116751
A survey shows that \(63 \%\) of the Indians like tea whereas \(76 \%\) like coffee. If \(x \%\) of the Indians like both tea and coffee, then
1 \(x=39\)
2 \(x=63\)
3 \(39 \leq \mathrm{x} \leq 63\)
4 none of these
Explanation:
C Given, number of the Indians like tea - \(\mathrm{n}(\mathrm{T})=63\) Number of the Indians like coffee \(\mathrm{n}(\mathrm{C})=76\) And number of the Indians like both tea and coffee \(\mathrm{n}(\mathrm{T} \cap \mathrm{C})=\mathrm{x}\) Then, \(n(T \cup C) =n(T)+n(C)-n(T \cap C)\) \(100 =63+76-x\) \(x =139-100\) \(x =39\) Also, \(\mathrm{n}(\mathrm{T} \cap \mathrm{C}) \leq \mathrm{n}(\mathrm{T})\) \(\mathrm{x} \leq 63\)So, \(\quad 39 \leq \mathrm{x} \leq 63\) Ans: a: Given, \(\mathrm{U}=\) Set of all days Ans: c : Given, \(\mathrm{n}(\mathrm{C})=20, \mathrm{n}(\mathrm{P})=30 \text { and } \mathrm{n}(\mathrm{C} \cap \mathrm{P})=10\) Where \(\mathrm{C}\) and \(\mathrm{P}\) be the number of students in chemistry and physics class. Find, \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=\) ? Here are two conditions - Condition no I : - When both classes meet at the same time, then \(\mathrm{n}(\mathrm{C} \cap \mathrm{P})=\phi\) Then, \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=\mathrm{n}(\mathrm{C})+\mathrm{n}(\mathrm{P})\) \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=20+30=50\) Condition no II :- When both classes meet at different hours. Then \(\mathrm{n}(\mathrm{C} \cap \mathrm{P})=10\) (Given) So, \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=\mathrm{n}(\mathrm{C})+\mathrm{n}(\mathrm{P})-\mathrm{n}(\mathrm{C} \cap \mathrm{P})\) \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=20+30-10\) \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=50-10\) \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=40\)Hence, both (a) and (b) correct
BITSAT-2007
Sets, Relation and Function
116752
Which of the following is an empty set?
1 \(\left\{x \mid x\right.\) is a realnumber and \(\left.x^2-1=0\right\}\)
2 \(\left\{x \mid x\right.\) is a realnumber and \(\left.x^2+3=0\right\}\)
3 \(\left\{x \mid x\right.\) is a real number and \(\left.x^2-9=0\right\}\)
4 \(\left\{x \mid x\right.\) is a real number and \(\left.x^2=x+2\right\}\)
Explanation:
B We check the following is an empty set by options - By option a: \(x^2-1=0\) \(x^2=1\) \(x= \pm 1 \in R\) This is not empty set. By option b: \(x^2+3=0\) \(x^2=-3\) \(x=\sqrt{-3} \in C\) This is a empty set. By option c: \(x^2-9=0\) \(x^2=9\) \(x= \pm 3 \in R\) This is not a empty set. By option d: \(x^2=x+2\) \(x^2-x-2=0\) \(x^2-2 x+x-2=0\) \(x(x-2)+1(x-2)=0\) \((x-2)(x+1)=0\) \(x=-1,2\)This is not empty set.
