Question 607242: Charlie was buying things for his holiday party. He spent one-third of what he had plus $3 on plates and tablecloths, then one-third of what was left plus $2 on food, then half of what was left plus $10 on soda. He then had $10 left for potato chips and other snacks. How much did he start with?
Answer by KMST(5328) (Show Source):
You can put this solution on YOUR website! THE ALGEBRA WAY:
Charlie started with (in $ units).
He spent $ on plates and tablecloths.
After buying the plates and tablecloths he had
$ left, and went to buy food.
He spent 1/3 of the amount above plus $2 on food.
After buying the food he had 1/3 of the amount above minus $2.
He had $ left, and went to buy soda.
He spent 1/2 of the amount above plus $10 from the other 1/2 on soda.
So, after buying the soda he had 1/2 of the amount above minus $10.
He had $ and that was the $10 left for chips and snacks.

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Charlie started with $99.
VERIFICATION:
One third of $99 plus $3 is $$33+$3=$36 spent on plates and tablecloths.
That leaves $99-$36=$63.
One third of $63 plus $2 is $21+$2=$23 spent on food.
That leaves $63-$23=$40.
One half of $40 plus $10 is $20+$10=$30 spent on soda.
That left $40-$30=$10 for chips and snacks.
THE FIFTH GRADER SOLUTION:
The final $10 left after buying the soda is $10 less that half what Charlie had before, so $20 is half what Charlie had before, and $40 is hat Charlie had before buying the soda.
The $40 amount left before buying the soda, but after buying the food is $2 less than two thirds of what Charlie had before, so $42 is two thirds of what Charlie had before. That means that $42/2=$21 is one third of what Charlie had before, and 3($21)=$63 is what Charlie had before buying the food.
The $63 amount left before buying the food, but after buying the plates and tablecloths is $3 less than two thirds of what Charlie had originally, so $63+$3=$66 is two thirds of what Charlie had originally. That means that $66/2=$33 is one third of what Charlie had originally, and 3($33)=$99 is the amount Charlie started with.
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