Savings on Gasoline
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.
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 your problem): 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/
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
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