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The better buy problem

  1. Sep 9, 2009 #1
    1. The problem statement, all variables and given/known data
    Bob is deciding whether to buy Car A which is $4000 and 20mpg, or Car B which is $5500 and 30mpg. Bob guesstimates that the gas will be about $1.25 per gallon. Repair costs will not factor in to this problem. How many miles would Bob need to drive, before Car B becomes the better buy?


    2. Relevant equations



    3. The attempt at a solution
    Well, I thought about subtracting 5500-4000 to get $1500. I guessed that this would be the amount that Bob would have to make-up. So if Car B drives 100 miles, and the car gets 30mpg, I calculated that the total gas would cost $4.16. So I made a proportion:

    100 = X
    $4.16 $1500

    So, 4.16X = 150,000
    X=150,000/4.16
    X=36057miles

    I got 36,057 miles. Is this correct? Is there another way to complete this problem? This section of my chapter is velocity, position, and time.

    Thanks
     
  2. jcsd
  3. Sep 9, 2009 #2

    berkeman

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    Staff: Mentor



    I don't think that is the correct answer. You need to set up the equation for the total cost of ownership for each car, based on the initial price plus the cost to drive X miles.
     
    Last edited: Sep 9, 2009
  4. Sep 9, 2009 #3

    berkeman

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    Staff: Mentor

    One thing that may help you visualize how to set up the problem....

    Draw an x-y graph, with the vertical axis labeled with total cost in $, and the horizontal axis labeled with distance travelled in miles. The $4000 car starts out as a point on the vertical axis at $4000, and the more expensive one starts out at $5500. There is a line for each car, starting at their initial point, and based on their gas consumption rate in mpg. The line up and to the right for the 4000 car has a steeper slope, because of the worse mileage rating. The two lines intersect at a point where the total cost of ownership is equal. Past that intersection point, the $5500 car is cheaper to own...
     
  5. Sep 9, 2009 #4
    I got 72,000 miles for where they are equal. So, after 72,001 miles would make Car B the best choice.

    Do you know if there is a faster way to complete this problem?
     
  6. Sep 9, 2009 #5

    berkeman

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    Staff: Mentor

    That's the same answer I got. I don't know about a faster way. I initially tried just figuring out how to get the $1500 difference, but then did the sketch/plot, and realized that I was neglecting the initial investment cost. If you think about it some, and keep checking your alternate methods' answers, you may find a good trick. Good luck!
     
  7. Sep 9, 2009 #6
    I found that if I plugged in the formula:
    CostA = ((1.25x)/20) + 4000
    CostB = ((1.25x)/30) + 5500

    into my calculator, and then made a table, plugging in values of x; in factors of 10000; it sped it up a little bit.
     
  8. Sep 9, 2009 #7
    Conceptually one can approach the problem in much the same way as the OP. Car B costs $1500 more than A but gets better gas milleage. What is important to consider is the relative savings between the cars. The cost per mile for A is (1.25)/20 while B is (1.25)/30. The difference is 1/48. So it boils down to how many many miles at $1/48 per mile savings must one drive in order to save $1500? The answer is 48 x 1500 = 72,000 miles.

    Whats really happening here is solving the system

    [tex]\left\{ \begin{array}{c} y=4000 + \frac{1.25}{20}x \\ \\
    y=5500 + \frac{1.25}{30}x \end{array}[/tex]

    which results in the equation and solution:

    [tex]5500 + \frac{1.25}{30}x = 4000 + \frac{1.25}{20}x[/tex]

    [tex]5500 - 4000 = \frac{1.25}{20}x - \frac{1.25}{30}x[/tex]

    [tex]1500 = \left( \frac{1}{16} - \frac{1}{24} \right) x[/tex]

    [tex]1500 = \frac{1}{48}x[/tex]

    [tex]x= 48 \cdot 1500 = 72000[/tex]

    --Elucidus
     
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