48 months. She does not
have a down payment so she plans to borrow all $35 000.
a. Calculate the regular monthly payment.
b. What is the total amount repaid for the loan?
c. How much interest is paid?
Solution
PV = $35 000
0.087
12
n = 48
i=
a.
Now use the formula
R=
R=
PVi
1 − (1 + i)
-n
35 000
1− (1+
( )
)
0.087
12
0.087 – 48
12
R ≈ $866.00
Pamela’s monthly payment is $866.00 for 48 months.
b.
There were 48 payments made of $866.00
$866.00 x 48 = $41 568
Pamela paid a total of $41 568 for the car.
c.
The difference between what Pamela paid and what she originally borrowed is
the amount of interest that she paid.
$41 568 – $35 000 = $6 568
Pamela paid $6 568 in interest for the car.
Copyright © 2007, Durham Continuing Education
Page 22 of 42
MFB3C – Foundations for College Mathematics
Unit 2 – Lesson 9
Support Questions
1.
Krista needs a laptop computer for college. She can purchase one for 36
monthly payments of $68. The interest is 10.5% compounded monthly. What is
the equivalent cash price of the laptop?
2.
Anna is planning to buy a stereo. She can afford to deposit $50 biweekly for the
next 4 years. If the interest rate is 7.6% compounded biweekly, what is the
maximum amount Anna can afford to pay for the stereo?
3.
Justin is taking out a personal loan of $12 000. He will be charged 14.35%
interest compounded monthly. The loan is to be repaid monthly for the next 5
years. Determine Justin’s monthly payment.
4.
James buys a motorcycle for $6575. He makes a down payment of $650 and
finances the rest. He plans to make monthly payments for the next 3 years at
12.8% compounded monthly. How much interest does James pay on the loan?
5.
Due to their credit ratings, Jake can borrow money at 9.45% compounded
monthly but Emily must pay 11.3% compounded monthly. How much more
interest would Emily pay than Jake on a $9 380 loan with monthly payments for 5
years?
RRSP’s
RRSP stands for Registered Retirement Savings Plan
It is a type of savings plan for people who earn income, where funds
contributed and interest earned are not taxed until the funds are
withdrawn.
An RRSP can be used to save money for your retirement. You can only contribute to
an RRSP if you earn an income. Once you stop working, you can no longer contribute
to an RRSP. Once you retire, you can set up a system where you withdraw regular
equal amounts from your RRSP.
Copyright © 2007, Durham Continuing Education
Page 23 of 42
MFB3C – Foundations for College Mathematics
Unit 2 – Lesson 9
If you plan to contribute to an RRSP, use the Annuity Formula:
R (1 + i)n − 1
A=
i
or
R=
Ai
(1+ i)n − 1
If you want to determine the monthly pension, use the Present Value of an Annuity
Formula:
R 1 − (1 + i)-n
PV =
i
or
R=
PVi
1 − (1 + i)-n
The next couple of examples will show you the advantages of early deposits into an
RRSP.
Example 1: Nick began to contribute to his RRSP at age 25. He made yearly
contributions that averaged $2000. His RRSP earned interest at an
average rate of 7.5% compounded annually until his 65th birthday.
Determine the amount in Nick’s RRSP on his 65th birthday.
Solution
R = $2000
i = 0.075
n = 65 – 25 = 40 years
Now use the annuity formula:
R (1 + i) − 1
A=
i
n
2000 (1 +0.075 )
A=
0.075
40
−1
A ≈ $454 513.04
Nick will have about $454 513.04 in his RRSP on his 65th birthday.
Copyright © 2007, Durham Continuing Education
Page 24 of 42
MFB3C – Foundations for College Mathematics
Unit 2 – Lesson 9
Example 2: Suppose Nick began to contribute to his RRSP at age 40 and his yearly
contributions averaged $5000. His RRSP earned the same average
interest rate of 7.5% compounded annually until his 65th birthday.
Determine the amount in Nick’s RRSP on his 65th birthday.
Solution
R = $5000
i = 0.075
n = 65 – 40 = 15 years
Now use the annuity formula:
A=
A=
R (1 + i) − 1
i
n
5000 (1 +0.075 )
0.075
15
−1
A ≈ $ 130 591.82
Nick will have about $ 130 591.82 in his RRSP on his 65th birthday.
