In a college,150 students of MBA are enrolled | vedclass

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(A) Total students $= 150$. Ratio of boys to girls $= 7:8$. Number of boys $= (7/15) 	imes 150 = 70$. Number of girls $= (8/15) 	imes 150 = 80$. Let

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$10:11$

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(A) Total students $= 150$. Ratio of boys to girls $= 7:8$. Number of boys $= (7/15) 	imes 150 = 70$. Number of girls $= (8/15) 	imes 150 = 80$. Let $M, H, F$ denote Marketing,$HR$,and Finance. For boys: $n(M) = 0.4 	imes 70 = 28$,$n(H) = 0.3 	imes 70 = 21$,$n(F) = 0.3 	imes 70 = 21$. Intersections: $n(H \cap M) = 7$,$n(H \cap F) = 6$,$n(M \cap F) = 5$,$n(M \cap H \cap F) = 2$. Boys only in $HR = n(H) - [n(H \cap M) + n(H \cap F) - n(M \cap H \cap F)] - n(M \cap H \cap F) = 21 - [7 + 6 - 2] - 2 = 21 - 11 - 2 = 8$. Wait,recalculating: Boys only in $HR = 21 - (7-2) - (6-2) - 2 = 21 - 5 - 4 - 2 = 10$. For girls: $n(M) = 0.5 	imes 80 = 40$,$n(H) = 0.3 	imes 80 = 24$,$n(F) = 0.2 	imes 80 = 16$. Intersections: $n(H \cap M) = 9$,$n(H \cap F) = 7$,$n(M \cap F) = 8$,$n(M \cap H \cap F) = 3$. Girls only in $HR = 24 - (9-3) - (7-3) - 3 = 24 - 6 - 4 - 3 = 11$. Ratio of boys to girls only in $HR = 10:11$.

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