Samacheer Kalvi 7th Maths (Term 2) Algebra Exercise Solutions

How to Use This Page

This page provides complete, step-by-step solutions for every exercise in Chapter 3 — Algebra of the Samacheer Kalvi 7th Maths Term 2 textbook. Work through each problem on your own first, then check your working against the solutions below. Focus on understanding why each step is taken, not just the final answer.

Key Concepts (Quick Recap)

Product rule of exponents: am × an = am+n (same base, add exponents)
Quotient rule of exponents: am ÷ an = am−n, a ≠ 0 (same base, subtract exponents)
Power rule of exponents: (am)n = am×n (multiply exponents)
Zero exponent rule: a0 = 1, a ≠ 0
Power of a product: (ab)n = an × bn
Standard form: Express numbers as a × 10n where 1 ≤ a < 10
Unit digit patterns: Numbers ending in 0, 1, 5, 6 always end in the same digit; numbers ending in 2, 3, 7, 8 cycle every 4; numbers ending in 4, 9 cycle every 2
Degree of a term: Sum of exponents of all variables in that term
Degree of an expression: Highest degree among its terms
Like terms: Terms with the same variable(s) raised to the same power(s) (coefficients can differ)
Unlike terms: Terms with different variables or different powers

Exercise 3.1 — Laws of Exponents

Q1. Simplify the following:

(i) 32 × 34 × 38

Solution

When multiplying powers with the same base, we add the exponents.

32 × 34 × 38 = 32+4+8 = 314

Answer314

(ii) 615 ÷ 610

Solution

When dividing powers with the same base, we subtract the exponents.

615 ÷ 610 = 615−10 = 65

Answer65

(iii) a3 × a2

Solution

Same base, add exponents.

a3 × a2 = a3+2 = a5

Answera5

(iv) 7x × 72

Solution

Same base, add exponents (even when one exponent is a variable).

7x × 72 = 7x+2

Answer7x+2

(v) (52)3 ÷ 53

Solution

First, apply the power rule: (52)3 = 52×3 = 56

Now divide: 56 ÷ 53 = 56−3 = 53

Answer53 = 125

Q2. Express as product of prime factors in exponential form:

(i) 108 × 192

Solution

First, factorise each number:

108 = 2 × 54 = 2 × 2 × 27 = 22 × 33

192 = 2 × 96 = 22 × 48 = 23 × 24 = 24 × 12 = 25 × 6 = 26 × 31

108 × 192 = 22 × 33 × 26 × 31 = 22+6 × 33+1 = 28 × 34

Answer28 × 34

(ii) 270

Solution

270 = 2 × 135 = 2 × 3 × 45 = 2 × 32 × 15 = 2 × 33 × 5

Answer2 × 33 × 5

(iii) 729 × 64

Solution

729 = 36 (since 36 = 729)

64 = 26

729 × 64 = 36 × 26

Answer36 × 26

(iv) 768

Solution

768 = 2 × 384 = 22 × 192 = 23 × 96 = 24 × 48 = 25 × 24 = 26 × 12 = 27 × 6 = 28 × 3

Answer28 × 3

Q3. Which is greater?

(i) 53 or 35?

Solution

53 = 5 × 5 × 5 = 125

35 = 3 × 3 × 3 × 3 × 3 = 243

Since 243 > 125, 35 > 53

Answer35 is greater

(ii) 28 or 82?

Solution

28 = 256

82 = 64

Since 256 > 64, 28 > 82

Answer28 is greater

(iii) 1002 or 2100?

Solution

1002 = 10,000

2100 is an extremely large number (approximately 1.27 × 1030).

Clearly, 2100 > 1002

Answer2100 is greater

(iv) 210 or 102?

Solution

210 = 1024

102 = 100

Since 1024 > 100, 210 > 102

Answer210 is greater

Q4. Simplify using laws of exponents:

(i) (23)2 × 24

Solution

First, apply the power rule: (23)2 = 23×2 = 26

Now multiply: 26 × 24 = 26+4 = 210

Answer210

(ii) (34)2 ÷ 33

Solution

First, apply the power rule: (34)2 = 34×2 = 38

Now divide: 38 ÷ 33 = 38−3 = 35

Answer35

Q5. Express in standard form:

(i) 25,000

Solution

25,000 = 2.5 × 10,000 = 2.5 × 104

Answer2.5 × 104

(ii) 0.0034

Solution

0.0034 = 3.4 × 0.001 = 3.4 × 10−3

Answer3.4 × 10−3

Exercise 3.2 — Unit Digits

Q1. Find the unit digit of the following:

(i) 255223

Solution

The base 255 ends in the digit 5.

