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Q1. Which of the following is a finite set?

• Set of natural numbers
• The set of numbers which are multiples of 5
• The set of lines parallel to y - axis
• Set of roots of the equation x² - 25 = 0

Solution

A set which is empty or consists of a definite number of elements is called a finite set. The roots of the equation x² - 25 = 0, are -5, 5. Hence the set of roots of the equation x² - 25 = 0 is finite.

Q2. The roaster form of the set B = {x : x is a vowel of English alphabet }

• B = {a, e, y, o, u}
• B = {a, e, i, o, u}
• B = {a, e, w, y, u}
• B = { a, i, y, o, u}

Solution

a, e, i, o and u are the vowels in English alphabet. So B = {a, e, i, o, u}

Q3. A die is thrown. The event A is that a number less than   or equal to 3 appears Event B is that a number between 3 and 5 appears Event C is that a number greater 3   appears. So what are A, B and C  called ?

• disjoint events
• complementary events
• Mutually exclusive events
• exhaustive events

Solution

Events A,B  and C are called exhaustive events if their union is the  sample space  S. In this case, S={1, 2, 3, 4, 5, 6}A={1, 2, 3}; B={4}; C={4, 5, 6} So, we see that,

Q4. A = {1, 2, 3, 4, 5} and B = {2, 3, 7} So A - B=?

• {1,2,3}
• {2,3,4}
• {2,3,7}
• {1,4,5}

Solution

A - B = those elements that are in A but not in B = {1, 4, 5}

Q5. Identify the finite set from the following:

• Set of prime numbers less than 10
• {x : 2 < x < 3, x Ε R}
• Set of positive integers > 50
• The set of lines passing through a point

Solution

The prime numbers less than 10 are 2,3,5,7. These are finite in number. Hence the set of prime numbers less than 10 is a finite set. All the other sets have infinite number of elements in it.  So option A is correct.

Q6. A set has 7 elements. The number of elements in its power set is:

• 64
• 32
• 128
• 24

Solution

If n(A) = m , then number of elements in its power set = 2^m Hence, number of elements in power set of A = 2⁷ = 128

Q7. How many elements does P(A) have, If A = Φ

• 1
• 0
• 2
• 4

Solution

Here A = Φ i.e. n(A) = 0 , therefore n [P(A)] = 2⁰ = 1

Q8. A is a set with 6 elements. So, the number of subsets is:

• 12
• 24
• 32
• 64

Solution

If n(A)= m , then the number of subsets of A is 2^m.So, in this case , the number of subsets == 64.

Q9. Out of 500 modules tested, 400 modules are from mathematics and 200 are from physics, how many modules are tested from both the subjects:

• 500
• 600
• 200
• 100

Solution

Let M denote the set of modules tested from mathematics and P denote the set of modules tested from physics. In the statement of the problem, the word ‘both’ gives us a clue of intersection. So, we have to determine n((MP). Now, n(M) = 400, n(P) = 200, n( MP) = 500. By using the result n( MP) = n(M) + n(P) – n(MP), we have n(MP) = 100

Q10. If in class XI of a certain school, 24 students got distinction in mathematics, 19 students got distinction in physics and 15 students got distinction in both the subjects,then the number of students in class XI is :

• 58
• 43
• 28
• 39

Solution

Let M denote the set of students got distinction in mathematics and P denote the set of students got distinction in physics. In the statement of the problem, we have n(M) = 24, n(P) = 19, n( MP) = 15.We have to determine n(MP). By using the result, we get n( MP) = n(M) + n(P) -  n( MP), we obtain n( MP) = 28

Q11. If A = {1, 4, 9, 16} then n (A) = _____

•   1
•   2
•   3
•    4

Solution

  n (A): No of elements in A = 4

Q12. In a group of people 50 speak Hindi, 20 speak Bengali and 10 speak both the languages. How many speak at least one language assuming that each person speaks at least one of the languages?

• 70
• 60
• 80
• 30

Solution

n (P): Number of people who speak Hindi = 50                                                                n (Q) : Number of people who speak Bengali = 20  n(PQ):  Number of people who speak both the languages = 10   We have,

Q13. Given the sets P = {2,4,6}, Q = {3,5,7} and  R = {1,3,5,7,9}, which of the following may be considered as universal set for all the three sets P, Q and R?

• {1,3,5,7,9}
• {1,2,3,4,5,6,7,9}
• Φ
• {2,3,4,5,6,7}

Solution

The universal set is a super set of all given sets under consideration. Hence, {1,2,3,4,5,6,7,9} is universal set for P, Q and R as it contains all the elements in P, Q and R.

Q14. The set P = {x:x is a two digit number and sum of whose digits is 6},written in roster form is:

• P ={60, 61, 62 ----- 69}
• P ={15, 51}
• P = {24, 42}
• P = {15, 24, 33, 42, 51, 60}

Solution

In roster form, all the elements of a set are listed, the elements are being separated by commas and are enclosed within braces {  }. All two digit numbers, sum of whose digits is 6, are 15, 24, 33, 42, 51 and 60. So option D is correct.

