Question

plzzzzzz solve this.....i will mark him braniliest....​

Answer 1

Given :-

• 2/5th of the students Play badminton.
• Twice of the number that means 2*(2/5) = 4/5 students play table Tennis .
• 15 students play both game.
• 5 play neither .

Solution :-

Let us Assume That, Total Number of Students in The class are X.

Now , Refer to Image for Venn Diagram.

From image now we can say that,

→ Play on Badminton + Play only TT + Play Both + Play Neither = Total Students = x

→ [ (2x/5) - 15 ] + [ (4x/5) - 15 ] + 15 + 5 = x

→ [ (2x/5) + (4x/5) ] - 30 + 20 = x

→ (6x/5) = (x + 10)

→ (6x/5) - x = 10

→ (x/5) = 19

→ x = 50.

Hence , Total Students In the class are 50.

_____________________________

Now, Given That, 76% of The students play Cricket.

So,

→ Played cricket = (50*76)/100 = 38 Students..

Hence,

→ Maximum Number of Students Playing all Three games = 50 - 38 = 12 Students. (Ans).

And,

→ Minimum Number of Students who might be playing all three games = 15 - 12 = 3 (Ans).

______________________________

Answer 2

${\boxed{\boxed{\mathtt{Given}}}}$ :-

• 2/5 th part of students play badminton
• Twice of the badminton players are table tennis player
• 15 students play both games
• 76% students play cricket .
• 5 play neither game .

${\boxed{\boxed{\mathtt{To\: Find}}}}$

• Total number of student in class
• Maximum number of student playing all the three games
• Minimum number of students playing the three games.

${\boxed{\boxed{\mathtt{Solution}}}}$

Firstly refer to the attachment .

Let's assume tha the class consists X number of students .

x = ( Badminton union tennis ) + 15

⇝ Badminton union tennis = n ( B ) + n ( T ) - Badminton intersection tennis .

⇝ $\frac{2x}{5}$ + $\frac{4x}{5}$ - 15

⇝ $\frac{2x \:+\:4x }{5}$ - 15

⇝ $\frac{6x}{5}$ - 15

x = $\frac{6x}{5}$ - 15 + 5

x = $\frac{6x}{5}$ - 10

x = $\frac{6x\: - \:50}{5}$

5x = 6x - 50

50 = X

So total total number of student in the class is 50.

Student playing cricket :-

⇝ 76% of the total students play cricket .

⇝ $\frac{76\times\: 50}{100}$

⇝ $\frac{3800}{100}$

⇝ 38

So number of student playing cricket is 38 .

Maximum number of student playing all the three games . :-

50 - 38 = 1 2

Minimum number of students playing all the three games :-

15 - 12 = 3