Financial Algebra 1st Edition Chapter 5 Automobile Ownership
Page 260 Problem 1 Answer
Given a car is traveling at R miles per hour for M minutes.
Need to write an algebraic expression for the distance traveled.
As the distance travelled is the product of speed and time and there are 60 minutes in one hour.
Then, the distance travelled in miles in one hour isR×1=Rmiles.
The distance travelled in miles in one minute is R÷60=R/60miles.
Thus, the distance travelled in miles in M minutes is R×M/60=RM/60miles.
Given a car is traveling at R miles per hour for M minutes.
Then, an algebraic expression for the distance traveled is RM/60 miles.
Page 261 Problem 2 Answer
Given: Danielle drove from Atlanta, Georgia, and Denver, Colorado, which is a distance of 1,401 miles.
D=1,401 R=58
T=?
To find: If she averaged 58 miles per hour on her trip, how long is her driving time to the nearest minute?
Using the formula D=R×T
Dividing throughout by R ∴T×R/R=D/R
Simplify ∴T=D/R
Substitute D and R ∴T=1,401/58
Calculate ∴T=24.155
The answer is a non-terminating, repeating decimal as indicated by the bar over the digit 6.
The time rounded to the nearest tenth of an hour is 24.1 hours.
If you are using a calculator and the display reads 24.1551724, the calculator has rounded the last digit, but it stores the repeating decimal in its memory.
Because you know that the exact time is between 24 and 25 hours, use only the decimal portion of the answer.
Once the answer is on the calculator screen, subtract the whole number portion.
24.1551724−24
=0.1551724
The number of sixes displayed will depend upon the accuracy of your calculator.
There are 60 minutes in an hour, so multiply by 60.
0.1551724×60
=9.310
The decimal portion of the hour is 9 minutes. Danielle will be driving for 24 hours and 9 minutes.
Therefore Danielle drove for 24 hours and 9 minutes.
Page 262 Problem 3 Answer
Given: The distance from the Canadian border to Montreal, Quebec, is approximately 65 kilometers.
If the entire trip took her about 3×3/4 hours, was her average speed for the trip?
To explain: In Example 3 above, could Kate’s km/h have been calculated by multiplying her miles per hour by the conversion factor?
Kate’s average speed can be reported in miles per hour or kilometers per hour.
To report her speed in miles per hour, convert the entire distance to miles.
To change 65 kilometers to miles, multiply by the conversion factor0.621371.
65×0.621371=40.389115
The distance from the Canadian border to Montreal is approximately 40.4 miles.
Kate’s total driving distance is the sum of the distances from Albany to the Canadian border and from the Canadian border to Montreal.
176+40.4=216.4miles
D=216.4 and T=3.75.
D=R×T
D/T=R×T/T
D/T=R
216.4/3.75=R
57.7≈R
When the conversion factor take place we can conclude the following reason to determine her speed in kilometers per hour.
To change the portion of the trip reported in miles to kilometers, multiply 176 by the conversion factor 1.60934.
176×1.60934≈283.2
There are approximately 283.2 kilometers in 176 miles.
The distance from Albany to Montreal is 283.2+65,or348.2 kilometers.
Let D=348.2 and T=3.75 in the distance formula.
348.2/3.75=R
92.853=R
Kate traveled approximately 93 kilometers per hour.
Hence form the above conclusion we can say that
Kate traveled at approximately 58 miles per hour or when the conversion take place we can write the same thing in diffrent way in different parameter
Kate traveled approximately 93 kilometers per hour.
Page 263 Problem 4 Answer
Given: Lily drove a total of 500 miles on g gallons of gas.
To Express her fuel economy measurement in miles per gallon as an algebraic expression.
Using the formula D=M×G
We have D=500
G=g
Therefore substitute the above value we get
500=M×g
Simplify M=500/g
Therefore her fuel economy measurement in miles per gallon as an algebraic expression is M=500/g
Page 263 Problem 5 Answer
Given: A person begins a trip with an odometer reading of A miles and ends the trip with an odometer reading of B miles.
To find: If the car gets C miles per gallon and the fi ll-up of gas for this trip cost D dollars, write an algebraic expression that represents the price per gallon.
Begin by computing the distance Barbara traveled. Find the difference between her ending and beginning odometer readings.
Let X be the difference between her ending and beginning odometer readings.
i.e.A−B=X
Therefore the person has travelled X miles .
