Complete solution of Math class 10(ten) Exercise 1.2 Solutions chapter-1 Sets/Venn Diagrams (All Questions 1-7).

Complete Exercise 1.2 Solutions

Set Theory and Venn Diagrams - Mathematics Grade 10

Table of Contents

• Question 1: Venn Diagram Analysis
• Question 2: Set Relations and Formulas
• Question 3: Set Calculations
• Question 4: Word Problems
• Question 5: Examination Results
• Question 6: Newspaper Readership
• Question 7: Film Language Preference
• Project Work
• Answer Summary

Question 1: Venn Diagram Analysis

1. In the given Venn-diagram, find the values of the following sets

(a) n(P) = 7

(b) n(Q) = 6

(c) n(P ∪ Q ∪ R) = 14

(d) n₀(P) = 4

(e) n₀(R) = 3

(f) n(P ∩ R) = 2

(g) n(P ∪ Q ∪ R) = 14

(h) n₀(P ∩ Q) = 1

(i) n(P ∩ Q ∩ R) = 1

Question 2: Set Relations and Formulas

2. Set Relations and Formula Verification

Given:

U = {positive integers less than 30} = {1, 2, 3, ..., 29}

P = {multiples of 2 less than 30} = {2, 4, 6, ..., 28}

Q = {multiples of 3 less than 30} = {3, 6, 9, ..., 27}

R = {multiples of 5 less than 30} = {5, 10, 15, ..., 25}

Venn Diagram Representation:

Sets P, Q, and R Relationship

P

Q

R

First, let's calculate the cardinalities:

n(P) = 14 (multiples of 2: 2,4,6,...,28)

n(Q) = 9 (multiples of 3: 3,6,9,...,27)

n(R) = 5 (multiples of 5: 5,10,15,20,25)

n(P∩Q) = 4 (multiples of 6: 6,12,18,24)

n(Q∩R) = 1 (multiples of 15: 15)

n(R∩P) = 2 (multiples of 10: 10,20)

n(P∩Q∩R) = 0 (no number less than 30 is a multiple of 2,3,5 i.e., 30)

n(P∪Q∪R) = 21

(a) Verify: n(P∪Q) = n(P) + n(Q) - n(P∩Q)

LHS: n(P∪Q) = 19 (elements in P or Q)

RHS: n(P) + n(Q) - n(P∩Q) = 14 + 9 - 4 = 19

∴ LHS = RHS, verified.

(b) Verify: n(P∪Q∪R) = n(P) + n(Q) + n(R) - n(P∩Q) - n(Q∩R) - n(R∩P) + n(P∩Q∩R)

LHS: n(P∪Q∪R) = 21

RHS: 14 + 9 + 5 - 4 - 1 - 2 + 0 = 28 - 7 = 21

∴ LHS = RHS, verified.

(c) Verify: n(P∪Q∪R) = n(P-Q) + n(Q-R) + n(R-P) + n(P∩Q∩R)

LHS: n(P∪Q∪R) = 21

RHS: n(P-Q) + n(Q-R) + n(R-P) + n(P∩Q∩R)

n(P-Q) = elements in P but not in Q = {2,4,8,10,14,16,20,22,26,28} → 10 elements

n(Q-R) = elements in Q but not in R = {3,6,9,12,18,21,24,27} → 8 elements

n(R-P) = elements in R but not in P = {5,15,25} → 3 elements

n(P∩Q∩R) = 0

RHS = 10 + 8 + 3 + 0 = 21

∴ LHS = RHS, verified.

Answer Summary

Answer Summary

Question 1

(a) 7

(b) 6

(c) 14

(d) 4

(e) 3

(f) 2

(g) 4

(h) 1

(i) 1

Question 3

(a) 80

(b) 9

(c) 15

(d) 5

Question 4

(a) (ii) 10

(b) (ii) 85

(c) (ii) 3

(c) (iii) 1

Question 5

(a) 5%

(b) 60%

(c) 30%

(d) 95%

Question 6

(b) 64

(c) 17

(d) 30

Question 7

(a) 13

(b) 59

(c) 24

Mathematics Grade 10 - Set Theory and Venn Diagrams

Complete Exercise 1.2 Solutions