Question

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 before it is possible that Romeo's ratio of correct solutions to attempted problems will be greater than Juliet's.

Answer 1

Juliet has already done 213 problems and got 210 correct. The ratio is 210/213.

Romeo has attempted 4 problems and got 2 right.

Now they are attempting x problems until both their ratios are same.

Each one attempting x problems

the ratios of Juliet will be

\[\frac{ 210 +x }{ 213 + x }\]
The ratio of Romeo will be

\[\frac{ 2 + x }{ 4 + x }\]
Now these ratios should be the same

\[\frac{ 210 + x }{213 + x } = \frac{ 2 + x }{ 4 + x }\]
Now cross multiply and solve for x

Answer 2

@sri_maths

Answer 3

answer is wrong i suppose

Answer 4

@rajee_sam