CAT 2020 Question Paper | Quants Slot 2
CAT Previous Year Paper | CAT Quants Questions | Question 7
CAT Questions It is vital for your CAT preparation online that you practice questions from CAT previous year paper so that you are well versed with the type of questions you can expect in your CAT exam. The CAT Question Paper is known to throw googlies at students and hence it is best that you prepare in a manner that'll help you tackle these googlies in the best possible way! Don't get caught off guard, start your preparation with this simple yet tricky question from CAT 2020 question paper. Make sure you do not peep at the solution before attempting it yourself!
🎉 Ace the Final Stretch with our Last Mile Excellence – Your Ultimate CAT 2024 Boost!
Click here!
Video Explanation
Best CAT Coaching in Chennai
CAT Coaching in Chennai - CAT 2022
Limited Seats Available - Register Now!
Explanatory Answer
To find the minimum possible value keep x and y as close as possible
So, x = y = 51
= 2601(1 + \\frac{1}{51}))(1 + \\frac{1}{51}))
= 2601(\\frac{52}{51}))(\\frac{52}{51}))
= 52 × 52
= 2704
The question is "If x and y are positive real numbers satisfying x + y = 102, then the minimum possible value of 2601(1 + \\frac{1}{x}))(1 + \\frac{1}{y})) is"
