Juliet has attempted 213 problems on Brilliant and solved 210 of them correctly. Her friend Romeo has just joined Brilliant, and has attempted 4 problems and solved 2 correctly. From now on, Juliet and Romeo will attempt all the same new problems. Find the minimum number of problems they must attempt such that it is possible that Romeo`s ratio of correct solutions to attempted problems will be strictly greater than Juliet`s.

Ajay Verma
10 years ago
solution:

for minimum no. of probs for rotio of correct solutions to attempted probs.. R's ratio > J's ratio.
assume X problems are attempted by both and R does all correctly and J does all wrong..

so for R: total no. of correct qus = 210
totel no. of attempted qus = 213+ X

for J: total no. of correct qus = 2+X
totel no. of attempted qus = 4+ X

given:
2+X / 4+x > 210/(213+ X)
after solving..

X2+ 5X - 414 > 0

(X - 18) (X +23) > 0

bcoz X is no. of qus.. so it wil be positive
so X > 18

Thanks and Regards,
Ajay verma,

shubham jain
24 Points
10 years ago
method was right but its not a right answer.
Ketan Jain
8 Points
10 years ago
then what is the correct answer?