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10 September, 20:56

A wallet contains $460 in $5, $10, and $20 bills. The number of $5 bills exceeds twice the number of $10 bills by 4, and the number of $20 bills is 6 fewer than the number of $10 bills. How many of each type are there?

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Answers (2)
  1. 10 September, 21:22
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    There are 14 $ 10 bills, 8 $ 20 bills and 32 $ 5 bills.

    Step-by-step explanation:

    Let x be the number of $10 bills,

    ∵ The number of $20 bills is 6 fewer than the number of $10 bills,

    So, the number of $ 20 bills = x - 6,

    Also, the number of $5 bills exceeds twice the number of $10 bills by 4,

    So, the number of $ 5 bills = 2x + 4

    Thus, the total amount = $ 10 * x + $ 20 (x - 6) + $ 5 (2x + 4)

    = 10x + 20x - 120 + 10x + 20

    = (40x - 100) dollars

    According to the question,

    40x - 100 = 460

    40x = 460 + 100

    40x = 560

    ⇒ x = 14

    Hence, the number of $ 10 bills is 14, $ 20 bills is 8 and $ 5 bills is 32.
  2. 10 September, 21:22
    0
    Number of $5 bills = 32

    Number of $10 bills = 14

    Number of $20 bills = 8

    Step-by-step explanation:

    Let x number of $5, y number of $10 and z number of $20

    The number of $5 bills exceeds twice the number of $10 bills by 4.

    Therefore, x = 2y + 4

    The number of $20 bills is 6 fewer than the number of $10 bills.

    Therefore, z = y - 6

    A wallet contains $460 in $5, $10, and $20 bills.

    Therefore,

    5x + 10y + 20z = 460

    Substitute x and y into equation

    5 (2y+4) + 10y + 20 (y-6) = 460

    10y + 20 + 10y + 20y - 120 = 460

    40y - 100 = 460

    40y = 460 + 100

    40y = 560

    y = 14

    Put the value of y into x = 2y + 4 and solve for x

    x = 2 (14) + 4

    x = 32

    Put the value of y into z = y - 6 and solve for z

    z = 14 - 6

    z = 8

    Hence, the each type of bills,

    Number of $5 bills = 32

    Number of $10 bills = 14

    Number of $20 bills = 8
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