Practice Set theory - quantitative aptitude Online Quiz (set-2) For All Competitive Exams
Q-1) Let x ε {2, 3, 4} and y ε {4, 6, 9, 10}. If A be the set of all order pairs (x, y) such that x is a factor of y. Then, how many elements does the set A contain?
(a)
(b)
(c)
(d)
Given that
x ε {2, 3, 4}
and
y ε {4, 6, 9, 10}
A = x × y
But, A is set of pairs in which $1^{st}$ number is factor of second number.
A = {2, 3, 4} × {4, 6, 9, 10}
= {(2, 4); (2, 6); (2, 10); (3, 6); (3, 9); (4, 4)}
Total number of elements = 6
Q-2) Let S be a set of first ten natural numbers. What is the possible number of pairs (a, b) where a, b E S and a ≠ b such that the product ab (> 12) leaves remainder 4 when divided by 12 ?
(a)
(b)
(c)
(d)
Since numbers which and leave 4 as Remainder when devided by 12 are
16, 28, 40, 52, 64, 76, 88, 100 and 124
16 = 2 × 8, 8 × 2
28 = 4 × 7, 7 × 4
40 = 4 × 10, 10 × 4, 5 × 5, 8 × 5
All remaining numbers doesn't meet the requirement
Answer is 8.
Q-3) If A and B are any two non-empty subsets of a set E, then what is A ∪ (A ∩ B) equal to?
(a)
(b)
(c)
(d)
A and B are non-empty subsets of E.
∴ A ∪ (A ∩ B) = A ∪ (Shaded portion) = A
Q-4) In a school there are 30 teachers who teach Mathematics or Physics. Of these teachers, 20 teach Mathematics and 15 teach Physics, 5 teach both Mathematics and Physics. The number of teachers teaching only Mathematics is
(a)
(b)
(c)
(d)
Total number of teachers = 30.
Number of teachers who teaches only Math
= 20 – 5 = 15.
Q-5) Out of 105 students taking an examination English and Mathematics, 80 students pass in English, 75 students pass in Mathematics 10 students fail in both the subjects. How many students fail in only one subject?
(a)
(b)
(c)
(d)
Number of students failing in Mathematics
= 105 – 75 = 30
Number of students failing in English
= 105 – 80 = 25
∴ Number of students failing in 1 subject
= (25 + 30) – 10 = 45
Q-6) In an examination, 35% students failed in Hindi, 45% students failed in English and 20% students failed in both the subjects. What is the percentage of students passing in both the subjects?
(a)
(b)
(c)
(d)
Let the total number of students be 100%
Number of students failed in Hindi = 35%
Number of students failed in English = 45%
Number of students failed in both the subjects = 20%
Total number of students failed = (35 + 45 – 20)% = 60%
Number of students passing in both the subjects = (100 – 60)% = 40%
Q-7) In a class of 60 boys, there are 45 boys who play chess and 30 boys who plays carrom. If every boy of the class plays at least one of the two games, then how many boys play carrom only?
(a)
(b)
(c)
(d)
Total number of boys in the class
n(A ∪ B) = 60
Number of boy play chess n(A) = 45
Number of boy play carrom n(B) = 30
∴ Number of boy play both chess and Carrom
n(A ∩ B) = n(A) + n(B) – n(A ∪ B) = 45 + 30 – 60 = 15
Number of boys plays only carrom n(B) – n(A ∩ B)
= 30 – 15 = 15
42.
Q-8) Consider the following for the next 04 (four) items that follow : In an examination of Class XII, 55% students passed in Biology, 62% passed in Physics, 60% passed in Chemistry, 25% passed in Physics and Biology, 30% passed in Physics and Chemistry, 28% passed in Biology and Chemistry. Only 2% failed in all the subjects.If the number of students is 360, then how many passed in at least two subjects?
(a)
(b)
(c)
(d)
Passed in at least two subject
= ${25 - x + 30 - x + 28 - x + x}/{100}$ × 360
= ${83 - 2x}/{100}$ × 360 = 270
Q-9) In a competitive examination, 250 students have registered. Out of these, 50 students have registered for Physics, 75 students for Mathematics and 35 students for both Mathematics and Physics. What is the number of students who have registered neither for Physics nor for Mathematics?
(a)
(b)
(c)
(d)
Number of students registered for
Physics n(P) = 50
Mathematics n(M) = 75
Number of students registered for both subjects n(P ∩ M) = 35.
Number of students, registered for either physics or mathematics
n(P ∪ M) = n(P) + n(M) – n(P ∩ M).
= 50 + 75 – 35 = 90
∴ Number of students registered neither for physics nor for mathematics
n($\ov{P ∪ M}$) = 250 – 90 = 160.
Q-10) In an examination, 50% of the candidates failed in English, 40% failed in Hindi and 15% failed in both the subjects. The percentage of candidates who passed in both English and Hindi is
(a)
(b)
(c)
(d)
Percent of candidates, who failed in either Hindi or English
= (50 + 40 – 15)% = 75%
∴ Percent of candidates who passed in both subject
= 100 – 75 = 25%
Q-11) Consider the following for the next three (03) items : In a certain town of population size 1,00,000 three types of newspapers (I, II and III) are available. The percentages of the people in the town who read these papers are as follows : Newspaper Proportion of readers I 10 % II 30 % III 5 % Both I and II 8 % Both II and III 4 % Both I and III 2 % All the three (I, II and III) 1 %
What is the number of people who read at least two newspapers?