Nick’s scenarios show that contributing smaller amounts for longer periods
of time is better in the long run.
Example 3: Suppose Lucy retires on her 60th birthday. The amount of all investments
in her RRSP is about $750 000. She decides that $650 000 of the funds
in the RRSP will be invested in an annuity that pays an average rate of 8%
compounded monthly for the next 25 years. What monthly pension will
she receive?
Solution
PV = $650 000
i=
0.08
12
n = 25 x 12 = 300
Now use the present value formula (rearranged):
PVi
R=
-n
1 − (1 + i)
Copyright © 2007, Durham Continuing Education
Page 25 of 42
MFB3C – Foundations for College Mathematics
R=
650000
1− (1+
Unit 2 – Lesson 9
( )
)
0.08
12
0.08 -300
12
R ≈ $5016.81
Lucy will receive a monthly pension of $5016.81.
Support Questions
6.
Mary is converting her RRSP into an income fund. She wishes to receive $1500
every 6 months for the next 20 years. She is guaranteed an interest rate of
6.25% compounded semi-annually. How much must Mary deposit now to pay for
the annuity?
7.
Josh is 27 years old and wants to start putting money into an RRSP on a
biweekly basis. If he wants to have $1,000,000 by his 65th birthday, how much
should he deposit regularly if he can earn an average interest rate of 6.75%
compounded biweekly? Does this seem reasonable? Justify your answer.
Key Question #9
1.
Luke wants to buy an Ipod. Determine the equivalent cash price if Luke makes
18 monthly payments of $31.48 at an interest rate of 15.2% compounded
monthly.
2.
Nancy purchased a hot tub at a total price of $6995. She made a $1200 down
payment and financed the rest at 10.4% compounded monthly. Nancy can repay
the loan in 36 months or 48 months. How much interest will Nancy save is she
repays the loan in 36 months instead of 48 months?
3.
Greg has a school debt of $9 875. He wants to make biweekly payments and
pay off the loan in 4 years. If the interest rate is 6.3% compounded biweekly,
how much are his payments?
4.
The Smith family refinished their basement. They borrowed $22 500 at 8.62%
compounded monthly to finance the project. They will pay back the loan with
monthly payments for the next 12 years.
a. How much are the monthly payments?
b. How much interest will they pay over the 12 years?
Copyright © 2007, Durham Continuing Education
Page 26 of 42
MFB3C – Foundations for College Mathematics
Unit 2 – Lesson 9
Key Question #9
5.
Susan borrowed $5000. The terms of the loan were equal monthly payments at
12% compounded monthly for 3 years. After making payments for 1 year, Susan
decided to pay off the balance of the loan.
a. What was Susan’s monthly payment?
b. How much must Susan pay at the end of 1 year to pay off the balance of the
loan?
c. How much interest did Susan save by repaying the loan in 1 year?
6.
Darcy began making monthly contributing to his RRSP at the age of 24. His
average monthly contribution, starting 1 month after his 24 th birthday, was $125
and his RRSP earned interest at an average rate of 7.5% compounded monthly.
Darcy stopped making contributions when he retired at the age of 50 and started
to withdraw a pension from his RRSP.
a. Determine the amount in Darcy’s RRSP on his 50th birthday.
b. Darcy decides to re-invest the total amount of his RRSP to provide an annuity
for his monthly income during his retirement. Suppose Darcy was able to
obtain the same average interest rate and chose a 20-year term for his
annuity. What monthly pension could he withdraw for the next 20 years?
Copyright © 2007, Durham Continuing Education
Page 27 of 42
MBF3C
Foundations for College
Mathematics, Grade 11, College
Preparation
Lesson 10
Buying New/Used Vehicles and Vehicle Costs
MFB3C – Foundations for College Mathematics
Unit 2 – Lesson 10
Lesson Ten Concepts
➢
➢
➢
➢
➢
➢
➢
Identify the procedures, costs, advantages, and disadvantages involved in buying
a new vehicle and a used vehicle
Determine, through investigation, the cost of purchasing or leasing a chosen new
vehicle or purchasing a chosen used vehicle, including financing
Describe a decision involving a choice between alternatives
Collect relevant information related to the alternatives to be considered in making
a decision
Summarize the advantages and disadvantages of the alternatives to a decision,
using lists and organization charts
Explain the process used in making a decision and justify the conclusions
reached.
