Set Theory Questions for CAT with Answers PDF

Total number of students studying one of the three subjects = 200 - 20
i.e. n(M ∪ P ∪ C) = 180
Now, n(M ∪ P ∪ C) = n(M) + n(P) + n(C) - n(M ∩ P) - n(P ∩ C) - n(M ∩ C) + n(M ∩ P ∩ C)
∴ 180 = 120 + 90 + 70 - 40 - 30 - 50 + n(M ∩ P ∩ C)
∴ n(M ∩ P ∩ C) = 20

Let BI, BW and BT represent the readers of the three newspapers Business India, Business World and Business Today, respectively.
Now, 5 read all three, so let us put 5 at the centre.
Since 10 read both BI and BW, we shall put 5 more in BI ∩ BW.
BW ∩ BT = 12 = 5 + 7. So, put 7 more in BT ∩ BW.
BI ∩ BT = 9 = 5 + 4, so put 4 more in BT ∩ BI.
Since we know that 25 read BI, only 25 - (5 + 5 + 4) = 11 read BI alone.
Similarly, only 13 read BW alone and only 8 read BT alone.
The required number = 11 + 13 + 8 = 32

Let C, F and V denote the sets of number of students who play cricket, football and volleyball, respectively.
n(C) = 20, n(F) = 25 and n(V) = 18
n(C ∩ F) = 15, n(F ∩ V) = 12, n(C ∩ V) = 10
Let 'x' be the number of students who play all the 3 games.
Number of students who play both cricket and football, but not volleyball = (15 - x)
Similarly, number of students who play both football and volleyball, but not cricket = (12 - x)
Number of students who play both cricket and volleyball, but not football = (10 - x)
Now, we can find the number of students who play cricket only, football only and volleyball only as follows:
n(C) only = 20 - (15 - x + x + 10 - x) = x - 5
n(V) only = 18 - (10 - x + x + 12 - x) = x - 4
And n(F) only = 25 - (15 - x + x + 12 - x) = x - 2
33 = (x - 5) + (15 - x) + x + (10 - x) + (12 - x) + (x - 4) + (x - 2)
33 = x + 26
x = 7
Number of students who play only cricket = 7 - 5 = 2

Let C, F and V denote the sets of number of students who play cricket, football and volleyball, respectively.
n(C) = 20, n(F) = 25 and n(V) = 18
n(C ∩ F) = 15, n(F ∩ V) = 12 and n(C ∩ V) = 10
Let 'x' be the number of students who play all the 3 games.
The number of students who play cricket and football, but not volleyball = (15 – x)
Similarly, the number of students who play football and volleyball, but not cricket = (12 – x)
The number of students who play cricket and volleyball, but not football = (10 – x)
Now, we can find the number of students who play cricket only, football only and volleyball only.
n(C) only = 20 – (15 – x + x + 10 – x) = x – 5
n(V) only = 18 – (10 – x + x + 12 – x) = x – 4
n(F) only = 25 – (15 – x + x + 12 – x) = x – 2
33 = (x – 5) + 15 – x + x + 10 – x + 12 – x + x – 4 + x – 2
33 = x + 26
x = 7
Number of students who play any 2 games = Total - [Students who play all 3 games + Students who play only 1 game]
= 33 - [7 + 2 + 5 + 3] = 33 - 17 = 16

B only = 50
Both A and C, but not B = 25
Required ratio = 50 : 25 = 2 : 1

Let A, B and C denote the sets of families buying newspapers A, B and C, respectively.
n(A) = 4000, n(B) = 2000 and n(C) = 1000
Also, n(A ∩ B) = 500, n(B ∩ C) = 300 and n(C ∩ A) = 400
n(A ∩ B ∩ C) = 200
Number of families who buy A only = Number of families who buy A - (Number of families who buy A and B, A and C and (A and B and C))
= 4000 - (300 + 200 + 200) = 3300
Alternative method:
We have, N = 10000, n(A) = 40% of 10000 = 4000, n(B) = 2000, n(C) = 1000, n(A ∩ B) = 500, n(B ∩ C) = 300, n(C ∩ A) = 400, n(A ∩ B ∩ C) = 200
We have to find

= n(A) - n((A ∩ B) ∪ (A ∩ C))
= n(A) - {n(A ∩ B) + n(A ∩ C) - n(A ∩ B ∩ C)}
= 4000 - (500 + 400 - 200) = 3300

Percentage of people who watch Chitrahaar

Percentage of people who watch Hindi films 

Required difference = 27% - 16% = 11%

Let C, F and V denote the sets of number of students who play cricket, football and volleyball, respectively.
n(C) = 20, n(F) = 25 and n(V) = 18
n(C ∩ F) = 15, n(F ∩ V) = 12 and n(C ∩ V) = 10
Let 'x' be the number of students who play all the 3 games.
The number of students who play cricket and football, but not volleyball = (15 – x)
Similarly, the number of students who play football and volleyball, but not cricket = (12 – x)
The number of students who play cricket and volleyball, but not football = (10 – x)
Now, we can find the number of students who play cricket only, football only and volleyball only.
n(C) only = 20 – (15 – x + x + 10 – x) = x – 5
n(V) only = 18 – (10 – x + x + 12 – x) = x – 4
n(F) only = 25 – (15 – x + x + 12 – x) = x – 2
33 = (x – 5) + 15 – x + x + 10 – x + 12 – x + x – 4 + x – 2
33 = x + 26
x = 7
The number of students who play all the 3 games = 7

Let C, F and V denote the sets of number of students who play cricket, football and volleyball, respectively.
n(C) = 20, n(F) = 25 and n(V) = 18
n(C ∩ F) = 15, n(F ∩ V) = 12 and n(C ∩ V) = 10
Let 'x' be the number of students who play all the 3 games.
The number of students who play cricket and football, but not volleyball = (15 – x)
Similarly, the number of students who play football and volleyball, but not cricket = (12 – x)
The number of students who play cricket and volleyball, but not football = (10 – x)
Now, we can find the number of students who play cricket only, football only and volleyball only.
n(C) only = 20 – (15 – x + x + 10 – x) = x – 5
n(V) only = 18 – (10 – x + x + 12 – x) = x – 4
n(F) only = 25 – (15 – x + x + 12 – x) = x – 2
33 = (x – 5) + 15 – x + x + 10 – x + 12 – x + x – 4 + x – 2
33 = x + 26
x = 7
Number of students who play only one game = Number who play (C only + V only + F only) = 2 + 3 + 5 = 10

Draw the rugs in the above manner, where a + b + c represents the total area of the floor covered by exactly two rugs and k represents the area of the floor covered by all three rugs. We are told that a + b + c = 24 m².
Since the total area of the floor covered when the rugs do not overlap is 200 m² and the total area covered when they do overlap is 140 m², 60 m² of rug is either double or triple layered.
Thus, a + b + c + 2k = 60 m²
2k = 36 or k = 18
Thus, the area of the floor covered by all three layers of rug is 18 m².