COMEDK 2014
Sets, Relation and Function
116753
In a class of students, 25 students play cricket, 20 student play tennis and 10 students play both the games. Then the number of students who play tennis only is
1 25
2 10
3 15
4 None of these
Explanation:
B From question, Let 'C' class of students play cricket and ' \(\mathrm{T}\) ' class of student play tennis respectively. Then, given - \(\mathrm{n}(\mathrm{C})=25, \mathrm{n}(\mathrm{T})=20\) \(\mathrm{n}(\mathrm{C} \cap \mathrm{T})=10\) \(\therefore \quad \mathrm{n}(\mathrm{C} \cup \mathrm{T})=\mathrm{n}(\mathrm{C})+\mathrm{n}(\mathrm{T})-\mathrm{n}(\mathrm{C} \cap \mathrm{T})\) \(\mathrm{n}(\mathrm{C} \cup \mathrm{T})=25+20-10=35\) So, number of student who play tennis only - \(=n(C \cup T)-n(C)\) \(=35-25\) \(=10 .\)
JCECE-2019
Sets, Relation and Function
116754
In a certain town, \(25 \%\) families own a cell phone, \(15 \%\) families own a scooter and \(65 \%\) families own neither a cell phone nor a scooter. If 1500 families own both a cell phone and a scooter, then the total number of families in the town is
1 10000
2 20000
3 30000
4 None of these
Explanation:
C Given, \(25 \%\) families own a cell phone \(15 \%\) families own a scooter \(65 \%\) families own neither cell phone nor a scooter. And, 1500 families own both a cell phone and a scooter. Let, the total number of families in the town is \(\mathrm{x}\). Then, \(\frac{25 \mathrm{x}}{100}+\frac{15 \mathrm{x}}{100}-1500+\frac{65 \mathrm{x}}{100}=\mathrm{x}\) \(\frac{105 \mathrm{x}}{100}-\mathrm{x}=1500\) \(\frac{5 \mathrm{x}}{100}=1500\) So, \(\mathrm{x}=\frac{1500 \times 100}{5}\) \(\mathrm{x}=30000\)
BCECE-2014
Sets, Relation and Function
116755
If \(A\) and \(B\) are two events associated to some experiment \(E\) such that \(P(A)=0.5 \mathrm{P}(B)=0.4\), \(\mathbf{P}(\mathbf{A} \cap \mathbf{B})=\mathbf{0 . 3}\) then \(\mathbf{P}\left(\mathrm{A}^{\mathrm{C}} / \mathbf{B}^{\mathrm{C}}\right)\) is equal to
116751
A survey shows that \(63 \%\) of the Indians like tea whereas \(76 \%\) like coffee. If \(x \%\) of the Indians like both tea and coffee, then
1 \(x=39\)
2 \(x=63\)
3 \(39 \leq \mathrm{x} \leq 63\)
4 none of these
Explanation:
C Given, number of the Indians like tea - \(\mathrm{n}(\mathrm{T})=63\) Number of the Indians like coffee \(\mathrm{n}(\mathrm{C})=76\) And number of the Indians like both tea and coffee \(\mathrm{n}(\mathrm{T} \cap \mathrm{C})=\mathrm{x}\) Then, \(n(T \cup C) =n(T)+n(C)-n(T \cap C)\) \(100 =63+76-x\) \(x =139-100\) \(x =39\) Also, \(\mathrm{n}(\mathrm{T} \cap \mathrm{C}) \leq \mathrm{n}(\mathrm{T})\) \(\mathrm{x} \leq 63\)So, \(\quad 39 \leq \mathrm{x} \leq 63\) Ans: a: Given, \(\mathrm{U}=\) Set of all days Ans: c : Given, \(\mathrm{n}(\mathrm{C})=20, \mathrm{n}(\mathrm{P})=30 \text { and } \mathrm{n}(\mathrm{C} \cap \mathrm{P})=10\) Where \(\mathrm{C}\) and \(\mathrm{P}\) be the number of students in chemistry and physics class. Find, \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=\) ? Here are two conditions - Condition no I : - When both classes meet at the same time, then \(\mathrm{n}(\mathrm{C} \cap \mathrm{P})=\phi\) Then, \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=\mathrm{n}(\mathrm{C})+\mathrm{n}(\mathrm{P})\) \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=20+30=50\) Condition no II :- When both classes meet at different hours. Then \(\mathrm{n}(\mathrm{C} \cap \mathrm{P})=10\) (Given) So, \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=\mathrm{n}(\mathrm{C})+\mathrm{n}(\mathrm{P})-\mathrm{n}(\mathrm{C} \cap \mathrm{P})\) \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=20+30-10\) \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=50-10\) \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=40\)Hence, both (a) and (b) correct
BITSAT-2007
Sets, Relation and Function
116752
Which of the following is an empty set?
