Current Affairs updated daily
KalviGavaniKalviGavani

Algebra Exercise Solutions | KalviGavani

Samacheer Kalvi 11th Business Maths Algebra exercise solutions — sets, relations, functions, sequences, series, permutations, combinations, binomial theorem, worked examples, MCQs, full answers.

Updated 15 June 202617 min read

Exercise Solutions Available

Step-by-step answers for all textbook questions in this chapter

View Solutions →

How to Use This Page

1. Solve first, then verify. Attempt every exercise problem on your own before checking the solutions here. Struggling through the problem is how you learn. 2. Compare your method with the step-by-step working shown. If your answer differs, rework from the step where you diverged — do not simply copy the correct answer. 3. Memorise the formulas in Key Concepts first — they are the foundation for every problem type in this chapter. 4. Pay close attention to the Common Mistakes section — these are the exact errors examiners mark down for in TN Board exams. 5. Use the Exam Checklist before your test to confirm you have covered every topic in the chapter.

Key Concepts (Quick Recap)

Sets

ConceptDefinition / Formula
SetA well-defined collection of distinct objects
SubsetA ⊆ B if every element of A is also in B
Proper subsetA ⊂ B if A ⊆ B and A ≠ B
Universal setU — the set containing all elements under consideration
UnionA ∪ B = {x : x ∈ A or x ∈ B}
IntersectionA ∩ B = {x : x ∈ A and x ∈ B}
DifferenceA − B = {x : x ∈ A and x ∉ B}
ComplementA′ = U − A
Power setP(A) = set of all subsets of A; \P(A)\= 2^n where n = \A\
Cardinality formula\A ∪ B\= \A\+ \B\− \A ∩ B\

De Morgan's Laws:

(A ∪ B)′ = A′ ∩ B′ (A ∩ B)′ = A′ ∪ B′

Relations

TypeCondition
Reflexive(a, a) ∈ R for every a ∈ A
Symmetric(a, b) ∈ R ⇒ (b, a) ∈ R
Transitive(a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R
EquivalenceReflexive + Symmetric + Transitive

Functions

TypeCondition
One-one (Injective)f(a) = f(b) ⇒ a = b (distinct inputs give distinct outputs)
Onto (Surjective)For every y ∈ B, ∃ x ∈ A such that f(x) = y
BijectiveBoth one-one and onto
Constant functionf(x) = c for all x in the domain
Identity functionf(x) = x

Sequences and Series — Arithmetic Progression (AP)

General term: Tn = a + (n − 1)d Sum of n terms: Sn = n/2 [2a + (n − 1)d] Sn = n/2 [a + l] where l = last term

Sequences and Series — Geometric Progression (GP)

General term: Tn = ar^(n−1) Sum of n terms: Sn = a(r^n − 1)/(r − 1) when r > 1 Sn = a(1 − r^n)/(1 − r) when 0 < r < 1 Sum to infinity: S∞ = a/(1 − r) when |r| < 1

Permutations and Combinations

Factorial: n! = n × (n−1) × (n−2) × … × 1; 0! = 1 Permutation: nPr = n!/(n−r)! Combination: nCr = n!/[r!(n−r)!] Key property: nCr = nC(n−r) Pascal's rule: nCr + nC(r−1) = (n+1)Cr

Binomial Theorem

(a + b)^n = Σ(nCr) a^(n−r) b^r for r = 0 to n General term: T(r+1) = nCr a^(n−r) b^r Middle term (n even): (n/2 + 1)th term Middle terms (n odd): [(n+1)/2]th and [(n+3)/2]th terms Number of terms: n + 1

Logarithms

log_a(xy) = log_a x + log_a y log_a(x/y) = log_a x − log_a y log_a(x^n) = n · log_a x log_a a = 1; log_a 1 = 0 Change of base: log_b a = log_c a / log_c b

I. Choose the Best Answer (MCQs)

1. If A = {1, 2, 3} and B = {3, 4, 5}, then A ∪ B is
Answer

(A) {3} &emsp; (B) {1, 2, 3, 4, 5} &emsp; (C) {1, 2, 4, 5} &emsp; (D) {3, 4}

Answer(B) {1, 2, 3, 4, 5}
ExplanationA ∪ B includes every element that belongs to A or B (or both). Combining both sets: {1, 2, 3} ∪ {3, 4, 5} = {1, 2, 3, 4, 5}. The element 3 appears in both but is listed only once in the union.
2. The number of subsets of a set with 5 elements is
Answer

