WEEK_2_HOMEWORK_SOLUTIONS[New_Version][1]

WEEK_2_HOMEWORK_SOLUTIONS[New_Version][1] - Threaded...

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Threaded Solutions For Week #2 1) Solution for Problem #5-12 Question: In 1955 only 435 Kaiser-Darrins were built, because Kaiser-Frazer bailed out of the car business. Only 435 of these fantastic cars were ever built, they sold for $3,668 according to an article in the Chicago Sun-Times March 5, 2007 edition. The Kaiser-Darrin ended up being the most prized of Henry J. Kaiser’s cars. It’s valued today at $62,125 if in excellent condition, which is 1 ¾ times as much as a car in very nice condition-if you can find an owner willing to part with one for any price. What would be the value of the car in very nice condition? Solution: Value = V 1 ¾ V = 62,125 Isolate the variable ‘V’ V = 62, 125 ÷ 1 ¾ V = $35,500 2) Solution for Problem #5-13 Question: Joe Sullivan and Hugh Key sell cars for a Ford dealer. Over the past year, they sold 300 cars. Joe sells 5 times as many cars as Hugh. How many cars did each sell? Solution: As in the question above there is a relationship between the “unknowns” of this problem, namely, the number of cars sold by Joe and Hugh . Therefore we need to determine a variable to represent the number of cars sold by Joe and Hugh respectively. . As stated in sentence #3 above ” Joe sells 5 times as many cars as Hugh” therefore we will use a variable that will represent the quantity of unknown cars sold by the individual who sold the least amount, namely, Hugh. So if C = the number of cars sold by Hugh, then Joe would have sold 5 times that amount or 5C. We start with the following equation: C + 5C = 300 This problem has less for us to do since value is not a consideration. We just want to know how many cars did Joe an Hugh sell. Our next equation is the result of combining like terms from above, thus we get: 6C = 300 Now divide both sides of the equal sign by 6 and you get: C = 50
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NOTE: When a relationship does not exist between two variables and by relationship I mean where one is in multiples of the other, i.e., twice as many as another, where one would be “x” and the other would be “2x”. We now substitute the 50 into our original equation of: 50 + 5(50) = 50 + 250 = 300 Since it totals what we had as the total in the original question we can state that Joe sold 250 cars and Hugh sold 50! 3) Solution for Problem
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This note was uploaded on 12/13/2010 for the course GM 400 taught by Professor - during the Spring '10 term at Keller Graduate School of Management.

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WEEK_2_HOMEWORK_SOLUTIONS[New_Version][1] - Threaded...

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