1 \(\left\{x \mid x\right.\) is a realnumber and \(\left.x^2-1=0\right\}\)
2 \(\left\{x \mid x\right.\) is a realnumber and \(\left.x^2+3=0\right\}\)
3 \(\left\{x \mid x\right.\) is a real number and \(\left.x^2-9=0\right\}\)
4 \(\left\{x \mid x\right.\) is a real number and \(\left.x^2=x+2\right\}\)
Explanation:
B We check the following is an empty set by options - By option a: \(x^2-1=0\) \(x^2=1\) \(x= \pm 1 \in R\) This is not empty set. By option b: \(x^2+3=0\) \(x^2=-3\) \(x=\sqrt{-3} \in C\) This is a empty set. By option c: \(x^2-9=0\) \(x^2=9\) \(x= \pm 3 \in R\) This is not a empty set. By option d: \(x^2=x+2\) \(x^2-x-2=0\) \(x^2-2 x+x-2=0\) \(x(x-2)+1(x-2)=0\) \((x-2)(x+1)=0\) \(x=-1,2\)This is not empty set.
COMEDK 2014
Sets, Relation and Function
116753
In a class of students, 25 students play cricket, 20 student play tennis and 10 students play both the games. Then the number of students who play tennis only is
1 25
2 10
3 15
4 None of these
Explanation:
B From question, Let 'C' class of students play cricket and ' \(\mathrm{T}\) ' class of student play tennis respectively. Then, given - \(\mathrm{n}(\mathrm{C})=25, \mathrm{n}(\mathrm{T})=20\) \(\mathrm{n}(\mathrm{C} \cap \mathrm{T})=10\) \(\therefore \quad \mathrm{n}(\mathrm{C} \cup \mathrm{T})=\mathrm{n}(\mathrm{C})+\mathrm{n}(\mathrm{T})-\mathrm{n}(\mathrm{C} \cap \mathrm{T})\) \(\mathrm{n}(\mathrm{C} \cup \mathrm{T})=25+20-10=35\) So, number of student who play tennis only - \(=n(C \cup T)-n(C)\) \(=35-25\) \(=10 .\)
JCECE-2019
Sets, Relation and Function
116754
In a certain town, \(25 \%\) families own a cell phone, \(15 \%\) families own a scooter and \(65 \%\) families own neither a cell phone nor a scooter. If 1500 families own both a cell phone and a scooter, then the total number of families in the town is
1 10000
2 20000
3 30000
4 None of these
Explanation:
C Given, \(25 \%\) families own a cell phone \(15 \%\) families own a scooter \(65 \%\) families own neither cell phone nor a scooter. And, 1500 families own both a cell phone and a scooter. Let, the total number of families in the town is \(\mathrm{x}\). Then, \(\frac{25 \mathrm{x}}{100}+\frac{15 \mathrm{x}}{100}-1500+\frac{65 \mathrm{x}}{100}=\mathrm{x}\) \(\frac{105 \mathrm{x}}{100}-\mathrm{x}=1500\) \(\frac{5 \mathrm{x}}{100}=1500\) So, \(\mathrm{x}=\frac{1500 \times 100}{5}\) \(\mathrm{x}=30000\)
BCECE-2014
Sets, Relation and Function
116755
If \(A\) and \(B\) are two events associated to some experiment \(E\) such that \(P(A)=0.5 \mathrm{P}(B)=0.4\), \(\mathbf{P}(\mathbf{A} \cap \mathbf{B})=\mathbf{0 . 3}\) then \(\mathbf{P}\left(\mathrm{A}^{\mathrm{C}} / \mathbf{B}^{\mathrm{C}}\right)\) is equal to
116751
A survey shows that \(63 \%\) of the Indians like tea whereas \(76 \%\) like coffee. If \(x \%\) of the Indians like both tea and coffee, then
1 \(x=39\)
2 \(x=63\)
3 \(39 \leq \mathrm{x} \leq 63\)
4 none of these
Explanation:
