Describe A Simple Process For Using Rates And Unit Prices That Might Help Someone Who Is Having Difficulty Understanding These Concepts. Include Either An Example From The Text Or One Of Your Own Examples To Explain The Solution Process?


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For problems of both types, I find it convenient to keep the units with the numbers. Rate has to do with how much useful activity is accomplished in a time period. The activity generally involves production or motion, but is not limited to such things. Sometimes the problems only involve one rate, such as travel speed (mile per hour). Sometimes the problems involve two or more rates (John, Jane, and Pat can paint a house in different amounts of time).  Here's an example of how carrying the units along provides a check of the work.   John can paint a house in 6 hours; Jane can paint the same house in 4 hours. If they work together, how long does it take?   John's rate is (1 house/(6 hours)) = (1/6 house/hour).    (note that time is in the denominator of rate)   Jane's rate is (1 house/(4 hours)) = (1/4 house/hour).   We add the two rates together to determine their total production rate in houses per hour.   (1/6 house/hour) + (1/4 house/hour) = (2/12 + 3/12) house/hour = 5/12 house/hour. Note that the problem asks "hours per house", so we need to invert this result to find the painting time.   1/(5/12 house/hour) = 12/5 hours/house = 2 2/5 hours per house John and Jane can paint the house together in 2 hours and 24 minutes.  Unit Prices are something we see every time we go to a supermarket. These are ratios that have a price in the numerator and a quantity in the denominator. For example, if 150 feet of fence costs $900 to install, the unit price is ($900)/(150 ft) = (900/150) $/ft = $6/ft.  When we compare unit prices, we are comparing fractions. This is most easily done if the same quantity is in the denominator, but it can also be done if the same price is in the numerator. For example, we know that $2 per quart is more expensive than $1.50 per quart. We also know that $3 per gallon (4 quarts) is less expensive (per quart) than $3 per quart.  Problems involving unit prices are sometimes mixture problems (find the total cost of a mix of items at different unit prices). Clearly, to find the cost of an item, you must multiply the unit price times the number of units. For example, if one liter of gas is $1.49, then 10 liters of gas will be (10 liters)*($1.49/liter) = $14.90.  (Note the units become liters/liter, which is 1. We say the units "cancel.")  Here's an example of a mixture problem.   I can buy peanuts in a 20 ounce can for $2. I can buy almonds in a 15 ounce can for $3. If I want to make a mixture that is half-and-half almonds and peanuts by weight, what do I need to sell it for to pay the cost of the ingredients?   The unit price of peanuts is $2/(20 ounces) = $0.10 per ounce.   The unit price of almonds is $3/(15 ounces) = $0.20 per ounce.   If I mix an ounce of peanuts and an ounce of almonds, their cost will be ($0.10 + $0.20)/(2 ounces) = $0.30/(2 ounces) = $0.15 per ounce.   This particular problem can also be worked by considering the least common multiple of the can sizes (60 ounces). 60 ounces of peanuts will cost (60 ounces)*($2/(20 ounces)) = 3*($2) = $6.  60 ounces of almonds will cost (60 ounces)*($3/(15 ounces)) = 4*$3 = $12.  Thus the total cost of 120 ounces of mix is ($6 + $12) = $18. The price per ounce must be at least $18/(120 ounces) = $0.15 per ounce.

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