Set Theory Questions for CAT with Answers PDF

Total number of students = n(H ∪ S ∪ F)
= n(H) + n(S) + n(F) - n(H ∪ S) - n(S ∪ F) - n(H ∪ F) + n(H ∩ S ∩ F)
= 100 + 70 + 70 - 17 - 25 - 11 + 5 = 192
Or adding the various numbers in different regions in the Venn diagram, total number of students
= 63 + 12 + 47 + 25 + 6 + 39 = 192

The Venn diagram for the situation is as shown above.
In the Venn diagram, H denotes the set of people who speak Hindi and E denotes the set of people who speak English.
So, a + x = 750 ... (i)
x + b = 400 ... (ii)
Adding equations (i) and (ii), we get a + x + b + x = 1150 ... (iii)
But a + x + b = 1000 ... (iv)
Putting this value of a + x + b from equation (iv) into equation (iii), we get
1000 + x = 1150
Or x = 150 ... (v)
Putting this value of x from equation (v) into equation (i) and solving for a, we get

Total students = Physics + Mathematics – Both 'Mathematics' and 'Physics' + Neither 'Mathematics' nor 'Physics'
Here's what we know:
Total = 100
Physics = 40
Mathematics = 70
Both (B) = Unknown
Neither (N) = Unknown
or we have
100 = 40 + 70 - Both + Neither
100 = 110 - B + N (Let's isolate 'B' since we are solving for possible values of 'B')
B = N + 10
Now, the minimum value of N is 20 and the maximum value can be 30.
Thus, the minimum value of B is 30 and the maximum value of B is 40.

n(E ∪ H ∪ T) = n(E) + n(H) + n(T) - n(E ∩ H) - n(E ∩ T) - n(H ∩ T) + n(E ∩ H ∩ T)
= 205 + 210 + 120 - 100 - 80 - 35 + 20
= 340
Number of people who cannot speak English, Hindi, or Tamil = 450 - 340 = 110

There are a total of 435 bags and 210 bags contain almonds.
From the diagram,
20y = 5x
y = x/4
Now, 435 = {Almonds} + 10x + y + x
435 = 210 + 10x + (x/4) + x
x = 20
Hence, number of bags that contain only one kind of item = 10x + x + 5x = 16x
= 320

We can calculate by using the formula,
n(a) + n(b) + n(c) - n(a ∩ b) - n(b ∩ c) - n(c ∩ a) + n(a ∩ b ∩ c)
n(a ∪ b ∪ c) = 180 + 190 + 190 - 80 - 90 - 50 + 30 = 370
Number of People did not come to know about the sale from any of these sources = (400 - 370) = 30
Thus, 30 people did not come to know about the sale from any of these sources.

n (A ∩ B) = n(A) + n(B) - n (A ∪ B)

For min n (A ∩ B), n (A ∪ B) should be maximum, which is 80.
Hence, n(A ∩ B) min = 50 + 70 - 80 = 40
Also, A ∩ B is maximum when all desktop users are also laptop users. Hence, 50.

Number of people who liked at least one drink = 12 + 15 + 20 + 10 + 20 + 5 + 8 = 90
Number of people who liked none of the drinks = 100 - 90 = 10

Let number of students who failed in all the subjects be x.
1000 = 658 + 372 + 590 - 166 - 126 - 434 + x
x = 106
Number of students who failed in Physics but not in Chemistry or Maths = 658 - 60 - 106 - 328 = 164

Total = 200
Neither having car nor bike = 80
Only bike = 60
Only car = 3 (using both car and bike)
Only bike + Only car + Neither of both + Both car and bike = Total
60 + 3 x {Both car and bike} + 80 + {Both car and bike} = 200
4 x {Both car and bike} = 60
Therefore, having both car and bike = 15

