Trick or Treat?

Discussion in 'Off-Topic' started by jpanhalt, Oct 16, 2009.

  1. jpanhalt

    Thread Starter AAC Fanatic!

    Jan 18, 2008
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    Instead of handing out candy this Halloween, hand out this problem. I dare you. Assume the price of each candy remains constant.

    Daniel bought 1 pound of jelly beans and 2 pounds of chocolates for $2.00. A week later, he bought 4 pounds of caramels and 1 pound of jelly beans, paying $3.00. The next week, he bought 3 pounds of licorice, 1 pound of jelly beans, and 1 pound of caramels for $1.50. How much would he have to pay on his next trip to the candy store if he bought 1 pound of each of the 4 kinds of candy?

    John
     
  2. mik3

    Senior Member

    Feb 4, 2008
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    Are you sure the problem statement is correct? :p
     
  3. mik3

    Senior Member

    Feb 4, 2008
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    He paid 2$ because the sales assistant forgot to include the licorice in the price. :rolleyes:
     
  4. jpanhalt

    Thread Starter AAC Fanatic!

    Jan 18, 2008
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    The problem is solvable.

    John
     
  5. mik3

    Senior Member

    Feb 4, 2008
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    There are 5 unknowns and 4 equations. How you can solve it?

    So you say that there is not trick but just mathematics.
     
  6. jpanhalt

    Thread Starter AAC Fanatic!

    Jan 18, 2008
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    There is no trick to the question. I believe there is only one answer.

    John
     
  7. mik3

    Senior Member

    Feb 4, 2008
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    Do you know it?
     
  8. jpanhalt

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    I just PM'd you the answer.

    John
     
  9. Ratch

    New Member

    Mar 20, 2007
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    jpanhalt,

    The trick to solving this problem is not to try to find the price per pound of all four different candies. That would be trying to solve 4 unknowns with only 3 equations. Instead, find what the sum of all 4 candies rates are. To do that, find out what each rate is in terms of a single reference rate. I used jelly beans as the reference rate. Without trying to post all the algebra involved, the results are:

    Using Cm for carmel,
    Ch for chocolate
    Jb for Jellybeans
    Lq for licorice


    Cm = -0.25*Jb + 0.75
    Ch = -0.50*Jb + 1.00
    Jb = Jb
    Lq = -.025*Jb + 0.25
    ---------------------------- summing the left and right sides
    Cm + Ch + Jb + Lq = 2.00

    Notice that 3 equations are sufficient if three unknowns are solved in terms of a single selected unknown. So the price paid for a pound of each type of candy is $2.00, and mik3 was correct. Congratulation to mik3 for solving it fast and first!

    Ratch
     
  10. mik3

    Senior Member

    Feb 4, 2008
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    Well, I didn't really solved it like you. I omitted the last equation

    Cm + Ch + Jb + Lq = ?

    and solved for the first three equations (three unknowns and three equations).

    Then I said that he forgot to include the licorice in the prize and the prize of the others came out to be 2%. It was just luck. :p
     
  11. Ratch

    New Member

    Mar 20, 2007
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    mik3,

    Being lucky is often better than being clever. Mr. Gates did not get to be a multi-BILLionaire just by being smart and clever. He had a incredible streak of good luck at the beginning of his career.

    Ratch
     
    Last edited: Oct 16, 2009
  12. jpanhalt

    Thread Starter AAC Fanatic!

    Jan 18, 2008
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    My solution was a little different. Once it is realized that the jelly beans are common to each purchase, it doesn't matter what they cost -- it is math puzzle after all. I set a sensible cost of $0 for them, which makes getting the total cost trivial.

    If you are stimulated to answer a question that wasn't asked, namely the price of each candy, one can do that by assigning another value to the jelly beans. If they are $1 per pound, then licorice is free. I like licorice, so that is my favorite non-solution. There are many more. For interest, try jelly beans at $0.50 per pound.

    John
     
  13. steveb

    Senior Member

    Jul 3, 2008
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    I just noticed this problem, so am too late. However, I'll mention my method to solve which is very simple minded.

    Total all 3 purchases first:

    3 JB + 2 Choc + 5 Caram + 3 Lic = $6.50

    Now add the third purchase:

    4 JB + 2 Choc + 6 Caram + 6 Lic = $8.00

    Now add 2 times the first purchase

    6 JB + 6 Choc + 6 Caram + 6 Lic = $12.00

    Now divide by 6

    1 JB + 1 Choc + 1 Caram + 1 Lic = $2.00
     
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