math

posted by .

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.

Respond to this Question

 First Name School Subject Your Answer

Similar Questions

1. Probability

A jar contains four red, four yellow, and three green jelly beans. If Joan and Jim take one jelly bean each, the probability that they both take a red jelly bean is?
2. math

The ratio of red jelly beans to yellow jelly beans in a dish is 2:3. If I eat 3 red jelly beans and 6 yellow ones the ratos is 4:5. How many yellow jelly beans were orginally in the dish?
3. Math

The ratio of red jelly beans to yellow jelly beans in a dish is 3:4. if greg eats 3 red jelly beans and 6 yellow ones, the ratio is 4;5. how many jelly beans were originally in the dish?
4. Math

The ratio of red jelly beans to yellow jelly beans in a dish is 3:4. If Greg eats 3 red jelly beans and 6 yellow ones, the ratio is 4:5. How many yellow jelly beans were originally in the dish?
5. Fractions

Juan ate 1/3 of the jellybeans. Maria then ate 3/4 of the remaining jelly beans, which left 10 jelly beans. How many jelly beans were there to begin with?
6. Math

Tory has 4 blue jelly beans, 6 red jelly beans, 5 yellow jelly beans, and 1 green jelly bean in his bag. What fraction of the jelly beans are A)Blue B)Yellow C)Red D)Green
7. Math

In a bag of 400 jelly beans, 25% of the jelly beans are red in color. If you randomly pick a jelly bean from the bag, what is the probability that the jelly bean picked is NOT one of red jelly beans?
8. math

A jar of jelly beans contains 50 red gumballs , 45 yellow gumballs, and 30 green gumballs. You reach into the jar and randomly select a jelly bean, then select another without putting the first jelly bean back. What is the probability …
9. Math

You have a jar with 60 jelly beans. You know that 1/3 of the jelly beans are yellow, 1/4 of them are white and the rest is red. You are not allowed to look. How many beans do you have to take out to be absolutely sure to get at least …
10. math

In a bag there are 8 red jelly beans and 7 green jelly beans. The 4 jelly beans are randomly chosen from the bag without replacement. a)create a probability distribution table for choosing red jelly beans b)what is the probability …

More Similar Questions