Question

In a class of 200 students, 70 played Cricket, 60 played hockey and 80 played football. 30 played Cricket and football, 30 played hockey and football, 40 played Cricket and hockey. Find the maximum number of people playing all the three games and also the minimum number of people playing at least one game?​

Answer 1

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

Let Assume that

• A denotes Cricket

• B denotes Hockey

• C denotes Football

According to given question,

$\rm :\longmapsto\:n(A) = 70$

$\rm :\longmapsto\:n(B) = 60$

$\rm :\longmapsto\:n(C) = 80$

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

$\rm :\longmapsto\:n(B\cap C) = 30$

$\rm :\longmapsto\:n(A\cap C) = 30$

Now, we know that

$\rm :\longmapsto\:n(A\cup B\cup C)$

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

Now, on Substituting the values from given data, we get

$\rm :\longmapsto\:n(A\cup B\cup C)$

$\rm \: = \: 70 + 60 + 80 - 30 - 30 - 40 + n(A\cap B\cap C)$

$\rm \: = \: 210 - 100 + n(A\cap B\cap C)$

$\rm \: = \: 110 + n(A\cap B\cap C)$

$\bf\implies \:n(A\cup B\cup C) = \: 110 + n(A\cap B\cap C)$

So, in order to find the value of the minimum number of people playing at least one game,

$\rm :\longmapsto\:n(A\cap B\cap C) = 0$

So,

$\bf\implies \:n(A\cup B\cup C) = \: 110$

Hence,

• The minimum number of people playing at least one game = 110.

Now,

We have to find the maximum number of people playing all the three games.

For this,

We have from given question,

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

$\rm :\longmapsto\:n(B\cap C) = 30$

$\rm :\longmapsto\:n(A\cap C) = 30$

So, to find the maximum value of people, who play all three games should be the smallest value of above intersection part.

So, smallest value is 30.

So,

$\bf\implies \:n(A\cap B\cap C) \leqslant 30$

Hence,

• The maximum number of people who play all three games = 30.

Additional Information :-

$\rm :\longmapsto\:(A\cup B)' = A'\cap B'$

$\rm :\longmapsto\:(A\cap B)' = A'\cup B'$

$\rm :\longmapsto\:A\cap A' = \: \phi$

$\rm :\longmapsto\:A\cup A' = U$

$\rm :\longmapsto\:U' = \: \phi$

$\rm :\longmapsto\: \phi ' = \: U$

$\rm :\longmapsto\:A - B = A\cap B'$

Answer 2

Answer:

The maximum number of people playing all the three games and also the minimum number of people playing at least one game = 30.

Step-by-step explanation:

Let us Assume that

A means Cricket

B means Hockey

C means Football

As indicated by given question,

n(A) = 70:⟼n(A)=70

n(B) = 60:⟼n(B)=60

n(C) = 80:⟼n(C)=80

⟼n(A∩B)=40

⟼n(B∩C)=30

⟼n(A∩C)=30

Presently, that's what we know

⟼n(A∪B∪C) = n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(C∩A)+n(A∩B∩C)

Presently, on Substituting the qualities from given information, we get

⟼n(A∪B∪C)

= 70+60+80−30−30−40+n(A∩B∩C)

= 210−100+n(A∩B∩C)

= 110+n(A∩B∩C)

⟹n(A∪B∪C)=110+n(A∩B∩C)

In this way, to find the worth of the base number of individuals playing something like one game,

⟼n(A∩B∩C)=0

In this way,

⟹n(A∪B∪C)=110

Subsequently,

The base number of individuals playing somewhere around one game = 110.

Presently,

We need to find the greatest number of individuals playing every one of the three games.

For this,

We have from given question

:⟼n(A∩B)=40

⟼n(B∩C)=30

⟼n(A∩C)=30

Thus, to find the most extreme worth of individuals, who play each of the three games ought to be the littlest worth of above crossing point part.

In this way, littlest worth is 30.

In this way,

⟹n(A∩B∩C)⩽30

Subsequently,

The greatest number of individuals who play every one of the three games = 30.

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