Unit 2: Algebra — Sets, Relations, Logic & Sequences — Class 11 Applied Maths | Free MCQs & Solved Examples | Boundless Maths
Unit 2 of 7 18 Marks CBSE 2026–27 Includes Mathematical Logic

Unit 2: Algebra
Sets, Relations, Logic & Sequences

CBSE Class 11 Applied Mathematics · Unit 2 · Free MCQs, AR Questions, Solved Examples & Case Studies

Unit 2 carries 18 marks — the highest-weightage unit in Class 11. Complete free resources: 15 MCQs, 5 Assertion-Reason questions, 13 solved examples and 4 case studies. Covers Sets & Venn Diagrams, Relations, Mathematical Logic (Syllogism, Blood Relations, Coding-Decoding), and AP & GP — fully aligned to CBSE 2026-27.

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Class 11 Applied Maths Unit 2: Algebra — Complete Free Resources (CBSE 2026-27)

This page covers all topics in Unit 2 of CBSE Class 11 Applied Mathematics — the highest-weightage unit in the syllabus, carrying 18 marks out of 80. You'll find 15 MCQs and 5 Assertion-Reason questions with step-by-step answers, 13 solved examples, and 4 case studies based on real-world contexts. It covers Sets (Venn Diagrams, De Morgan's Laws, Intervals), Relations and Ordered Pairs (Cartesian product, domain, range), Mathematical Logic (Syllogism, Blood Relations, Coding-Decoding), and Arithmetic and Geometric Progressions. Free CBSE 2026-27 aligned practice on De Morgan's laws with examples, Venn diagram word problems, syllogism reasoning questions, blood relations puzzles, and AP & GP nth term and sum formulas explained step by step. Note: P&C has moved to Unit IV in the 2026-27 syllabus.

Sets & Venn Diagrams Relations Mathematical Logic AP & GP 18 Marks
Unit 2 · 18 Marks

Topics & Key Formulas

Two sub-units, five topic areas. Highest-weightage unit in Class 11 Applied Maths. Note: P&C has moved to Unit IV in 2026-27.

Sub-Unit A: Sets & Relations

1. Sets

Well-defined collection of objects. Representation, types, operations, Venn diagrams and intervals.

  • Representation: Roster form {1,2,3} and Set-builder form {x : x < 5}
  • Types: Finite, Infinite, Empty (∅), Singleton, Equal, Equivalent sets
  • Universal Set (U): set containing all elements under consideration
  • Union A∪B, Intersection A∩B, Difference A−B, Complement A'
  • De Morgan's Laws: (A∪B)'=A'∩B' and (A∩B)'=A'∪B'
  • Total subsets = 2ⁿ  ·  Proper subsets = 2ⁿ−1
  • Intervals: open (a,b), closed [a,b], semi-open (a,b]
n(A∪B) = n(A) + n(B) − n(A∩B)

2. Relations & Ordered Pairs

Ordered pairs, Cartesian product, domain and range of a relation.

  • Ordered pair: (a,b) ≠ (b,a) unless a = b
  • Cartesian product: A×B = {(x,y) : x∈A, y∈B}
  • n(A×B) = n(A) × n(B)
  • Domain = set of all first elements; Range = set of all second elements
A×B ≠ B×A (not commutative)

3. Mathematical Logic

Reasoning-based — no formula. Read carefully and think logically step by step.

  • Odd Man Out: identify the element that doesn't fit the pattern
  • Syllogism: draw valid conclusions from All/Some/No premises
  • Blood Relations: build family tree step by step from each clue
  • Coding-Decoding: find the shift or pattern, apply or reverse it
No formula — logical thinking only

Sub-Unit B: Sequences & Series

4. Arithmetic Progression (AP)

Sequences with constant common difference d.

