# CAT Quantitative Aptitude Questions | CAT Ratios, Mixtures, Alligations and Averages Questions

###### CAT Questions | Averages | Averages - Maximum

The question is from CAT Ratio and Proportions, Mixtures and Averages. We need to maximize the average based on the given details. CAT exam is known to test on basics rather than high funda ideas. A range of CAT questions can be asked from Ratios and Proportions, Mixtures, Alligations and Averages. Make sure you master the topics. 2IIMs CAT questions bank provides you with CAT questions that can help you gear for CAT Exam CAT 2020.

Question 1: Consider a class of 40 students whose average weight is 40 kgs. m new students join this class whose average weight is n kgs. If it is known that m + n = 50, what is the maximum possible average weight of the class now?

1. 40.18 kgs
2. 40.56 kgs
3. 40.67 kgs
4. 40.49 kgs

## Best CAT Coaching in Chennai

#### CAT Coaching in Chennai - CAT 2021Online Batches Available Now!

##### Method of solving this CAT Question from Ratio, Mixtures and Averages: This question gives us a sense of how averages change when entities are added to or taken away from the system. One needs to learn how the system behaves without having to go through the onerous algebraic route.

If the overall average weight has to increase after the new people are added, the average weight of the new entrants has to be higher than 40.
So, n > 40

Consequently, m has to be < 10 (as n + m = 50)
Working with the â€œdifferencesâ€? approach, we know that the total additional weight added by â€œmâ€? students would be (n - 40) each, above the already existing average of 40. m(n - 40) is the total extra additional weight added, which is shared amongst 40 + m students.
So, m * $$frac{$n-40$}{(m+40)}$ has to be maximum for the overall average to be maximum. At this point, use the trial and error approach$or else, go with the answer options) to arrive at the answer.

The maximum average occurs when m = 5, and n = 45

And the average is 40 + (45 â€“ 40) * $$frac{5}{45}\\$ = 40 + $\frac{5}{9}\\$ = 40.56 kgs The question is "what is the maximum possible average weight of the class now?" ##### Hence, the answer is "40.56 kgs". Choice B is the correct answer. Â ###### Best CAT Online Coaching Try upto 40 hours for free Learn from the best! ###### Already have an Account? ###### CAT Coaching in ChennaiCAT 2021Enroll at 49,000/- 44,000/- Online Classroom Batches Starting Now! ###### Best CAT Coaching in Chennai Introductory offer of 5000/- Attend a Demo Class ###### Best Indore IPM & Rohtak IPM CoachingSignup and sample 9 full classes for free. Register now! ##### Where is 2IIM located? 2IIM Online CAT Coaching A Fermat Education Initiative, 58/16, Indira Gandhi Street, Kaveri Rangan Nagar, Saligramam, Chennai 600 093 ##### How to reach 2IIM? Phone:$91) 44 4505 8484
Mobile: (91) 99626 48484
WhatsApp: WhatsApp Now
Email: prep@2iim.com