SOLUTION: Tim had some storybooks for sale at the start of the day. School A bought 2/5
of the storybooks and received 30 books for free. School B bought 7/9 of the
remaining books and rec
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Question 1186833: Tim had some storybooks for sale at the start of the day. School A bought 2/5
of the storybooks and received 30 books for free. School B bought 7/9 of the
remaining books and received 38 books for free. Tim then had 54 storybooks
left. How many storybooks did Tim have at first?
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39626) (Show Source): You can put this solution on YOUR website!
Tim started with b number of books.
School A distribution
Tim still has .
School B distribution
Tim still has
Use your algebra arithmetic skills to solve for b.
740
Answer by greenestamps(13203) (Show Source): You can put this solution on YOUR website!
Let's take the equation shown by the other tutor and go through the process of solving it to answer the question.
Now let's work the problem backwards; we will see that we do the same calculations that we made in the solution above.
He finished with 54 books.
Before that, he gave away 38 books; so before that he had 54+38=92 books.
Before that, he sold 7/9 of the books he had, so what he had left was 2/9 of his books. So the number he had before he sold 7/9 of his books was 92*(9/2)=414.
Before that, he gave away 30 books; so before that he had 414+30=444 books.
Before that, he sold 2/5 of his books, so what he had left was 3/5 of his books. So the number he had before he sold 2/5 of his books was 444*(5/3)=740.
We do the same calculations using either method -- so why show two different methods?
Imagine a similar problem where, instead of selling some books and giving away some books twice, he does that four or five times. Solving the problem by writing and solving an equation similar to the one the other tutor used would get VERY messy; but working the problem backwards one step at a time would be much easier.
So practice solving problems like this by both methods, knowing that for more complicated problems the solution will be easier if you work the problem backwards.
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