  • nᵗʰ term: aₙ = a + (n−1)d
  • Sum of n terms: Sₙ = n/2 × [2a + (n−1)d]
  • If last term l known: Sₙ = n/2 × (a + l)
  • Arithmetic Mean: AM = (a+b)/2  ·  AM ≥ GM
aₙ = a+(n−1)d  ·  Sₙ = n/2[2a+(n−1)d]

5. Geometric Progression (GP)

Sequences with constant common ratio r.

  • nᵗʰ term: aₙ = arⁿ⁻¹
  • Sum: Sₙ = a(rⁿ−1)/(r−1) when r ≠ 1
  • Infinite GP: S∞ = a/(1−r) only when |r| < 1
  • Geometric Mean: GM = √(ab) for positive a, b
S∞ = a/(1−r)  ·  valid only when |r| < 1
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Interactive Practice

Practice MCQs — Unit 2: Algebra

Click Show Answer after each question to check with a full step-by-step explanation. Five Assertion-Reason questions follow after the MCQs.

Sets
Q1 Sets · Difference
If A = {1, 2, 3} and B = {2, 3, 4, 5}, then A − B is:
(a) {2, 3}
(b) {4, 5}
(c) {1}
(d) {1, 2, 3, 4, 5}
Answer: (c) {1}
A − B = elements in A but NOT in B. {1,2,3} − {2,3,4,5} = {1}. The elements 2 and 3 are removed because they are in B.
Q2 De Morgan's Law
According to De Morgan's Second Law, (A ∩ B)' is equal to:
(a) A' ∩ B'
(b) A ∪ B'
(c) A' ∪ B'
(d) A' ∩ B
Answer: (c) A' ∪ B'
De Morgan's Second Law: (A∩B)' = A'∪B'. First Law: (A∪B)' = A'∩B'. Complement flips ∩ to ∪ and vice versa.
Q3 Subsets
The number of proper subsets of A = {a, b, c, d} is:
(a) 16
(b) 14
(c) 15
(d) 8
Answer: (c) 15
Total subsets = 2⁴ = 16. Proper subsets = 2⁴ − 1 = 15 (excludes the set itself).
Q4 Venn Diagram
In a survey of 60 students, 25 play cricket, 20 play football, 10 play both. How many play neither?
(a) 15
(b) 20
(c) 25
(d) 30
Answer: (c) 25
n(C∪F) = 25+20−10 = 35. Neither = 60−35 = 25.
Relations
Q5 Cartesian Product
If A = {1, 2, 3} and B = {x, y}, then n(A × B) is:
(a) 5
(b) 8
(c) 6
(d) 9
Answer: (c) 6
n(A×B) = n(A) × n(B) = 3 × 2 = 6. Pairs: {(1,x),(1,y),(2,x),(2,y),(3,x),(3,y)}.
Q6 Domain & Range
For R = {(1,2),(2,3),(3,4),(4,5)}, the range of R is:
(a) {1,2,3,4}
(b) {2,3,4,5}
(c) {1,2,3,4,5}
(d) {1,3,5}
Answer: (b) {2,3,4,5}
Range = set of all second elements of the ordered pairs = {2,3,4,5}.
Mathematical Logic
📌 About Mathematical Logic

Four types of questions — no formula for any: Odd Man Out (find the element that breaks the pattern), Syllogism (draw valid conclusions from All/Some/No premises — visualise Venn diagrams), Blood Relations (build family tree step by step), and Coding-Decoding (find the shift or pattern, apply or reverse it).

