In a class of $40$ students, $12$ enrolled for both English and German. $22$ enrolled for German. If the students of the class enrolled for at least one of the two subjects, then how many students enrolled for English but not German ?

10

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20

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18

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22

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Solution

The correct option is C $18$
Let $E$ be the set of students enrolled in English, $G$ be the set of students enrolled in German
Given: $n (E \cup G) = 40, n (E \cap G) = 12, n (G) = 22$
Now, $n (E \cap G^{'}) = n (E) - n (E \cap G)$
Also, $n (E \cup G) = n (E) + n (G) - n (E \cap G)$

$\Rightarrow 40 = n (E) + 22 - 12 \Rightarrow n (E) = 30$

$∴ n (E - G) = n (E) - n (E \cap G) = 30 - 12 = 18$

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In a class of 40 students, 12 enrolled for both English and German. 22 enrolled for German. If the students of the class enrolled for at least one of the two subjects, then how many students enrolled for English but not German ?

The correct option is C 18Let E be the set of students enrolled in English, G be the set of students enrolled in German Given: n(E∪G)=40,n(E∩G)=12,n(G)=22 Now, n(E∩G′)=n(E)−n(E∩G) Also, n(E∪G)=n(E)+n(G)−n(E∩G) ⇒40=n(E)+22−12⇒n(E)=30 ∴n(E–G)=n(E)–n(E∩G)=30−12=18

In a class of $40$ students, $12$ enrolled for both English and German. $22$ enrolled for German. If the students of the class enrolled for at least one of the two subjects, then how many students enrolled for English but not German ?