Q15. The Roster form of the set C = {x : x is a letter of the word TOPPER}

• C = {T, O, P, E, R}
• C = {T, P, E, R}
• C = {T, O, P, R}
• C = {T, R, P}

Solution

The letters of the word are T, O, P, E and R.  The repetition of the letters does not affect the set.  So, C = {T, O, P, E, R}

Q16. In a complex there are 25 families, 12 take coffee, 8 take coffee but not the tea. Find the number of families who take tea but not coffee assuming that each of them take atleast one of the drinks

• 13
• 4
• 8
• 5

Solution

n (P): Number of families who take coffee = 12 n(P-Q):Number of families who take coffee but not tea=8 n (PQ):  Number of families who take both tea and coffee =n(P)-n(P-Q)= 12 - 8 = 4                      n (Q):  Number of families who take tea n (PUQ) : Number of families who take at least one of the drink = 25                   n(PUQ) = n (P) + n(Q) - n (P ∩Q)                         25 = 12 + n (Q) - 4                        n (Q) = 25 - 8 = 17 No. of families who take tea but not coffee=n(Q)-n(PQ)                         = No. of families who take coffee - No of families who take both tea and coffee                         = 17 - 4 = 13 

Q17. U= set of all teachers in a school B= Set of all Mathematics teachers in a school So B’=?

• Set of all science teachers
• set of all non science teachers
• set of all language teachers
• set of all non mathematics teachers

Solution

set of all non mathematics teachers B’={x |  x∈  U and x  ∉B}= set of all non mathematics teachers.    

Q18. If n(A) + n(B) + n(C) = n(AUBUC) then what are the sets A, B, C are called?

• equal
• equivalent 
•  disjoint
• joint

Solution

If n(A) + n(B)+n(C) = n(AUBUC) then the sets A, B, C are called disjoint sets.

Q19. If A = {2, 4, 6, 8} and U = {1, 2, 3, 4, 5, 6, 7, 8, 9,}, then A'=

• {1,3,5,7}
• {1,3,5,7,9}
• {1,3,5,6,9}
• {1,3,5,8,9}

Solution

Q20. If U, the universal set, is the set of all students of a school which are in classes from I to X and B = set of all students of class X. What is B'?

• set of all students of class IX
• set of all students from classes I to IX
• set of all students of class VIII
• set of all students of class I to VIII

Solution

U = set of all students from classes I to IX The complement of a subset B of universal set U is the set of all elements of U which are not the elements of B . Hence B' = set of all students from classes I to IX

Q21. From the sets given below, select equal sets : A = { 2, 4, 8, 12}, B = { 1, 2, 3, 4}, C = { 4, 8, 12, 14}, D = { 3, 1, 4, 2}

• A and C
• A and B
• B and D
• B and C

Solution

A and C are unequal as they do not contain same elements.  B and D are equal as they contain same elements, namely, 1, 2, 3 and 4.

Q22. For the set of all natural numbers the universal set can be

• Set of all even numbers
• Set of all prime numbers
• Set of all odd numbers
• Set of all integers

Solution

The universal set is a super set of all given sets under consideration. Here, the set of all natural numbers is a subset of set of all integers.

Q23. The set builder form of the set B = {2, 3, 5, 7, 11} is

• B = {x : x is an odd natural number less than 12}
• B = {x : x is a prime number less than 12}
• B = {x : x is a cube of a natural number less than 12}

Solution

2 is the smallest prime number. All the other numbers are also prime less than 12. Hence, option B is correct.

Q24. If n (P) = 5, n(Q) = 12 and n(PUQ) = 14 then n(P∩Q) =

• 3
• 4   
• 5
• 7

Solution

We have, n(PUQ) = n (P) + n(Q) - n (PQ) n (PQ) = n (P) + n(Q) - n (PQ) = 5 + 12 -14 = 3

Q25. Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}, which of the following may be considered as universal set  for all the three sets A, B and C?

• {0,1,2,3,4,5,6,7,9,10}
• {1,2,3,4,5,6,7,8}
• Φ
• {0, 1, 2, 3, 4, 5, 6, 8}

Solution

Universal set contains every element that is relevant to the  sets under consideration. So the universal set here is {0,1,2,3,4,5,6,8}  

Q26. If A={2,4,6,8,10}B={3,4,5,6,7}So A-B=?

• {2,6,8,10} 
• {2,4,5,6,7} 
• {2,8,10} 
• {2,3,4,5,6,7,8,10} 

Solution

{2,8,10} The difference of two sets A and B is the set of elements which belongs to A but  not to B. i.e  A-B={X: X A And X B} Hence the answer.

Q27. The Roster form of the set A = {x : x is a natural number and a perfect square less than 50} is

• A = {4, 9, 16, 25}
• A = {1, 4, 9, 16, 25, 36}
• A = { 1, 4, 9, 16, 25, 36, 49}

Solution

Square of first natural number 1 is 1. Similarly, 2² = 4, 3² = 9, 4² = 16, 5² = 25, 6² = 36, 7² = 49 8² = 64, which is more than 50. So, A = {1, 4, 9, 16, 25, 36, 49} Option C is correct.

