SOLUTION: A worm is at the bottom of a 12 ft. wall. Every day it crawls up 3 ft., but at night it slips down 2 ft. How many days does it take the worm to get to the top of the wall? How do

Click here to see ALL problems on Miscellaneous Word Problems

Question 40587: A worm is at the bottom of a 12 ft. wall. Every day it crawls up 3 ft., but at night it slips down 2 ft. How many days does it take the worm to get to the top of the wall? How do I solve this?
Found 3 solutions by rajagopalan, JoelSchwartz, MathTherapy:
Answer by rajagopalan(174)   (Show Source):
You can put this solution on YOUR website!
Height for climbing 12
Daily climbing rate = 3
daily sipping rate = 2
Net height climbed per day = 1
Like this, on day 9 it will on the 9 foot marking
on day 10 it will reach the top ( 9 + 3) = 12
At the top it need not slip !!
Bravo..

Answer by JoelSchwartz(130)   (Show Source):
You can put this solution on YOUR website!

Answer by MathTherapy(10551)   (Show Source):
You can put this solution on YOUR website!

A worm is at the bottom of a 12 ft. wall. Every day it crawls up 3 ft., but at night it slips down 2 ft. How many days does it take the worm to get to the top of the wall? How do I solve this?

days
You can do a check!!
If you need a complete and detailed solution, let me know!!
Send comments, �thank-yous,� and inquiries to �D� at MathMadEzy@aol.com.
Further help is available, online or in-person, for a fee, obviously.
For FREE info and answers to questions about the ASVAB exam, the NYS 3 � 8 city/state wide exams,GENERAL MATH QUESTIONS, MATH HOMEWORK QUESTIONS, or MATH QUESTIONS related to the Regents Integrated Algebra,
Regents Geometry, Regents Algebra 2/Trigonometry, SHSAT, COOP/HSPT/TACHS, PSAT, SAT, ACT, SSAT/ISEE,
GRE, CLEP,and the TASC/GED, you can visit: http://asvabstudyzone.freeforums.net/.