(A) 5 &emsp; (B) 10 &emsp; (C) 25 &emsp; (D) 32

Answer(D) 32
ExplanationThe number of subsets of a set with n elements is 2^n. For n = 5: 2^5 = 32. This includes the empty set and the set itself.
3. The 10th term of the AP 3, 7, 11, 15, … is
Answer

(A) 39 &emsp; (B) 43 &emsp; (C) 35 &emsp; (D) 47

Answer(A) 39
ExplanationHere a = 3, d = 4, n = 10.
T10 = a + (n − 1)d = 3 + (10 − 1) × 4 = 3 + 36 = 39
4. 7P3 equals
Answer

(A) 210 &emsp; (B) 35 &emsp; (C) 120 &emsp; (D) 720

Answer(A) 210

Explanation:

7P3 = 7! / (7 − 3)! = 7! / 4! = (7 × 6 × 5 × 4!) / 4! = 7 × 6 × 5 = 210
5. 8C3 equals
Answer

(A) 56 &emsp; (B) 336 &emsp; (C) 28 &emsp; (D) 24

Answer(A) 56

Explanation:

8C3 = 8! / [3! × (8 − 3)!] = 8! / (3! × 5!) = (8 × 7 × 6) / (3 × 2 × 1) = 336/6 = 56
6. In the expansion of (a + b)^6, the coefficient of the middle term is
Answer

(A) 10 &emsp; (B) 15 &emsp; (C) 20 &emsp; (D) 6

Answer(C) 20
ExplanationFor (a + b)^6, n = 6 (even). There is one middle term at position (n/2 + 1) = 4th term. The coefficient is:
6C3 = 6! / (3! × 3!) = (6 × 5 × 4) / (3 × 2 × 1) = 120/6 = 20
7. A relation R on a set A = {1, 2, 3} is given by R = {(1,1), (2,2), (1,2), (2,1)}. R is
Answer

(A) Reflexive only &emsp; (B) Symmetric only &emsp; (C) Transitive only &emsp; (D) Both reflexive and symmetric

Answer(D) Both reflexive and symmetric
Explanation(1,1) and (2,2) are in R, so R is reflexive on {1, 2}. Since (1,2) ∈ R and (2,1) ∈ R, R is symmetric. However (3,3) ∉ R, so R is not reflexive on the full set A = {1, 2, 3}. Among the given options, (D) is the best description — R is symmetric and reflexive on its domain.

II. Fill in the Blanks

1. If A = {a, b, c}, then the number of proper subsets of A is 7 (2^3 − 1 = 7; exclude the set itself).

2. If f(x) = 2x + 3, then f(4) = 11 (2 × 4 + 3 = 11).

3. The common difference of the AP 5, 8, 11, 14, … is 3 (8 − 5 = 3).

4. The sum of the first 20 natural numbers is 210 (Sn = n(n + 1)/2 = 20 × 21/2 = 210).

5. The common ratio of the GP 2, 6, 18, 54, … is 3 (6/2 = 3).

6. The value of 5P0 is 1 (nPr = n!/(n−r)!; 5P0 = 5!/5! = 1).

7. The value of nC0 is 1 for any positive integer n (choosing 0 items from n items can be done in exactly 1 way).

8. The 5th term in the expansion of (a + b)^7 is 35a^3 b^4 (T5 = 7C4 · a^3 · b^4 = 35 a^3 b^4).

9. log_10 1000 = 3 (10^3 = 1000).

10. If A ∩ B = ∅, then A and B are disjoint sets (they have no common elements).

11. The sum to infinity of the GP 1, 1/2, 1/4, 1/8, … is 2 (S∞ = a/(1 − r) = 1/(1 − 1/2) = 2).

12. The number of combinations of n objects taken r at a time where r > n is 0 (you cannot choose more objects than are available).

III. True or False

1. Statement: Every subset of a set is a proper subset.

False. A set is a subset of itself (A ⊆ A), but it is not a proper subset. A proper subset must be strictly contained: A ⊂ B means A ⊆ B and A ≠ B.

2. Statement: Every one-one function is onto.

False. A one-one (injective) function maps distinct inputs to distinct outputs, but the range may be smaller than the codomain. For example, f: R → R defined by f(x) = e^x is one-one but not onto, because negative values are never attained.

3. Statement: The sum of the first n terms of an AP is always given by Sn = n/2 [2a + (n − 1)d].

True. This is the standard formula for the sum of an AP. Here a is the first term and d is the common difference. It can also be written as Sn = n/2 (a + l) where l is the last term.

4. Statement: nCr + nC(r−1) = (n+1)Cr.

True. This is Pascal's identity, a fundamental property of combinations. It is used in constructing Pascal's triangle and in many combinatorial proofs.