Q7 Syllogism
Statements: All cats are animals. All animals are living beings.
Conclusion I: All cats are living beings.   Conclusion II: Some living beings are cats.
Which conclusions follow?
(a) Only I
(b) Only II
(c) Both I and II
(d) Neither
Answer: (c) Both I and II
All cats → animals → living beings ∴ All cats are living beings ✓ (I). Since all cats are living beings, some living beings are cats ✓ (II — valid conversion). Both follow.
Q8 Syllogism
Statements: No doctor is a teacher. All teachers are engineers.
Conclusion I: No engineer is a doctor.   Conclusion II: Some engineers are teachers.
(a) Only I
(b) Only II
(c) Both I and II
(d) Neither
Answer: (b) Only II
All teachers → engineers ∴ some engineers are teachers ✓ (II). We cannot conclude no engineer is a doctor — there may be engineers who are not doctors but some could be. I does NOT follow.
Q9 Blood Relations
Ravi says, "She is the daughter of my grandfather's only son." How is the girl related to Ravi?
(a) Mother
(b) Aunt
(c) Sister
(d) Cousin
Answer: (c) Sister
Grandfather's only son = Ravi's father. Daughter of Ravi's father = Ravi's Sister.
Q10 Blood Relations
A is B's sister. C is B's mother. D is C's father. How is A related to D?
(a) Grandmother
(b) Granddaughter
(c) Daughter
(d) Great-granddaughter
Answer: (b) Granddaughter
A and B share mother C. D is C's father → D is A's grandfather. So A is D's Granddaughter.
Q11 Coding-Decoding
In a code, MANGO is written as NBOHP. Using the same rule, how is APPLE written?
(a) BQQMF
(b) BOOMF
(c) CQQMF
(d) BPQMF
Answer: (a) BQQMF
Rule: each letter shifts +1. M→N, A→B, N→O, G→H, O→P ✓. APPLE: A→B, P→Q, P→Q, L→M, E→F = BQQMF.
Q12 Coding-Decoding
In a code: 35 = "good weather", 58 = "weather forecast", 589 = "good weather forecast". What does digit 9 stand for?
(a) good
(b) weather
(c) forecast
(d) Cannot be determined
Answer: (a) good
5 is common to 35 and 58 → 5 = weather. 8 is in 58 and 589 → 8 = forecast. 589 adds "good" compared to 58 → 9 = good.
Sequences & Series — AP & GP
Q13 AP
The 15th term of the AP: 3, 7, 11, 15, … is:
(a) 55
(b) 59
(c) 63
(d) 67
Answer: (b) 59
a = 3, d = 4. a₁₅ = a + 14d = 3 + 14×4 = 3 + 56 = 59.
Q14 GP
In a GP, the 2nd term is 6 and the 4th term is 54. The common ratio is:
(a) 2
(b) 3
(c) 6
(d) 9
Answer: (b) 3
ar = 6 …(i)   ar³ = 54 …(ii). Dividing: r² = 9 → r = 3.
Q15 Infinite GP
Sum of the infinite GP: 1, 1/2, 1/4, 1/8, … is:
(a) 1/2
(b) 3/2
(c) 2
(d) 4
Answer: (c) 2
a = 1, r = 1/2 (|r| < 1 ✓). S∞ = a/(1−r) = 1/(1−1/2) = 2.
Assertion-Reason Questions (AR 1–5)

📋 Assertion-Reason Questions

Statement I is Assertion (A) and Statement II is Reason (R). Choose the correct option:

  • (a) Both A and R are True and R is the correct explanation of A
  • (b) Both A and R are True but R is not the correct explanation of A
  • (c) A is True but R is False
  • (d) A is False but R is True
AR 1 Sets
Assertion (A): The total number of subsets of a set with n elements is 2ⁿ.

Reason (R): For each element there are exactly 2 choices — include or exclude — and these choices are independent for every element.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is NOT the correct explanation
(c) A is true, R is false
(d) A is false, R is true
Answer: (a)   A true: 2ⁿ subsets. R true and correctly explains A: n independent binary choices (include/exclude) give 2ⁿ by the multiplication principle.
AR 2 Relations
Assertion (A): If A = {1,2,3} and B = {a,b}, then n(A×B) = 6.

