Question

In a mathematics class of 100 students, at least one of the language french, german and russian have to be opt by students. If 65 study french, 45 students study german, 42 study russian, 20 study french and german, 25 study french and russian, 15 study german and russian, then find the number of students studying all the subjects solve by set theory

Answer 1

Answer with Explanation :

in a Mathematics class total no of students are 100 ,

No of students studying French language = 65 ,

let, ∴ n(A)=65.

No of students studying German language = 45 ,

let, ∴ n(B)=45.

Here 42 students studying Russian,

let, ∴ n(C)=42.

20 students studying French and German, ∴ n (A∩B) =20

25 studying French and Russian,

∴ n (A∩C)=25

15 students study german and russian,

∴ n(B∩C) = 15

No of students studying all the subjects = n(A ∩ B ∩ C)

we know,

n(A ∪ B ∪ C) = n(A)+n(B)+n(C)-n(A∩B) - n(B∩C) -n(A∩C) + n( A ∩ B ∩ C )

Here, n(A ∪ B ∪ C) = total no of students in a Mathematics class

100 = 65 + 45 + 42 - 20 - 15 -25 + n( A ∩ B ∩ C )

100 = 152 -20 - 15 - 25 + n( A ∩ B ∩ C )

100 = 117 - 25 + n( A ∩ B ∩ C )

100 = 92 + n( A ∩ B ∩ C )

100 - 92 = n( A ∩ B ∩ C )

8 = n( A ∩ B ∩ C )

Therefore 8 students studying all the subjects.

Answer 2

$\huge{ \green{ \mathfrak{ \underline{Question}}}}$

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In a mathematics class of 100 students, at least one of the language french, german and russian have to be opt by students. If 65 study french, 45 students study german, 42 study russian, 20 study french and german, 25 study french and russian, 15 study german and russian, then find the number of students studying all the subjects solve by set theory.$</font></i>$

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$\huge{ \green{ \mathfrak{ \underline{Answer}}}}$

$\boxed{\red{\bold{Given:}}}$

$French \: students \: = n(A) = 65$

$German \: students \: = n(B) = 45$

$Russian \: students \: = n(C) = 42$

$French \:and \: German \:= n(A \cap B) = 20$

$German\: and \:Russian \:= n(B \cap C) = 15$

$French\: and\: Russian \:= n(A \cap C) = 25$

$\boxed{\red{\bold{By\:set \:theory:}}}$

$n(A \cup B \cup C) = n(A) + n(B) + n(C) \\ -n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C)$

$\implies 100 = 65 + 45 + 42 - 20 - 15 - 25 + \\ n(A \cap B \cap C)$

$\implies 100 = 152 - 60 + n(A \cap B \cap C)$

$\implies n(A \cap B \cap C) = 100 - 92$

$\implies n(A \cap B \cap C) = 8$

$\boxed{\red{\bold{Hence:}}}$

$Studying \: all \: the \: subjects = n(A \cap B \cap C) = 8$

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