SOLUTION: Jim gave one-third of his jellybeans to Jack, ate two-thirds of the remainder, and had 12 left over. How many jellybeans did he have originally?
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Question 193650: Jim gave one-third of his jellybeans to Jack, ate two-thirds of the remainder, and had 12 left over. How many jellybeans did he have originally? Found 2 solutions by josmiceli, Edwin McCravy:Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website! Let = the number he had originally
given:
The amount he gave to Jack =
The remainder =
Two-thirds of the remainder =
------------------
The amount he ate plus the left over beans has to
equal , so
Multiply both sides by
He originally had 54 jelly beans
check answer:
He gave Jack
He had left
He ate
There were left over
Jack's + what he ate + left overs +
OK
You can put this solution on YOUR website! Edwin's solution:
Jim gave one-third of his jellybeans to Jack, ate two-thirds of
the remainder, and had 12 left over. How many jellybeans did he
have originally?
The 12 he had left over at the end was the other one-third of
what he had before eating the two-thirds. So 12 was 1/3 of
those he had before starting eating. So he had 3x12 or 36 before
starting to eat.
So those 36 he had before starting eating was the other 2/3 of what
he had before giving the one-third to Jack. So since 1/3 is half
of 2/3, then he gave Jack half of 36 or 18. So before giving those
18 to Jack, he must have had 36+18 or 54.
---
Now we check:
Jim gave one-third of his jellybeans to Jack,
So Jim started with 54 and gave 54*1/3 or 18 of them to Jack.
That means he only had 54-18 or 36 left.
Then Jim ate two-thirds of the remainder,
That means he ate 2/3 of 36 or 24.
So he had 36-24 or 12 left.
So 54 is the correct answer.
No algebra necessary!
Edwin