Since the person car get C mpg,you can determine the number of gallons of gas used on the trip with the formula D=M×G
Where D is the distance traveled, M is the miles per gallon, and G is the number of gallons used.
Substitute X in D and C in M
we get X=C×G
Simplify G=X/C
Therefore the person used X/C gallons of gas in this trip.
If the person’s total gas bill was D dollars divide this total amount by the number of gallons used to get the price per gallon paid.
DX/C=price per gallons
Therefore the person paid DX/C per gallon for this fill-up.
Therefore the price per gallons is DX/C
Page 264 Problem 6 Answer
Given: The average price of gas per liter was 1.28 Canadian dollars
To find: what is the equivalent gas price per gallon in U.S. currency?
Angie must find the current currency exchange rate.
The currency exchange rate is a number that expresses the price of one country’s currency calculated in another country’s currency. Up-to-date
exchange rates are available on the Internet.
Angie needs to know what 1 U.S. dollar (USD) is worth in candian dollars
For the time of his travel, 1 USD = 1.07 Canadian dollars
Here we know that 1 gallon to liter = 3.78541 liter.
1.28/1.07
=1.19
so the total cost is 1.19×3.78=4.49
Hence we conclued that the equivalent gas price per gallon in U.S. currency will be 4.49 us dollar
Page 264 Problem 7 Answer
Given: Angie knew that the price of gas in her home town was 2.50 per gallon
To find: What is the equivalent price in Canadian dollars per liter?
Angie needs to express the U.S. gas price as a price in USD per liter.
There are approximately 3.8 liters in a gallon.
Divide the price per gallon by 3.8 to determine the price per liter in USD.
2.50÷3.8≈0.76
Her home town gas price is equivalent to about 0.76 USD per liter.
So gas is less expensive in Canada, $0.64<$0.76.
To compare the prices in Canada, multiply the USD amount by the exchange rate
Exchange rate was 13.3
0.76×13.3=10.11
The gas in her home town would sell for about 10.11 Mexican pesos.
Just as the comparison in USD showed, the comparison in Canada shows that gas is less expensive in Canada,8.50<10.11.
Therefore $10.11 is the equivalent price in Canadian dollars per liter.
Page 265 Problem 8 Answer
To find: How might the quote apply to what you have learned?
If you want to drive with a car, then knowing how to drive the car is not sufficient and you then also need to know other important information from the car (namely, what is “under the hood” of the car).
For example, it is important to know the location of the oil, tire pressure, brakes, antifreeze/coolant, etc. such that you can regularly check if they need to be fixed/replaced and such that you then can take action to fix the problem.
To drive with a car, you need to know important information from the car beside the knowledge on how to drive.
Page 265 Problem 9 Answer
Given: Arthur travels for 3 hours on the freeway. His average speed is 55 mi/hr
To find: How far does he travel?
Using the formula D=R×T
D=55×3
D=165
Therefore Arthur Travels 165 miles.
Page 265 Problem 10 Answer
Given: Steve’s SUV has a17 -gallon gas tank. The SUV gets an estimated 24 miles per gallon.
To find: Approximately how far can the SUV run on half a tank of gas?
Total capacity is 17 gallon but it is half-full.
Amount of gas in the tank of SUV=17/2=8.5gallon
Distance=miles per gallon× gallons
=24×8.5=204
Distance =204 miles .
Therefore The SUV can run 204 miles on half a tank of gas.
Page 265 Problem 11 Answer
Given: Becky is planning a 2,100-mile trip to St. Louis to visit a college. Her car averages 30 miles per gallon.
TO find out: About how many gallons will her car use on the trip?
Using the formula D=M×G
Substitute the values.2100=30×G
SimplifyG=2100/30
=70
Therefore Becky’s car will use 70 gallons of gas on the trip.
Page 265 Problem 12 Answer
Given: Robbie’s car gets M miles per gallon.
To Write: an algebraic expression that represents the number of gallons he would use when traveling 270 miles.
Substitute the given values in the formula D=M×G
We get 270=M×G
Simplify further G=270/M
Therefore G=270/M is the required solution.
Page 265 Problem 13 Answer
Given: Michael used his car for business last weekend.
When he reports the exact number of miles he traveled, the company will pay him 52 cents for each mile.
At the beginning of the weekend, the odometer in Michael’s car read 74,902.6 miles.
At the end of the weekend, it read 75,421.1 miles.
