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

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.

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.

To solve this problem, we can use algebraic equations. Let's assume the number of red jelly beans in the jar is represented by the variable 'r' and the number of yellow jelly beans is represented by the variable 'y'.

Given that if the child ate 1 red jelly bean, 1/7 of the remaining candies would be red, we can write the equation:

(r-1) = (1/7)(r-1+y)

Similarly, given that if the child ate 5 yellow jelly beans, 1/6 of the remaining candies would be red, we can write another equation:

(r) = (1/6)(r+y-5)

Now, we can solve these two equations simultaneously to find the values of 'r' and 'y'.

Let's start by solving the first equation:
Multiply both sides of the equation by 7 to eliminate the fraction:
7(r-1) = r-1+y

Simplifying:
7r - 7 = r - 1 + y

Moving the variables to one side and the constants to the other side:
6r - y = 6

Now let's solve the second equation:
Multiply both sides of the equation by 6 to eliminate the fraction:
6(r) = r+y-5

Simplifying:
6r = r + y - 5

Moving the variables to one side and the constants to the other side:
5r - y = 5

Now we have a system of equations:
6r - y = 6 ----(1)
5r - y = 5 ----(2)

To solve this system of equations, we can use the elimination method. Subtract equation (2) from equation (1):
(6r - y) - (5r - y) = 6 - 5

Simplifying:
6r - y - 5r + y = 1

The 'y' terms cancel out:
6r - 5r = 1

Simplifying further:
r = 1

Now, substitute the value of 'r' back into equation (1) or (2) to find the value of 'y'. Let's substitute it into equation (2):
5(1) - y = 5

Simplifying:
5 - y = 5

Move the constant to the other side:
-y = 5 - 5

Simplifying:
-y = 0

Multiply both sides by -1 to isolate 'y':
y = 0

Therefore, the number of red jelly beans (r) is 1 and the number of yellow jelly beans (y) is 0.

So, in the jar, there is 1 red jelly bean and 0 yellow jelly beans.