# Samacheer Kalvi Class 7 Maths Term 2 Chapter 3 Solutions

## Samacheer Kalvi Class 7 Maths Term 2 Chapter 3 Algebra Solutions

Welcome to NCTB Solutions. Here with this post we are going to help 7th class students by providing Solutions for Samacheer Kalvi Class 7 Maths Term 2 chapter 3 Algebra. Here students can easily find all the solutions for Algebra Exercise 3.1, 3.2, 3.3 and 3.4. Also here our Expert Maths Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 3 solutions. Here all the solutions are based on Tamil Nadu State Board latest syllabus.

#### Algebra Exercise 3.1 Solutions :

(1) Fill in the blanks.

(i) The exponential form 149 should be read as ____.

(ii) The expanded form of p3q2 is ____.

(iii) When base is 12 and exponent is 17, its exponential form is ___.

(iv) The value of (14  × 21)0 is ____.

Solution :

(i) 14 power 9

(ii) p × p × p × q × q

(iii) 1217

(iv) 1

(2) Say true or false

(i) 23 × 32 = 65

(ii) 29 × 32 = (2 × 3)9 × 2

(iii) 34 × 37 = 311

(iv) 20 = (1000)0

(v) 23 < 32

Solution :

(i) → True

Because, in multiplication power is plus.

(ii) → False

This is because in multiplication power is plus.

(iii)  → True,

Because in multiplication power is plus.

(iv) → True,

This is because when the power is 0 of any number, then the answer is 1.

(4) Express the following in exponential form

(i) 6 × 6 × 6 × 6

(ii) t × t

(iii) 5 × 5 × 7 × 7 × 7

(iv) 2 × 2 × a × a

Solution :

(i) 6×6×6×6 = 61+1+1+1 = 6⁴

(ii) t×t = t²

(iii) 5×5×7×7×7 = 5² ×7³

(iv) 2×2 × a×a = 2² × a² = (2×a)²

(5) Express each of the following numbers using exponential form.

(i) 512

(ii) 343

(iii) 729

(iv) 3125

Solution :

(i) 512 = 8 × 64

= 2³ × 8²

= 2³ × (2³) ²

= 2³ × 2⁶

= 2⁹

(ii) 343 = 7 × 49

= 7×7²

= 7³

(iii)729 = 9 × 81

= 3² × 9²

= 3² × (3²) ²

= 3²×3⁴

=3⁶

(iv) 3125 = 5 × 625

= 5 × 25²

= 5 × (5²)²

= 5 × 5⁴

= 5⁵

(8) Find the value of the following

(i) (-4)²

(ii) (–3) × (–2)³

(iii) (-2)³ × (-10)³

Solution :

(i) (-4)²

= – 4 × – 4

= 16

(ii) (–3) × (–2)³

= (–3) × (–2)³ = (–3) × (–2)× (–2)²

= 6 × 4

= 24

(iii) (-2)³ × (-10)³

= (-2)³ × (-10)³ = (-2) × (-2)² × (-10) × (-10) ²

= 20 × 4 × 100

= 80 × 100

= 8000

(10) If a = 3 and b = 2, then find the value of the following

(i) a^b + b^a

(ii) a^a – b^b

(iii) (a+b)^b

(iv) (a-b)^a

Solution :

(i) a^b + b^a

Given, a = 3 and b = 2

= a^b + b^a

= 3² +2³

= 9 + 8

= 17

(ii) a^a – b^b

a^a – b^b = 3³ – 2²

= 27 – 4

= 23

(iii) (a+b)^b

= (a + b) ^b = (3 + 2)²

= 5²

= 2⁵

(iv) (a-b)^a

= (a-b)^a

= (3-2)³

= 1³

= 1

(11) Simplify and express each of the following in exponential form

(i) 4⁵ × 4² × 4⁴

(ii) (3² × 3³)⁷

(iii) (5² × 5⁸) ÷ 5⁵

(iv) 2°× 3° ×4°

(v) 4⁵ × a⁸ × b³/4³ × a⁵ × b²

Solution :