To Find: How many miles did Michael drive during the weekend?
Given: Reading beginning weekend:74,902.6 miles
Reading end weekend:75,421.1 miles
The amount that Michael drove during the weekend is then the difference between the two readings of the odometer.
Amount driven in weekend = Reading end weekend−Reading beginning weekend
=75,421.1−74,902.6
=518.5
Thus Micheal drove 518.5 miles.
Therefore Micheal drove 518.5 miles during the weekend.
Page 265 Problem 14 Answer
Given: Michael used his car for business last weekend.
When he reports the exact number of miles he traveled, the company will pay him 52 cents for each mile.
At the beginning of the weekend, the odometer in Michael’s car read 74,902.6 miles.
At the end of the weekend, it read 75,421.1 miles.
To Find: How much money should his company pay him for the driving?
Given: Reading beginning weekend=74,902.6 miles
Reading end weekend =74,902.6 dollars
Price per mile=52 cents=0.52 dollars
The amount that Michael drove during the weekend is then the difference between the two readings of the odometer.
Amount driven in weekend= Number of miles= Reading end weekend− Reading beginning
=75,421.1−74,902.6
=518.5
The total price of the trip is then the product of the price per mile and the number of miles driven:
Total price=Price per mile ×Number of miles
=0.52×518.5
=269.62
Thus the company should then pay him the total price of $269.62.
Therefore $269.62 money should his company pay him for the driving.
Page 265 Problem 15 Answer
Given: Lenny’s car gets approximately 20 miles per gallon.
He is planning a 750-mile trip.
To Find a. About how many gallons of gas should Lenny plan to buy?
Using the formula D=M×G
Substitute the values. 750=20×G
Simplify G=750/20
=37.5
Therefore Lenny should buy 37.5 gallons of gas.
Page 260 Problem 16 Answer
Given: from previous exercise No. of gallons of gas to be brought =37.5
To find: At an average price of4.10 per gallon, how much should Lenny expect to spend for gas
Multiply the price per gallon by no. of gallons of gas to be bought.
4.10×37.5=153.75
Therefore Lenny will have to spend 153.75.
Page 265 Problem 17 Answer
Given: Francois’ car gets about 11 kilometers per liter. She is planning a 1,200-kilometer trip.
To find: About how many liters of gas should Francois plan to buy? Round your answer to the nearest liter.
Using the formula D=K×L
Substitute the values. 1200=11×L
Simplify L=1200/11
=109.09
Therefore Francois should buy about 109 liters of gas.
Page 265 Problem 18 Answer
Given : From previous exercise we have No. of gallons to be =109
To find: At an average price of$1.45 per liter, how much should Francois expect to spend for gas?
Multiply the price per liter by no. of liters of gas to be bought.1.45×109=158.05.
Francois will have to spend158.05
Therefore Francois will have to spend 158.05.
Page 265 Problem 19 Answer
Given: Nola’s car gets approximately 42 miles per gallon. She is planning to drive x miles to visit her friends.
To find: What expression represents the number of gallons of gas she should expect to buy?
Using the formula D=M×G
Substitute the values. x=42×G
Simplify G=x/42
Therefore G=x/42 is the required expression
Page 265 Problem 20 Answer
Given: From previous exercise G=x/42
To find: At an average price of $2.38 per gallon, write an expression for the amount that Nola will spend for gas.
Multiply the price per gallon by no. of gallons of gas to be bought. 2.38×x/42
2.38×x/42
=0.057x
She will have to spend0.057x
2.38×x/42
=0.057x miles
Therefore the amount that Nola will spend for gas is 0.057x miles.
Page 266 Exercise 1 Answer
Given, Initial odometer reading42876.1
Final odometer reading43156.1
Mileage 35 per gallon Cost$34.24
To find the price pay per gallon of gas fill up.
Distance travelled will be difference of odometer reading
=43156.1−42876.1
=280 miles
From tip,280=35×no. of gallons of gallons=8
Eight gallons cost$34.24
Therefore, per gallon cost$4.28
He pay per gallon price of $4.28 on filling gas.
