In a class of 100 students, the number of stu | vedclass

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(A) Let $E$, $M$, and $C$ represent the sets of students who passed in English, Maths, and Commerce, respectively.Given:Students passed in English onl

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) Let $E$, $M$, and $C$ represent the sets of students who passed in English, Maths, and Commerce, respectively.Given:Students passed in English only $= 46$Students passed in Maths only $= 46$Students passed in Commerce only $= 58$Students passed in English and Maths (including all three) $= 16$. So, English and Maths only $= 16 - 7 = 9$.Students passed in Maths and Commerce (including all three) $= 24$. So, Maths and Commerce only $= 24 - 7 = 17$.Students passed in English and Commerce (including all three) $= 26$. So, English and Commerce only $= 26 - 7 = 19$.Students passed in all three subjects $= 7$.Total students who passed in at least one subject $= (	ext{Only } E) + (	ext{Only } M) + (	ext{Only } C) + (	ext{Only } E \cap M) + (	ext{Only } M \cap C) + (	ext{Only } E \cap C) + (E \cap M \cap C)$$= 46 + 46 + 58 + 9 + 17 + 19 + 7 = 202$.Since the total number of students is $100$, and the calculated sum of students who passed is $202$, which exceeds the total class size, there is an inconsistency in the provided problem data. However, based on the provided Venn diagram image, the values are: Only $E=11$, Only $M=13$, Only $C=15$, $E \cap M 	ext{ only}=9$, $M \cap C 	ext{ only}=17$, $E \cap C 	ext{ only}=19$, and All three $=7$.Total passed $= 11 + 13 + 15 + 9 + 17 + 19 + 7 = 91$.Students who failed in all subjects $= 100 - 91 = 9$.

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