# PE 1 - Venn Diagram | LR - Venn Diagram

**Answer the next 4 questions based on the information given below.**

In a school, some students out of 400 are trained for 3 sports i.e., Cricket, Football and Hockey. Students may opt to receive training for more than one of these 3 sports.

- The number students training for Cricket is 40 less than those who are not training for Cricket.
- Number of students who train in more than one of these sports is 180.
- Number of students who are not training for any of the sports is twice the number of students who are training for all three sports.
- Number of students who are training for Football and exactly one other sport is 120.
- Number of students are training for only Football is twice that of the number of students training for only Hockey.
- Number of students who train for Cricket but not for Football is 70.
- Number of students training for both Football and Hockey is 90.

**PE 1 - Venn Diagram | LR - Venn Diagram**

How many students do not train for any of the 3 sports?

- (a)
80

- (b)
40

- (c)
60

- (d)
Cannot be determined

Answer: Option A

**Explanation** :

The given information can be represented through a Venn Diagram.

The number students training for Cricket is 40 less than those who are not training for Cricket.

Let the number of students training for cricket = x and not training for cricket will be 400 - x.

∴ x + 40 = 400 – x

⇒ x = 180

From the figure we can see that,

Number of people training for only Football and Hockey = 90 – y.

∴ Number of people who train for only Cricket and Football = 120 – (90 - y) = 30 + y.

⇒ Number of people training for Cricket = 180 = 70 + (30 + y) + y

⇒ y = 40

Now, total number of people = 400 = 180 + 2x + (90 - y) + x + 2y

⇒ 130 = 3x + y = 3x + 40

⇒ x = 30

Now,

Number of students who train in more than one of these sports is 180 = y + 120 + (students training for only Cricket and hockey)

⇒ Students training for only Cricket and Hockey = 60 – y = 20

⇒ Number of students training for Only Cricket = 70 – 20 = 50.

Hence, we get the following diagram.

∴ Number of students who do not train for any of the 3 sports = 2y = 80.

Hence, option (a).

Workspace:

**PE 1 - Venn Diagram | LR - Venn Diagram**

How many students train for only Football?

- (a)
90

- (b)
50

- (c)
60

- (d)
30

Answer: Option C

**Explanation** :

Consider the solution to first question of this set.

Number of students training only for Football = 60.

Hence, option (c).

Workspace:

**PE 1 - Venn Diagram | LR - Venn Diagram**

The percentage point difference between the number of students who train for exactly one sport and those who train for exactly three sports is:

- (a)
20

- (b)
15

- (c)
25

- (d)
Cannot be determined

Answer: Option C

**Explanation** :

Consider the solution to first question of this set.

Difference in number of students who train for exactly one and exactly three sports = (50 + 60 + 30) – 40 = 100

∴ % difference = 100/400 × 100 = 25%

Hence, option (c).

Workspace:

**PE 1 - Venn Diagram | LR - Venn Diagram**

How many students train for both Cricket and Hockey but not in Football?

- (a)
20

- (b)
40

- (c)
70

- (d)
30

Answer: Option A

**Explanation** :

Consider the solution to first question of this set.

Number of students training for both Cricket and Hockey but not Football = 20.

Hence, option (a).

Workspace:

**Answer the next 2 questions based on the information given below.**

Of the 580 members of a sports club, 224 play Cricket, 195 play Football, 196 play Hockey. Each member plays at least one game out of these 3 games. The members who play Football and Hockey both must play Cricket also. For every four members who play at least two games, there are three members who play all the three games.

**PE 1 - Venn Diagram | LR - Venn Diagram**

Find the number of members who play all the three games.

Answer: 15

**Explanation** :

Given that members who play Football and Hockey both must play Cricket also. It means there is no player who plays only Football and Hocket.

For every four members who play at least two games, there are three members who play all the three games.

Let the number of players playing all three sports be 3x

∴ Number of players playing at least two games = 4x

⇒ Number of players playing exactly two + exactly three games = 4x.

⇒ Number of players playing exactly two + 3x = 4x

∴ Number of players playing exactly two games = 4x – 3x = x.

All players play at least one game.

We know:

A ∪ B ∪ C = A + B + C – (Those who play exactly two games) - 2 × (Those who play all three games)

⇒ 580 = 224 + 195 + 196 – x – 2 × 3x

⇒ 7x = 35

∴ x = 5

∴ Number of players playing all three games = 3x = 15.

Hence, 15.

Workspace:

**PE 1 - Venn Diagram | LR - Venn Diagram**

Find the number of members who play only Cricket.

Answer: 204

**Explanation** :

Consider the solution to first question of this set.

Number of players playing only Cricket = 224 - x - 3x = 224 – 4x = 224 – 20 = 204

Hence, 204.

Workspace:

**Answer the next 4 questions based on the information given below.**

A survey was conducted among 500 people regarding their likings for Candy Crush, Chess, Temple Run. 280 people did not like Candy Crush, 260 people did not like Chess while 310 people did not like Temple Run. 20 people like all the three games while 80 people did not like any of the three games. 100 people like both Candy Crush and Chess. 90 people like both Chess and Temple Run while 60 people like both Candy Crush and Temple Run.

**PE 1 - Venn Diagram | LR - Venn Diagram**

Find the number of people who like Candy Crush but do not like Chess.

Answer: 120

**Explanation** :

Number of people who like Candy Crush = 500 – 280 = 220

Number of people who like Chess = 500 – 260 = 240

Number of people who like Temple Run = 500 – 310 = 190

20 people like all the three games

100 people like both Candy Crush and Chess

∴ Number of people liking only Candy Crush and Chess = 100 – 20 = 80

90 people like both Chess and Temple Run

∴ Number of people liking only Chess and Temple Run = 90 – 20 = 70

60 people like both Candy Crush and Temple Run

∴ Number of people liking only Candy Crush and Temple Run = 60 – 20 = 40

We can make the following Venn Diagram:

∴ Number of people who like Candy Crush but do not like Chess = 220 – 80 – 20 = 120

Hence, 120.

Workspace:

**PE 1 - Venn Diagram | LR - Venn Diagram**

Find the number of people who like either Chess or Temple Run.

Answer: 340

**Explanation** :

Consider the solution to first question of this set.

Number of people who like either Chess or Temple Run = 240 + 60 + 40 = 340

Hence, 340.

Workspace:

**PE 1 - Venn Diagram | LR - Venn Diagram**

Find the number of people who like at most one game.

Answer: 290

**Explanation** :

Consider the solution to first question of this set.

The number of people who like at most one game = 500 – (20 + 80 + 70 + 40) = 500 – 210 = 290

Hence, 290.

Workspace:

**PE 1 - Venn Diagram | LR - Venn Diagram**

Find the number of people who like exactly one of Candy Crush and Chess.

Answer: 150

**Explanation** :

Consider the solution to first question of this set.

The number of people who like exactly one of Candy Crush and Chess = 80 + 70 = 150

Hence, 150.

Workspace:

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