5. Statement: The binomial expansion of (a + b)^n has n + 2 terms.

False. The expansion of (a + b)^n has exactly n + 1 terms, corresponding to r = 0, 1, 2, …, n.

IV. Match the Following

Column AColumn B
1. A ∪ A′(d) Universal set U
2. nPr / r!(f) nCr
3. Tn of AP(a) a + (n − 1)d
4. S∞ of GP (when \r\< 1)(e) a/(1 − r)
5. General term in binomial expansion(c) nCr a^(n−r) b^r
6. log_a a(b) 1

Matches: 1 → d, 2 → f, 3 → a, 4 → e, 5 → c, 6 → b

V. Answer Briefly (Two-Mark Questions)

1. Define a set and give an example. Distinguish between a finite and an infinite set.
Answer

A set is a well-defined collection of distinct objects. For example, A = {1, 2, 3} is a set of the first three natural numbers. A finite set has a countable number of elements (e.g., {a, b, c} has 3 elements), while an infinite set has an unending number of elements (e.g., the set of all natural numbers N = {1, 2, 3, …}).

2. State De Morgan's laws with an example.
Answer

De Morgan's laws relate the complement of unions and intersections:

  • (A ∪ B)′ = A′ ∩ B′
  • (A ∩ B)′ = A′ ∪ B′

Example: Let U = {1, 2, 3, 4, 5}, A = {1, 2}, B = {2, 3}. Then A′ = {3, 4, 5}, B′ = {1, 4, 5}. A ∪ B = {1, 2, 3}, so (A ∪ B)′ = {4, 5}. And A′ ∩ B′ = {3, 4, 5} ∩ {1, 4, 5} = {4, 5}. Hence (A ∪ B)′ = A′ ∩ B′ — verified.

3. Distinguish between permutations and combinations with one example each.
Answer

Permutations are arrangements where order matters. Combinations are selections where order does not matter. For example, from 3 letters {A, B, C}: the number of arrangements of 2 letters is 3P2 = 6 (AB, BA, AC, CA, BC, CB), but the number of selections of 2 letters is 3C2 = 3 (AB, AC, BC). The relationship is nPr = r! × nCr.

4. What is the condition for the sum to infinity of a GP to exist? Find S∞ for the GP 4, 2, 1, 1/2, ….
Answer
The sum to infinity of a GP exists only when the common ratio satisfiesr< 1. Here, r = 2/4 = 1/2 and1/2< 1, so the infinite sum converges.
S∞ = a / (1 − r) = 4 / (1 − 1/2) = 4 / (1/2) = 8

VI. Answer in Detail (Five-Mark Questions)

1. Worked AP/GP Sum Problem

Question: Find the sum of the first 15 terms of the AP 5, 9, 13, 17, … Also, find the sum of the first 6 terms of the GP 3, 6, 12, 24, … Hence find the combined total.

Solution — AP part
Given AP: 5, 9, 13, 17, … First term a = 5 Common difference d = 9 − 5 = 4 Number of terms n = 15 S15 = n/2 [2a + (n − 1)d] = 15/2 [2(5) + (15 − 1)(4)] = 15/2 [10 + 56] = 15/2 × 66 = 15 × 33 = 495
Solution — GP part
Given GP: 3, 6, 12, 24, … First term a = 3 Common ratio r = 6/3 = 2 Number of terms n = 6 S6 = a(r^n − 1)/(r − 1) = 3(2^6 − 1)/(2 − 1) = 3(64 − 1)/1 = 3 × 63 = 189

Combined total:

Total = 495 + 189 = 684

2. Worked Permutation/Combination Counting Problem

Question: A committee of 5 is to be formed from 7 men and 6 women. In how many ways can this be done if (i) there is no restriction, (ii) the committee must include exactly 3 women, (iii) the committee must include at least 2 men?

(i) No restriction:

Total people = 7 men + 6 women = 13 Committee of 5 from 13 people: Number of ways = 13C5 = 13! / [5! × 8!] = (13 × 12 × 11 × 10 × 9) / (5 × 4 × 3 × 2 × 1) = 154440 / 120 = 1287

(ii) Exactly 3 women (and therefore 2 men):

Select 3 women from 6: 6C3 = 20 Select 2 men from 7: 7C2 = 21 Number of ways = 6C3 × 7C2 = 20 × 21 = 420

(iii) At least 2 men (i.e., 2, 3, 4, or 5 men):

2 men + 3 women: 7C2 × 6C3 = 21 × 20 = 420 3 men + 2 women: 7C3 × 6C2 = 35 × 15 = 525 4 men + 1 woman: 7C4 × 6C1 = 35 × 6 = 210 5 men + 0 women: 7C5 × 6C0 = 21 × 1 = 21 ───── Total = 420 + 525 + 210 + 21 = 1176

3. Worked Binomial Expansion Problem

Question: Expand (2x + 3y)^5 using the binomial theorem and find the coefficient of x^3 y^2.

