anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
rajee_sam
  • rajee_sam
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
rajee_sam
  • rajee_sam
@sri_maths
anonymous
  • anonymous
answer is wrong i suppose

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
@rajee_sam

Looking for something else?

Not the answer you are looking for? Search for more explanations.