=if(condition1,responce1,If(condition2,responce2,r esponce3)
or
=if(and(condition1,condition2),responce1,reponce2)

-----Original Message-----

.

Here's is the formula that I am trying to use, I want to get an answer
using a variable (B13) based on whether the cell is 1,2,3,4,5,6 I'm
fairly new to this so any help would be awesome.

=IF(C22=1,"=0.2",IF(C22=2,"=((B13-0.4)/5)+0.2",IF(C22=3,"=(((B13-0.4)/5)*2)+0.2",IF(C22=4,"=(((B13-0.4)/5)*3)+0.2",IF(C22=5,"=(((B13-0.4)/5)*4)+0.2",IF(C22=6,"=B13-0.2"))))))


--
Wazza
------------------------------------------------------------------------
Wazza's Profile: http://www.excelforum.com/member.php...o&userid=16225
View this thread: http://www.excelforum.com/showthread...hreadid=276384

So where is my original formula wrong? When I punch in the random number
1-6 it just puts the corresponding formula on the screen and does not
calcualte it i.e when I put it 1 D22 results in =0.2 instead of just
0.2


--
Wazza
------------------------------------------------------------------------
Wazza's Profile: http://www.excelforum.com/member.php...o&userid=16225
View this thread: http://www.excelforum.com/showthread...hreadid=276384

Hi
try:
=IF(C22=1,0.2,IF(C22=2,(B13-0.4)/5+0.2,IF(C22=3,((B13-0.4)/5)*2+0.2,IF(
C22=4,((B13-0.4)/5)*3+0.2,IF(C22=5,((B13-0.4)/5)*4+0.2,IF(C22=6,B13-0.2
))))))

--
Regards
Frank Kabel
Frankfurt, Germany

"Wazza" schrieb im Newsbeitrag
...

Here's is the formula that I am trying to use, I want to get an
answer
using a variable (B13) based on whether the cell is 1,2,3,4,5,6 I'm
fairly new to this so any help would be awesome.


=IF(C22=1,"=0.2",IF(C22=2,"=((B13-0.4)/5)+0.2",IF(C22=3,"=(((B13-0.4)/5
)*2)+0.2",IF(C22=4,"=(((B13-0.4)/5)*3)+0.2",IF(C22=5,"=(((B13-0.4)/5)*4
)+0.2",IF(C22=6,"=B13-0.2"))))))


--
Wazza
---------------------------------------------------------------------
---
Wazza's Profile:
http://www.excelforum.com/member.php...o&userid=16225
View this thread:
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Hi

=0.2+(C22-1)*(B13-0.4)/5


Arvi Laanemets


"Wazza" wrote in message
...

Here's is the formula that I am trying to use, I want to get an answer
using a variable (B13) based on whether the cell is 1,2,3,4,5,6 I'm
fairly new to this so any help would be awesome.


=IF(C22=1,"=0.2",IF(C22=2,"=((B13-0.4)/5)+0.2",IF(C22=3,"=(((B13-0.4)/5)*2)+
0.2",IF(C22=4,"=(((B13-0.4)/5)*3)+0.2",IF(C22=5,"=(((B13-0.4)/5)*4)+0.2",IF(
C22=6,"=B13-0.2"))))))


--
Wazza
------------------------------------------------------------------------
Wazza's Profile:
http://www.excelforum.com/member.php...o&userid=16225
View this thread: http://www.excelforum.com/showthread...hreadid=276384

Hi
don't enclose the return values in apostrophes nor use the equation
sign

--
Regards
Frank Kabel
Frankfurt, Germany

"Wazza" schrieb im Newsbeitrag
...

So where is my original formula wrong? When I punch in the random
number
1-6 it just puts the corresponding formula on the screen and does not
calcualte it i.e when I put it 1 D22 results in =0.2 instead of just
0.2


--
Wazza
---------------------------------------------------------------------
---
Wazza's Profile:
http://www.excelforum.com/member.php...o&userid=16225
View this thread:
http://www.excelforum.com/showthread...hreadid=276384

Thanks Frank :)


--
Wazza
------------------------------------------------------------------------
Wazza's Profile: http://www.excelforum.com/member.php...o&userid=16225
View this thread: http://www.excelforum.com/showthread...hreadid=276384

lol
that is a "little bit" simpler than my formula

--
Regards
Frank Kabel
Frankfurt, Germany

"Arvi Laanemets" schrieb im Newsbeitrag
...
Hi

=0.2+(C22-1)*(B13-0.4)/5


Arvi Laanemets


"Wazza" wrote in message
...

Here's is the formula that I am trying to use, I want to get an
answer
using a variable (B13) based on whether the cell is 1,2,3,4,5,6
I'm
fairly new to this so any help would be awesome.



=IF(C22=1,"=0.2",IF(C22=2,"=((B13-0.4)/5)+0.2",IF(C22=3,"=(((B13-0.4)/5
)*2)+

0.2",IF(C22=4,"=(((B13-0.4)/5)*3)+0.2",IF(C22=5,"=(((B13-0.4)/5)*4)+0.2
",IF(
C22=6,"=B13-0.2"))))))


--
Wazza

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