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Math Help - Can I use inequalities?

  1. #1
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    Can I use inequalities?

    I have a problem:
    A man walks in a store and spend half of his money. After he got out he saw that he has as much cents, as he had dollars in the beginning and that he has dollars as half of the cents he had before he entered the store.

    So I have for variables and only three equations.

    Let's say something like that:
    100*X1 + X2 = 200*Y1 - 2*Y2
    Y2 = X1
    2*Y1 = X2
    where X1 is the dollars he had, X2 -cents he got, Y1 - dollars he has now, Y2 -cents now.

    But I also know that in one dollar there is no more than 100 cents.
    Can I use that to solve the problem?

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by Cherufe View Post
    I have a problem:
    A man walks in a store and spend half of his money. After he got out he saw that he has as much cents, as he had dollars in the beginning and that he has dollars as half of the cents he had before he entered the store.

    So I have for variables and only three equations.

    Let's say something like that:
    100*X1 + X2 = 200*Y1 - 2*Y2
    Y2 = X1
    2*Y1 = X2
    where X1 is the dollars he had, X2 -cents he got, Y1 - dollars he has now, Y2 -cents now.

    But I also know that in one dollar there is no more than 100 cents.
    Can I use that to solve the problem?

    Thanks in advance.
    1. The man has x \$ + y\ cts = (x + \frac y{100})\ \$.

    2. When he has spent half of his money he has:

    \dfrac12 (x + \frac y{100}) = \dfrac12 y + \dfrac x{100}

    Solve for y:

    3. y = \dfrac{98}{99} x

    Since there aren't any fractions of cents possible the smallest amount of cents is 98 if x = 99.

    That means the man had 99.98 $ and when he left the store he had 49.99 $ which is exactly the half of 99.98 $.
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  3. #3
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    Quote Originally Posted by Cherufe View Post
    I have a problem:
    A man walks in a store and spend half of his money. After he got out he saw that he has as much cents, as he had dollars in the beginning and that he has dollars as half of the cents he had before he entered the store.

    So I have for variables and only three equations.

    Let's say something like that:
    100*X1 + X2 = 200*Y1 - 2*Y2
    Y2 = X1
    2*Y1 = X2
    where X1 is the dollars he had, X2 -cents he got, Y1 - dollars he has now, Y2 -cents now.
    Your first equation is wrong but probably a typo. It should be 200Y1+ 2Y2, not "-".

    You can use the last two equations to write the first as
    100X1+ X2= 200*(X2/2)+ 2X1
    which is the same as
    98X1= 99X2

    so that X1= 99X2/98

    That is now one equation with two unknowns but, since X1 is the number of dollars and X2 is the number of cents, they must be integers. Since X1 must be an integer, X2 must be a multiple of 98.

    But I also know that in one dollar there is no more than 100 cents.
    Can I use that to solve the problem?

    Thanks in advance.
    Yes, you can use that. X2 must be a multiple of 98 and less than 100. X2 must be equal to 98.
    That means that X1= 99 so he must have started with $99.98.

    Check: half of $99.98 is $49.99.
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  4. #4
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    Yeah, I saw that after the post. I got to the end equation, but I was thinking about the 100 cents in a dollar thing and did not thought about that they need to be a whole number.
    Thanks any ways, you are great
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