116773
In a class of 60 students, 25 students play cricket and 20 students play tennis and 10 students play both the games, then the number of students who play neither is
1 45
2 0
3 25
4 35
Explanation:
C Given, \(\mathrm{n}(\mathrm{C})=25\) \(\mathrm{n}(\mathrm{T})=20\) \(\mathrm{n}(\mathrm{C} \cap \mathrm{T})=10\) Where, \(\mathrm{C}=\) Number of students play cricket \(\mathrm{T}=\) Number of students play tennis Then, \(\mathrm{P}(\mathrm{C} \cup \mathrm{T})=\mathrm{P}(\mathrm{C})+\mathrm{P}(\mathrm{T})-\mathrm{P}(\mathrm{C} \cap \mathrm{T})\) \(\mathrm{P}(\mathrm{C} \cup \mathrm{T})=25+20-10\) \(\mathrm{P}(\mathrm{C} \cup \mathrm{T})=45-10\) \(\mathrm{P}(\mathrm{C} \cup \mathrm{T})=35\) So, the number of student who play neither is - \(=60-35\) \(=25 .\)
Karnataka CET 2014
Sets, Relation and Function
116774
If \(U\) is the universal set with 100 elements; \(A\) and \(B\) are two sets such that \(n(A)=50, n(B)=\) \(60, n(A \cap B)=20\) then \(n\left(A^{\prime} \cap B^{\prime}\right)=\)
116773
In a class of 60 students, 25 students play cricket and 20 students play tennis and 10 students play both the games, then the number of students who play neither is
1 45
2 0
3 25
4 35
Explanation:
C Given, \(\mathrm{n}(\mathrm{C})=25\) \(\mathrm{n}(\mathrm{T})=20\) \(\mathrm{n}(\mathrm{C} \cap \mathrm{T})=10\) Where, \(\mathrm{C}=\) Number of students play cricket \(\mathrm{T}=\) Number of students play tennis Then, \(\mathrm{P}(\mathrm{C} \cup \mathrm{T})=\mathrm{P}(\mathrm{C})+\mathrm{P}(\mathrm{T})-\mathrm{P}(\mathrm{C} \cap \mathrm{T})\) \(\mathrm{P}(\mathrm{C} \cup \mathrm{T})=25+20-10\) \(\mathrm{P}(\mathrm{C} \cup \mathrm{T})=45-10\) \(\mathrm{P}(\mathrm{C} \cup \mathrm{T})=35\) So, the number of student who play neither is - \(=60-35\) \(=25 .\)
Karnataka CET 2014
Sets, Relation and Function
116774
If \(U\) is the universal set with 100 elements; \(A\) and \(B\) are two sets such that \(n(A)=50, n(B)=\) \(60, n(A \cap B)=20\) then \(n\left(A^{\prime} \cap B^{\prime}\right)=\)
116773
In a class of 60 students, 25 students play cricket and 20 students play tennis and 10 students play both the games, then the number of students who play neither is
1 45
2 0
3 25
4 35
Explanation:
C Given, \(\mathrm{n}(\mathrm{C})=25\) \(\mathrm{n}(\mathrm{T})=20\) \(\mathrm{n}(\mathrm{C} \cap \mathrm{T})=10\) Where, \(\mathrm{C}=\) Number of students play cricket \(\mathrm{T}=\) Number of students play tennis Then, \(\mathrm{P}(\mathrm{C} \cup \mathrm{T})=\mathrm{P}(\mathrm{C})+\mathrm{P}(\mathrm{T})-\mathrm{P}(\mathrm{C} \cap \mathrm{T})\) \(\mathrm{P}(\mathrm{C} \cup \mathrm{T})=25+20-10\) \(\mathrm{P}(\mathrm{C} \cup \mathrm{T})=45-10\) \(\mathrm{P}(\mathrm{C} \cup \mathrm{T})=35\) So, the number of student who play neither is - \(=60-35\) \(=25 .\)
Karnataka CET 2014
Sets, Relation and Function
116774
If \(U\) is the universal set with 100 elements; \(A\) and \(B\) are two sets such that \(n(A)=50, n(B)=\) \(60, n(A \cap B)=20\) then \(n\left(A^{\prime} \cap B^{\prime}\right)=\)