Calculate the fixed and variable costs involved in owning and operating a vehicle
(e.g., the license fee, insurance, maintenance) Collect relevant information
related to the alternatives to be considered in making a decision
Buying a New Vehicle
When buying a vehicle, it is important to understand the terminology dealerships use
and the additional costs that are involved. Not all dealerships and vehicles have the
same additional costs but it is still important to know what they are so there are no
surprises. The following are some definitions that may help you when purchasing a
vehicle.
Manufacturers Suggested Retail Price (MSRP): the starting point for the price of a
new vehicle. Usually it is only a suggested price and you can
negotiate a lower price to pay.
Freight Charge:
a charge paid to transport the vehicle from where it was
manufactured to the dealership.
Tire/air tax:
an environmental tax to be used to dispose of tires and clean up
the air from the chemicals used in the air conditioning system.
a charge on vehicles with larger engines because they are not fuel
efficient.
Gas tax:
Administration fee: a charge to do all the paper work and calculations in a transaction.
License plate fee: the cost of obtaining a license plate.
Gas fee:
a charge for supplying gas with the vehicle (so you can actually
drive off the lot with it!)
Lien payout:
money still owed on a vehicle used for a trade-in (trading an used
vehicle for a newer one)
Copyright © 2007, Durham Continuing Education
Page 29 of 42
MFB3C – Foundations for College Mathematics
Deal Review:
Unit 2 – Lesson 10
a computer printout of all the costs and financial arrangements
involved in purchasing a vehicle.
Manufacturer’s rebate (MFG): money returned to the buyer by the manufacturer when
the vehicle is purchased (usually for inventory reasons, the GM
credit card is considered a rebate, etc.)
Depreciates:
the amount the value of the car goes down each year.
When financing a new vehicle, we use the following annuity formula:
R=
PVi
-n
1 − (1 + i)
There are many advantages and disadvantages to buying a new vehicle. It is important
to research what you want and ask lots of questions before buying.
When buying new, a vehicle’s costs include the base price (what you and the dealership
decide on), the options (sunroof, spoiler, air-conditioning, etc), and taxes. There is also
a fee to register and license the vehicle, and additional costs such as freight, and
delivery charges.
Example 1: A car costs $21 500. How much will it cost, including 15% taxes
(remember – tax amounts are dependent upon in which province you purchase the car
– tax rates vary from province to province)?
Solution
$21 500 x 0.15 = $3225 in taxes
Another way to figure out the total
cost is to multiply by 1.15.
$21 500 x 1.15 = $24 725
$21 500 + $3225 = $24 725
The cost of the car including taxes is $24 725.
Example 2: At the end of 2004, a car dealership was clearing out any unsold new
vehicles by advertising a discount of 10% off. If a new car costs $18 930,
how much is the car before taxes?
Solution
If there is a 10% discount, then the car is worth 100% – 10% = 90% the
original value.
$18 390 x .90 = $16 551
The car is worth $16 551 before taxes.
Copyright © 2007, Durham Continuing Education
Page 30 of 42
MFB3C – Foundations for College Mathematics
Unit 2 – Lesson 10
The following example is a complete deal, including all taxes and additional fees.
Example 3: Liam purchases a new car with an MSRP of $27 600. He negotiates a
discount of $700. Liam pays these additional costs: $750 freight charge,
$100 tire/air tax, $75 gas tax, $80 administration fee, $74 license plate
fee, and a $20 gas fee. Liam does not have a vehicle to trade-in but he
does make a down payment of $1500 and finances the rest through the
dealership at 4.9% compounded monthly for 48 months. The finance fee
is $52. Complete a deal review for Liam’s purchase:
Solution
Step 1:
Calculate the purchase price:
MSRP – Discount = Purchase Price
$27 600 – $700 = $26 900
Step 2:
Calculate the taxable total (all the costs that will be taxed added together
then subtract the trade-in allowance)
**In this case, there isn’t a trade-in so subtract 0.
Taxable Total = Purchase price + freight charge + tire/air tax +
administration fee + gas tax – trade-in
= $26 900 + $750 + $100 + $75 + $80 – 0
= $27 905
Step 3:
Calculate the taxes.
