Question 1194891
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In MBA 45th batch, 40% of the girls are fair and the {{{highlight(cross(reminder))}}} <U>remainder</U> is brown. 
Half of the girl are beautiful and half are moderate. 
If 10% of the girls are fair and beautiful, and 40 girls are brown and moderates, 
how many girls are fair and moderate?
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<pre>
In the set of all girls in MBA 45th batch, we have 4 subsets:

    - F (fair, 40%);

    - R (brown, 60%);

    - B (beautiful, 50%);

    - M (moderate, 50%).


We also are given in-pair intersections

    - FB (fair and beautiful, 10%);

    - RM (brown and moderate, 40 girls).


From it, the union 

    F U B = 40% + 50% - 10% = 80%  (fair or beautiful).


The complement for (F U B) is the set RM = (R and M) = (brown and moderate); so it contains 100%-80% = 20%.

From the other side, the set (brown and moderate) is 40 girls;

So, the total girls in MBA 45th batch is  40/0.2 = 200.


    +-----------------------------------------------+
    |    OK.  Half of the problem is just solved.   |
    |          Now let' solve the rest.             |
    +-----------------------------------------------+


Notice, that the sought set FM = (fair and moderate)  is the complement 
of the set FB (fair and beautiful) to F (fair).


In other words,  FM = F \ FB.


Hence,  FM is 40% minus 10% :  FM = 40% - 10% = 30%.


30% of 200 is 0.3*200 = 60.


<U>ANSWER</U>.  In MBA 45th batch, there are 60 girls fair and moderate.
</pre>

Solved.



Really good entertainment problem.