Rule: Any power of a number ending in 5 always ends in 5 (since 5 × 5 = 25, 25 × 5 = 125, and so on — the unit digit is always 5).

AnswerThe unit digit is 5

(ii) 81111000

Solution

The base 8111 ends in the digit 1.

Rule: Any power of a number ending in 1 always ends in 1 (since 1 × 1 = 1, always).

AnswerThe unit digit is 1

(iii) 4866431

Solution

The base 4866 ends in the digit 6.

Rule: Any power of a number ending in 6 always ends in 6 (since 6 × 6 = 36, 36 × 6 = 216, and so on — the unit digit is always 6).

AnswerThe unit digit is 6

Q2. Find the unit digit:

(i) 1844671

Solution

The base 1844 ends in 4.

The pattern for powers of 4 is:

41 = 4 (unit digit 4)
42 = 16 (unit digit 6)
43 = 64 (unit digit 4)
44 = 256 (unit digit 6)

The cycle is: 4, 6 (length 2). Odd powers give 4, even powers give 6.

Since 671 is odd, the unit digit is 4.

AnswerThe unit digit is 4

(ii) 1564100

Solution

The base 1564 ends in 4.

Using the same cycle as above: 4, 6 (length 2). Odd powers give 4, even powers give 6.

Since 100 is even, the unit digit is 6.

AnswerThe unit digit is 6

Q3. Find the unit digit:

(i) 999222

Solution

The base 999 ends in 9.

The pattern for powers of 9 is:

91 = 9 (unit digit 9)
92 = 81 (unit digit 1)
93 = 729 (unit digit 9)
94 = 6561 (unit digit 1)

The cycle is: 9, 1 (length 2). Odd powers give 9, even powers give 1.

Since 222 is even, the unit digit is 1.

AnswerThe unit digit is 1

(ii) 1549777

Solution

The base 1549 ends in 9.

Using the same cycle: 9, 1 (length 2). Odd powers give 9, even powers give 1.

Since 777 is odd, the unit digit is 9.

AnswerThe unit digit is 9

Q4. Find the unit digit of 1549101 + 654120

Solution

Step 1: Find the unit digit of 1549101.

The base 1549 ends in 9. The cycle for 9 is: 9, 1 (length 2). Since 101 is odd, the unit digit is 9.

Step 2: Find the unit digit of 654120.

The base 6541 ends in 1. Any power of a number ending in 1 ends in 1. So the unit digit is 1.

Step 3: Add the unit digits.

Unit digit of (1549101 + 654120) = Unit digit of (9 + 1) = Unit digit of 10 = 0.

AnswerThe unit digit is 0

Exercise 3.3 — Degree and Algebraic Expressions

Q1. Find the degree of each expression:

(i) x5 − x4 + 3

Solution

The degrees of the individual terms are: x5 (degree 5), x4 (degree 4), 3 (degree 0).

The highest degree is 5.

AnswerDegree = 5

(ii) 2 − y5 − y3 + 2y8

Solution

The degrees of the individual terms are: 2 (degree 0), y5 (degree 5), y3 (degree 3), 2y8 (degree 8).

The highest degree is 8.

AnswerDegree = 8

(iii) 2

Solution

2 is a constant term. The degree of any non-zero constant is 0.

AnswerDegree = 0

(iv) 5x3 + 4x2 + 7x

Solution

The degrees of the individual terms are: 5x3 (degree 3), 4x2 (degree 2), 7x (degree 1).

The highest degree is 3.

AnswerDegree = 3

(v) 4xy + 7x2y + 3xy3

Solution

The degrees of the individual terms are:

4xy: degree = 1 + 1 = 2
7x2y: degree = 2 + 1 = 3
3xy3: degree = 1 + 3 = 4

The highest degree is 4.

