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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.
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 = 210totel no. of attempted qus = 213+ Xfor J: total no. of correct qus = 2+Xtotel no. of attempted qus = 4+ Xgiven:2+X / 4+x > 210/(213+ X)after solving..X2+ 5X - 414 > 0(X - 18) (X +23) > 0bcoz X is no. of qus.. so it wil be positiveso X > 18Thanks and Regards,Ajay verma,askIITians faculty,IIT HYDERABAD
method was right but its not a right answer.
then what is the correct answer?
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