(i) 45 × 4² × 4⁴

= 4⁵ × 4² × 4⁴

= 4^5+2+4

= 4^11

(ii) (3² × 3³)⁷

= (3² × 3³)⁷

= (3^2+3)7

= (3⁵)7

= 3^35

(iii) (5² × 5⁸) ÷ 55

= (5² × 5⁸) ÷ 5⁵

= (5^2+8) ÷ 5⁵

= 5^10 ÷ 5⁵

= 5^10-5

= 5⁵

(iv) 2° × 3° ×4°

= 2° × 3° ×4°

= 1 × 1 × 1

= 1

(v) 4⁵ × a⁸ × b³/4³ × a⁵ × b²

= 4⁵ × a⁸ × b³/4³ × a⁵ × b²

= 4^5-3 × a^8-5 × b^3-2

= 4² × a³ × b

= 16a³b

Objective Type Question Solutions :

(12) a × a × a × a × a is equal to

(i) a⁵

(ii) 5^a

(iii) 5a

(iv) a + 5

Solution :

Correct Option → (i)

a × a × a × a × a is equal to a⁵.

(13) The exponential form of 72 is

(i)72

(ii) 27

(iii) 22 × 33

(iv) 23 × 32

Solution :

Correct Option → (iv)

The exponential form of 72 is 2³ × 3²

(15) How many zeros are there in 100^10

(i) 2

(ii) 3

(iii) 10

(iv) 20

Solution :

Correct Option → (iv)

There will be 20 zeros.

(16) 2^40 + 2^40 is equal to

(i) 4^40

(ii) 2^80

(iii) 2^41

(iv) 4^80

Solution :

Correct Option → (iii)

2^40 + 2^40 is equal to 2^41.

#### Algebra Exercise 3.2 Solutions :

(1) Fill in the blanks.

(i) Unit digit of 124 ×  36 × 980 is ___.

(ii) When the unit digit of the base and its expanded form of that number is 9, then the exponent must be ____ power.

Solution :

(i) 0

(ii) odd

(2) Match the following:

 Group A Exponential form Group B  Unit digit of the number (i) 20^10 (a) 6 (ii) 121^11 (b) 4 (iii) 444^41 (c) 0 (iv) 25^100 (d) 1 (v) 716^83 (e) 9 (vi) 729^725 (f) 5

Solution :

(i) → (c) 0

(ii) → (d) 1

(iii) → (b) 4

(iv) → (f) 5

(v) → (a) 6

(vi) → (e) 9

Objective Type Question Solutions :

(5) Observe the equation (10 + y)⁴ = 50625 and find the value of y

(i) 1

(ii) 5

(iii) 4

(iv) 0

Solution :

Correct Option → (ii)

The value of y will be 5.

(6) The unit digit of (32 × 65)° is

(i) 2

(ii) 5

(iii) 0

(iv) 1

Solution :

Correct Option → (iv)

The unit digit of (32 × 65)° is 1

(7) The unit digit of the numeric expression 10^71 + 10^72 + 10^73 is

(i) 0

(ii) 3

(iii) 1

(iv) 2

Solution :

Correct Option → (i)

The unit digit of the numeric expression 10^71 + 10^72 + 10^73 is 0.

#### Algebra Exercise 3.3 Solutions :

(1) Fill in the blanks.

(i) The degree of the term a3b2c4d2 is ___.

(ii) Degree of the constant term is ___.

(iii) The coefficient of leading term of the expression 3z2y + 2x – 3 is ___.

Solution :

(i) → 11

(ii) → 3

(2) Say True or False.

(i) The degree of m²n and mn² are equal

(ii) 7a2b and -7ab2 are like terms

(iii) The degree of the expression -4x2yz is – 4

(iv) Any integer can be the degree of the expression.