Q28. In a certain class 15 students play basket-ball, 13 play chess and 5 play both the games. How many students are there in the class assuming that each students play at least one of the game?

• 23
• 33
• 43
• 48

Solution

 n (P) : Number of students who play basket ball = 15  n (Q) : Number of students who play chess = 13   n (PQ):  Number of students who play both the games = 5                    n(PUQ) = n (P) + n(Q)  - n (PQ)                                 = 15 + 13 -5 = 23

Q29. Consider the set A of all divisors of 30. How many subsets of A contains even divisors only?

• 4
• 16
• 2
• 2⁸

Solution

{1, 2, 3, 5, 6, 10, 15, 30} The set of even divisors is B = {2, 6, 10, 30} n(B) = 4 Hence the number of subsets of B is 2⁴=16.

Q30. Choose the incorrect statement:

• Φ is not a subset of any set.
• Every set is a subset of itself.
• If a set has only one element, we call it a singleton set.
• Set of all even prime numbers is a subset of set of all natural numbers.

Solution

Since the empty set Φ has no elements, we say that Φ is a subset of every set. So the statement in option A is incorrect.

Q31. The set builder form of the set A = {1, 8, 27, 64} is

• A = {x : x is cube of a natural number less than 5}
• A = {x : x is square of a natural number less than 8}
• A = {x : x is sum of any two consecutive odd natural numbers less than 10}
• A = {x : x is a multiple of 4 less than 64}

Solution

1 = 1³ 8 = 2³ 27 = 3³ 64 = 4³ So set A has cubes of all natural numbers less than 5. Option A is correct.

Q32. If A = {a, b, c}, then the number of proper subsets of A is:

• 8
• 3
• 6
• 10

Solution

There are 2ⁿ-2 proper subsets for every set with n elements. So power set of A = 2³-2 = 6

Q33. A = set of all triangles B = set of all right trianglesA - B=?

• set of all equilateral triangles
• set of all isosceles triangles
• set of all triangles which do not have one right angle
• set of all triangles which have all three angles different

Solution

set of all triangles which do not have one right angle set of all triangles which are not right triangles    

Q34. In a class of 50 students, 35 students play cricket and 30 students play football; identify the students playing both the games.

• 65
• 15
• 20
• 115

Solution

Let C be the set of students who play cricket, and F be the set of students who play Football. Here we have to determine n(C F). So, we have n( CF) = n(C) + n(F) – n(C F). By using the above equation, we get n(C F) = 15 

Q35. If P = {1, 2, 3, 4}, then the number of subsets in its power set will be

• 16
• 4
• 8
• 6

Solution

If A is a set with n(A) = m, then n [P(A)] = 2^m .Here P has 4 elements,so the number of elements in its power set will be 2⁴ = 16

Q36. The set builder form of the set C = {2, 4, 6, 8, 10} is

• C = {x :x is an even natural number less than 12}
• C = {x : x is an odd natural number less than 12}
• C = {x : x is a prime number less than 12}
• C = {x : x is an even natural number less than 10}

Solution

All the numbers are even natural numbers less than 12.  So option A is correct.

Q37. A group of 120 students, 90 take mathematics and 72 take economics. If 10 students take neither of the two, how many students take both:

• 30
• 62
• 20
• 52

Solution

Let X denote the set of all students. Let M denote the set of students from mathematics and E denote the set of students from physics and 10 students take neither of the two subjects. We have, n(M) = 90,  n(M∩E)'= 10 n(ME) = 120 - n(M∩E)' =120 - 10 = 110 By using the result n(ME) = n(M) + n(E) - n(ME), we have n(ME) = 90 + 72 - 110 = 52

Q38. Which of the following is a set ?

• The collection of all the months of a year
• The collection of all rich persons of Delhi
• The collection of all interesting novels
• A collection of 10 best writers of India

Solution

A set is a well defined collection of objects. Since the months of a year are defined, so the collection given in option A is a set. The sets in options B, C and D have adjectives 'rich', 'interesting' and 'best writers' which are not well-defined. So thoose collections do not form a set.

Q39. In a class of 40 students 14 take physics and 29 take chemistry. If 5 students take both, how many students take neither of the subjects?

• 5
• 38
• 14
• 2

Solution

n (P): Number of students who study physics = 14   n (PQ):  Number of students who study both physics and chemistry = 5                      n (Q):  Number of students who study chemistry = 29  n (PUQ): No of students who study at least one of the subject   n(PUQ) = n (P) + n(Q) - n (P∩Q)                               = 14 + 29 -5 = 38  No of students who study neither of the subjects                                      = Total number of students - No of students who study at least one subject                                    = 40 -38 = 2

Q40. If A = {3, 6, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20}, then A - B is

• {3, 15, 18, 21}
• {3, 6, 12, 18, 21}
• {3, 6, 15, 18, 21}
• Φ

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