If the ordered pairs (x, -1)and (5, y) belong to set {(a, b) : b = 2a – 3}, find the value of x and y.

If Ordered Pairs (x, −1) and (5, y) Belong to the Set {(a, b) : b = 2a − 3}, Find x and y If the Ordered Pairs \( (x,-1) \) and \( (5,y) \) Belong to the Set \( \{(a,b): b=2a-3\} \), Find \(x\) and \(y\) Question If the ordered pairs \[ (x,-1) \] […]

If the ordered pairs (x, -1)and (5, y) belong to set {(a, b) : b = 2a – 3}, find the value of x and y. Read More »

Let S = the set of points inside the square, T = the set of points inside the triangle and C = the set of points inside the circle. If the triangle and circle intersect each other and are contained in a square. Then,(a) S ∩ T ∩ C = ϕ(b) S ∪ T ∪ C = C(c) S ∪ T ∪ C = S(d) S ∪ T = S ∩ C

Let \(S\) = the set of points inside the square, \(T\) = the set of points inside the triangle and \(C\) = the set of points inside the circle. If the triangle and circle intersect each other and are contained in a square. Then, (a) \(S\cap T\cap C=\phi\) (b) \(S\cup T\cup C=C\) (c) \(S\cup T\cup

Let S = the set of points inside the square, T = the set of points inside the triangle and C = the set of points inside the circle. If the triangle and circle intersect each other and are contained in a square. Then,(a) S ∩ T ∩ C = ϕ(b) S ∪ T ∪ C = C(c) S ∪ T ∪ C = S(d) S ∪ T = S ∩ C Read More »

If (a/3 + 1, b – 2/3 ) = (5/3, 1/3) find the value of a and b.

If (a/3 + 1, b − 2/3) = (5/3, 1/3), Find the Values of a and b If \( \left(\frac{a}{3}+1,\ b-\frac{2}{3}\right)=\left(\frac{5}{3},\ \frac{1}{3}\right) \), Find the Values of \(a\) and \(b\) Question If \[ \left(\frac{a}{3}+1,\ b-\frac{2}{3}\right)=\left(\frac{5}{3},\ \frac{1}{3}\right) \] find the values of \(a\) and \(b\). Solution Since two ordered pairs are equal, their corresponding components are

If (a/3 + 1, b – 2/3 ) = (5/3, 1/3) find the value of a and b. Read More »

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(a) 0(b) 25(c) 35(d) 45

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 (a) 0 (b) 25 (c) 35 (d) 45 Solution Let \[ n(C)=25 \] \[ n(T)=20 \] \[ n(C\cap T)=10 \] Using the formula, \[

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(a) 0(b) 25(c) 35(d) 45 Read More »

In a town of 840 persons, 450 persons read Hindi, 300 read English and 200 both. Then the number of persons who read neither is(a) 210(b) 290(c) 180(d) 260

In a town of 840 persons, 450 persons read Hindi, 300 read English and 200 both. Then the number of persons who read neither is (a) 210 (b) 290 (c) 180 (d) 260 Solution Let \[ n(H)=450 \] \[ n(E)=300 \] \[ n(H\cap E)=200 \] Using the formula, \[ n(H\cup E)=n(H)+n(E)-n(H\cap E) \] \[ =450+300-200

In a town of 840 persons, 450 persons read Hindi, 300 read English and 200 both. Then the number of persons who read neither is(a) 210(b) 290(c) 180(d) 260 Read More »

Let S = {x : x is a positive multiple of 3 less than 100}, P = {x : x is a prime less than 20}. Then, n(S) + n(P) is(a) 34(b) 31(c) 33(d) 30

Let \(S=\{x:x \text{ is a positive multiple of }3 \text{ less than }100\}\), \[ P=\{x:x \text{ is a prime less than }20\} \] Then, \(n(S)+n(P)\) is (a) 34 (b) 31 (c) 33 (d) 30 Solution Positive multiples of \(3\) less than \(100\): \[ 3,6,9,\ldots,99 \] Number of elements in \(S\): \[ \frac{99}{3}=33 \] Therefore, \[

Let S = {x : x is a positive multiple of 3 less than 100}, P = {x : x is a prime less than 20}. Then, n(S) + n(P) is(a) 34(b) 31(c) 33(d) 30 Read More »

The set (A ∪ B ∪ C) ∩ (A ∩ B′ ∩ C′)′ ∪ C′ is equal to(a) B ∩ C′(b) A ∩ C(c) B ∪ C′(d) A ∩ C′

The set \((A \cup B \cup C) \cap (A \cap B′ \cap C′)′ \cup C′\) is equal to (a) \(B \cap C’\) (b) \(A \cap C\) (c) \(B \cup C’\) (d) \(A \cap C’\) Solution \[ (A\cup B\cup C)\cap(A\cap B’\cap C’)’\cup C’ \] Using De Morgan’s law, \[ (A\cap B’\cap C’)’=A’\cup B\cup C \] Therefore,

The set (A ∪ B ∪ C) ∩ (A ∩ B′ ∩ C′)′ ∪ C′ is equal to(a) B ∩ C′(b) A ∩ C(c) B ∪ C′(d) A ∩ C′ Read More »

Two finite sets have m and n elements respectively. The total number of subsets of first set is 56 more than the total number of subsets of the second set. The value of m and n respectively are:(a) 7, 6(b) 5, 1(c) 6, 3(d) 8, 7

Two finite sets have \(m\) and \(n\) elements respectively. The total number of subsets of first set is 56 more than the total number of subsets of the second set. The value of \(m\) and \(n\) respectively are: (a) \(7,6\) (b) \(5,1\) (c) \(6,3\) (d) \(8,7\) Solution Number of subsets of a set having \(m\)

Two finite sets have m and n elements respectively. The total number of subsets of first set is 56 more than the total number of subsets of the second set. The value of m and n respectively are:(a) 7, 6(b) 5, 1(c) 6, 3(d) 8, 7 Read More »