Dell Set Theory Questions

Number of candidates who had none of the three
= Total number of candidates - number of candidates
who had at least one of three devices.

Total number of candidates = 200.

Number of candidates who had at least one
of the three = A U B U C, where A is the set
of those who have a two wheeler, B the set of
those who have a credit card and C the set of
those who have a mobile phone.

We know that AUBUC = A + B + C -
{A n B + B n C + C n A} + A n B n C
Therefore, AUBUC = 100 + 70 + 140 - {40 + 30 + 60} + 10
Or AUBUC = 190.

As 190 candidates who attended the interview had
at least one of the three gadgets, 200 - 190
= 10 candidates had none of three.

We need to find out the number of students who
took at least one of the three subjects and subtract
that number from the overall 120 to get the number
of students who did not opt for any of the three subjects.

Number of students who took at least one of the three
subjects can be found by finding out A U B U C, where
A is the set of those who took Physics, B the set of those
who took Chemistry and C the set of those who opted for Math.

Now, AUBUC =
A + B + C - (A n B + B n C + C n A) + (A n B n C)
A is the set of those who opted for Physics = 120/2
= 60 students
B is the set of those who opted for Chemistry = 120/5
= 24
C is the set of those who opted for Math = 120/7
= 17.

The 10th, 20th, 30th…… numbered students would
have opted for both Physics and Chemistry.
Therefore, A n B = 120/10 = 12

The 14th, 28th, 42nd…. Numbered students would
have opted for Physics and Math.
Therefore, C n A = 120/14 = 8

The 35th, 70th …. Numbered students would have
opted for Chemistry and Math.
Therefore, A n B = 120/35 = 3

And the 70th numbered student would have opted
for all three subjects.

Therefore, AUBUC
= 60 + 24 + 17 - (12 + 8 + 3) + 1 = 79.

Number of students who opted for none
of the three subjects = 120 - 79 = 41

Let A = Set of people who like cold drinks B = Set of people
who like hot drinks Given,
(A ∪B) = 60 n(A) = 27 n(B) = 42 then;
n(A ∩ B) = n(A) + n(B) - n(A ∪ B)
= 27 + 42 - 60
= 69 - 60 = 9
= 9
Therefore, 9 people like both tea and coffee

Let A = set of persons who got medals in dance.
B = set of persons who got medals in dramatics.
C = set of persons who got medals in music.
Given,
n(A) = 36
n(B) = 12
n(C) = 18
n(A ∪ B ∪C) = 45
n(A ∩ B ∩ C) = 4
We know that number of elements belonging to
exactly two of the three sets A, B, C
= n(A ∩ B) + n(B ∩ C) + n(A ∩ C) - 3n(A ∩ B ∩ C)
= n(A ∩ B) + n(B ∩ C) + n(A ∩ C) - 3 × 4 ……..(i)
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C)
- n(A ∩ C) + n(A ∩ B ∩C)
Therefore, n(A ∩ B) + n(B ∩ C) + n(A ∩ C) =
n(A) + n(B) + n(C) + n(A ∩ B ∩ C) - n(A ∪ B ∪ C)
From (i) required number
= n(A) + n(B) + n(C) + n(A ∩ B ∩ C) - n(A ∪ B ∪ C) - 12
= 36 + 12 + 18 + 4 - 45 - 12
= 70 - 67
= 3

Let A be the set of people who speak English.
B be the set of people who speak French.
A - B be the set of people who speak English and not French.
B - A be the set of people who speak French and not English.
A ∩ B be the set of people who speak both French and English.
Given,
n(A) = 72
n(B) = 43
n(A ∪ B) = 100
Now, n(A ∩ B) = n(A) + n(B) - n(A ∪ B)
= 72 + 43 - 100
= 115 - 100
= 15
Therefore, Number of persons who speak both French and English = 15

Using the formula n(A∪B) = n(A - B) + n(A ∩ B) + n(B - A)

70 = 18 + 25 + n(B - A)

70 = 43 + n(B - A)

n(B - A) = 70 - 43

n(B - A) = 27

Now n(B) = n(A ∩ B) + n(B - A)

= 25 + 27

= 52

Let A = Set of people who like cold drinks.

B = Set of people who like hot drinks.

Given

(A ∪ B) = 60 n(A) = 27 n(B) = 42 then;
n(A ∩ B) = n(A) + n(B) - n(A ∪ B)

= 27 + 42 - 60

= 69 - 60 = 9

= 9

Therefore, 9 people like both tea and coffee.

Using the formula n(A ∪ B) = n(A) + n(B) - n(A ∩ B).
then n(A ∩B) = n(A) + n(B) - n(A ∪B)
= 20 + 28 - 36
= 48 - 36
= 12

% students enrolled only for Maths=40-15=25, Economics=70-15=55
% Total enrolled students=15+25+55=95,
so % students not enrolled either of the two=100-95=5

Let F, H and C represent the set of students who
play football, hockey and cricket respectively.
From the given information, we have
n(F) = 65 , n(H) = 45, n(C) = 42,
n(FnH) = 20, n(FnC) = 25, n(HnC) = 15
n(FnHnC) = 8
From the basic stuff, we have
Total number of students in the group = n(FuHuC)
= n(F) + n(H) + n(C) - n(FnH) - n(FnC) - n(HnC) + n(FnHnC)
= 65 + 45 + 42 -20 - 25 - 15 + 8
=100
Hence, the total number of students in the group is 100