C Given, number of the Indians like tea - \(\mathrm{n}(\mathrm{T})=63\) Number of the Indians like coffee \(\mathrm{n}(\mathrm{C})=76\) And number of the Indians like both tea and coffee \(\mathrm{n}(\mathrm{T} \cap \mathrm{C})=\mathrm{x}\) Then, \(n(T \cup C) =n(T)+n(C)-n(T \cap C)\) \(100 =63+76-x\) \(x =139-100\) \(x =39\) Also, \(\mathrm{n}(\mathrm{T} \cap \mathrm{C}) \leq \mathrm{n}(\mathrm{T})\) \(\mathrm{x} \leq 63\)So, \(\quad 39 \leq \mathrm{x} \leq 63\) Ans: a: Given, \(\mathrm{U}=\) Set of all days Ans: c : Given, \(\mathrm{n}(\mathrm{C})=20, \mathrm{n}(\mathrm{P})=30 \text { and } \mathrm{n}(\mathrm{C} \cap \mathrm{P})=10\) Where \(\mathrm{C}\) and \(\mathrm{P}\) be the number of students in chemistry and physics class. Find, \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=\) ? Here are two conditions - Condition no I : - When both classes meet at the same time, then \(\mathrm{n}(\mathrm{C} \cap \mathrm{P})=\phi\) Then, \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=\mathrm{n}(\mathrm{C})+\mathrm{n}(\mathrm{P})\) \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=20+30=50\) Condition no II :- When both classes meet at different hours. Then \(\mathrm{n}(\mathrm{C} \cap \mathrm{P})=10\) (Given) So, \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=\mathrm{n}(\mathrm{C})+\mathrm{n}(\mathrm{P})-\mathrm{n}(\mathrm{C} \cap \mathrm{P})\) \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=20+30-10\) \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=50-10\) \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=40\)Hence, both (a) and (b) correct
BITSAT-2007
Sets, Relation and Function
116752
Which of the following is an empty set?
1 \(\left\{x \mid x\right.\) is a realnumber and \(\left.x^2-1=0\right\}\)
2 \(\left\{x \mid x\right.\) is a realnumber and \(\left.x^2+3=0\right\}\)
3 \(\left\{x \mid x\right.\) is a real number and \(\left.x^2-9=0\right\}\)
4 \(\left\{x \mid x\right.\) is a real number and \(\left.x^2=x+2\right\}\)
Explanation:
B We check the following is an empty set by options - By option a: \(x^2-1=0\) \(x^2=1\) \(x= \pm 1 \in R\) This is not empty set. By option b: \(x^2+3=0\) \(x^2=-3\) \(x=\sqrt{-3} \in C\) This is a empty set. By option c: \(x^2-9=0\) \(x^2=9\) \(x= \pm 3 \in R\) This is not a empty set. By option d: \(x^2=x+2\) \(x^2-x-2=0\) \(x^2-2 x+x-2=0\) \(x(x-2)+1(x-2)=0\) \((x-2)(x+1)=0\) \(x=-1,2\)This is not empty set.
COMEDK 2014
Sets, Relation and Function
116753
In a class of students, 25 students play cricket, 20 student play tennis and 10 students play both the games. Then the number of students who play tennis only is
1 25
2 10
3 15
4 None of these
Explanation:
B From question, Let 'C' class of students play cricket and ' \(\mathrm{T}\) ' class of student play tennis respectively. Then, given - \(\mathrm{n}(\mathrm{C})=25, \mathrm{n}(\mathrm{T})=20\) \(\mathrm{n}(\mathrm{C} \cap \mathrm{T})=10\) \(\therefore \quad \mathrm{n}(\mathrm{C} \cup \mathrm{T})=\mathrm{n}(\mathrm{C})+\mathrm{n}(\mathrm{T})-\mathrm{n}(\mathrm{C} \cap \mathrm{T})\) \(\mathrm{n}(\mathrm{C} \cup \mathrm{T})=25+20-10=35\) So, number of student who play tennis only - \(=n(C \cup T)-n(C)\) \(=35-25\) \(=10 .\)
JCECE-2019
Sets, Relation and Function
116754
In a certain town, \(25 \%\) families own a cell phone, \(15 \%\) families own a scooter and \(65 \%\) families own neither a cell phone nor a scooter. If 1500 families own both a cell phone and a scooter, then the total number of families in the town is
1 10000
2 20000
3 30000
4 None of these
Explanation:
