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Algebra II question?

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A jar contains 41 coins consisting of nickels,dimes,and quarters and having a total value of $4.75. How many of each kind of coins are there if there are twice as many dimes as quarters

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  1. Let n = # of nickels and q = # of quarters

    Since there are as many dimes as quarters, # of dimes = 2q

    n + 2q + q = 41 (eq1) --> total # of coins = 41

    0.05n + 0.10(2q) + 0.25(q) = 4.75 (eq2)

    From eq1, n + 3q = 41 ==> n = 41 - 3q

    From eq2, 0.05n + 0.20q + 0.25q = 4.75

    0.05n + 0.45q = 4.75

    Sub n = 41 - 3q into eq2,

    0.05(41-3q) + 0.45q = 4.75

    2.05 - 0.15q + 0.45q = 4.75

    0.30q = 2.7

    q = 9

    # of quarters = 9

    # of dimes = 2q = 2x9 = 18

    # of nickels = 41 - 3q = 41 - 27 = 14

    Your answer is correct.


  2. quarter 9

    dime  18  

    nickel 14

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