AnswerDegree = 4

Q2. Identify like and unlike terms:

(i) 1, 100

Solution

Both are constants (no variables). They have the same degree (0) and the same variables (none).

AnswerLike terms (both are constants with degree 0)

(ii) −7x, (5/2)x

Solution

Both terms have the same variable x raised to the same power (x1). Only the coefficients differ (−7 and 5/2).

AnswerLike terms

(iii) 4m2p, 4mp2

Solution

4m2p: variable part is m2p (powers m2, p1)
4mp2: variable part is mp2 (powers m1, p2)

The powers of the variables are different.

AnswerUnlike terms

(iv) 12xz, 12x2z2

Solution

12xz: variable part is xz (powers x1, z1)
12x2z2: variable part is x2z2 (powers x2, z2)

The powers of the variables are different.

AnswerUnlike terms

Q3. Subtract and find the degree: (3ab − 2a2 − 2b2) − (5a2 − 7ab + 5b2)

Solution

Step 1: Remove parentheses, changing signs for the second expression.

= 3ab − 2a2 − 2b2 − 5a2 + 7ab − 5b2

Step 2: Group like terms.

= (3ab + 7ab) + (−2a2 − 5a2) + (−2b2 − 5b2)

= 10ab − 7a2 − 7b2

Step 3: Find the degree.

10ab: degree 2 (1 + 1)
−7a2: degree 2
−7b2: degree 2

Answer10ab − 7a2 − 7b2, Degree = 2

Q4. Add and find the degree: (x2 − y2 − 1) + (y2 − 1 − x2) + (1 − x2 − y2)

Solution

Step 1: Remove parentheses and group like terms.

x2 − y2 − 1 + y2 − 1 − x2 + 1 − x2 − y2

Step 2: Combine coefficients.

= (x2 − x2 − x2) + (−y2 + y2 − y2) + (−1 − 1 + 1)

= −x2 − y2 − 1

Step 3: Find the degree.

−x2: degree 2
−y2: degree 2
−1: degree 0

Answer−x2 − y2 − 1, Degree = 2

Q5. Find the degree of each term:

(i) x2

Solution

The only variable is x with exponent 2.

AnswerDegree = 2

(ii) 4xyz

Solution

Exponents of variables: x1, y1, z1.

Degree = 1 + 1 + 1 = 3

AnswerDegree = 3

(iii) 7x2y4/(xy)

Solution

Simplify first: 7x2y4 ÷ xy = 7x2−1y4−1 = 7xy3

Degree = 1 + 3 = 4

AnswerDegree = 4

(iv) x2y2/(xy2)

Solution

Simplify first: x2y2 ÷ xy2 = x2−1y2−2 = x1y0 = x

Degree = 1

AnswerDegree = 1

Multiple Choice Questions (MCQs)

Q1. The value of 23 × 25 is:

(a) 215 &emsp; (b) 45 &emsp; (c) 28 &emsp; (d) 23

Solution

Same base, add exponents: 23 × 25 = 23+5 = 28

Answer(c) 28

Q2. The unit digit of 750 is:

(a) 7 &emsp; (b) 9 &emsp; (c) 3 &emsp; (d) 1

Solution

Powers of 7 cycle: 71=7, 72=49, 73=343, 74=2401. Unit digits: 7, 9, 3, 1 (cycle length 4).

50 ÷ 4 = 12 remainder 2. The 2nd element in the cycle is 9.

Answer(b) 9

Q3. The degree of 3x2y + 5xy is:

(a) 3 &emsp; (b) 2 &emsp; (c) 1 &emsp; (d) 5

Solution

3x2y: degree = 2 + 1 = 3
5xy: degree = 1 + 1 = 2

Highest degree = 3.

Answer(a) 3

Q4. Which of the following are like terms?

(a) 3x and 3y &emsp; (b) 3x2 and −5x2 &emsp; (c) 3x2 and 3x &emsp; (d) 3xy and 3x2y2

Solution

Like terms have the same variable(s) with the same power(s). Only (b) has the same variable x2.

Answer(b) 3x2 and −5x2

Q5. (a0) equals:

(a) 0 &emsp; (b) a &emsp; (c) undefined &emsp; (d) 1

Solution

By the zero exponent rule, any non-zero number raised to the power 0 equals 1.

