SOLUTION: Hi-I was trying to help my 6th grader with his math homework. Many years since I have done any, the problem was from a work sheet. A Frog is in a 10 foot well, climbs 3 feet during

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Question 160158: Hi-I was trying to help my 6th grader with his math homework. Many years since I have done any, the problem was from a work sheet. A Frog is in a 10 foot well, climbs 3 feet during the day and slides back 2 feet at night. At what point will the frog make it to the 10 foot level and be able to crawl out? I thought the answer would be 8 days, but I did not have a formula to use. Would you please help out a frustrated dad?? Thanks for your time.
Found 2 solutions by stanbon, gonzo:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
It really is more of a logic problem.
Each day the frog gains 1 ft.
When he gets to the 7 ft. level (7 days)
his next 3 ft. jump will get him to the
10 ft.level. That will occur on the 8th day.
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You have the correct answer.
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Cheers,
Stan H.

Answer by gonzo(654) (Show Source):
You can put this solution on YOUR website!
goes up 3 in the day and goes down 2 in the night.
at the end of the first day there's a gain of 3.
at the end of each succeeding day there's a gain of 1 (loss of 2 during the night).
end of day 1 he's up 3
end of day 2 he's up 4
end of day 3 he's up 5
end of day 7 he's up 9
end of day 8 he's up 10
you're correct.
number of days is 8.
now the formula.
let x be the number of days needed to get out of the well.
formula looks like it will be
(x-1)*(3-2) + 3 = 10
what this says is that you have x-1 days where the net gain was only 1.
that would be the loss of 2 the night before plus the gain of 3 during the day.
the + 3 means you got a net gain of 3 the first day.
so (x-1)*(3-2) + 3 = 10
solving you get
(x-1)(1) + 3 = 10
x - 1 + 3 = 10
x - 1 = 7
x = 8
i must admit i didn't think about it correctly the first time i looked at it.
congratulations on good logic.