Page 266 Exercise 2 Answer
Given,
| Price per gallon | Total gas cost | Number of people in car pool | Gas cost per person |
| $3.99 | a. | 4 | g. |
| $4.08 | b. | 5 | h. |
| $4.15 | c. | 3 | i. |
| $4.30 | d. | 6 | j. |
| D | e. | 4 | k. |
| P | f. | C | I. |
To find the values of variables−l
From tip,
a=10×3.99=$39.9
b=12×4.08=$48.96
c=17×4.15=$70.55
d=26×4.30=$111.8
e=$15D
f=$GP
g=a/4=$9.975
h=b/5=$9.792
i=c/3=$23.52
j=d/6=$18.63
k=e/4=$15D/4
l=f/C=GP/C
The value of variables are
a=$39.9
b=$48.96
c=$70.55
d=$111.8
e=$15D
f=$GP
g=$9.975
h=$9.972
i=$23.52
j=$18.63
k=$15D/4
l=$GP/C
Page 266 Exercise 3 Answer
Given, Distance from Burlington to Canadian border is approximately 42 miles.
Canadian border to Ottawa is approximately280km=173.98mi
Time=4.3hr
To find the average speed in miles per hour.
Total distance=42+173.98
=219.98
Average speed=219.98/4.3
=50.23
≈50
The average speed in miles per hour is 50 approximately.
Page 266 Exercise 4 Answer
Given, A car averages 56mi/h on a trip.
To write an equation that shows the relationship between distance, rate, and time for this situation.
From tip, Distance=56×time
The equation between distance, rate and time is Distance=56×time
Page 266 Exercise 5 Answer
Given, on a trip.
A car averages 56mi/h on a trip
For time be the independent variable and distance be the dependent variable, to draw and label the graph of this equation.
The equation is distance=56×time
y=distance
x=time
The graph is
Page 266 Exercise 6 Answer
Given, A car averages 56mi/h on a trip.
To determine approximately how far this car would travel after 14 hours.
From the graph, After 14 hours,
distance=784mi
After 14 hours, the car travels 784 miles.
Page 266 Exercise 7 Answer
Given, on a trip.
A car averages 56mi/h
To use the graph to determine the approximate length of time a 500-mile trip would take
Graph is
About nine hours
The approximate time a 500-mile trip would take is 9 hours.
Page 266 Exercise 8 Answer
To write a formula to calculate the speed of the car for the trip in cellC1
Distance will be ending odometer reading−initial odometer reading
Speed=(A2−A1)/A4
A formula to calculate the speed of the car for the trip in cell C1 is C1=(A2−A1)/A4
Page 266 Exercise 9 Answer
To write a formula to calculate the number of gallons of gas used in cellC2.
Distance is(A2−A1)
From tip, No. of gallons required is (A2−A1)/A3
A formula to calculate the speed of the car for the trip in cell C2 is (A2−A1)/A3
Page 266 Exercise 10 Answer
To write a formula to calculate the total cost of gas for the trip in cellC3
Distance is(A2−A1)
Total cost is (A2−A1)×A5/A3
=C2×A5
A formula to calculate the speed of the car for the trip in cell C3 is =C2×A5
Page 267 Exercise 11 Answer
Given,1USD=1.07Canadiandollars(CAD)1USD=89.85Japaneseyen(JPY)
1USD=1.16Australiandollars(AUD)1USD=1.00Swissfranc(CHF)
1USD=0.69Euros(EUR)1USD=7.34SouthAfricanrand(ZAR)
To complete the chart.
By using spreadsheet
| CAD | EUR | AUD |
| 4.01 | 2.62 | 4.41 |
| 16.85 | 10.87 | 18.27 |
| 21.4 | 13.8 | 23.2 |
| 191 | 123.16 | 207.06 |
| 267.5 | 172.5 | 290 |
| 5885 | 3795 | 6380 |
The final chart is
| USD | CAD | EUR | AUD |
| 3.8 | 4.01 | 2.62 | 4.41 |
| 15.75 | 16.85 | 10.87 | 18.27 |
| 20 | 21.4 | 13.8 | 23.2 |
| 178.5 | 191 | 123.16 | 207.06 |
| 250 | 267.5 | 172.5 | 290 |
| 5500 | 5885 | 3795 | 6380 |
Page 267 Exercise 12 Answer
Given, 1USD=1.07Canadiandollars(CAD)1USD=89.85Japaneseyen(JPY)
1USD=1.16Australiandollars(AUD)1USD=1.00Swissfranc(CHF)
1USD=0.69Euros(EUR)1USD=7.34SouthAfricanrand(ZAR)
To complete the chart.
| x | 1 | 2 | 3 |
| y | 56 | 112 | 168 |
By using spreadsheet
a=79.44
b=123.19
c=73.28
d=85
e=11.58
f=0.95
The values are
a=79.44
b=123.19
c=73.28
d=85
e=11.58
f=0.95
Page 267 Exercise 13 Answer
Average price of gas in Spain is approximately 1.12 euros per liter.
