Posted by Diana on .
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 
Reiny,
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 = x1
number of yellows = y
sum = x+y1
given: 1/7 of remaining are red
(1/7)(x+y1) = x1
x+y1 = 7x  7
y = 6x  6 (equation #1)
case 2: child eats 5 yellow
number of reds = x
number of yellows = y5
sum = x+y5
given:
(1/6)(x+y5) = x
x+y5 = 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 
tchrwill,
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.