Reason (R): A×B is the set of all ordered pairs (x,y) where x∈B and y∈A.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is NOT the correct explanation
(c) A is true, R is false
(d) A is false, R is true
Answer: (c)   A true: n(A×B) = 3×2 = 6 ✓. R false: definition has x and y swapped — correct definition is x∈A and y∈B, not the other way around.
AR 3 AP
Assertion (A): The sum of first n natural numbers is n(n+1)/2.

Reason (R): Natural numbers form an AP with a = 1, d = 1. Applying Sₙ = n/2×[2a+(n−1)d] gives n(n+1)/2.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is NOT the correct explanation
(c) A is true, R is false
(d) A is false, R is true
Answer: (a)   Both true. Sₙ = n/2×[2+(n−1)] = n(n+1)/2 ✓. R correctly derives A.
AR 4 GP · Infinite Sum
Assertion (A): Sum of infinite GP 1 + 1/3 + 1/9 + … is 3/2.

Reason (R): Infinite GP sum exists only when |r| > 1, given by S∞ = a/(1−r).
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is NOT the correct explanation
(c) A is true, R is false
(d) A is false, R is true
Answer: (c)   A true: r=1/3, S∞=1/(1−1/3)=3/2 ✓. R false: the condition must be |r| < 1, not |r| > 1. The formula itself is correct but the stated condition is wrong.
AR 5 Syllogism
Assertion (A): Given "All birds are animals" and "All animals are living beings", the conclusion "All birds are living beings" is valid.

Reason (R): In a universal affirmative syllogism, if All A are B and All B are C, then All A are C follows by the transitive property.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is NOT the correct explanation
(c) A is true, R is false
(d) A is false, R is true
Answer: (a)   A true: All birds→animals→living beings ✓. R true and correctly explains A: transitive property of universal affirmative statements.

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Step-by-Step Solutions

Short Answer Questions

Click Show Solution to reveal complete working for each question.

