75% Right
A Brilliant member has solved 500 problems and got 350 of them right.
If he now starts getting every problem right, after how many problems will he have got 75% of all his problems right?
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5 solutions
When I solved this question, it said 75% of people got this right.
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So now one could ask how many of the next people should get it right to make the percentage exactly 75% again. Of course, if the next person gets it wrong, it will still be over 75%, but after two wrong people it might go under 75%. Interesting observation
It’s 73% now
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Unfortunately, it has been like this for quite some time. I don't think it will eventually stabilize at 75%, which would be funny.
I tried this with 3/4, not 0.75 and got x=-50. I just did trial and error after that
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You did something wrong then.
Simplifying (350 + x)/(500 + x) = 3/4 results in:
4(350 + x) = 3(500 + x)
Which further simplifies down to:
1400 + 4x = 1500 + 3x
From this, we get x = 100.
The member in question's current score is 5 0 0 3 5 0 , so the score after he gets p problems correct is 5 0 0 + p 3 5 0 + p . Then, we set this new expression equal to 75% or .75: 5 0 0 + p 3 5 0 + p = . 7 5 ⟹ 3 5 0 + p = 3 7 5 + . 7 5 p ⟹ . 2 5 p = 2 5 ⟹ p = 1 0 0 β ⌈ ∣ ⌉
At my point of submission 75% of people got this right
Lol, just saying - on a somewhat unrelated note - 75% of people also got this right,
Actually, this is the third time somone is pointing that out. This brings me to the question, when the percentage of people is at 75% what is the exspected number of people who solve or don't solve it until the percentage reaches 75% again. Interesting!
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Keep in mind that if enough people solve this problem, it may stay at 75% for example 97/129 and 97/130 to the nearest whole percent are both 75%
(Incidentally, I got it right but it currently says 74%)
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Oh right, of course that makes it difficult
So at the moment, 301 out of 405 people solved it, which is – as you said – about 74%. But all 17 fractions 4 0 5 + x 3 0 1 + x , 3 ≤ x ≤ 1 9 , x ∈ N get rounded to 75%.
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@Henry U – I actually submitted a related problem a few months ago: https://brilliant.org/problems/possible-percent-problem/?ref_id=1491623
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@Jeremy Galvagni – Interesting problem!
Do you know a generalization to it? I'd imagine the solution to be almost the same for all values of the percentage because you're always looking for a fraction that lies or doesn't lie inside this intervall of lenght 1.
@Jeremy Galvagni – I tried to find a pattern in the values of x (the smallest possible denominator), but it's difficult to make predictions about specific numbers. Now, I have also posted a problem on this topic.
Now it says 73%
(350+x)/(500+x)*100=75
Therefore x=100
Let x be the problems the person does :
5 0 0 + x 3 5 0 + x 3 5 0 + x 0 . 2 5 x = 0 . 7 5 = 3 7 5 + 0 . 7 5 x = 2 5 x = 1 0 0
The percentage of right problems after solving x more is given by
5 0 0 + x 3 5 0 + x .
To get a percentage of 75% we set this equal to 0.75 and solve the resulting equation.
5 0 0 + x 3 5 0 + x = 0 . 7 5 ⇔ 3 5 0 + x = 0 . 7 5 ⋅ 5 0 0 + 0 . 7 5 x ⇔ 0 . 2 5 x = 2 5 ⇔ x = 1 0 0