# IF STATEMENT not working

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4 - Data Explorer

Hi,

I am trying to create a simple IF Statement to produce the result PASS or FAIL if the value in another column is >= 80%:  In Column 1 ('% germination rate on second check') are % figures with values ranging from 0% to 100%. The purpose of the Column 2 ('PASS/FAIL'), is to produce the word 'PASS' if the value of Column 1 is >= 80%, or 'FAIL' if otherwise. Currently all cells in Column 2 are showing 'FAIL', irrespective of the % figure showing in Column 1. Is this some kind of syntax or formatting error on my end?

Here is my formula: IF({% germination rate on second check} >= 80, "PASS", "FAIL")

Thanks for the help! 🙂

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11 - Venus

80% is only an apparent number formatted as the amount per hundred.
Internally, it has a value of 0.8.

``IF({% germination rate on second check} >= 0.8, "PASS", "FAIL")``

Multiplying the "percent" by 100 is also a good idea.

5 Replies 5
11 - Venus

Percentage field, right?
Then 100% is 1 and 80% is 0.8.

8000% would be a PASS!

4 - Data Explorer

Hi @Sho ,

Thanks for you reply. I am a beginner here and I don't follow what you're saying. Would you be able to explain? Or offer an alternative formula that may work if your logic is correct?

Thanks so much for the support!

Z

4 - Data Explorer

I think I've figured it out by changing my formula to:

IF({% germination rate on second check}*100 >= 80, "PASS", "FAIL")
but I don't fully understand why?!
11 - Venus

80% is only an apparent number formatted as the amount per hundred.
Internally, it has a value of 0.8.

``IF({% germination rate on second check} >= 0.8, "PASS", "FAIL")``

Multiplying the "percent" by 100 is also a good idea.

4 - Data Explorer

Ok - I understand. Cool - thanks for your help @Sho