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Find the value of a percentage
Hi,
I tried my inquiry once but I did not get a reply. Maybe I was not clear and explicit with my question, therefor I will try again. Please anyone with a reply. I have column "A" 1 thru 200 numbers. I need to create column "B" showing the total of an increase of 35% (value plus the 35%). Now I have to create a column "C" with the value or a number that if you subtract 40% will give me the number in colmn "B". I hope I made myself clear and comprehesible this time. Please do reply, anyone. Thank you -- smile |
#2
Posted to microsoft.public.excel.newusers
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Find the value of a percentage
Hi,
to apply 35 % in column B enter =a1*1.35 copy the formula down in column C enter =a1/0.6 if this helps please click yes thanks "israel" wrote: Hi, I tried my inquiry once but I did not get a reply. Maybe I was not clear and explicit with my question, therefor I will try again. Please anyone with a reply. I have column "A" 1 thru 200 numbers. I need to create column "B" showing the total of an increase of 35% (value plus the 35%). Now I have to create a column "C" with the value or a number that if you subtract 40% will give me the number in colmn "B". I hope I made myself clear and comprehesible this time. Please do reply, anyone. Thank you -- smile |
#3
Posted to microsoft.public.excel.newusers
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Find the value of a percentage
You received several replies yesterday.
Quit waiting for a personal email reply and start searching for your posts which have the replies. Gord Dibben MS Excel MVP On Tue, 22 Dec 2009 09:45:02 -0800, israel wrote: Hi, I tried my inquiry once but I did not get a reply. Maybe I was not clear and explicit with my question, therefor I will try again. Please anyone with a reply. I have column "A" 1 thru 200 numbers. I need to create column "B" showing the total of an increase of 35% (value plus the 35%). Now I have to create a column "C" with the value or a number that if you subtract 40% will give me the number in colmn "B". I hope I made myself clear and comprehesible this time. Please do reply, anyone. Thank you |
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