Sets
Q1 Sets · Operations
If A = {x : x is an even natural number ≤ 10} and B = {x : x is a prime natural number ≤ 10}, find A∪B, A∩B and A−B.
A = {2, 4, 6, 8, 10}    B = {2, 3, 5, 7}
A∪B = {2, 3, 4, 5, 6, 7, 8, 10}
A∩B = {2} (only element common to both)
A−B = {4, 6, 8, 10} (even numbers that are not prime)
✓ A∪B = {2,3,4,5,6,7,8,10}  |  A∩B = {2}  |  A−B = {4,6,8,10}
Q2 De Morgan's Law
Let U = {1,…,10}, A = {1,3,5,7,9}, B = {2,3,5,7}. Verify: (A∪B)' = A'∩B'.
A∪B = {1,2,3,5,7,9} → (A∪B)' = {4,6,8,10} … LHS
A' = {2,4,6,8,10}    B' = {1,4,6,8,9,10}
A'∩B' = {4,6,8,10} … RHS
✓ LHS = RHS = {4,6,8,10} — De Morgan's First Law verified
Q3 Intervals
Write as intervals: (a) {x∈ℝ : −3 < x ≤ 5}   (b) {x∈ℝ : 0 ≤ x ≤ 4}   (c) {x∈ℝ : −1 < x < 2}
(a) (−3, 5] — Semi-closed. Left endpoint excluded, right included.
(b) [0, 4] — Closed interval. Both endpoints included.
(c) (−1, 2) — Open interval. Both endpoints excluded.
✓ (a) (−3,5] semi-closed  |  (b) [0,4] closed  |  (c) (−1,2) open
Q4 Venn Diagram
In a group of 75 students, 45 study Maths, 30 study Biology, 15 study both. Find: (i) at least one, (ii) only Maths, (iii) only Biology, (iv) neither.
n(M)=45, n(B)=30, n(M∩B)=15, Total=75
(i) n(M∪B) = 45+30−15 = 60
(ii) Only Maths = 45−15 = 30
(iii) Only Biology = 30−15 = 15
(iv) Neither = 75−60 = 15
✓ (i) 60  |  (ii) 30  |  (iii) 15  |  (iv) 15
Q5 Power Set
If A = {p, q, r}, find: (i) all subsets, (ii) P(A), (iii) proper subsets count, (iv) 2-element subsets.
(i) 8 subsets: ∅, {p}, {q}, {r}, {p,q}, {p,r}, {q,r}, {p,q,r}
(ii) P(A) = {∅, {p}, {q}, {r}, {p,q}, {p,r}, {q,r}, {p,q,r}}
(iii) Proper subsets = 2³−1 = 7 (excludes {p,q,r} itself)
(iv) ³C₂ = 3: {p,q}, {p,r}, {q,r}
✓ (i) 8 subsets  |  (iii) 7 proper  |  (iv) 3 two-element subsets
Relations
Q6 Cartesian Product
If A = {2,3} and B = {1,4,9}, write A×B and B×A. Are they equal? Find domain and range of A×B.
A×B = {(2,1),(2,4),(2,9),(3,1),(3,4),(3,9)}
B×A = {(1,2),(1,3),(4,2),(4,3),(9,2),(9,3)}
A×B ≠ B×A — Cartesian product is not commutative
Domain of A×B = {2,3}    Range of A×B = {1,4,9}
✓ A×B≠B×A  |  Domain={2,3}  |  Range={1,4,9}
Mathematical Logic
Q7 Syllogism
Statements: (1) All mangoes are fruits. (2) Some fruits are sweet. Can we conclude that some mangoes are sweet? Justify.
All mangoes ⊂ fruits. Some fruits are sweet — but we don't know which ones.
The sweet fruits may or may not include mangoes — cannot be determined from the given statements.
✓ No — "some mangoes are sweet" does NOT follow. The premises are insufficient.
Q8 Blood Relations
P is the mother of Q. Q is the sister of R. R is the father of S. S is the brother of T. How is P related to T?
P is Q's mother. Q is R's sister → P is also R's mother.
R is S's father. S is T's brother → R is also T's father.
P is R's mother and R is T's father → P is T's Grandmother.
✓ P is the Grandmother of T
Q9 Coding-Decoding
PENCIL is written as RGPEKN. Using the same rule, decode JQTUG back to the original word.
P→R(+2), E→G(+2), N→P(+2), C→E(+2), I→K(+2), L→N(+2). Rule: each letter shifts +2.
Decode JQTUG by applying −2: J→H, Q→O, T→R, U→S, G→E.
✓ JQTUG decodes to HORSE
Sequences & Series — AP & GP
Q10 AP
The 4th term of an AP is 11 and the 12th term is 35. Find first term, common difference and S₂₀.
a+3d=11 …(i)    a+11d=35 …(ii)
(ii)−(i): 8d=24 → d=3
From (i): a = 11−9 = 2
S₂₀ = 20/2 × [4+57] = 10×61 = 610
✓ a=2, d=3, S₂₀=610
Q11 AP · Sum Given
The sum of first n terms is Sₙ = 3n²+4n. Find first term, common difference and 10th term.
a₁ = S₁ = 3+4 = 7
S₂ = 12+8 = 20 → a₂ = 20−7 = 13 → d = 13−7 = 6
a₁₀ = 7+9×6 = 61
✓ a=7, d=6, a₁₀=61
Q12 GP
3rd and 9th terms of a GP are 4 and 256 respectively. Find first term, common ratio and S₆.
ar²=4 …(i)    ar⁸=256 …(ii). Dividing: r⁶=64 → r=2
From (i): 4a=4 → a=1
S₆ = 1×(2⁶−1)/(2−1) = 63
✓ a=1, r=2, S₆=63
Q13 Infinite GP & AM-GM
(a) Find S∞ for GP: 3, 1, 1/3, 1/9, …    (b) Verify AM ≥ GM for a=9, b=4.
(a) a=3, r=1/3, |r|<1 ✓. S∞ = 3/(1−1/3) = 3/(2/3) = 4.5
(b) AM = (9+4)/2 = 6.5
GM = √(9×4) = √36 = 6
6.5 ≥ 6 ✓ AM ≥ GM verified
✓ (a) S∞=4.5  |  (b) AM=6.5, GM=6 — AM≥GM verified

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4-Mark Questions

Case Studies

Board-pattern case questions across all Unit 2 topics. Click Show Answers for each case.