C Given, \(25 \%\) families own a cell phone \(15 \%\) families own a scooter \(65 \%\) families own neither cell phone nor a scooter. And, 1500 families own both a cell phone and a scooter. Let, the total number of families in the town is \(\mathrm{x}\). Then, \(\frac{25 \mathrm{x}}{100}+\frac{15 \mathrm{x}}{100}-1500+\frac{65 \mathrm{x}}{100}=\mathrm{x}\) \(\frac{105 \mathrm{x}}{100}-\mathrm{x}=1500\) \(\frac{5 \mathrm{x}}{100}=1500\) So, \(\mathrm{x}=\frac{1500 \times 100}{5}\) \(\mathrm{x}=30000\)
BCECE-2014
Sets, Relation and Function
116755
If \(A\) and \(B\) are two events associated to some experiment \(E\) such that \(P(A)=0.5 \mathrm{P}(B)=0.4\), \(\mathbf{P}(\mathbf{A} \cap \mathbf{B})=\mathbf{0 . 3}\) then \(\mathbf{P}\left(\mathrm{A}^{\mathrm{C}} / \mathbf{B}^{\mathrm{C}}\right)\) is equal to
116751
A survey shows that \(63 \%\) of the Indians like tea whereas \(76 \%\) like coffee. If \(x \%\) of the Indians like both tea and coffee, then
1 \(x=39\)
2 \(x=63\)
3 \(39 \leq \mathrm{x} \leq 63\)
4 none of these
Explanation:
C Given, number of the Indians like tea - \(\mathrm{n}(\mathrm{T})=63\) Number of the Indians like coffee \(\mathrm{n}(\mathrm{C})=76\) And number of the Indians like both tea and coffee \(\mathrm{n}(\mathrm{T} \cap \mathrm{C})=\mathrm{x}\) Then, \(n(T \cup C) =n(T)+n(C)-n(T \cap C)\) \(100 =63+76-x\) \(x =139-100\) \(x =39\) Also, \(\mathrm{n}(\mathrm{T} \cap \mathrm{C}) \leq \mathrm{n}(\mathrm{T})\) \(\mathrm{x} \leq 63\)So, \(\quad 39 \leq \mathrm{x} \leq 63\) Ans: a: Given, \(\mathrm{U}=\) Set of all days Ans: c : Given, \(\mathrm{n}(\mathrm{C})=20, \mathrm{n}(\mathrm{P})=30 \text { and } \mathrm{n}(\mathrm{C} \cap \mathrm{P})=10\) Where \(\mathrm{C}\) and \(\mathrm{P}\) be the number of students in chemistry and physics class. Find, \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=\) ? Here are two conditions - Condition no I : - When both classes meet at the same time, then \(\mathrm{n}(\mathrm{C} \cap \mathrm{P})=\phi\) Then, \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=\mathrm{n}(\mathrm{C})+\mathrm{n}(\mathrm{P})\) \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=20+30=50\) Condition no II :- When both classes meet at different hours. Then \(\mathrm{n}(\mathrm{C} \cap \mathrm{P})=10\) (Given) So, \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=\mathrm{n}(\mathrm{C})+\mathrm{n}(\mathrm{P})-\mathrm{n}(\mathrm{C} \cap \mathrm{P})\) \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=20+30-10\) \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=50-10\) \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=40\)Hence, both (a) and (b) correct
BITSAT-2007
Sets, Relation and Function
116752
Which of the following is an empty set?
1 \(\left\{x \mid x\right.\) is a realnumber and \(\left.x^2-1=0\right\}\)
2 \(\left\{x \mid x\right.\) is a realnumber and \(\left.x^2+3=0\right\}\)
3 \(\left\{x \mid x\right.\) is a real number and \(\left.x^2-9=0\right\}\)
4 \(\left\{x \mid x\right.\) is a real number and \(\left.x^2=x+2\right\}\)
Explanation:
B We check the following is an empty set by options - By option a: \(x^2-1=0\) \(x^2=1\) \(x= \pm 1 \in R\) This is not empty set. By option b: \(x^2+3=0\) \(x^2=-3\) \(x=\sqrt{-3} \in C\) This is a empty set. By option c: \(x^2-9=0\) \(x^2=9\) \(x= \pm 3 \in R\) This is not a empty set. By option d: \(x^2=x+2\) \(x^2-x-2=0\) \(x^2-2 x+x-2=0\) \(x(x-2)+1(x-2)=0\) \((x-2)(x+1)=0\) \(x=-1,2\)This is not empty set.
