Question

In a group of students, 50 students passed in English, 60 students passed in Mathematics and 40 students passed in both the subjects. Find the number of students who passed either in English or in Mathematics. How many students passed only in English ? How many students passed only in Mathematics ?​

Answer 1

Step-by-step explanation:

in Mathematics = 60 students passed

in English = 50 students passed

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Answer 2

$\large\underline{\sf{Solution-}}$

Given that

• 50 students passed in English.

• 60 students passed in Mathematics.

• 40 students passed in both the subjects.

Let assume that

• A represents the set of students passed in English

• B represents the set of students passed in Mathematics

So,

$\red{\rm :\longmapsto\:n(A) = 50}$

$\red{\rm :\longmapsto\:n(B) = 60}$

$\red{\rm :\longmapsto\:n(A\cap B) = 40}$

Now, Number of students who passed either in English or in Mathematics are

$\rm \:  =  \: n(A\cup B)$

$\rm \:  =  \: n(A) + n(B) - n(A\cap B)$

$\rm \:  =  \: 60 + 50 - 40$

$\rm \:  =  \: 70$

Now, Number of students passed in English only are

$\rm \:  =  \: n(A - B)$

$\rm \:  =  \: n(A) - n(A\cap B)$

$\rm \:  =  \: 50 - 40$

$\rm \:  =  \: 10$

Now, Number of students who passed in Mathematics only are

$\rm \:  =  \: n(B - A)$

$\rm \:  =  \: n(B) - n(A\cap B)$

$\rm \:  =  \: 60 - 40$

$\rm \:  =  \: 20$

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More to Learn :-

1. Commutative Law

$\boxed{\tt{ A\cup B = B\cup A \: }}$

$\boxed{\tt{ A\cap B = B\cap A \: }}$

2. Associative Law

$\boxed{\tt{ A\cup (B\cup C) = (A\cup B)\cup C \: }}$

$\boxed{\tt{ A\cap (B\cap C) = (A\cap B)\cap C \: }}$

3. Distributive Law

$\boxed{\tt{ A\cup (B\cap C) = (A\cup B)\cap (A\cup C) \: }}$

$\boxed{\tt{ A\cap (B\cup C) = (A\cap B)\cup (A\cap C) \: }}$