To find the U.S. dollar equivalent of this From tip,1.12
euro=1.12/0.69 U.S. dollars
=1.62 U.S. dollars
The amount equivalent to1.12
Euro is 1.62 U.S. dollars.
Page 267 Exercise 14 Answer
Average price of gas in Spain is approximately 1.12 euros per liter.
To find the U.S. dollar equivalent per one gallon.
From tip, and 1.12US. dollar=1.62 Euro Per gallon price is 1.62×3.8
=6.16
The rate equivalent of this in U.S. Dollars per gallon is$6.16.
Page 267 Exercise 15 Answer
The average price of gas in Johannesburg is about 19.24ZAR per liter.
To find the U.S. dollar equivalent of this
From tip,19.24ZAR=19.24/7.34 USD
=2.62USD
The amount equivalent to19.24Z AR is 2.62USD.
Page 267 Exercise 16 Answer
Average price of gas in Johannesburg is approximately19.24ZAR
To find the U.S. dollar equivalent per one gallon.
From tip, and2.62USD=19.24ZAR
Per gallon price is 2.62×3.8=9.96USD
The rate equivalent of this in U.S. Dollars per gallon is9.96USD.
Page 267 Exercise 17 Answer
Given that “Willie traveled in India, he paid an average of 87.42 Indian rupees for a liter of gas.”
Exchange rate = x that means 1 USD =x INR
Here we will use this relation to represent the given price in US Dollars.
We have 1 USD =x INR
Thus 1/x
USD =1 INR
Now we have 87.42 Indian rupees.
Thus 87.42/x
USD=87.42INR
We then not that 87.42 rupees for a liter of gas is equal to 87.42/x USD/L
Price of gas in US Dollars is 87.42/x USD/L
Page 267 Exercise 18 Answer
We have “Willie traveled in India, he paid an average of 87.42 Indian rupees for a liter of gas.”
And result from last part we have 1 USD =x INR
And we know that 1 gallon ≈3.8 liters
We will use these two relation to find the rate equivalent to in U.S. dollars per gallon.
We have 1 gallon ≈3.8 liters
And 1USD=xINR
We know that 87.42 Indian rupees per liter is equal to 87.42/x USD per liter.
Per liter gas is equivalent to 87.42/x USD
Such that 87.42 USD
xL×3.8L 1 gallons
=87.4×3.8/x USD/gallon =332.12/x USD/gallon
The rate equivalent to in U.S. dollars per gallon is 332.12/x USD/gallon
Page 267 Exercise 19 Answer
We have “Willie traveled in India, he paid an average of 87.42 Indian rupees for a liter of gas.”
And Willie spent about$115 From previous results we have 87.42 Indian rupees per liter is equal to 87.42/x USD per liter.
We will use these results to find the number of gallons of gas.
We know that 87.42 Indian rupees per liter is equal to 87.42/x USD per liter.
And rate equivalent to in U.S. dollars per gallon is 332.12/x USD/gallon
Next we know that the total price is $115 (115 USD). The total price is also known to be the product of the price per gallon and the number of gallons.
Let number of gallons is y
Then Total price = Price per gallon × Number of gallons
115 USD =332.12/x USD/gallon ×y gallons
Thus 115 USD
115x/332.12
=332.12y/x USD
=332.12y
=y
Thus the number of gallons of gas is 115x/332.12 =y
Page 266 Exercise 20 Answer
We have “Willie traveled in India, he paid an average of 87.42 Indian rupees for a liter of gas.”
And “Willie spent about $115.”
We also have 1 gallon ≈3.8 liters And 87.42 Indian rupees per liter is equal to 87.42/x USD per liter.
The rate equivalent to in U.S. dollars per gallon is 332.12/x USD/gallon ( previous results )
Number of gallons =115x/332.12
We know that 1 gallon ≈3.8 liters
Thus amount of gas in liters is
Total price = Price per liter × Number of liters
115 USD =87.42/x USD/liter ×y liters
115x=87.42y
115x/87.42=y
Thus the amount of gas in liters is 115x/87.42