This isn't a homework problem or anything, simply a Math problem from real-life that I can't figure out how to solve without resorting to brute force checking.
Let n= the number of attempts and x=the number of successes. The fraction of success is therefore x/n, simple so far. Given that the original percentage of success was 78% and after another failed attempt the percentage dropped to 70%, what was the original number of successes, x, and attempts, n?
Furthermore, the percentages were given as 2 digit numbers, i.e. it was rounded to the nearest integer percentage, e.g. Both 77.5% and 78.49% would show as 78%, so there is some flexibility with the values.
This problem "feels" to me that it should be solvable algebraically somehow, but I can't figure it out for the life of me. I've graduated university a year ago and haven't done math since then, so I might just be having a brain fart. I was able to solve this with brute force checking, the answer being 7/9 = x/n, but I would like to know if there was a more intelligent, faster way of solving this, preferably by hand. I realize that given we know the difference in percentage before and after a failed attempt, from n -> (n+1), this should lead us to a single answer. That's as far as I got. It's just frustrating that I can't figure out a way to solve this in a smarter way.
Thanks for any help!
Let n= the number of attempts and x=the number of successes. The fraction of success is therefore x/n, simple so far. Given that the original percentage of success was 78% and after another failed attempt the percentage dropped to 70%, what was the original number of successes, x, and attempts, n?
Furthermore, the percentages were given as 2 digit numbers, i.e. it was rounded to the nearest integer percentage, e.g. Both 77.5% and 78.49% would show as 78%, so there is some flexibility with the values.
This problem "feels" to me that it should be solvable algebraically somehow, but I can't figure it out for the life of me. I've graduated university a year ago and haven't done math since then, so I might just be having a brain fart. I was able to solve this with brute force checking, the answer being 7/9 = x/n, but I would like to know if there was a more intelligent, faster way of solving this, preferably by hand. I realize that given we know the difference in percentage before and after a failed attempt, from n -> (n+1), this should lead us to a single answer. That's as far as I got. It's just frustrating that I can't figure out a way to solve this in a smarter way.
Thanks for any help!