Sets and Relation questions
172 published questions with worked solutions.
- Let A and B be two sets, then (A B)' (A' B) is equal to
- For any two sets A and B, if A X=B X= and A X=B X for some set X, then (AMU 2009)
- The shaded region in the figure represents
- A class has 175 students. The following data shows the number of students opting one or more subjects: Mathematics 100; Physics...
- Let S= 1,2,3,4 . The total number of unordered pairs of disjoint subsets of S is equal to
- Suppose A_1,A_2, ,A_ 30 are thirty sets, each having 5 elements, and B_1,B_2, ,B_n are n sets, each with 3 elements. Let _ i=1 ^...
- Let Z denote the set of all integers and A= (a,b):a^2+3b^2=28, a,b Z and B= (a,b):a>b, a,b Z . Then, the number of elements in A...
- If A= 1,2,3 and B= a,b , then A B is (DCE 2004)
- The finite sets A and B have m and n elements respectively. If the total number of subsets of A is 112 more than the total number...
- In a class of 30 pupils, 12 take needlework, 16 take physics and 18 take history. If all the 30 pupils take at least one subject...
- If A= 1,2,3,4,5 , B= 2,4,6 , C= 3,4,6 , then (A B) C is (OJEE 2004)
- If two sets A and B are having 99 elements in common, then the number of elements common to each of the sets A B and B A is
- For any two sets A and B, A-(A-B) equals
- If n(A)=4, n(B)=3 and n(A B C)=24, then n(C) is equal to
- Let Y= 1,2,3,4,5 , A= 1,2 , B= 3,4,5 and denote the null set. If A B denotes the Cartesian product of the sets A and B, then (Y...
- If A= 1,2,3 , B= 3,4 and C= 4,5,6 , then A (B C) is (OJEE 2008)
- If sets A and B are defined as A= (x,y):y= 1 x , 0 x R and B= (x,y):y=-x, x R , then (Guj CET 2007)
- 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...
- Let the universal set be U= x:x^5-6x^4+11x^3-6x^2=0 , A= x:x^2-5x+6=0 and B= x:x^2-3x+2=0 . Then, (A B)' is equal to (BITSAT 2006)
- An investigator interviewed 100 students to determine their preference for three drinks: milk, coffee and tea. The investigator...
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