Taxes = 13% x taxable total
= 0.13 x $27 905
= $3522.35
Step 4:
Calculate the delivery price.
Delivery price = taxable total + taxes + license plate fee + gas fee – lien
payout
= $27 905 + $3522.35 + $74 + $20 – 0
= $31 521.35
Step 5:
Calculate the amount to be financed.
Amount Financed = Delivery price – MFG rebate – down payment +
finance fee
= $31521.35 – 0 – $1500 + $52
= $30 073.35
Copyright © 2007, Durham Continuing Education
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MFB3C – Foundations for College Mathematics
Step 6:
Unit 2 – Lesson 10
Calculate the monthly payment (using the annuity formula)
( )
)
30 073.35 0.049
PVi
12
R=
=
-n
1 − 1 + 0.049 -48
1 − (1 + i)
12
(
= $691.21
Liam will have a monthly payment of $691.21
A deal review looks something like the chart below. All the information in the question
can be filled in right on the review and the bolded terms are the steps requiring the
calculations in the 6 steps above on the previous two pages.
Deal Review
MSRP/List Price
Discount
Purchase Price
Freight
Tire/Air Tax
Gas Tax
Administration Fee
Trade-in Allowance
Taxable Total
Taxes
License Plate Fee
Gas Fee
Lien Payout
Delivery Price
MFG Rebate
Down Payment
Finance Fee
Amount Financed
Interest Rate
Loan Term
Monthly Payment
Support Questions
1.
Marty is buying a new van for $16 550, plus an options package costing $1175.
How much is the van, including 13% taxes?
2.
Richard wants to buy a new SUV. It is on sale at a discount of 4.5%. The SUV
costs $31 690. How much will is cost after the discount (not including taxes)?
3.
Patty is planning the purchase a new car with an MSRP of $19 995. She adds a
few options, including air-conditioning ($1000), automatic transmission ($1200),
and a roof-rack ($450). The salesperson informs her of the following additional
costs: $459 freight charge, $100 tire/air tax, $65 gas tax, $55 administration fee,
$74 license plate fee, and a $30 gas fee. Patty does not have a vehicle to trade-
Copyright © 2007, Durham Continuing Education
Page 32 of 42
MFB3C – Foundations for College Mathematics
Unit 2 – Lesson 10
in but she does make a down payment of $3000 and finances the rest through
the dealership at 3.9% compounded monthly for 5 years. The finance fee is $45.
Complete a deal review for Patty’s purchase.
Buying a Used Vehicle
The big difference in buying a used vehicle from a dealership compared to a new one is
that you do not have to pay many of the additional costs (because they have already
been paid). For instance, there isn’t a freight charge, air/tire tax, and gas tax. As well,
because vehicles depreciate in value, especially in their first year, used vehicles can be
a lot less expensive.
When you buy a used vehicle, the price you pay is called the resale value. This price
takes into account how much the vehicle depreciated.
The formula for the resale value of a vehicle is:
V = P(1 – r)n
Where
V = the resale value in dollars
P = the MSRP in dollars
r = the annual depreciation rate, written as a decimal
n = the age of the car in years
Example 4: A car has an MSRP of $23 500. It depreciates at a rate of 16% per year.
Estimate the resale value of the car after 5 years.
Solution
P = $23 500
R = 0.16
N=5
Use the resale formula:
V = P(1 – r)n
V = 23 500(1 – 0.16)5
V = $9 827.98
After 5 years, the car is worth $9 827.98
Example 5: A 3-year old car is worth $15 750. The next year it is worth $11 800. By
what percent has the value of the car depreciated?
Solution
$15 750 – $11 800 = $3 950
Copyright © 2007, Durham Continuing Education
Page 33 of 42
MFB3C – Foundations for College Mathematics
Percent depreciation =
Unit 2 – Lesson 10
amount of depreciation
x100
original price
$3 950
x100
$15 750
≈ 25.08
=
The car’s depreciation value is about 25%
Support Questions
4.
What are 2 advantages and 2 disadvantages of buying a used vehicle versus a
new vehicle?
5.
Determine the resale value of a 2-year old car with an MSRP of $30 450 and an
annual depreciation rate of 20%.
6.
Determine the monthly payment on an $8 790 vehicle financed at 4.9%
compounded monthly for 48 months.
