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Phrases Sets And Relations PYQ



The Set of Intelligent Students in a class is





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Solution

Since, intelligency is not defined for students in a class i.e., Not a well defined collection.


If A is a subset of B and B is a subset of C, then cardinality of A ∪ B ∪ C is equal to





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Let R be reflexive relation on the finite set a having 10 elements and if m is the number of ordered pair in R, then





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Number of real solutions of the equation  is





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Inverse of the function f(x)=10x10x10x+10x is 





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Solution

Let f(x)=y, then 

⇒ 10x10x10x+10x=y

⇒ 102x1102x+1=y

⇒ 102x=1+y1y   By Componendo Dividendo Rule 

⇒ x=12log10(1+y1y)

⇒ f1(y)=12log10(1+y1y)

⇒ f1(x)=12log10(1+x1x)



Suppose A1 , A2 , A3 , …..A30 are thirty sets each having 5 elements with no common elements across the sets and B1 , B2 , B3 , ..... , Bn are





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Find the cardinality of the set C which is defined as C={x|sin4x=12forx(9π,3π)}.





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Solution

We are given:

sin(4x)=12,x(9π, 3π)

Step 1: General solutions for \sin(θ) = \frac{1}{2}

θ = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad θ = \frac{5\pi}{6} + 2n\pi

Let θ = 4x , so we get:

  • x = \frac{\pi}{24} + \frac{n\pi}{2}
  • x = \frac{5\pi}{24} + \frac{n\pi}{2}

✅ Step 2: Count how many such x fall in the interval (-9\pi, 3\pi)

By checking all possible n values, we find:

  • For x = \frac{\pi}{24} + \frac{n\pi}{2} : 24 valid values
  • For x = \frac{5\pi}{24} + \frac{n\pi}{2} : 24 valid values

? Total distinct values = 24 + 24 = 48

✅ Final Answer: \boxed{48}



A survey is done among a population of 200 people who like either tea or coffee. It is found that 60% of the pop lation like tea and 72% of the population like coffee. Let x be the number of people who like both tea & coffee. Let m{\leq x\leq n}, then choose the correct option.





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Let Z be the set of all integers, and consider the sets X=\{(x,y)\colon{x}^2+2{y}^2=3,\, x,y\in Z\} and Y=\{(x,y)\colon x{\gt}y,\, x,y\in Z\}. Then the number of elements in X\cap Y is:





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Solution

Given: x^2 + 2y^2 = 3 \text{ and } x > y \text{ with } x, y \in \mathbb{Z}

Solutions to the equation are: \{(1,1), (1,-1), (-1,1), (-1,-1)\}

Among them, only (1, -1) satisfies x > y .

Answer: \boxed{1}



Out of a group of 50 students taking examinations in Mathematics, Physics, and Chemistry, 37 students passed Mathematics, 24 passed Physics, and 43 passed Chemistry. Additionally, no more than 19 students passed both Mathematics and Physics, no more than 29 passed both Mathematics and Chemistry, and no more than 20 passed both Physics and Chemistry. What is the maximum number of students who could have passed all three examinations?





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Solution

? Maximum Students Passing All Three Exams

Given:

  • Total students = 50
  • |M| = 37 , |P| = 24 , |C| = 43
  • |M \cap P| \leq 19 , |M \cap C| \leq 29 , |P \cap C| \leq 20

We use the inclusion-exclusion principle:

|M \cup P \cup C| = |M| + |P| + |C| - |M \cap P| - |M \cap C| - |P \cap C| + |M \cap P \cap C|

Let x = |M \cap P \cap C| . Then:

50 \geq 37 + 24 + 43 - 19 - 29 - 20 + x \Rightarrow 50 \geq 36 + x \Rightarrow x \leq 14

✅ Final Answer: \boxed{14}



There are two sets A and B with |A| = m and |B| = n. If |P(A)| − |P(B)| = 112 then choose the wrong option (where |A| denotes the cardinality of A, and P(A) denotes the power set of A)





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In a survey where 100 students reported which subject they like, 32 students in total liked Mathematics, 38 students liked Business and 30 students liked Literature. Moreover, 7 students liked both Mathematics and Literature, 10 students liked both Mathematics and Business. 8 students like both Business and Literature, 5 students liked all three subjects. Then the number of people who liked exactly one subject is






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If A={1,2,3,4} and B={3,4,5}, then the number of elements in (A∪B)×(A∩B)×(AΔB)





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Suppose A_1,A_2,\ldots,A_{30} are 30 sets each with five elements and B_1,B_2,B_3,\ldots,B_n are n sets (each with three elements) such that  \bigcup ^{30}_{i=1}{{A}}_i={{\bigcup }}^n_{j=1}{{B}}_i=S\, and each element of S belongs to exactly ten of the A_i's and exactly 9 of the B^{\prime}_j's. Then n=





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The number of elements in the power set P(S) of the set S = [2, (1, 4)] is





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If X and Y are two sets, then X∩Y ' ∩ (X∪Y) ' is 





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In a class of 50 students, it was found that 30 students read "Hitava", 35 students read "Hindustan" and 10 read neither. How many students read both: "Hitavad" and "Hindustan" newspapers?





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Solution

P(x)=50,
P(A ∩ B)' = 10 
So P(A ∪ B) = 50 - 10 = 40. 
So P(A ∩ B) = P(A) + P(B) - P(A ∪ B) 
= 30 + 35 - 40 = 25

Solution Contribution by Priyanka Soni


If A = {4x - 3x - 1 : x ∈ N} and B = {9(x - 1) : x ∈ N}, where N is the set of natural numbers, then





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Solution

A = {0,9,54...}
B = {0,9,18,27...}
So, A ⊂ B


If A = { x, y, z }, then the number of subsets in powerset of A is





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If the sets A and B are defined as A = {(x, y) | y = 1 / x, 0 ≠ x ∈ R}, B = {(x, y)|y = -x ∈ R} then





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





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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 R) = 3, n(P \cap R) = 3, n(Q \cap R) = 3, n(Q \cap R) = 6, 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 compilers is





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Let A and B be sets. A\cap X=B\cap X=\phi and A\cup X=B\cup X for some set X, relation between A & B





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