Let the total number of people who filled the questionnaire be x. 
If 120 didn't enjoy; then x - 120 enjoyed. 
Also (3x/4) = those who enjoyed and understood 
Total = x - 120 + x - 120 - (3x/4) + 120 = x 
⇒ (5x/4) - 120 = x 
⇒ (x/4) = 120 
⇒ x = 480

{Total} = {Movie X} + {Movie Y} + {Movie Z} - {Exactly 2 movies} - 2 × {All 3 movies} + {No movies}
Let 'n' denote the number of people who had seen only two of the movies.
Therefore,
100 = 55 + 45 + 50 - n - (2 × 20) + 10
Or, n = 20
Thus, number of people who had seen only two of the movies is 20.

n(A ∩ B) = n(A) + n(B) - n(A ∪ B)
= 410 + 220 - n(A ∪ B)
To minimise intersection, we need to maximise union, which could be most 500. Putting the value,
We get n(A ∩ B) = 410 + 220 - 500 = 130

If total = 1, then:
Last year
B = 1/2
C = 1/3
P = 1/6
This year
B = (1/2) x (2/3) = 1/3
C = (1/3) x (3/4) = 1/4
Total = (1/3) + (1/4) = 7/12

Students left this year:
B = (1/2) x (1/3) = (1/6)
C = (1/3) x (1/4) = (1/12)
Total = (1/6 + 1/12) = (1/4)

Students at the school this year
= Number of students last year - Number of students left this year
= 1 - (1/4)
= (3/4)
Fraction of all students who took biology and chemistry = (7/12) ÷ (3/4) = 7/9

Let the total number of people surveyed be 100.
Number of people in favour of at least one proposal = 78
From Venn diagram:
Number of people in favour of exactly two proposals = x + y + z
Number of people in favour of all three proposals = 5
Number of people in favour of only proposal I = 50 - (x + y + 5)
Number of people in favour of only proposal II = 30 - (y + z + 5)
Number of people in favour of only proposal III = 20 - (x + z + 5)
According to the question:
78 = 50 - (x + y + 5) + 30 - (y + z + 5) + 20 - (x + z + 5) + (x + y + z) + (5)
⇒ 78 = 100 - (x + y + z) - 10
⇒ (x + y + z) = 12
Number of people who favoured more than one proposal = Number of people in favour of exactly two proposals +
Number of people in favour of all three proposals
= x + y + z + 5
= 12 + 5 = 17
Required percentage = 17%

We have: 8ⁿ - 7n - 1 = (7 + 1)ⁿ - 7n - 1 = (ⁿC₂ 72 + ⁿC₃ 7³ + … ⁿC_n 7ⁿ)
= 49(ⁿC₂ + ⁿC₃ 7 + … + ⁿC_n 7^{n - 2}) for n ≥ 2. For n = 1, 8n - 7n - 1 = 0.
Thus, 8ⁿ - 7n - 1 is a multiple of 49 for n ≥ 2 and 0 for n = 1.
Hence, X consists of all positive integral multiples of 49 of the form 49 K_n, where K_n = ⁿC₂ + ⁿC₃ 7 + … + ⁿC_n 7^{n - 2} together with zero.
Also, Y consists of all positive integral multiple of 49, including zero. Therefore, X ⊂ Y.

The Venn diagram for the same is as shown.
Percentage of consumers who buy all the 3 grades 

Number of customers who buy A or B, but not C = 90 + 50 + 0 = 140

Ganashakti ∩ Anand Bazar Patrika = Ganashakti + Anand Bazar Patrika - [Ganashakti ∩ Anand Bazar Patrika]
Ganashakti ∩ Anand Bazar Patrika = 121 + 117 - 200 = 38

Ganashakti ∩ Anand Bazar Patrika = Ganashakti + Anand Bazar Patrika - [Ganashakti ∩ Anand Bazar Patrika]
Ganashakti ∩ Anand Bazar Patrika = 33 + 32 - 52 = 13
So, the number of persons in Dighoshpur who read only Ganashakti = 83