COMEDK 2014
Sets, Relation and Function
116753
In a class of students, 25 students play cricket, 20 student play tennis and 10 students play both the games. Then the number of students who play tennis only is
1 25
2 10
3 15
4 None of these
Explanation:
B From question, Let 'C' class of students play cricket and ' \(\mathrm{T}\) ' class of student play tennis respectively. Then, given - \(\mathrm{n}(\mathrm{C})=25, \mathrm{n}(\mathrm{T})=20\) \(\mathrm{n}(\mathrm{C} \cap \mathrm{T})=10\) \(\therefore \quad \mathrm{n}(\mathrm{C} \cup \mathrm{T})=\mathrm{n}(\mathrm{C})+\mathrm{n}(\mathrm{T})-\mathrm{n}(\mathrm{C} \cap \mathrm{T})\) \(\mathrm{n}(\mathrm{C} \cup \mathrm{T})=25+20-10=35\) So, number of student who play tennis only - \(=n(C \cup T)-n(C)\) \(=35-25\) \(=10 .\)
JCECE-2019
Sets, Relation and Function
116754
In a certain town, \(25 \%\) families own a cell phone, \(15 \%\) families own a scooter and \(65 \%\) families own neither a cell phone nor a scooter. If 1500 families own both a cell phone and a scooter, then the total number of families in the town is
1 10000
2 20000
3 30000
4 None of these
Explanation:
C Given, \(25 \%\) families own a cell phone \(15 \%\) families own a scooter \(65 \%\) families own neither cell phone nor a scooter. And, 1500 families own both a cell phone and a scooter. Let, the total number of families in the town is \(\mathrm{x}\). Then, \(\frac{25 \mathrm{x}}{100}+\frac{15 \mathrm{x}}{100}-1500+\frac{65 \mathrm{x}}{100}=\mathrm{x}\) \(\frac{105 \mathrm{x}}{100}-\mathrm{x}=1500\) \(\frac{5 \mathrm{x}}{100}=1500\) So, \(\mathrm{x}=\frac{1500 \times 100}{5}\) \(\mathrm{x}=30000\)
BCECE-2014
Sets, Relation and Function
116755
If \(A\) and \(B\) are two events associated to some experiment \(E\) such that \(P(A)=0.5 \mathrm{P}(B)=0.4\), \(\mathbf{P}(\mathbf{A} \cap \mathbf{B})=\mathbf{0 . 3}\) then \(\mathbf{P}\left(\mathrm{A}^{\mathrm{C}} / \mathbf{B}^{\mathrm{C}}\right)\) is equal to
116751
A survey shows that \(63 \%\) of the Indians like tea whereas \(76 \%\) like coffee. If \(x \%\) of the Indians like both tea and coffee, then
1 \(x=39\)
2 \(x=63\)
3 \(39 \leq \mathrm{x} \leq 63\)
4 none of these
Explanation:
C Given, number of the Indians like tea - \(\mathrm{n}(\mathrm{T})=63\) Number of the Indians like coffee \(\mathrm{n}(\mathrm{C})=76\) And number of the Indians like both tea and coffee \(\mathrm{n}(\mathrm{T} \cap \mathrm{C})=\mathrm{x}\) Then, \(n(T \cup C) =n(T)+n(C)-n(T \cap C)\) \(100 =63+76-x\) \(x =139-100\) \(x =39\) Also, \(\mathrm{n}(\mathrm{T} \cap \mathrm{C}) \leq \mathrm{n}(\mathrm{T})\) \(\mathrm{x} \leq 63\)So, \(\quad 39 \leq \mathrm{x} \leq 63\) Ans: a: Given, \(\mathrm{U}=\) Set of all days Ans: c : Given, \(\mathrm{n}(\mathrm{C})=20, \mathrm{n}(\mathrm{P})=30 \text { and } \mathrm{n}(\mathrm{C} \cap \mathrm{P})=10\) Where \(\mathrm{C}\) and \(\mathrm{P}\) be the number of students in chemistry and physics class. Find, \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=\) ? Here are two conditions - Condition no I : - When both classes meet at the same time, then \(\mathrm{n}(\mathrm{C} \cap \mathrm{P})=\phi\) Then, \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=\mathrm{n}(\mathrm{C})+\mathrm{n}(\mathrm{P})\) \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=20+30=50\) Condition no II :- When both classes meet at different hours. Then \(\mathrm{n}(\mathrm{C} \cap \mathrm{P})=10\) (Given) So, \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=\mathrm{n}(\mathrm{C})+\mathrm{n}(\mathrm{P})-\mathrm{n}(\mathrm{C} \cap \mathrm{P})\) \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=20+30-10\) \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=50-10\) \(\mathrm{n}(\mathrm{C} \cup \mathrm{P})=40\)Hence, both (a) and (b) correct
BITSAT-2007
Sets, Relation and Function
116752
Which of the following is an empty set?
