conversion formula

  • Thread starter Thread starter Guest
  • Start date Start date
G

Guest

hi I work with Excel 2000. I have a spreadsheet that holds the pass marks
for students displayed like so
27-Apr-07 El Health & Safety 13 93/130 72%
29-May-07 First Aid 18 124/180 69%
08-Jun-07 Geography 19 156/190 82%
this shows date, name of course, how many students sat the exam, what their
pass mark was and then the pass mark as a persentage.
The person who set up this spreadsheet obviously has better math skills than
me as she was just typint in the pass mark persentage.

I need to know if there is a formula that will automatically show me what
the persentage rate is. (my fingers are crossed in the hope that there is)
Thank you for reading my post. Hopefully you can answer my querie
 
Ok sorry about this but i have managed to figure it out. Now i am wondering
if there is any way that i can get the percentage value to show up
automaticaly into a different cell without me having to type it in as i am
not allowed to break up the pass mark into 2 different cells.
 
=LEFT(D5,FIND("/",D5)-1)/MID(D5,FIND("/",D5)+1,99)

--
HTH

Bob

(there's no email, no snail mail, but somewhere should be gmail in my addy)
 
Try in column E (assuming the ratio 93/130 etc is in column D):

=ROUND(LEFT(D2,FIND("/",D2)-1)/RIGHT(D2,LEN(D2)-FIND("/",D2)),2)

Format cell as % with zero (0) decimal places

HTH
 
That only works because co-incidentally the / is in position 3 and there are
3 numbers after it

Neither 101/130 or 81/90 will work with that formula.

--
HTH

Bob

(there's no email, no snail mail, but somewhere should be gmail in my addy)
 
Did you mean to say
=LEFT(J51, FIND("/", J51, 1)-1)/RIGHT(J51, LEN(J51) - FIND("/", J51, 1)) ?
 
You are right Bob
we can use this also:
=LEFT(I19,FIND("/",I19,1)-1)/RIGHT(I19, LEN(I19)-LEN(RIGHT(I19, FIND("/",
I19,1))))

it is a bit lenghty. I couldn't understand why did you take 99 in Mid
function. Could you please explain that. thank a lot
 
I wanted all of the remaining characters after the /. I could have used
RIGHT subtracting the position of the / from the LEN of the string. I chose
to use MID with a character count of 99 because if you use a number of
characters in the MID function greater than the possible number of
characters it doesn't choke, it just returns as many as it can. Same result
as the RIGHT function, less code.
--
HTH

Bob

(there's no email, no snail mail, but somewhere should be gmail in my addy)
 
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