Solution :

(i) → True

(ii) → False

(iii) → False

(iv) → True

(4) Find the degree of the following expressions

(i) x³ – 1

(ii) 3x²+ 2x+1

(iii) 3t⁴ – 5st² + 7s³t²

(iv) 5 – 9y + 15y² – 6y³

(v) u⁵ + u⁴v + u³v² + u²v³ + uv⁴

Solution :

(i) x³ – 1

The terms of the given expression are x³, -1

Degree of each of the terms : 3, 0

Terms with highest degree : x³

Therefore, degree of the expression is 3.

(ii) 3x² + 2x + 1

The terms of the given expression are 3x², 2x, 1

Degree of each of the terms : 2, 1, 0

Terms with highest degree : 3x²

Therefore, degree of the expression is 2.

(iii) 3t⁴ – 5st² + 7s³t²

The terms of the given expression are 3t⁴, -5st² , 7s³t²

Degree of each of the terms : 4, 3, 5

Terms with highest degree : 7s³t²

Therefore, degree of the expression is 5.

(iv) 5 – 9y + 15y² – 6y³

The terms of the given expression are 5, -9y , 15y², -6y³

Degree of each of the terms : 0, 1, 2, 3

Terms with highest degree : -6y³

Therefore, degree of the expression is 3.

(v) u⁵ + u⁴v + u³v² + u²v³ + uv⁴

The terms of the given expression are u⁵, u⁴v, u³v², u²v³, uv⁴

Degree of each of the terms : 5, 5, 5, 5, 5

Terms with highest degree : u⁵, u⁴v, u³v², u²v³, uv⁴

Therefore, degree of the expression is 5.

(6) Add and find the degree of the following expressions

(i) (9x+3y) and (10x – 9y)

(ii) (k² – 25k + 46) and (23 – 2k²+ 21k)

(iii) (3m²n + 4pq²) and ( 5nm²- 2q²p)

Solution :

(i) (9x + 3y) + (10x – 9y)

= 9x + 10 x + 3y – 9y

= 19x – 6y

Hence, the degree of the expression is 1.

(ii) (k² – 25k + 46) + (23 – 2k² + 21k)

= k² – 2k² – 25k + 21k + 46 + 23

= -k² – 4k + 69

Hence, the degree of the expression is 2.

(iii) (3m²n + 4pq²) + ( 5nm² – 2q²p)

= 3m²n + 5nm² + 4pq² – 2q²p

= 8nm² + 2pq²

Hence, the degree of the expression is 3.

(7) Simplify and find the degree of the following expressions :

(i) 10x² – 3xy + 9y² – (3x² – 6xy – 3y²)

(ii) 9a⁴ – 6a³ – 6a⁴ – 3a² + 7a3 + 5a²

(iii) 4x² – 3x – [8x -(5x² -8)]

Solution :

(i) 10x² – 3xy + 9y² – (3x² – 6xy – 3y²)

= 10x² – 3xy + 9y² – 3x² + 6xy + 3y²

= 7x² + 3xy + 12y²

Hence, the degree of the expression is 2.

(ii) 9a⁴ – 6a³ – 6a⁴ – 3a² + 7a³ + 5a²

= 9a⁴ – 6a⁴ -6a³ +7a³ -3a² + 5a²

= 3a⁴ + a³ + 2a²

Hence, the degree of the expression is 4.

(iii) 4x² – 3x – [8x -(5x² -8) ]

= 4x² – 3x – 8x + 5x² – 8

= 9x² -11x -8

Hence, the degree of the expression is 2.

Objective Type Question Solutions :

(9) 3p² – 5pq + 2q² + 6pq – q² + pq is a

(i) Monomial

(ii) Binomial

(iii) Trinomial

Solution :

Correct Option → (iii)

3p² – 5pq + 2q² + 6pq – q² + pq is a Trinomial.

