Ajay Verma
Last Activity: 11 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,
askIITians faculty,
IIT HYDERABAD