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Sasa had some 20-cent and 50-cent coins.
7/8 of the coins were 20-cent coins and the rest were 50-cent coins.After Sasa had spent $72.50 worth of 50-cent coins and 5/7 of the 20-cent coins, she had 2/7 of the coins left. Find the total amount of money Sasa left.

  • math -

    let there be:
    c coins
    t 20¢ coins
    f 50¢ coins

    7/8 c = t
    1/8 c = f
    so, t = 7f

    72.50 is 145 of the 50¢ coins
    (f + t) - 145 - (5/7 t) = 2/7 (f+t)

    since t = 7f,

    f + 7f - 145 - 5f = 2/7 * 8f
    f = 203
    so, t = 1421
    and 1624 coins in all

    check:
    1624 - 145 - 5/7 (1421)
    = 1624 - 145 - 1015
    = 464

    464 is 2/7 of 1624

  • math -

    Thank you, Steve.

  • math -

    The first answer is wrong. What is being asked is the amount. 464 is too much compared to $72.50 which is the combination of 50c and 5/7 of 20c.

    Listing method is used to find the answer.

    The total fraction for the 20c and 50c is 8/8.
    7/8 were 20c
    1/8 were 50c

    Singaporean Math is applied using the block method.
    7 units = 20c
    1 unit = 50c
    8 units in all

    If the denominator is 8, find a multiple that will match the given amount of $72.50. The multiple is equivalent to 1 unit.

    The multiple is 48.
    Each unit is 48.
    $72.50 is 5units of 20c and 1unit of 5c.

    48 x 5 x .20 = $48
    48 x .50 = $24
    $48 + $24 = $72
    Add .50 to make it $72.50

    The question is the total amount of money left which 2/7.
    So that is 48 x 2 x .20 = $19.20

    48 48 48 48 48 48 48 49
    20c 20c 20c 20c 20c 20c 20c 50c

    There is a total of 336pcs 20c coins amounting to $67.20 and
    49pcs 50c coins amounting to $24.50
    =$91.70 - $19.20 = $72.50

    All 50c were spent. Only 2/7 of 20c were left.

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