Vehicle Costs
There are many costs involved in owning and operating a vehicle. The regular
expenses include fuel costs, maintenance costs, insurance, and licensing costs. These
expenses fall under two types of costs: fixed costs and variable costs.
Fixed costs:
Variable costs:
are the expected costs such as insurance and licensing. These
costs do not depend on how the vehicle is used.
are the costs the owner has little control over. These costs include
gas, repairs, and maintenance. The frequency and amount of
repairs needed depend on the way the car is driven and taken care
of.
Right now with the raising gas prices, fuel consumption can be an important factor when
buying a vehicle.
Example 1: Doug’s car has a fuel consumption rating of 6.2 L/100km. Doug estimates
that he drives 30 000 km per year and the average cost of fuel during the
year is $0.82/L. Estimate Doug’s fuel cost for the year.
A fuel consumption rating of 6.2 L/100 km means that Doug can
drive 100 km on 6.2L of gas.
Copyright © 2007, Durham Continuing Education
Page 34 of 42
MFB3C – Foundations for College Mathematics
Unit 2 – Lesson 10
Solution
If 6.2L/100 km, then fuel for 1 km is
6.2
L
100
Doug drives about 30 000 km in 1 year so
The amount of fuel used is 30 000 x
6.2
L = 1860L
100
The cost of 1 L of gas is $0.82. The total cost of gas for the year is
1860 L x $0.82 = $1525.20
Doug spends approximately $1525.20 on fuel in 1 year.
Example 2: One week gas was selling for 89.9¢/L. Tre put $25 worth of gas into his
car. How many litres of gas did he buy? Round to one decimal place.
Solution
89.9¢/L means $0.899 for 1 L
Number of litres bought for $25
=
25
0.899
≈ 27.8 L
Tre bought approximately 27.8 L of gas.
Example 3: Christy drives to work everyday and estimates that she drives 30 000 km
each year. In general, it is recommended to get a lube, oil, and filter
change every 5000 km or 3 months and costs approximately $22.95 each
time. It is also recommended to get a $39.99 brake inspection and tire
rotation every 10 000 km. Estimate how much Christy will spend on
maintenance this year.
Solution
A lube, oil, and filter change should be done every 5000 km.
30 000 ÷ 5 000 = 6 times in 1 year
6 x $22.95 = $137.70
A brake inspection and tire rotation should be done every 10 000 km.
30 000 ÷ 10 000 km = 3 times in 1 year
3 x $39.99 = $119.97
$137.70 + $119.97 = $257.67
It will cost Christy about $257.67 in maintenance this year.
Copyright © 2007, Durham Continuing Education
Page 35 of 42
MFB3C – Foundations for College Mathematics
Unit 2 – Lesson 10
Support Questions
7.
Determine the cost to drive each distance at the given fuel consumption rating
and average cost of fuel.
a. 50 500 km at 5.6 L/100 km and $0.78/L
b. 21 485 km at 6.7 L/100 km and 92.4¢/L
8.
Andrea drives about 27 000 km per year. Her car has a fuel consumption rating
of 6.9L/100 km. The average cost of fuel is $0.91/L. Andrea gets regular $31.95
lube, oil and filter services done every 5000 km, tire rotation and brake inspection
for $45.50 every 10 000 km, and a wheel alignment inspection every 25 000 km
for $72.99. She also had to replace all four tires at $105 each plus an additional
$20 per tire installation and balancing charge. Determine Andrea’s average
monthly fuel and maintenance cost.
Key Question #10
1.
How much is a $19 825 car including 13% taxes?
2.
Sean has ordered a new sporty car listed at $25 500. The only option that Sean
ordered is the power package, which includes power windows, steering, and
braking. This option costs $1275. The dealer charges $500 for additional costs.
How much will Sean pay in total, including taxes?
3.
Sue and Frank plan to buy their son a truck as a graduation present. They can
afford to spend no more than $15 000. A dealership has a sale on a truck for
15% off the listed price of $17 500. Can Sue and Frank get the truck for their
son? Explain.
4.
Scott purchases a new car with an MSRP of $32 800. He negotiates a discount
of $1200. Scott wants a few options, including air-conditioning ($800), automatic
transmission ($1250), and a power package ($1450). The salesperson informs
him of the following additional costs: $625 freight charge, $100 tire/air tax, $75
gas tax, $45 administration fee, $74 license plate fee, and a $45 gas fee. Scott
does not have a vehicle to trade-in but he does make a down payment of $2500
and finances the rest through the dealership at 1.9% compounded monthly for 5
years. The finance fee is $54. Complete a deal review for Scott’s purchase.
