Question 1

If A = {a, d}, B = {b, c, e} and C = {b, c, f}, then A × (B ∪ C) =

Accepted Answer

As per given question, (B ∪ C) = {b, c, e} ∪ {b, c, f} = {b, c, e, f} ∴ A × (B ∪ C) = {a, d} × {b, c, e, f} = {(a, b), (a, c), (a, e), (a, f), (d, b), (d, c), (d, e), (d, f)} …...........(1) Also, (A × B) = {(a, b), (a, c), (a, e), (d, b), (d, c), (d, e)} and, (A × C) = {(a, b), (a, c), (a, f), (d, b), (d, c), (d, f)} …............(2) ∴ (A × B) ∪ (A × C) = {(a, b), (a, c), (a, e), (a, f), (d, b), (d, c), (d, e), (d, f)} From (1) and (2), we have A × (B ∪ C) = (A × B) ∪ (A × C).

Question 2

Out of 100 families in the neighborhood, 50 have radios, 75 have TVs and 25 have VCR. Only 10 families have all three and each VCR owner also has a TV. If some families have radio only, how many have only TV?

Accepted Answer

Let us assume that the N number of people have only TV. According to question, ∴ (40 – P) + P + 10 + 15 + (50 – P) = 100 ⇒ 115 – 2P + P = 100 ∴ P = 115 – 100 = 15 ∴ Only TV = 75 – 15 – 10 – 15 = 35

Question 3

72% of the students of a certain class took Biology and 44% took Mathematics. If each student took at least one of Biology or Mathematics and 40 students took both of these subjects, the total number of students in the class is:

Accepted Answer

n(A) = Biology students = 72% n(B) = Mathematics students = 44% Total students = n(A ∪ B) = 100% ∴ 100% = 72% + 44% - n(A ∩ B) = 116% - n(A ∩ B) ∴ n(A ∩ B) = 16% ⇒ number of students studying both subjects = 40 ∴ Let the total number of students be P Now, according to the question, 16P/100 = 40 ∴ P = 250

Question 4

In a class of 50 students, 35 opted for mathematics and 37 opted for Biology. How many have opted for both Mathematics and Biology? How many have opted for only Mathematics? (Assume that each student has to opt for at least one of the subjects).

Accepted Answer

Given in the question, n(M ∪ B) = 50, n(M) = 35, n(B) = 37, n(M ∩ B) =? use the below formula n(M ∪ B) = n(M) + n(B) – n(M ∩ B) We get 50 = 35 + 37 – n(M ∩ B) ⇒ n(M ∩ B) = 35 + 37 – 50 = 72 – 50 = 22 ∴ 22 students have opted for both Mathematics and Biology. Again number of students who have opted for only Mathematics = n(M) – n(M ∩ B) = 35 – 22 = 13.

Sets and Functions Questions