1 \(\left\{x \mid x\right.\) is a realnumber and \(\left.x^2-1=0\right\}\)
2 \(\left\{x \mid x\right.\) is a realnumber and \(\left.x^2+3=0\right\}\)
3 \(\left\{x \mid x\right.\) is a real number and \(\left.x^2-9=0\right\}\)
4 \(\left\{x \mid x\right.\) is a real number and \(\left.x^2=x+2\right\}\)
Explanation:
B We check the following is an empty set by options - By option a: \(x^2-1=0\) \(x^2=1\) \(x= \pm 1 \in R\) This is not empty set. By option b: \(x^2+3=0\) \(x^2=-3\) \(x=\sqrt{-3} \in C\) This is a empty set. By option c: \(x^2-9=0\) \(x^2=9\) \(x= \pm 3 \in R\) This is not a empty set. By option d: \(x^2=x+2\) \(x^2-x-2=0\) \(x^2-2 x+x-2=0\) \(x(x-2)+1(x-2)=0\) \((x-2)(x+1)=0\) \(x=-1,2\)This is not empty set.
COMEDK 2014
Sets, Relation and Function
116753
In a class of students, 25 students play cricket, 20 student play tennis and 10 students play both the games. Then the number of students who play tennis only is
1 25
2 10
3 15
4 None of these
Explanation:
B From question, Let 'C' class of students play cricket and ' \(\mathrm{T}\) ' class of student play tennis respectively. Then, given - \(\mathrm{n}(\mathrm{C})=25, \mathrm{n}(\mathrm{T})=20\) \(\mathrm{n}(\mathrm{C} \cap \mathrm{T})=10\) \(\therefore \quad \mathrm{n}(\mathrm{C} \cup \mathrm{T})=\mathrm{n}(\mathrm{C})+\mathrm{n}(\mathrm{T})-\mathrm{n}(\mathrm{C} \cap \mathrm{T})\) \(\mathrm{n}(\mathrm{C} \cup \mathrm{T})=25+20-10=35\) So, number of student who play tennis only - \(=n(C \cup T)-n(C)\) \(=35-25\) \(=10 .\)
JCECE-2019
Sets, Relation and Function
116754
In a certain town, \(25 \%\) families own a cell phone, \(15 \%\) families own a scooter and \(65 \%\) families own neither a cell phone nor a scooter. If 1500 families own both a cell phone and a scooter, then the total number of families in the town is
1 10000
2 20000
3 30000
4 None of these
Explanation:
C Given, \(25 \%\) families own a cell phone \(15 \%\) families own a scooter \(65 \%\) families own neither cell phone nor a scooter. And, 1500 families own both a cell phone and a scooter. Let, the total number of families in the town is \(\mathrm{x}\). Then, \(\frac{25 \mathrm{x}}{100}+\frac{15 \mathrm{x}}{100}-1500+\frac{65 \mathrm{x}}{100}=\mathrm{x}\) \(\frac{105 \mathrm{x}}{100}-\mathrm{x}=1500\) \(\frac{5 \mathrm{x}}{100}=1500\) So, \(\mathrm{x}=\frac{1500 \times 100}{5}\) \(\mathrm{x}=30000\)
BCECE-2014
Sets, Relation and Function
116755
If \(A\) and \(B\) are two events associated to some experiment \(E\) such that \(P(A)=0.5 \mathrm{P}(B)=0.4\), \(\mathbf{P}(\mathbf{A} \cap \mathbf{B})=\mathbf{0 . 3}\) then \(\mathbf{P}\left(\mathrm{A}^{\mathrm{C}} / \mathbf{B}^{\mathrm{C}}\right)\) is equal to