| Newspaper | Proportion of readers |
| I | 10 % |
| II | 30 % |
| III | 5 % |
| Both I and II | 8 % |
| Both II and III | 4 % |
| Both I and III | 2 % |
| All the three (I, II and III) | 1 % |
(a)
(b)
(c)
(d)
As we can see from the above venn diagram the number of people who read two or more newspapers are 1% + 1% + 3% + 7% = 12% = 12/100 × 100000 = 12000
Q-12) Consider the following for the next three (03) items : In a certain town of population size 1,00,000 three types of newspapers (I, II and III) are available. The percentages of the people in the town who read these papers are as follows : Newspaper Proportion of readers I 10 % II 30 % III 5 % Both I and II 8 % Both II and III 4 % Both I and III 2 % All the three (I, II and III) 1 %
What is the number of people who read only one newspaper?
| Newspaper | Proportion of readers |
| I | 10 % |
| II | 30 % |
| III | 5 % |
| Both I and II | 8 % |
| Both II and III | 4 % |
| Both I and III | 2 % |
| All the three (I, II and III) | 1 % |
(a)
(b)
(c)
(d)
The number of people who read only 1, only II and only II are
1% + 19% + 0% = 20% of total population = 20/100 × 100000 = 20000
Q-13) If X= {a, {b}, c}, Y = {{a}, b, c} and Z = {a, b, {c}}, then (X ∩ Y) ∩ Z equals to
(a)
(b)
(c)
(d)
(X ∩ Y) = {a,{b},c} ∩ {{a},b,c} = c
Now, (X ∩ Y) ∩ Z
= c ∩ {a, b,{c}} = Φ
Q-14) Let A = {7, 8, 9, 10, 11, 12} and B = {7, 10, 14, 15}. What is the number of elements in (A–B) and (B–A) respectively ?
(a)
(b)
(c)
(d)
A = {7, 8, 9, 10, 11, 12}; B = {7, 10, 14, 15}
(A – B) = {8, 9, 11, 12} (B – A) = {14, 15}
n (A – B) = 4; n (B – A) = 2
Q-15) Consider the following for the next three (03) items : In a certain town of population size 1,00,000 three types of newspapers (I, II and III) are available. The percentages of the people in the town who read these papers are as follows : Newspaper Proportion of readers I 10 % II 30 % III 5 % Both I and II 8 % Both II and III 4 % Both I and III 2 % All the three (I, II and III) 1 %
What is the number of people who do not read any of these three newspapers?
| Newspaper | Proportion of readers |
| I | 10 % |
| II | 30 % |
| III | 5 % |
| Both I and II | 8 % |
| Both II and III | 4 % |
| Both I and III | 2 % |
| All the three (I, II and III) | 1 % |
(a)
(b)
(c)
(d)
Number of people who do not read any of these newspaper = total population – number of people who read atleast one of these newspapers.
number of people who read atleast one of these newspapers = 1% + 1% + 3% + 1% + 7% + 19% = 32% of total population = 32000
required number of people = 100000 – 32000 = 68000
Q-16) Consider the following statements:
I. Set of points of a given line is a finite set.
II. Intelligent students in a class is a set.
III. Good books in a school library is a set.
Which of the above statements is/are not correct?
(a)
(b)
(c)
(d)
I. The set of points of a given line is not a finite set.
II. Here, we cannot decide, which students are intelligent.
III. Here, we cannot decide, which books are good a school library.
Q-17) The set of natural numbers is closed under
I. addition
II. subtraction
III. multiplication
IV. division
Which of the above is/are correct?
(a)
(b)
(c)
(d)
Set of natural numbers,
N = {1, 2, 3, 4, 5, 6, ...
(i) Addition: 2 + 3 = 5 is also an element of N.
(ii) Subtraction: 2 – 3 = –1 ∉ N
(iii) Multiplication: 2 × 3 = 6 ε N
(iv) Division: $3/2$. = 1.5 ∉ N (since, N contains only positive integers)
Therefore, the set of natural numbers is closed under addition and multiplication.
Q-18) If A = {x : x is an even natural number},
B = {x : x is a natural number and multiple of 5} and
C = {x : x is a natural number and multiple of 10},
then what is the value of A ∩ (B ∪ C)?
(a)
(b)
(c)
(d)
We know that
A ∩ (B ∪ C) = {A ∩ B) ∪ (A ∩ C)
Example:
A = Set of an even natural number
A = {2, 4, 6, 8, 10, 12, ...}
B = Set of natural number and multiples of 5.
B = {5, 10, 15, 20, 25, ...}
C = Set of natural number and multiple of 10.
C = {10, 20, 30, 40, 50, ...}
A ∩ B = {2, 4, 6, 8, 10, 12, ...} ∩ {5, 10, 15, 20, 25, ...} = {10, 20, 30, ...}
A ∩ C = {2, 4, 6, 8, 10, 12, ...} ∩ {10, 20, 30, 40, 50, ...} = {10, 20, 30, 40, ...}
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
= {10, 20, 30, ...} ∪ {10, 20, 30, 40, ...}
= {10, 20, 30, 40, ...}
Q-19) Which one of the following is a correct statement?
(a)
(b)
(c)
(d)
In the given options, the correct statement is Φ ε P(Φ).
Q-20) Let A denote the set of quadrilaterals having two diagonals equal and bisecting each other. Let B denote the set of quadrilaterals having diagonals bisecting each other at 90°. Then A ∩ B denotes
(a)
(b)
(c)
(d)
Externality is a result of an economic activity which is realised by third one. It may be of two types - negative and positive. Pollution caused by a factory is negative one and increase in land price of a plot due to construction of a road.