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

Answer 1

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.

Answer 2

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.