In a group of 100 children, 32% play hoceky, 64% play football and 40% play cricket. 14% play hockey and football , 155 play football and cricket , 13% play hockey and cricket. Only 6% play all the three games .
(a) How many children play one game ?
(b) How many children play atleast two game ?
(c) How many children play hockey or football but not circket ?

Answer

To solve the problem step by step, we will use the information given about the percentages of children playing different games and apply set theory concepts.

### Given Data:
- Total children = 100
- Percentage of children playing:
- Hockey (H) = 32% of 100 = 32 children
- Football (F) = 64% of 100 = 64 children
- Cricket (C) = 40% of 100 = 40 children
- Percentage of children playing combinations of games:
- Hockey and Football (H ∩ F) = 14% of 100 = 14 children
- Football and Cricket (F ∩ C) = 15% of 100 = 15 children
- Hockey and Cricket (H ∩ C) = 13% of 100 = 13 children
- All three games (H ∩ F ∩ C) = 6% of 100 = 6 children

### Step 1: Define Variables
Let:
- A = Children playing only Hockey
- B = Children playing only Football
- C = Children playing only Cricket
- D = Children playing Hockey and Football but not Cricket
- E = Children playing Football and Cricket but not Hockey
- F = Children playing Hockey and Cricket but not Football
- G = Children playing all three games (H ∩ F ∩ C)

From the problem, we know:
- G = 6

### Step 2: Find Values for D, E, and F
Using the intersections:
1. For Hockey and Football (H ∩ F):
\[ D + G = 14 \implies D + 6 = 14 \implies D = 8 \]
2. For Football and Cricket (F ∩ C):
\[ E + G = 15 \implies E + 6 = 15 \implies E = 9 \]
3. For Hockey and Cricket (H ∩ C):
\[ F + G = 13 \implies F + 6 = 13 \implies F = 7 \]

### Step 3: Find Values for A, B, and C
Using the total number of players in each game:
1. For Hockey:
\[ A + D + F + G = 32 \implies A + 8 + 7 + 6 = 32 \implies A = 32 - 21 = 11 \]
2. For Football:
\[ B + D + E + G = 64 \implies B + 8 + 9 + 6 = 64 \implies B = 64 - 23 = 41 \]
3. For Cricket:
\[ C + E + F + G = 40 \implies C + 9 + 7 + 6 = 40 \implies C = 40 - 22 = 18 \]

### Summary of Values:
- A (only Hockey) = 11
- B (only Football) = 41
- C (only Cricket) = 18
- D (Hockey and Football only) = 8
- E (Football and Cricket only) = 9
- F (Hockey and Cricket only) = 7
- G (All three) = 6

### (a) How many children play one game?
To find the number of children who play only one game:
\[ \text{Total playing one game} = A + B + C = 11 + 41 + 18 = 70 \]

### (b) How many children play at least two games?
To find the number of children who play at least two games:
\[ \text{Total playing at least two games} = 100 - \text{Total playing one game} = 100 - 70 = 30 \]

### (c) How many children play hockey or football but not cricket?
To find the number of children who play Hockey or Football but not Cricket:
\[ \text{Total playing Hockey or Football but not Cricket} = A + B + D = 11 + 41 + 8 = 60 \]

### Final Answers:
(a) 70 children play one game.
(b) 30 children play at least two games.
(c) 60 children play hockey or football but not cricket.