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Math Help - Coin problem

  1. #1
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    Coin problem

    The value (in cents) of the change in a purse that contains twice as many nickels as pennies, four more dimes than nickels, and as many quarters as dimes and nickels combined; p = number of pennies?
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  2. #2
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    Coin problem

    Hello tmac11522
    Quote Originally Posted by tmac11522 View Post
    The value (in cents) of the change in a purse that contains twice as many nickels as pennies,
    Quote Originally Posted by tmac11522 View Post
    four more dimes than nickels, and as many quarters as dimes and nickels combined; p = number of pennies?
    Number of pennies = p. Each penny is worth one cent. So the value of the pennies = 1\times p = p cents

    Number of nickels = twice as many as number of pennies = 2p. Each nickel is worth 5 cents. So the value of the nickels = 5\times 2p = 10p cents

    Number of dimes = four more than the number of nickels = (2p + 4). Each dime is worth 10 cents. So the value of the dimes = ??? cents.

    Number of quarters = number of dimes plus number of nickels = ??. Each quarter is worth 25 cents, so the value of the quarters is ??? cents.

    Add together all four value totals; remove the brackets and simplify the result. You'll then arrive at an expression like Ap + B, where A and B are two whole numbers. That will be your final answer, if you don't have any more information.

    Can you complete it now?

    Grandad
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