(10) If p(x) and q(x) are two expressions of degree 3, then the degree of p(x) + q(x) is

(i) 6

(ii) 0

(iii) 3

(iv) Undefined

Solution :

Correct Option → (iii)

Then the degree of p(x) + q(x) is 3

#### Algebra Exercise 3.4 Solutions :

(1) 6² × 6^m = 6⁵, find the value of m

Solution :

Given, 6² × 6^m = ^65

We know a^m × a^n = a^m + n

Thus, 2 + m = 5

m = 5 – 2 = 3

m = 3

∴ The value of m will be 3

(4) Find the value of (-1)⁶ × (-1)⁷ × (-1)⁸ / (-1)³ × (-1)⁵

Solution :

(-1)⁶ × (-1)⁷ × (-1)⁸/(-1)³ × (-1)⁵

= (-1)^6 + 7 + 8/(-1)^3 + 5

= (-1)^21/(-1)⁸

= (-1)^21 – 8

= (-1)^13

= -1

Therefore, the value will be -1.

(5) Identify the degree of expression, 2a³bc + 3a³b + 3a³c – 2a²b²c²

Solution :

Given, 2a³bc + 3a³b + 3a³c – 2a²b²c²

The terms of the given expression are 2a³bc, 3a³b , 3a³c, – 2a²b²c²

Degree of each of the terms : 5, 4, 4,6

Terms with highest degree : – 2a²b²c²

Therefore, degree of the expression is 6.

(6) If p = -2, q = 1 and r=3, find the value of 3p²q²r

Solution :

Given, p = -2 , q = 1 and r = 3

Now, 3p²q²r = 3 × (-2)² × 1² × 3

= 3 × 4 × 1 × 3

= 12 × 3

= 36

Therefore, the value is 36.

Challenge Problem Solutions :

(8) Find x such that 3^x+2 = 3x + 216

Solution :

3^x + 2 = 3^x + 216

3^x + 2 = 3^x + 63

3^x + 2 = 3^x + (32)^3

(9) If X = 5x² + 7x + 8 and Y = 4x² – 7x + 3, then find the degree of X + Y

Solution :

Given, X = 5x² + 7x + 8 and Y = 4x² – 7x + 3

Now, X + Y = 5x² + 7x + 8 + 4x² – 7x + 3

= 9x² +11

Therefore, the degree of X + Y is 2.

(10) Find the degree of (2a² + 3ab – b²) – (3a² – ab – 3b²)

Solution :

(2a² + 3ab – b²) – (3a² – ab – 3b²)

= 2a² + 3ab – b² – 3a² + ab + 3b²

= -a² + 4ab + 2b²

Hence, the degree of expression is 2

(11) Find the value of w, given that x =3, y =4, z =−2 and w = x² – y² +z² – xyz

Solution :

Given, x = 3, y = 4, z = −2

Now, w = x² – y² + z² – xyz

w = 3² – 4² + (-2)² – (3 × 4 × (-2))

w = 9 – 16 + 4 + 24

w = 21

Therefore, the value of w will be 21.

(12) Simplify and find the degree of 6x² + 1 – [ 8x – {3x² – 7 – (4x² – 2x + 5x + 9) }]

Solution :

6x² + 1 – [ 8x – {3x² – 7 – (4x² – 2x + 5x + 9) }]

= 6x² + 1 – [ 8x – {3x² – 7 -4x² – 3x – 9 }]

= 6x² + 1 – [ 8x – { -x² – 3x – 16}]

= 6x² + 1 – [ 8x + x² + 3x + 16 ]

= 6x² + 1 – 11x – x² – 16

= 5x² – 11x – 15

Hence, the degree of expression is 2.

(13) The two adjacent sides of a rectangle are 2x² – 5xy + 3z² and 4xy – x² – z² . find the perimeter and the degree of the expression.

Solution :

Given, The two adjacent sides of a rectangle are 2x² – 5xy + 3z² and 4xy -x² -z² .

Now, perimeter of rectangle = 2 [2x² – 5xy + 3z² + 4xy -x² -z² ]

= 2 [x² – xy + 2z²]

= 2x² – 2xy + 4z²

Hence, the degree of expression is 2.

Next Chapter Solution :

👉 Geometry

Updated: July 28, 2023 — 3:58 pm