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In MBA 45th batch, 40% of the girls are fair and the remainder 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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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.
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| OK. Half of the problem is just solved. |
| Now let' solve the rest. |
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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.
ANSWER. In MBA 45th batch, there are 60 girls fair and moderate.
Solved.
Really good entertainment problem.