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Question 1186025: Sam and Lynn had 96 storybooks altogether. Sam gave 1/4 of his
storybooks to Lynn. Lynn then gave 1/3 of the total number of storybook
she had to Sam. In the end, they had the same number of storybooks.
How many storybooks did each of them have at first?
Found 2 solutions by EaazyGoarth, ikleyn:
Answer by EaazyGoarth(2)   (Show Source):
Answer by ikleyn(52752)  (Show Source):
You can put this solution on YOUR website! .
Dear student @EaazyGoarth (!)
In your post, I see some manipulations with some numbers, but I do not understand these numbers,
these manipulations and your thoughts behind these manipulations.
At this forum, we, the tutors, do not serve the visitors this way.
We make it differently . . .
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Sam and Lynn had 96 storybooks altogether. Sam gave 1/4 of his
storybooks to Lynn. Lynn then gave 1/3 of the total number of storybook
she had to Sam. In the end, they had the same number of storybooks.
How many storybooks did each of them have at first?
~~~~~~~~~~~~~~~~~
Sam = x books at the beginning.
Lynn = y books at the beginning.
After giving  of his book to Lynn, Sam has  books left; Lynn has  books.
After giving 1/3 of her  books to Sam, Lynn has  books left; Sam has +  .
At the end, each of them has 48 books, which gives you these two equations
+ = 48 (1) (the number of Sam's books at the end)
= 48 (2) (the number of Sam's books at the end)
To solve, multiply equations (1) and (2) by 12 each; then simplify
9x + 4*(4y+x) = 576 (1')
8*(4y+x) = 576 (2')
From (2'), 4y + x = 576/8 = 72. Substitute this value 72 into equation (1'), replacing 4y+x there. You will get
9x + 4*72 = 576
which implies
9x = 576 - 4*72 = 288, x = 288/9 = 32.
Thus Sam had 32 books at the beginning; Lynn had the rest 96 - 32 = 64 books. ANSWER
Your last step is to check / (to verify) that the manipulations as described in the problem, lead
to the correct result.
I did this check for myself; you should do it for yourself.
Solved.
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