MCA Entrance Previous Year Question Bank

If $A$ and $B$ are two sets, then $A^{\prime }-B^{\prime }$ is equal to:

In a class, 50 student play cricket, 20 play hockey, 10 play both the games, then the no. of students playing at least one game is:

Suppose $P_1,P_2,\dots ,P_{30}$ are thirty sets each having 5 elements and $Q_1,Q_2,\dots ,Q_n$ are $n$ sets with 3 elements each. Let ${\bigcup }_{i=1}^{30}{P}_i={\bigcup }_{j=1}^n{Q}_j=S$ and each element of $S$ belongs to exactly 10 of the $P_i$s and exactly 9 of the $Q_j$s. Then $n$ is equal to:

From 50 students taking examination in Mathematics, Physics and Chemistry, 37 passed Mathematics, 24 Physics and 43 Chemistry. At most 19 passed Mathematics and Physics, at most 29 Mathematics and Chemistry, and at most 20 Physics and Chemistry. The largest possible number that could have passed all three examinations is:

$A_1,A_2,A_3$ and $A_4$ are subsets of a set $U$ containing 75 elements with the following properties: Each subset contains 28 elements; the intersection of any two of the subsets contains 12 elements; the intersection of any three of the subsets contains 5 elements; the intersection of all four subsets contains 1 element. The number of elements belonging to none of the four subsets is:

Let $\overline{P}$ and $\overline{Q}$ denote the complements of two sets $P$ and $Q$. Then, the set $(P-Q)\cup (Q-P)\cup (P\cap Q)$ is:

A professor has 24 text books on computer science and is concerned about their coverage of the topics $(P)$ compilers, $(Q)$ data structures and $(R)$ operating systems. The following data gives the number of books that contain material on these topics : $n(P)=8,n(Q)=13,n(R)=13,n(P\cap Q)=3,n(P\cap Q\cap R)=2$, where $n(x)$ is the cardinality of the set $x$, then, the number of text books that have no material on compiler is:

In a class of $100$ students, $55$ students have passed in Mathematics and $67$ students have passed in Physics. Then, the number of students who have passed in Physics only, is:

In a college of 300 students, every student reads 5 newspapers and every newspaper is read by 60 students. The number of newspaper is:

Let $X$ be the universal set for sets $A$ and $B$. If $n(A)=200$, $n(B)=300$ and $n(A\cap B)=100$, then $n(A^{\prime }\cap B^{\prime })$ is equal to 300 provided $n(X)$ is equal to: