Test: Set Theory- 1 - CAT MCQ

If we carefully observe set A, then we find that 6²ⁿ - 35n-1 is divisible by 35. So, set A contains multiples of 35. However, not all the multiples of 35 are there in set A, for different values of n.

For n = 1, the value is 0, for n = 2, the value is 1225 which is the 35th multiple of 3.

If we observe set B, it consists of all the multiples of 35 including 0.

So, we can say that every member of set A will be in B while every member of set B will not necessarily be in set A.

P = {1,2,3,4} and  Q = {2,3,5,6,}
PΔQ = {1, 4, 5, 6}
R = {1,3,7,8,9} and S = {2,4,9,10}
RΔS = {1, 2, 3, 4, 7, 8, 10}
(PΔQ)Δ(RΔS) = {2, 3, 5, 6, 7, 8, 10}
Thus, there are 7 elements in (PΔQ)Δ(RΔS) .
hence, 7 is the correct answer.

It has been given that among 200 students, 105 like pizza and 134 like burger.
The question asks us to find out the number of students who can be liking only burgers among the given values.

The least number of students who like only burger will be obtained when everyone who likes pizza likes burger too.
In this case, 105 students will like pizza and burger and 134-105 = 29 students will like only burger. Therefore, the number of students who like only burger cannot be less than 29.

The maximum number of students who like only burger will be obtained when we try to separate the 2 sets as much as possible.
There are 200 students in total. 105 of them like pizza. Therefore, the remaining 95 students can like only burger and 134-95 = 39 students can like both pizza and burger. As we can see, the number of students who like burger cannot exceed 95.

The number of students who like only burger should lie between 29 and 95 (both the values are included).
93 is the only value among the given options that satisfies this condition and hence, option D is the right answer.

Let us draw a Venn diagram using the information present in the question.

It is given that the number of students studying H equals that studying E.

Let ‘x’ be the total number of students who studied H, and H and P but mot E.We can also say that the same will be the number of students who studied E, and E and P but not H.Therefore,

x + 20 + 10 + x = 74

x = 22

Hence, the number of students studying H = 22 + 10+ 20 = 52

The number of candidates registered for 3 of the 4 courses is 45, 55, and 70.
There are 2 cases.
70 can be the total number of students registered in an elective or the mandatory course.
Case (i):
70 students have registered in an elective.
Each student will choose 2 elective.
=> Total number of electives chosen = 2* total number of students.
45+55+70 = 2*number of students.
Number of students = 170/2 = 85.
In this case, at least 45 students would have selected an elective.

Case (ii):
70 students have registered in the mandatory class.
=> 45+55+x = 2*70
=> x =40.
As we can see, the number of students registered in an elective can be 40 or 45. Therefore, option D is the right answer.

Let ‘100x’ be the number of students who joined XCRI last year.

Let ‘a’, ‘b’, ‘c’ and d be the number of students who play 1 game, 2 games, 3 games and 4 games respectively.

Therefore,

a+b+c+d = 100x … (1)

a+2b+3c+4d = 70x+75x+80x+85x

a+2b+3c+4d = 310x … (2)

By equation (2) – (1)

b+2c+3d = 210x

We have to minimize ‘d’ for that we have to maximize c. But c ≤ 100x

At c_max = 90x, d_min = 10x

Therefore, we can say that the minimum percentage of students who play all four games = 10%.

Let us note down the information given:

No person can major in all 3 subjects. 6 students major in both Mathematics and Physics, 15 major in both Physics and Biology, but no one majors in both Mathematics and Biology. There are 60 students in total.

It has been given that average marks scored by students majoring in Maths in English is 45.
Average marks scored by students majoring in Biology in English is 60.
But the combined average marks scored by students majoring in Maths and Biology in English is 50.

To maximize b, we have to minimize ‘a’ and ‘c’. The least value that ‘c’ can take is 0.
The corresponding value of a is 24.
24+6+b+15 = 60
=> b = 15.
Therefore, the maximum number of students who could have majored only in physics is 15. 

22% of the families own agricultural land, 18% own a mobile phone, 1600 families own both and 68% families own none.

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

32 = 22 + 18 – x

x=8. Hence, there will be 8% families who own both.

8% =1600 (Given in question)

Total families in a village =>100% = 20000

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