5.
List 3 advantages and 3 disadvantages of buying a new vehicle versus a used
vehicle.
Copyright © 2007, Durham Continuing Education
Page 36 of 42
MFB3C – Foundations for College Mathematics
Unit 2 – Lesson 10
Key Question #10 (con’t)
6.
Determine the resale value of a 4-year old vehicle with an MSRP of $21 300 and
an annual depreciation rate of 18%.
7.
Eric wants to sell his 3-year old car. The car had an MSRP of $35 450. Eric
knows that the average annual depreciation of the make and model of his car is
23%. He does not want to sell it for less than $25 000. Is this realistic? Explain
your reasoning.
8.
Determine the cost to drive each distance at the given fuel consumption rating
and average cost of fuel.
a. 22 000 km at 7.1 L/100 km and $0.88/L
b. 49 900 km at 5.8 L/100 km and 97.3¢/L
9.
The Gather’s have 2 cars. Last year, one car was driven about 13 500 km and
has a fuel consumption rating of 8.1L/100 km and 55 625 km was put on the
other car which has a fuel consumption rating of 6.3L/100 km. The average cost
of fuel during the year was $0.87/L. Using the following maintenance table,
determine how much the Gather’s spent on all relevant variable costs last year.
Recommended Service
Lube, oil, filter
Tire rotation and brake inspection
Wheel alignment inspection
Tune-up and emission control
Cooling system
Brakes replaced
Frequency
(km)
5000
10 000
25 000
50 000
50 000
50 000
Copyright © 2007, Durham Continuing Education
Approximate
cost ($)
29.99
34.50
64.99
250.00
84.95
355.00
Page 37 of 42
MFB3C – Foundations for College Mathematics
Unit 2 –Support Questions Answers
Unit 2 – Support Question Answers
Lesson 6
1.
a.
I = ($485)(0.0275)(1.5)
b. P =
I = $20.01
2.
P = $1449.75
$18.25
($925)(0.0 4)
t = approx. 5 months
$23.84
($895)( 128 )
r = approx. 4%
c.
$16.98
90
(0.0475)( 365
)
r=
I = ($1236.90)(0.182)
d. t =
( )
80
365
I = $49.34
$26.25
($500)(3)
r = 1.75%
3.
r=
4.
P=
$75.75
(0.0875)( 14
)
12
P = $742.04
Lesson 7
1.
a.
A = 375(1 +
0.035
12
)48
b. A = 100 000(1 +
A = $431.26
c.
0.0525
12
)12
A = $105 378.19
A = 235(1 + 0.0768
)(20 x 365)
365
A = $1091.63
2.
a.
3.
a.
A = 2500(1 + 0.0425)7
A = $3345.59
Year
0
1
2
3
4
5
b. I = $3345.59 – $2500
I = $845.59
Amount ($)
350
350(1 + 0.0375)1 = 363.13
350(1 + 0.0375)2 ≈ 376.74
350(1 + 0.0375)3 ≈ 390.87
350(1 + 0.0375)4 ≈ 405.53
350(1 + 0.0375)5 ≈ 420.73
Copyright © 2007, Durham Continuing Education
Page 38 of 42
1100
1000
MFB3C – Foundations for College Mathematics
Unit 2 –Support Questions Answers
900
800
700
600
500
Total
Amount
($)
f( x) = 350( 1+ 0.0375) x
400
300
200
100
5
10
15
20
Time in years
b.
The investment is non-linear because it is not a straight line when graphed
and the rate of change is not constant.
Lesson 8
1.
P=
1500
(1+ ( ))
48
0.065
12
P = $1157.39
2.
P=
$1500 – $1157.39 = $342.61
Tricia can spend about $342.61 on the CD player.
1426.73
(1+ ( ))
0.125
12
36
P = $ 982.48
3.
2 = 1(1 +
0.065
52
)n
2 = (1.00125)n
n ≈ 555
555
≈ 10.7 years
52
4.
