Wasn't sure where to post this please help ASAP :-)

[IcantMath]

Ok, I've been through three formulas so far and each one works for a set equation and then breaks, so here's my dilemma:

I can receive a score from 1-6. Let's say I have four 6's and a two 2's. (6,6,6,6,2,2,) I need my scores to average a 5.5. I need to find out how many more 6's bring me to a 5.5. I'm using this in excel and need to be able to account for any number of scores. Please help

 

[IcantMath]

First of all thank you for your immediate response, holy cow this site rocks. Secondly, you mentioned the "6" can be changed to a variable. For the sake of clarity, does that (6-n) represent (c-n) or is it 6 based upon the fact that 6 is my highest score? Again thank you for your fast response and all of your help. WOW...

 

[IcantMath]

Well, I'll give you the coding in UBasic :cool:

c = current number of scores (6)
k = current sum of scores (28)
n = new average required (5.5)
a = additional scores @ 6 required (?)

GET c,k,n
a = (c*n - k) / (6 - n)
PRINT a

Note: the "6" can be changed to a variable,
in case the additional scores are not 6 each.
OK??

You just helped me solve something that made me lose sleep for a week...A solid week. There's no thank you button on this forum but if I could shake your hand I would. I'll buy you a beer or something. Thanks so much my friend.

 

[IcantMath]

Well, I'll give you the coding in UBasic :cool:

c = current number of scores (6)
k = current sum of scores (28)
n = new average required (5.5)
a = additional scores @ 6 required (?)

GET c,k,n
a = (c*n - k) / (6 - n)
PRINT a

Note: the "6" can be changed to a variable,
in case the additional scores are not 6 each.
OK??

Ok I found one small hiccup, if I have two scores that add up to 6, my equation divides by zero. I'm in excel and want to do some error checking, thoughts? IF, my desired average "new average" is 6 that is. So a = (2*6 - 6) / (6 - 6). Blows up on me. True answer is 10, how do i get there?

 

[IcantMath]

I'm trying to account for two scores: a one and a five. Since 6 is a perfect average I wanted to see what it took to get back there. This is a valid scenario

 

[DrPhil]

Ok I found one small hiccup, if I have two scores that add up to 6, my equation divides by zero. I'm in excel and want to do some error checking, thoughts? IF, my desired average "new average" is 6 that is. So a = (2*6 - 6) / (6 - 6). Blows up on me. True answer is 10, how do i get there?

If any score was ever not a 6, there is no way to get the average back to 6.

I'll try to think in terms of Excel.
A1 = N (number of scores already done)
column B = scores
A2 = Sum(B)
A3 = 5.5 (or desired average)
A4 = (A1*A3 - A2)/(6 - A3) --> number of 6s you have to score