A piggy bank contains some dimes and nickels. There are 8 more dimes than nickels in the bank. There is a total of $1.40. How many of each t

Question

A piggy bank contains some dimes and nickels. There are 8 more dimes than nickels in the bank. There is a total of $1.40. How many of each type of coin are in the bank?

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Charlie 1 week 2021-09-15T04:29:22+00:00 2 Answers 0

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    0
    2021-09-15T04:30:56+00:00

    This question is solved using a system of equations. I am going to say that:

    • x is the number of dimes.
    • y is the number of nickels.

    Doing this, we get that: There are 4 nickels and 12 dimes.

    There are 8 more dimes than nickels in the bank.

    This means that: y = x + 8

    There is a total of $1.40.

    • Nickel is worth $0.05.
    • Dime is worth $0.1.

    Thus, for the nickels:

    0.05x + 0.1y = 1.4

    Since y = x + 8

    0.05x + 0.1(x + 8) = 1.4

    0.05x + 0.1x + 0.8 = 1.4

    0.15x = 0.6

    x = \frac{0.6}{0.15}

    x = 4

    For the dimes:

    y = x + 8 = 4 + 8 = 12

    There are 4 nickels and 12 dimes.

    A similar question is found at https://brainly.com/question/24342899

    0
    2021-09-15T04:31:11+00:00

    Answer:

    4 nickels

    12 dimes

    Step-by-step explanation:

    Dimes are worth .1 each while nickels are .05 each.

    We have 8 more dimes than nickels. Let d represent number of dimes and n for number of nickels. This means we have d=8+n.

    If all our nickels and dimes together are worth 1.4 then we have another equation .1d+.05n=1.4

    Lets put our equations together:

    d=8+n

    .1d+.05n=1.4

    ‐—————–Plug first equation into second.

    .1(8+n)+.05n=1.4

    Distribute

    .8+.1n+.05n=1.4

    Combine like terms

    .8+.15n=1.4

    Subtract .8 on both sides

    .15n=.6

    Divide both sides by .15

    n=.6/.15=4

    Remember there are 8 more dimes so d=8+4=12.

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