Answer(d) 1

True or False

1. Any number raised to power 0 equals 1.

AnswerTrue (for any non-zero number, a0 = 1)

2. Like terms can have different coefficients.

AnswerTrue (like terms differ only in coefficient; e.g., 3x and −5x are like terms)

3. The degree of 5 is 1.

AnswerFalse (the degree of a non-zero constant is 0, not 1)

4. 34 × 32 = 38

AnswerFalse (34 × 32 = 34+2 = 36, not 38)

5. The unit digit of 25100 is 5.

AnswerTrue (any power of a number ending in 5 ends in 5)

Fill in the Blanks

1. am × an = ___

Answeram+n (Product rule: add exponents when bases are the same)

2. am ÷ an = ___ (a ≠ 0)

Answeram−n (Quotient rule: subtract exponents when bases are the same)

3. (am)n = ___

Answeram×n (Power rule: multiply exponents)

4. The degree of 7x3y2 is ___

Answer5 (degree = 3 + 2 = 5)

5. The unit digit of 6100 is ___

Answer6 (any power of a number ending in 6 ends in 6)

Common Mistakes

1. Adding exponents when bases are different. For example, 23 × 32 ≠ 65. You can only add exponents when the bases are the same. When bases differ, evaluate each power separately and then multiply.

2. Confusing the multiplication rule with the power rule. The product rule adds exponents (am × an = am+n), but the power rule multiplies them ((am)n = am×n). These are fundamentally different operations.

3. Forgetting that a0 = 1 (not 0). Any non-zero number raised to the power zero equals 1. This is one of the most commonly tested facts.

4. Mixing up like and unlike terms. Terms like 3x2y and 3xy2 are unlike terms because the variable powers are different, even though they have the same variables and coefficients.

5. Not following the cyclical pattern for unit digits correctly. For bases ending in 2, 3, 7, 8 the cycle length is 4; for 4, 9 it is 2; for 0, 1, 5, 6 the unit digit is constant. Always find the remainder when the exponent is divided by the cycle length.

6. Subtracting exponents in the wrong order for division. Remember: am ÷ an = am−n, not an−m. The exponent in the numerator comes first.

7. Forgetting to add exponents of ALL variables when finding degree of a term. The degree of 4xy3 is 1 + 3 = 4, not 3. Add up the exponents of every variable in the term.

Exam Checklist

Before your exam, make sure you can confidently:

[ ] Apply all five laws of exponents (product, quotient, power, zero exponent, power of a product)
[ ] Express large numbers in exponential/standard form
[ ] Compare numbers using their exponential forms
[ ] Find unit digits of large powers using cyclical patterns
[ ] Determine the degree of algebraic expressions and individual terms
[ ] Distinguish between like terms and unlike terms
[ ] Add and subtract algebraic expressions by grouping like terms
[ ] Simplify expressions involving exponents step by step

Frequently Asked Questions

Q1: What are the five laws of exponents?

Answer

The five fundamental laws of exponents are: 1. Product rule: am × an = am+n (add exponents when multiplying) 2. Quotient rule: am ÷ an = am−n (subtract exponents when dividing) 3. Power rule: (am)n = am×n (multiply exponents when raising a power to a power) 4. Zero exponent rule: a0 = 1 for a ≠ 0 5. Power of a product: (ab)n = an × bn

Q2: How do you find the unit digit of a large power?

Answer

Find the pattern of unit digits by raising the last digit of the base repeatedly. For numbers ending in 0, 1, 5, 6, the unit digit is constant for any power. For numbers ending in 2, 3, 7, 8, the cycle length is 4. For numbers ending in 4, 9, the cycle length is 2. Divide the exponent by the cycle length, find the remainder, and use it to locate the correct unit digit in the cycle.

Q3: What is the degree of an algebraic expression?

Answer

The degree of a term is the sum of the exponents of all variables in that term. For example, 3x2y has degree 2 + 1 = 3. The degree of an expression is the highest degree among all its terms. For example, in 3x2y + 5xy3, the degrees are 3 and 4, so the expression has degree 4. Constants have degree 0.

