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### Savings on Gasoline

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Date: 05/22/2001 at 22:03:27
From: Janet
Subject: Problem of the Week

This is the question: Your old vehicle averaged 22 mpg and your new
vehicle has an EPA sticker stating that it should average 37 mpg.
Suppose you drive about 12,500 miles per year and the cost of gasoline
averages \$1.85 per gallon. How much should you expect to save on
gasoline during the first year you own your new vehicle? ...the second
year? ...the third year?

I don't know what kind of math to use for this, and I have tried, but
I'm not sure if I have to divide, add, or multiply.
```

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Date: 05/23/2001 at 16:41:21
From: Doctor Rick
Subject: Re: Problem of the Week

Hi, Janet.

The best way to approach this problem is to calculate separately the
cost of driving each car for a year, then compare the costs.

The main calculation is a rate problem - actually, TWO rate problems,
one after the other. Just as you would use:

Distance = Speed * Time
Miles    = Mph   * Hours

to calculate how far you go in a certain time, you can use:

Miles = Mpg * Gallons

to calculate how far you can go on a quantity of gasoline, and:

Dollars = Dollars per gallon * Gallons

to calculate the cost of a certain amount of gasoline.

You don't want to calculate how far you can go on a certain amount of
gasoline, though: you want to calculate how many gallons it takes to
go a certain distance. This "backward problem" is the same sort of
thing as calculating how long it takes to get somewhere at a certain
speed. You use the inverted form of the rate equation (divide both
sides by mpg):

Gallons = Miles / Mpg

Once you have the gallons used in a year, you can use the last rate
equation to find how much that many gallons will cost.

Rate problems are easy when you use dimensional analysis. That sounds
tough, but it just means keeping the units with your numbers, and
treating them as if they were numbers. Here's an example (not from

How long does it take to go 50 miles at 40 mph?

That unit "mph" is "miles per hour." It means the number of miles you
go in one hour. We can write it as a ratio:

40 miles
40 miles/hour = --------
1 hour

We want to end up with a time. Its units will be hours (the unit of
time in the speed). The hours must be in the numerator. For the speed,
the hours are in the denominator. To get them to the numerator, we
take the reciprocal:

1 hour
--------
40 miles

We just want hours in our answer, so we need to get rid of the miles
in the denominator. How can we do that? We can multiply by miles (in
the numerator), and the miles will "cancel out." It happens that we
have a number of miles handy, namely the 50 miles that we're driving.
Do the multiplication:

1 hour
-------- * 50 miles = 50/40 hours
40 miles

I have canceled the miles/miles, leaving just the hours as our final
unit - just what we wanted. Of course, we can also simplify the
numbers:

50/40 = 5/4 = 1 1/4 hours

This method is useful, but it isn't the solution to every problem. You
need to know that you have a rate problem in order to use it. Let's
just go back to the rate equation and see how it gives the same
result.

Distance = Speed * Time

Another form of this equation is:

Time = Distance / Speed

Plugging in the numbers from the problem:

Time = 50 miles / 40 miles/hour = 5/4 hours

I hope this helps you. For further information, we have an item in our
Dr. Math Archives that discusses the ideas of rates and dimensional
analysis:

Problems on Rates and Unit Conversions
http://mathforum.org/library/drmath/view/58038.html

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/
```

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Date: 06/02/2001 at 00:12:18
From: Janet
Subject: Re: Problem of the Week

I just want to thank you for taking time to help me. So thank you a
lot.

Janet
```
Associated Topics:
High School Basic Algebra
High School Physics/Chemistry
Middle School Algebra
Middle School Word Problems

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