1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Can you figure this out? (word problem)

  1. Nov 27, 2006 #1
    Greetings,

    On the board today in my calculus class, the teacher had a word problem that has stumped some other teacher, and wanted to see if we could figure it out (I realize this is the PREcalculus board, but I'm sure calculus isn't required to solve the problem). I don't remember it word for word, but I know the gist of it...

    "You go into a store with a certain amount of dollars and cents. You spend half of your money. After spending half of your money, you now have half as many dollars as you had in cents, and the same number of cents as you had in dollars. How much money did you have to start with?"

    Just in case that isn't clear, here's an example of what an answer COULD look like... original: $48.66
    new: $33.48

    See how the $33 in the "new" amount is half of the cents that was in the original, and the cents in the "new" amount is exactly the amount of dollars in the "original"? Of course, $48.66 isn't the right answer, because 33.48 isn't half of $48.66. I imagine I could come up with an answer after some "brute force", but I'd rather not guess and check. Good luck!
     
  2. jcsd
  3. Nov 27, 2006 #2
    I hope the answer isn't $0.00
     
  4. Nov 27, 2006 #3
    Oh nevermind I got it. $99.98
     
  5. Nov 27, 2006 #4
    KoGs, how was the money split up? How many dollars and how many cents? Also, how was it spent? I'm kinda curious... I'm working this out right now as well.
     
  6. Nov 27, 2006 #5
    Originally the person started with $99.98 since. After spending exactly half, he/she was left with $49.99. Note that 49 is exactly half of 98, and 99 = 99.

    As for how this person spent it.......I dunno. Dinner maybe? :)
     
  7. Nov 27, 2006 #6
    Damnit, I was accidentally counting one cent as 100 dollars :rofl:

    I haven't glanced at your post (post #5), so I can solve it by myself first...

    EDIT: Looks like I have the same thing as you...

    For a second I didn't realize that she could get change from the cashier...lol

    So, Put a quark in it, set up the total amount and then try to set half of that equal to the new amount...
     
    Last edited: Nov 28, 2006
  8. Nov 28, 2006 #7
    I understand in theory how it should work, I just don't know how to go about doing it on paper.
     
  9. Nov 28, 2006 #8
    Use algebra and define 2 variables. One for cents and one for dollars.
    Now you have 2 equations with 2 unknowns. You also have a relationship between the first equation and the 2nd equation.
     
  10. Nov 30, 2006 #9

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Well, from what I can see, you can make a system of 4 equations in 5 unknowns:
    Let:
    D: Number of dollars before spending
    C: Number of cents before spending
    T: Total amount before spending, measured in cents
    D*: Number of dollars after spending
    C*: Number of cents after spending.

    Thus, I get the four equations:
    100D+C=T
    100D*+C*=T/2
    D*=C/2
    C*=D

    From what I see, this gives an infinity of whole numbered solutions.

    However, the reasonable, but by no means necessary, inequality C<100, along with the tacit assumption T>0 yields a unique solution.
     
    Last edited: Nov 30, 2006
  11. Dec 1, 2006 #10

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Here's how I did it. Let C be the number of cents initially, D the number of dollars. Of course, the way the problem is worded, as arildno pointed out, I might have, say, $5 in bills and another $2.50 in "cents"! In order to get a single solution we have to assume "a certain amount of dollars and cents" means $D.C whether the "dollars" are in bills or coins!

    In order to be able to spend half, C must be even. Assume, first, that D is also even. Then D*, how many dollars I have now, is D/2 and C*, how many cents I have now, is C/2. "You now have half as many dollars as you had in cents": D*= C/2 or D/2= C/2 so D= C. "The same number of cents as you had in dollars": C*= D or C/2= D so C= 2D. The only numbers satisfying both are D= C= 0.

    Now suppose D is odd. Then D*= (D-1)/2 and C*= 50+ C/2. "You now have half as many dollars as you had in cents": D*= C/2 so (D-1)/2= C/2 so C= D-1 or D= C+ 1. "The same number of cents as you had in dollars": C*= D so 50+ C/2= D or 50+ C/2= C+1. Then C/2= 49 so C= 98 and D= 98+ 1= 99.

    You must have started with $99.98. After spending half of it, you have $49.99 left. The number of dollars now, 49 is half the number of cents you initially had. The number of cents you have now, 99, is the same as the number of dollars you initially had.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?