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19 October, 12:42

A bag contains balls that are red, green, or blue. One third of the balls are red, and 2/7 are blue. The number of green balls is eight fewer than the twice the number of blue balls. How many green balls are in the bag?

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  1. 19 October, 14:45
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    16 green balls

    Explanation:

    x = total balls in bag

    Red = 1/3x

    Blue = 2/7x

    Green = - 8+2 (2/7x)

    Since these are balls, the answers have to be whole numbers. The equations representing the different colors contain fractions and have denominators 3 and 7. So the total number of balls, represented by x, needs to be a factor of 3 and 7. So I did a trial and error thing here:

    First possible common denominator = 21, which is 3x7

    We test that there are 21 balls in the bag. Substituting 21 into the equations for red, blue and green we get:

    red = 1/3 (21) = 7

    blue = 2/7 (21) = 6

    green = - 8+2 (2/7) (21) = 4

    But if you add up the colors: 7+6+4 = 14 which is < 21

    So we test another common denominator

    Next possible common denominator = 42, which is 3x7x2

    So we test that there are 42 balls in the bag. Substituting 42 into the equations for red, blue and green we get:

    red = 1/3 (42) = 14

    blue = 2/7 (42) = 12

    green = - 8+2 (2/7) (42) = 16

    If you add up the colors: 14+12+16 = 42
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