ZS Set Theory Quiz -1

Solution
(A\cap B) = { P, M }
(A\cap B)\cup C = { T, M, P, R }
So 4 elements

Solution
(P \times Q ) \cap ( P \times R) = ( P \times (Q \cap R))
n((P \times Q) \cap (P x R)) = 4
n(P \times n(Q \cap R) = 4
n(P) \times 2 = 4
So n(P) = 2

Solution

Let A = {1, 2, 3, 4} and B = {1, 2, 3, 4, 5, 6}

B – A = {5, 6}

So A ∩ (B – A) = ф

Solution
Let A = {Ronaldo fans}
B = {Dhoni fans}
n(A) = 35
n(B) = 55
We have total students = n(S) = 70
n(S) = n(A) + n(B) - n( A\cap B )
70 = 35 + 55 - n( A\cap B )
n( A\cap B ) = 20
We need to find
n(A) - n( A\cap B ) = 35 - 20 = 15

Solution
Let A = {set of 2 digit numbers divisible by 5} = { 10, 15, 20, 25, …, 95}

Let B = {set of 2 digit numbers divisible by 3} = { 12, 15, 18, 21, …, 99}

Let C = {set of 2 digit numbers divisible by 6} = { 12, 18, 24, 30, …, 96}

We need to find (A\cap B)-C

(A\cap B) = { 15, 30, 45, 60, 75, 90}

30, 60 and 90 are also elements of C

So, (A\cap B)-C = { 15, 45, 75 }

So, 3 elements are divisible by 5 and 3, but not 6

Solution
Let D represent the set of people with Debit card and C represent the set of people with Credit card.
Given n(D) = 60, n(C) = 45
Let S be set of people holding cards
One out of every 20 people does not hold any of these card, so 5% doesn't have any card.
So n(S) = 100 - 5 = 95
We have n(S) = n(C) + n(D) - n( c \cap D )
95 = 60 + 45 - n( C \cap D )
n( C \cap D ) = 60 + 45 - 95 = 10

Solution

(A\cup B) can have max 11 elements if all elements in A are distinct from that of B
(B\cup C) can have max 12 elements if all elements in C are distinct from that of B
So the intersection between 2 sets can’t have elements more than the number of elements of the smallest set
So here, the maximum number of possible elements is 11

Solution
A = {M. I, S, P }
B = {S, I, M, P, L, F, Y }
C= { S, I, M, P }
So words that can be formed using 4 alphabets = 4! = 24

Solution
A = { 2, 3, 5, 7, 11, 13, 17, 18}
B = { 1, 2, 3, 5, 8, 13}
Common elements = { 2, 3, 5, 13}
Four elements are common between A and B
So A X B and B X A will have 4^{2} elements common
i.e., 16 elements

Solution

Given: n(H) = 800, n(E) = 900

Let n(S) = number of people in the group = 1500

Number of people who speak both = 1500 = 800 + 900 – n(H ∩ E)

n(H ∩ E) = 1700 – 1500 = 200