Full expansion:

(2x + 3y)^5 = Σ (5Cr) (2x)^(5−r) (3y)^r r = 0: 5C0 (2x)^5 (3y)^0 = 1 × 32x^5 × 1 = 32x^5 r = 1: 5C1 (2x)^4 (3y)^1 = 5 × 16x^4 × 3y = 240x^4 y r = 2: 5C2 (2x)^3 (3y)^2 = 10 × 8x^3 × 9y^2 = 720x^3 y^2 r = 3: 5C3 (2x)^2 (3y)^3 = 10 × 4x^2 × 27y^3 = 1080x^2 y^3 r = 4: 5C4 (2x)^1 (3y)^4 = 5 × 2x × 81y^4 = 810x y^4 r = 5: 5C5 (2x)^0 (3y)^5 = 1 × 1 × 243y^5 = 243y^5 Therefore: (2x + 3y)^5 = 32x^5 + 240x^4 y + 720x^3 y^2 + 1080x^2 y^3 + 810x y^4 + 243y^5

Coefficient of x^3 y^2:

The term containing x^3 y^2 corresponds to r = 2: T3 = 5C2 (2x)^3 (3y)^2 = 10 × 8x^3 × 9y^2 = 10 × 72 x^3 y^2 = 720 x^3 y^2 Coefficient of x^3 y^2 = 720

Common Mistakes to Avoid

1. Confusing nPr with nCr. Permutations (nPr) count arrangements where order matters; combinations (nCr) count selections where order does not. Always ask: "Does the order matter?" before choosing the formula. Arranging people in a row = permutation; choosing a committee = combination.

2. Forgetting that 0! = 1. Many students write 0! = 0. This is wrong. By convention and definition, 0! = 1. This directly affects calculations in nPr and nCr.

3. Misapplying the AP sum formula. Remember Sn = n/2 [2a + (n − 1)d]. A common error is using n instead of (n − 1) in the formula, or forgetting to divide by 2. Double-check your substitution.

4. Using the wrong GP sum formula based on r. When r > 1, use Sn = a(r^n − 1)/(r − 1). When 0 < r < 1, use Sn = a(1 − r^n)/(1 − r). Mixing these up gives a negative denominator, leading to wrong answers.

5. **Ignoring the conditionr< 1 for infinite GP sum.** The formula S∞ = a/(1 − r) is valid only whenr< 1. Ifr≥ 1, the infinite sum diverges — do not apply this formula.

6. Wrong middle term in binomial expansion. When n is even, there is one middle term: the (n/2 + 1)th term. When n is odd, there are two middle terms. Students often forget to add 1 to n/2 for the even case.

7. Confusing the general term index. In the binomial expansion, T(r+1) = nCr · a^(n−r) · b^r. The index starts from r = 0 (the first term), not r = 1. So the 3rd term uses r = 2, not r = 3.

8. Forgetting De Morgan's laws in set problems. When asked to find the complement of a union or intersection, apply (A ∪ B)′ = A′ ∩ B′ or (A ∩ B)′ = A′ ∪ B′. Do not distribute the complement to only one set — it applies to both.

Exam Checklist

  • [ ] I can define sets, subsets, power sets, and universal sets and state their properties
  • [ ] I can perform union, intersection, difference, and complement operations on sets
  • [ ] I can state and apply De Morgan's laws to simplify set expressions
  • [ ] I can identify whether a relation is reflexive, symmetric, transitive, or an equivalence relation
  • [ ] I can classify functions as one-one, onto, or bijective and give examples of each type
  • [ ] I can find the nth term and sum of n terms of an AP using the correct formula
  • [ ] I can find the nth term and sum of n terms of a GP, and the sum to infinity when |r| < 1
  • [ ] I can distinguish between permutations and combinations and compute nPr and nCr correctly
  • [ ] I can expand (a + b)^n using the binomial theorem and identify any term or coefficient
  • [ ] I can apply logarithm properties to simplify expressions and solve logarithmic equations

Frequently Asked Questions

Permutations consider order (arrangements), while combinations do not (selections). Use nPr when order matters, nCr when it does not.