Internal Links

[Term 2 Number System Solutions](/school/tamil-nadu-state-board/class/7/maths/solutions/term-2-number-system-exercise-solutions)
[Term 2 Measurements Solutions](/school/tamil-nadu-state-board/class/7/maths/solutions/term-2-measurements-exercise-solutions)
[All Class 7 Maths Solutions](/school/tamil-nadu-state-board/class/7/maths/solutions)

Samacheer Kalvi 7th Maths (Term 2) Algebra Exercise Solutions

Samacheer Kalvi 7th Maths (Term 2) Algebra Exercise Solutions

How to Use This Page

Key Concepts (Quick Recap)

Exercise 3.1 — Laws of Exponents

Q1. Simplify the following:

Q2. Express as product of prime factors in exponential form:

Q3. Which is greater?

Q4. Simplify using laws of exponents:

Q5. Express in standard form:

Exercise 3.2 — Unit Digits

Q1. Find the unit digit of the following:

Q2. Find the unit digit:

Q3. Find the unit digit:

Q4. Find the unit digit of 1549<sup>101</sup> + 6541<sup>20</sup>

Exercise 3.3 — Degree and Algebraic Expressions

Q1. Find the degree of each expression:

Q2. Identify like and unlike terms:

Q3. Subtract and find the degree: (3ab − 2a<sup>2</sup> − 2b<sup>2</sup>) − (5a<sup>2</sup> − 7ab + 5b<sup>2</sup>)

Q4. Add and find the degree: (x<sup>2</sup> − y<sup>2</sup> − 1) + (y<sup>2</sup> − 1 − x<sup>2</sup>) + (1 − x<sup>2</sup> − y<sup>2</sup>)

Q5. Find the degree of each term:

Multiple Choice Questions (MCQs)

True or False

Fill in the Blanks

Common Mistakes

Exam Checklist

Frequently Asked Questions

Internal Links

Other question types — same chapter

Tamil Nadu Class 7 Maths: Algebra

Samacheer Kalvi 7th Maths (Term 1) Direct and Inverse Proportion Exercise Solutions

Samacheer Kalvi 7th Maths (Term 3) Percentage and Simple Interest Exercise Solutions

Samacheer Kalvi 7th Maths (Term 2) Algebra Exercise Solutions

Samacheer Kalvi 7th Maths (Term 2) Algebra Exercise Solutions

How to Use This Page

Key Concepts (Quick Recap)

Exercise 3.1 — Laws of Exponents

Q1. Simplify the following:

Q2. Express as product of prime factors in exponential form:

Q3. Which is greater?

Q4. Simplify using laws of exponents:

Q5. Express in standard form:

Exercise 3.2 — Unit Digits

Q1. Find the unit digit of the following:

Q2. Find the unit digit:

Q3. Find the unit digit:

Q4. Find the unit digit of 1549<sup>101</sup> + 6541<sup>20</sup>

Exercise 3.3 — Degree and Algebraic Expressions

Q1. Find the degree of each expression:

Q2. Identify like and unlike terms:

Q3. Subtract and find the degree: (3ab &minus; 2a<sup>2</sup> &minus; 2b<sup>2</sup>) &minus; (5a<sup>2</sup> &minus; 7ab + 5b<sup>2</sup>)

Q4. Add and find the degree: (x<sup>2</sup> &minus; y<sup>2</sup> &minus; 1) + (y<sup>2</sup> &minus; 1 &minus; x<sup>2</sup>) + (1 &minus; x<sup>2</sup> &minus; y<sup>2</sup>)

Q5. Find the degree of each term:

Multiple Choice Questions (MCQs)

True or False

Fill in the Blanks

Common Mistakes

Exam Checklist

Frequently Asked Questions

Internal Links

Other question types — same chapter

Tamil Nadu Class 7 Maths: Algebra

Samacheer Kalvi 7th Maths (Term 1) Direct and Inverse Proportion Exercise Solutions

Samacheer Kalvi 7th Maths (Term 3) Percentage and Simple Interest Exercise Solutions

Q3. Subtract and find the degree: (3ab − 2a<sup>2</sup> − 2b<sup>2</sup>) − (5a<sup>2</sup> − 7ab + 5b<sup>2</sup>)

Q4. Add and find the degree: (x<sup>2</sup> − y<sup>2</sup> − 1) + (y<sup>2</sup> − 1 − x<sup>2</sup>) + (1 − x<sup>2</sup> − y<sup>2</sup>)