🏫

Case Study 1: School Sports Survey (Sets)

A school surveyed 200 Class 11 students. 60% like Cricket, 45% like Football, 20% like both. Let C = set of students who like Cricket and F = set of students who like Football.
(i)

Find n(C), n(F) and n(C∩F).

(ii)

How many like at least one sport?

(iii)

How many like only Cricket and only Football?

(iv)

How many like neither? Verify De Morgan's Law: (C∪F)' = C'∩F'.

(i)n(C) = 120  |  n(F) = 90  |  n(C∩F) = 40
(ii)n(C∪F) = 120+90−40 = 170
(iii)Only Cricket = 120−40 = 80  |  Only Football = 90−40 = 50
(iv)Neither = 200−170 = 30. C'=80, F'=110. n(C'∩F') = 30 = n(C∪F)' ✓ De Morgan's Law verified.
🕵️

Case Study 2: Company Recruitment Test (Mathematical Logic)

A company uses Syllogism, Blood Relations and Coding-Decoding to shortlist candidates in their aptitude test.
(i) Syllogism

Statements: All pens are stationery. No stationery is food. Conclusion I: No pen is food. Conclusion II: Some stationery is pen. Which follow?

(ii) Blood Relations

A woman says "His mother is the only daughter of my father." How is she related to the man?

(iii) Coding-Decoding

STRONG is written as VWURQJ. (a) Find the rule. (b) Encode BRIGHT using the same rule.

(i)Both I and II follow. All pens→stationery + No stationery→food ∴ No pen is food ✓ (I). All pens are stationery → some stationery are pens ✓ (II).
(ii)Only daughter of her father = herself. Man's mother = the woman → she is his Mother.
(iii)(a) Each letter shifts +3.  (b) B→E, R→U, I→L, G→J, H→K, T→W = EULJKW
💼

Case Study 3: Salary Growth — Arithmetic Progression

Rajan joins a company with monthly salary ₹25,000 and annual increment ₹1,500. His salary each year forms an AP.
(i)

Write the AP and identify a and d.

(ii)

What will his monthly salary be in the 8th year?

(iii)

In which year will his salary first exceed ₹40,000?

(iv)

Find total monthly salary earned across the first 10 years.

(i)AP: 25000, 26500, 28000, …  |  a = ₹25,000, d = ₹1,500
(ii)a₈ = 25000+7×1500 = ₹35,500
(iii)25000+(n−1)×1500 > 40000 → n > 11 → 12th year.
(iv)S₁₀ = 10/2×[50000+13500] = 5×63500 = ₹3,17,500
🦠

Case Study 4: Virus Spread — Geometric Progression

During an outbreak, new daily cases follow a GP. Day 1: 5 cases. Each day, cases triple.
(i)

Identify a and r. Write the first five terms.

(ii)

How many new cases on Day 7?

(iii)

Total new cases over first 6 days?

(iv)

Find GM of cases on Day 1 and Day 3. Verify it equals Day 2 cases.

(i)a=5, r=3. Terms: 5, 15, 45, 135, 405
(ii)a₇ = 5×3⁶ = 5×729 = 3,645
(iii)S₆ = 5×(3⁶−1)/(3−1) = 5×364 = 1,820
(iv)GM = √(5×45) = √225 = 15 = a₂ ✓

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Score Full Marks

Exam Tips — Unit 2: Algebra

What separates 16-mark answers from 12-mark answers in this unit.

