Do you let them try a second time with the original number and when they get it right the second time you still take off for that? Before you have a formula you have to have the exact rules.

Anyway, let me try to answer your question even without knowing the exact rules, you can adapt it accordingly

let's say you give them 10 questions

you have a variable called mistake. Every time they got an answer wrong you add 1 to mistake.

10 = 100% so for each mistake you take off 10%. So your formula would be

100-(mistake*10)

Thanks for the help guys. Okay, I tried running the program, using both the formulas you supplied at the same time. When I use the program with no incorrect answers I get...

plusplus = 0

zmariow = 10

And when I have one incorrect answer (i.e. 11 attempts for 10 numbers)...

plusplus = -10

zmariow = 9.090909

From what I can work out plusplus is close to the mark (probably due to my ambiguity) but it's not as accurate as zmariow. So, because I want to show it as a percentage, here's what I came up with...

```
res2 =(success / attempt) * 100 * 10
lblresults.caption "Your accuracy is " and res2 & "%"
```

Is this right?

...When I use the program with no incorrect answers I get...

plusplus = 0

zmariow = 10

Since you're getting 10 using my formula I assume you're doing something wrong when counting the number of mistakes and the total number.

This is because you should have s and t both equal to 10 and the formula should give you: 100... that is 10/10 * 100 = 100...

Can you show us what are the values of s and t??? (or whatever names you're using for those variables)