Wikipedia:Reference desk/Archives/Mathematics/2014 June 24

Mathematics desk
< June 23 << May | June | Jul >> Current desk >
Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.


June 24

edit

How many steps

edit

How do I figure out in how many steps one number will overtake another if they grow by different rates and start at different points? Is there a basic formula I can just plug them into? I don't know much math, so please keep it as simple as possible for me little brain. Here are my numbers: 400+35+... and 1000+28+... but i would like to know more generally. Mingmingla (talk) 18:58, 24 June 2014 (UTC)[reply]

You divide the difference in the initial values by the difference in the growth rates (rounded up). In your case,  . To understand why that is, you'll need to learn about arithmetic progressions. -- Meni Rosenfeld (talk) 19:16, 24 June 2014 (UTC)[reply]
No you won't. The difference between the two numbers decreases at a rate given by the difference between the absolute rates of change of each number. — Preceding unsigned comment added by 86.132.39.95 (talk) 23:29, 24 June 2014 (UTC)[reply]
Since the OP asks for simplicity, I would start with a basic algebra setup, using the variable 'n' as the number of steps:
400 + 35n = 1000 + 28n
and solving for n gives the equation mentioned earlier, thus n = 85.714285... which would then be rounded up to 86 for answering the question. My equation theoretically states when the two sums will be the same, so we round up to get the next whole number of steps when the first sum 'overtakes' the second. (It may or may not be better/helpful to use an inequality instead: 400 + 35n > 1000 + 28n which would still be true for n=86 whereas the above equation wouldn't; this is a tricky conceptual issue in translating word problems to algebra, in this case with practical "steps" and rounding.) El duderino (abides) 05:02, 25 June 2014 (UTC)[reply]
I was playing with Wolfram Alpha but the plain language thing didn't quite get what I was asking. Close, but not quite. Thank you for your answers. Mingmingla (talk) 04:48, 26 June 2014 (UTC)[reply]