✅ Tip 1 — Sets & Venn Diagrams

Always draw and label the Venn diagram first. Even a rough sketch earns method marks. Label each region: only A, only B, A∩B, and neither. This prevents arithmetic errors and lets the examiner follow your working even if a final number is wrong.

✅ Tip 2 — AP & GP

Always write the formula before substituting. Write aₙ = a+(n−1)d or Sₙ = n/2×[2a+(n−1)d] explicitly — this earns the formula mark even if your arithmetic is wrong. Many students lose 1 mark by jumping straight to the calculation.

✅ Tip 3 — Syllogism

Visualise Venn diagrams mentally before answering. "All A are B" = circle A inside circle B. "No A are B" = two non-overlapping circles. This makes valid conclusions obvious and prevents the most common Syllogism errors on exam day.

✅ Tip 4 — Blood Relations

Always write out the chain step by step. Don't solve Blood Relation questions in your head — write each link: "X is Y's ___", then trace the full chain to the final answer. One missed step leads to a wrong answer every time.

❌ Common Mistakes to Avoid in Unit 2

  • Using n(A∪B) = n(A)+n(B) without subtracting n(A∩B)
  • Confusing proper subsets (2ⁿ−1) with total subsets (2ⁿ)
  • Writing A×B = B×A — Cartesian product is NOT commutative
  • Applying S∞ = a/(1−r) when |r| ≥ 1 — only valid when |r| < 1
  • Syllogism: concluding "Some A are C" when correct answer is "All A are C"
  • Blood Relations: confusing maternal/paternal when gender is unspecified
  • Not verifying AP/GP answers by substituting back into the original conditions
Common Questions

Frequently Asked Questions

Questions students ask most about Class 11 Applied Maths Unit 2 Algebra.

Unit 2 covers four sections: Sets (Types, Venn Diagrams, Union, Intersection, Difference, Complement, De Morgan's Laws, Intervals), Relations & Ordered Pairs (Cartesian product, domain, range), Mathematical Logic (Syllogism, Blood Relations, Coding-Decoding), and Sequences & Series (AP and GP with applications). Note: P&C has moved to Unit IV in 2026-27 — it is NOT part of Unit 2.
Unit 2 Algebra carries 18 marks — the highest-weightage unit in Class 11 Applied Maths (out of 80 marks total). Questions appear as 1-mark MCQs and Assertion-Reason, 2–3 mark short answers, and 4-mark case studies.
No. In CBSE Applied Maths 2026-27, P&C has moved to Unit IV — Combinatorics & Probability. If your notes or textbook include P&C in Unit 2, they may be based on an older syllabus. Always verify against the official CBSE document.
Mathematical Logic has three types: Syllogism (given two statements, which conclusions follow?), Blood Relations (decode family chains step by step), and Coding-Decoding (find the shift or pattern, apply or reverse it). No formula required. Preparation: practise 5–6 questions of each type. For Syllogism draw Venn diagrams. For Blood Relations always write each step. For Coding-Decoding always identify the rule first.
In an AP, you add the same number each time (common difference d). Example: 3,7,11,15 (add 4). In a GP, you multiply by the same number each time (common ratio r). Example: 2,6,18,54 (multiply by 3). To identify: check if consecutive differences are constant (AP) or consecutive ratios are constant (GP).
An infinite GP has a finite sum only when |r| < 1. Formula: S∞ = a/(1−r). If |r| ≥ 1 the terms don't shrink to zero and no finite sum exists. This is a common Assertion-Reason trap — AR4 on this page tests exactly this point.
De Morgan's Laws: (A∪B)' = A'∩B' and (A∩B)' = A'∪B'. In exams you typically verify one law by computing both sides separately and showing they are equal. Remember: complement of a union = intersection of complements, and complement of an intersection = union of complements.
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Unit 1
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Unit 7
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