Question 436522: Jane and Ken had 288 marbles altogether. Jane gave 1/5 of her marbles to Ken. Ken then gave 1/4 of his total amount of marbles to Jane. In the end, they each had the same amount of marbles.
(a) How many marbles had Jane at first?
(b) How much marbles had Ken at first?
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! Let x=number of marbles that Jane had at first
Then 288-x=number of marbles that Ken had at first
When Jane gave 1/5 of her marbles to Ken:
Ken had (288-x)+(1/5)x=288-(5/5)x+(1/5)x=288-(4/5)x
And Jane had x-(1/5)x=(4/5)x
When Ken gave 1/4 of his total amount of marbles to Jane, he gave her (1/4)(288-(4/5)x)=72-(1/5)x
After this, Jane had(4/5)x)+(72-(1/5)x)=72+(3/5)X
And Ken had left (288-(4/5)x)-(72-(1/5)x)=216-(3/5)x
Now we are told that the above two numbers are equal, so our equation to solve is:
72+(3/5)x=216-(3/5)x subtract 72 from and add (3/5)x to each side
(3/5)x+(3/5)x=216-72
(6/5)x=144
6x=720
x=120----number of marbles that Jane had at first
288-x=288-120=168----number of marbles that Ken had at first
CK
Jane gave (1/5)*120=24 to Ken leaving her with 96
Ken then had 168+24=192
Ken gave (1/4)*192=48 to Jane leaving him with 144
Then Jane has 96+48=144
They both have the same
Hope this helps---ptaylor
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