Mathematics (SNC) – Class 9th

Q1. Choose the correct option.

1. Consider $P=\{x\mid x\text{ is a prime number}\cap 0<x\le 10\}$P={x∣x is a prime number∩0<x≤10} and $Q=\{x\mid x\text{ is a divisor of 210}\cap 0<x\le 10\}$Q={x∣x is a divisor of 210∩0<x≤10}, then $P\cup Q$P∪Q is:
   (A) $\{1,2,3,5,6,7,10\}${1,2,3,5,6,7,10}
2. $A\cap B=B\cap A$A∩B=B∩A holds the property:
   (C) Commutative law of intersection
3. If $A=\mathbb{N}$A=N and $B=\mathbb{Z}$B=Z, then $A\cap B$A∩B is:
   (A) $\mathbb{N}$N
4. Associative law for sets $A,B,C$A,B,C is:
   (C) Both A and B
5. Universal set contains elements of:
   (A) All sets
6. If $A=\mathrm{\varnothing }$A=∅, then $P(A)$P(A) is:
   (D) $\{\mathrm{\varnothing }\}${∅}
7. The proper subset of $\mathbb{W}$W is:
   (A) $\mathbb{N}$N
8. Any two subsets of $\mathbb{N}$N are:
   (C) $\{1,2,3\},\{7,8,9\}${1,2,3},{7,8,9}
9. If $\mathbb{Z}=$Z= set of integers, then:
   (C) Both A and B
10. The power set of $\mathrm{\varnothing }$∅ contains number of elements:
    (B) 1

Q2. Short Answers

[i] $\{x\mid x=n^2,n\in \mathbb{N},1\le n\le 22\}${x∣x=n2,n∈N,1≤n≤22}
[ii] $\{x\mid x\in \mathbb{Z},-1000\le x\le 1000\}${x∣x∈Z,−1000≤x≤1000}
[iii] $\{x\mid x=6n,n\in \mathbb{N},1\le n\le 20\}${x∣x=6n,n∈N,1≤n≤20}
[iv] $\{x\mid x=3^{n-1},n\in \mathbb{N}\}${x∣x=3n−1,n∈N}
[v] $\{5,10,15,20,25,30,35\}${5,10,15,20,25,30,35}
[vii] $\{2,3,5,7,11\}${2,3,5,7,11}
[viii] $\mathrm{\varnothing }$∅
[viii] $\mathbb{N}$N
[x] Proper subsets of $\mathbb{Z}$Z: $\mathbb{N},\mathbb{W}$N,W
[x] Proper subsets of $\{x\mid x\in \mathrm{\Omega },0<x\le 2\}${x∣x∈Ω,0<x≤2}: $\{1\},\{2\}${1},{2}
[xi] $4$4
[xii] $\{\mathrm{\varnothing },\{+\},\{-\},\{\times \},\{÷\},\{+,-\},\{+,\times \},\{+,÷\},\{-,\times \},\{-,÷\},\{\times ,÷\},\{+,-,\times \},\{+,-,÷\},\{+,\times ,÷\},\{-,\times ,÷\},\{+,-,\times ,÷\}\}${∅,{+},{−},{×},{÷},{+,−},{+,×},{+,÷},{−,×},{−,÷},{×,÷},{+,−,×},{+,−,÷},{+,×,÷},{−,×,÷},{+,−,×,÷}}
[xiii] Verified: $A\cap U=A$A∩U=A
[xiv] $9$9 students
[xv] Intersection of Two Sets: The set of elements common to both sets. Example: $A=\{1,2,3\},B=\{2,3,4\}\Rightarrow A\cap B=\{2,3\}$A={1,2,3},B={2,3,4}⇒A∩B={2,3}

Q3. Detailed Answers (Answer any 2)

1. Participants with Laptops, Tablets, or Books

Let:

• $L$L: Laptops
• $T$T: Tablets
• $B$B: Books

Given:
$\mathrm{\mid }L\mathrm{\mid }=17,\mathrm{\mid }T\mathrm{\mid }=11,\mathrm{\mid }L\cap T\mathrm{\mid }=9,\mathrm{\mid }L\cap B\mathrm{\mid }=6,\mathrm{\mid }T\cap B\mathrm{\mid }=4,\mathrm{\mid }L\cap T\cap B\mathrm{\mid }=8,\mathrm{\mid }L\cup T\cup B\mathrm{\mid }=35$∣L∣=17,∣T∣=11,∣L∩T∣=9,∣L∩B∣=6,∣T∩B∣=4,∣L∩T∩B∣=8,∣L∪T∪B∣=35

Using inclusion-exclusion:$$\mathrm{\mid }L\cup T\cup B\mathrm{\mid }=\mathrm{\mid }L\mathrm{\mid }+\mathrm{\mid }T\mathrm{\mid }+\mathrm{\mid }B\mathrm{\mid }-\mathrm{\mid }L\cap T\mathrm{\mid }-\mathrm{\mid }L\cap B\mathrm{\mid }-\mathrm{\mid }T\cap B\mathrm{\mid }+\mathrm{\mid }L\cap T\cap B\mathrm{\mid }$$∣L∪T∪B∣=∣L∣+∣T∣+∣B∣−∣L∩T∣−∣L∩B∣−∣T∩B∣+∣L∩T∩B∣$$35=17+11+\mathrm{\mid }B\mathrm{\mid }-9-6-4+8$$35=17+11+∣B∣−9−6−4+8$$35=17+11+\mathrm{\mid }B\mathrm{\mid }-11$$35=17+11+∣B∣−11$$35=17+\mathrm{\mid }B\mathrm{\mid }$$35=17+∣B∣$$\mathrm{\mid }B\mathrm{\mid }=18$$∣B∣=18

✅ 18 participants have books.

2. Students in Both Swimming and Tug-of-War

Let:

• $S$S: Swimming club = 58
• $T$T: Tug-of-war club = 50
• Total students = 98

Using:$$\mathrm{\mid }S\cup T\mathrm{\mid }=\mathrm{\mid }S\mathrm{\mid }+\mathrm{\mid }T\mathrm{\mid }-\mathrm{\mid }S\cap T\mathrm{\mid }$$∣S∪T∣=∣S∣+∣T∣−∣S∩T∣$$98=58+50-\mathrm{\mid }S\cap T\mathrm{\mid }$$98=58+50−∣S∩T∣$$98=108-\mathrm{\mid }S\cap T\mathrm{\mid }$$98=108−∣S∩T∣$$\mathrm{\mid }S\cap T\mathrm{\mid }=10$$∣S∩T∣=10

✅ 10 students participated in both games.

3. Books in Both Islamic and Science Categories

Let:

• $I$I: Islamic books = 70
• $S$S: Science books = 90
• Neither = 15
• Total books = 150

Books in at least one category:$$\mathrm{\mid }I\cup S\mathrm{\mid }=150-15=135$$∣I∪S∣=150−15=135

Using:$$\mathrm{\mid }I\cup S\mathrm{\mid }=\mathrm{\mid }I\mathrm{\mid }+\mathrm{\mid }S\mathrm{\mid }-\mathrm{\mid }I\cap S\mathrm{\mid }$$∣I∪S∣=∣I∣+∣S∣−∣I∩S∣$$135=70+90-\mathrm{\mid }I\cap S\mathrm{\mid }$$135=70+90−∣I∩S∣$$135=160-\mathrm{\mid }I\cap S\mathrm{\mid }$$135=160−∣I∩S∣$$\mathrm{\mid }I\cap S\mathrm{\mid }=25$$∣I∩S∣=25

✅ 25 books are in both categories.

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