Question

c) In a hostel with 400 children, some opt for Football and 250 children opt for
Cricket and 160 opt for both Football and Cricket.
(4M)
i) How many children have opted for only Football? ii) How many children have
opted for only Cricket?​

Answer 1

Answer:

otp for children's are in 240, children have is- 490

Answer 2

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

Given that,

In a hostel with 400 children, some opt for Football and 250 children opt for Cricket and 160 opt for both Football and Cricket.

Let assume that,

A represents the set of students opted for football.

B represents the set of students opted for Cricket.

According to statement, we have

$\rm :\longmapsto\:n(A\cup B) = 400$

$\rm :\longmapsto\:n(B) = 250$

$\rm :\longmapsto\:n(A\cap B) = 160$

We know, that

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

$\rm :\longmapsto\:400 = n(A) + 250 - 160$

$\rm :\longmapsto\:400 = n(A) + 90$

$\rm :\longmapsto\:400 - 90= n(A)$

$\rm :\longmapsto\:n(A) = 310$

So, it implies 310 students opted for football.

Now,

(i) How many students opted for only Football?

$\rm :\longmapsto\:n(A - B)$

$\rm \:  =  \:n(A) - n(A\cap B)$

$\rm \:  =  \:310 - 160$

$\rm \:  =  \:150$

(ii) How many children opted only for Cricket?

$\rm :\longmapsto\:n(B - A)$

$\rm \:  =  \:n(B) - n(A\cap B)$

$\rm \:  =  \:250 - 160$

$\rm \:  =  \:90$

Additional Information :-

$\boxed{ \bf{ \: A\cap \phi \: = \: \phi}}$

$\boxed{ \bf{ \: A\cup \phi \: = \: A}}$

$\boxed{ \bf{ \: U' = \phi}}$

$\boxed{ \bf{ \: \phi \: ' = \: U}}$

$\boxed{ \bf{ \: A\cup A' = U}}$

$\boxed{ \bf{ \: A\cap A' = \phi}}$

$\boxed{ \bf{ \: {(A\cup B)}^{'} = A'\cap B'}}$

$\boxed{ \bf{ \: {(A\cap B)}^{'} = A'\cup B'}}$