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It is simply a mixture problem. Men's concentration last year is 3/5. We don't know men's concentration in new members this year. That will decide the overall concentration and whether it is greater than 3/5.
(1) More than half of the new members are male.
This means C1 can be anywhere as shown in the diagram and accordingly Cavg will be less than / equal to / more than 3/5.
If C1 is very close to 1/2 as shown, Cavg will be less than 3/5. If C1 is more than 3/5, Cavg will be more than 3/5.
This alone is not sufficient.
Attachment:
Screenshot 2025-01-03 at 2.41.47 PM.png [ 40.75 KiB | Viewed 196 times ]
(2) The number of members of the club this year is 6/5 the number of members last year.
Gives us the weights that last year : new members = 5 : 1 but not C1. We cannot find Cavg without C1.
Using both also, we don't have C1. The same diagram is applicable.
Answer (E)
Check these video on weighted averages and mixtures:
https://www.youtube.com/watch?v=_GOAU7moZ2Q
https://www.youtube.com/watch?v=VdBl9Hw0HBg
and these posts:
https://anaprep.com/arithmetic-weighted-averages/
https://anaprep.com/arithmetic-mixtures/
