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March 30, 2017

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A jar contains some red and some yellow jelly beans. If a child ate 1 red jelly bean, 1/7 of the remaining candies would be red. If instead the child ate 5 yellow jelly beans, 1/6 of the remaining candies would be red

  • math - ,

    let the number of original reds be x
    let the number of original yellows be y

    case 1: child eats 1 res
    number of reds = x-1
    number of yellows = y
    sum = x+y-1
    given: 1/7 of remaining are red
    (1/7)(x+y-1) = x-1
    x+y-1 = 7x - 7
    y = 6x - 6 (equation #1)

    case 2: child eats 5 yellow
    number of reds = x
    number of yellows = y-5
    sum = x+y-5
    given:
    (1/6)(x+y-5) = x
    x+y-5 = 6x
    y = 5x+5 (equation #2)

    #1 = #2
    6x - 6 = 5x + 5
    x = 11
    then y = 5(11) + 5 = 60

    so original had 11 reds and 60 yellow.

  • math - ,

    Same problem, same answer.

    Same problem, same answer.


    A jar contains some red and some yellow jelly beans. If a child ate 1 red jelly bean, 1/7 of the remaining candies would be red.

    If instead the child ate 5 yellow jelly beans, 1/6 of the remaining candies would be red.

    If there are R red beans and Y yellow beans, there are a total of (R + Y) beans.

    If one red bean was eaten, there would be (R - 1) red beans left, the total number of beans now being (R - 1 + Y).

    If 1/7th of the remaining beans are red, then (R - 1)= (R - 1 + Y)/7.


    If instead the child ate 5 yellow beans,
    R = (R + Y - 5)/6.

    Then, 7R - 7 = r - 1 + y or Y = 6R - 6 from the first and Y = 5R - 5 from the second.

    Equating, 6R - 6 = 5R + 5 making R = 11 and Y = 60 for a total of 71 jelly beans.

    With 1 red bean eaten, 10/(71 - 1)= 10/70 = 1/7 and with 5 yellows eaten, 11/(71 - 5) = 11/66 = 1/6.

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