***Use guess and check
349.82 = 200(1 + i)16
1.7491 = (1 + i)16
***Use guess and check
i ≈ 0.03555
The annual interest rate is 0.03555 x 2 x 100 ≈ 7.11%
Copyright © 2007, Durham Continuing Education
Page 39 of 42
25
MFB3C – Foundations for College Mathematics
Unit 2 –Support Questions Answers
Lesson 9
(
1.
PV =
2.
50 1 − 1 +
PV =
3.
4.
)
-36
68 1 − 1 + 0.105
12
0.105
12
(
= $2092.15
)
0.076 -104
26
0.076
26
R=
1− (1+
(
)
0.1435
12
0.1435 – 60
12 000
12
)
= $4478.36
= $281.40
$6575 – $650 = $5925
R=
( )
1− (1+ ) = $199.07
0.128
12
0.128 – 36
5 925
12
5.
Jake: R =
9380
1− (1+
(
)
0.0945
12
0.0945 – 60
12
)
= $196.77
$196.77 x 60 = 11806.20
$11 806.20 – $9380 = $2426.20
Emily:
( )
R=
1− (1+ ) = $205.35
9380
0.113
12
0.113 – 60
12
$205.35 x 60 = $12 321
$12 321 – $9380 = $2941
$2941 – $2426.20 = $514.80
Emily pays $514.80 more in interest than Jake.
6.
PV =
(
)
0.0625 -40
2
0.0625
2
1500 1 − 1 +
= $33 982.11
Copyright © 2007, Durham Continuing Education
Page 40 of 42
MFB3C – Foundations for College Mathematics
7.
Unit 2 –Support Questions Answers
65 – 27 = 38 years
R=
(
1 000 000 0.0675
26
(1+
)
0.0675 988
26
)
= $217.11
−1
$217.11every two weeks seems reasonable if Josh makes good money and
does not have much debt or other bills to pay. (This is less than a car payment)
Lesson 10
1.
($16 550 + $1175) x 1.13 = $20 029.25
2.
$31 690 X 0.045 = $1426.05
$31 690 – $1426.05 = $30 263.95
3.
MSRP/List Price
Discount
Purchase Price
Freight
Tire/Air Tax
Gas Tax
Administration Fee
Trade-in Allowance
Taxable Total
Patty’s Deal Review
$22 645
Gas Fee
Lien Payout
$22 645
Delivery Price
$459
MFG Rebate
$100
Down Payment
$65
Finance Fee
$55
Amount Financed
0
$23 324
Interest Rate
Taxes
License Plate Fee
$3032.12
$74
Loan Term
Monthly Payment
$30
0
$26460.12
0
$3000
$45
$23505.12
0.039
12
60 months
$431.83
4.
Two advantages for buying a used vehicle over a new one are it is cheaper and
there is less depreciation on a used vehicle. Two disadvantages are it is older so
you don’t know how much repair work will need to be done to the vehicle and
here won’t be a warranty if bought privately.
5.
V = $30 450(1 – 0.20)2
V = $19 488
6.
( ) = $202.03
R=
1− (1+ )
0.049
12
0.049 – 48
8 790
12
Copyright © 2007, Durham Continuing Education
Page 41 of 42
MFB3C – Foundations for College Mathematics
7.
8.
a.
5.6
= 0.056 L/1 km
100
b.
Unit 2 –Support Questions Answers
6.7
= 0.067 L/1 km
100
0.056 x 50 500 = 2828 L
0.067 x 21 485 = 1439.495 L
2828 x 0.78 = $2205.84
1439.495 x 0.924 = $1330.09
Gas:
6.9
= 0.069 L/1 km
100
0.069 x 27 000 = 1863 L
1863 x 0.91 = $1695.33
Lube, oil and filter:
27 000
= 5.4 (round to 5)
5000
5 x $31.95 = $159.75
Tire rotation and brake inspection:
27 000
= 2.7 (round to 2)
10000
2 x $45.50 = $91.00
Wheel alignment inspection:
27 000
= 1.08 (round to 1)
25000
1 x $72.99 = $72.99
Replace all four tires:
4 x $105 + 4 x $20 = $500
Total for the year: $1695.33 + $159.75 + $91.00 + $72.99 + $500 = 2519.07
Total for the month: $2519.07 ÷ 12 = $209.92
Copyright © 2007, Durham Continuing Education
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