Brilliant.org solvers

Probability Level 3

The percentage of people to solve a problem correctly on Brilliant.org is rounded to the nearest integer.

If 27 % 27 \% of the attempts solve my problem, what is the minimum amount of people that can get my problem wrong?


The answer is 8.

Chris Sapiano by Chris Sapiano , 19, United Kingdom

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Eric Scholz
Jul 15, 2019

8/11 = 0.727272... Is the fraction with smallest integers, that when rounded to 2 decimals results in 0.73 (0.27 rounded get it correct). Thus answer is 8

4 Helpful 1 Interesting 0 Brilliant 0 Confused
Pop Wong
Aug 18, 2020

Let x x be the number of correct answer, N N be the number of total attempts.

The rounded percentage of correct answer means

265 1000 ≤ x N < 275 1000 ⇒ 1000 265 ≥ N x > 1000 275 ⇒ 200 x 53 ≥ N > 200 x 55 ⇒ 3 x + 41 x 53 ≥ N > 3 x + 35 x 55 Goal: find min. x for which there exists an integer in between ⇒ 41 x 53 ≥ K > 35 x 55 looking for an integer K= N-3x in between ⇒ 41 x 53 ≥ K > 7 x 11 it is clear that x=1, 2 no solution, for x=3 ⇒ 41 × 3 53 ≥ K > 7 × 3 11 = 21 11 K=2, N=2+3(3) = 11 \begin{aligned} \cfrac{265}{1000} &\leq \cfrac{x}{N} < \cfrac{275}{1000} \\ \Rightarrow \cfrac{1000}{265} &\geq \cfrac{N}{x} > \cfrac{1000}{275} \\ \Rightarrow \cfrac{200x}{53} &\geq N > \cfrac{200x}{55} \\ \Rightarrow 3x + \cfrac{41x}{53} &\geq N > 3x + \cfrac{35x}{55} \hspace{5mm} \textcolor{#3D99F6}{ \text{Goal: find min. x for which there exists an integer in between}} \\ \Rightarrow \cfrac{41x}{53} &\geq K > \cfrac{35x}{55} \hspace{12mm} \textcolor{#3D99F6}{ \text{ looking for an integer K= N-3x in between } } \\ \Rightarrow \cfrac{41x}{53} &\geq K > \cfrac{7x}{11} \\ \text{ it is clear that x=1, 2 no solution, for x=3 } \Rightarrow \cfrac{41 \times 3}{53} &\geq K > \cfrac{7 \times 3}{11} = \cfrac{21}{11} \hspace{2mm} \textcolor{#3D99F6}{ \text{K=2, N=2+3(3) = 11} } \\ \end{aligned}

Minimum correct answer x = 3 \boxed{x = 3} with total attempts N = 11 \boxed{ N = 11} , therefore minimum amount of people that can get my problem wrong = 11 − 3 = 8 \boxed{= 11 -3 = 8}

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Kyle T
Jul 16, 2019 
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
 
<?php
for($a=1;$a<=20;$a++){ //a represents the number of people who got the problem correct 
    for($b=1;$b<=20;$b++){ //b represents the number of people who got the problem wrong
        if(round(($a/($a+$b))*100)==27){ //calculate percentage, compare against integer 27
            echo $b; //answer: 8
            exit;
        }
    }
}
?>

0 Helpful 0 Interesting 0 Brilliant 1 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...