Students can Download Maths Chapter 3 Algebra Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 3 Algebra Additional Questions

Additional Questions and Answers

Exercise 3.1

Question 1.

Simplify the following and write the answer in Exponential form.

(i) 3^{2} × 3^{4} × 3^{8}

(ii) 6^{15} ÷ 6^{10}

(iii) a^{3} × a^{2}

(iv) 7^{x} × 7^{2}

(v) (5^{2})^{3} ÷ 5^{3}

Solution:

(i) 3^{2} × 3^{4} × 3^{2} = 3^{2+4+8} = 3^{14} [∵ a^{m} × a^{n} = a^{m+n}]

So 3^{2} × 3^{4} × 3^{2} = 3^{14}

(ii) 6^{15} ÷ 6^{10} = 6^{15-10} = 6^{5} [∵ a^{m} × a^{n} = a^{m-n}]

(iii) a^{3} × a^{2} = a^{3+2} = a^{5}

(iv) 7^{x} × 7^{2} = 7^{x+2}

(v) (5^{2})^{3} ÷ 5^{3} = 5^{2×3} ÷ 5^{3} = 5^{6} ÷ 5^{3} [∵ (a^{m})^{n} = a^{m×n}]

Question 2.

Express the following as a product of factors only in exponential form

(i) 108 × 192

(ii) 270

(iii) 729 × 64

(iv) 768

Solution:

(i) 108 × 192

108 = 2 × 2 × 3 × 3 × 3

192 = 2 × 2 × 2 × 2 × 2 × 2 × 3

∴ 108 × 192 = (2 × 2 × 3 × 3 × 3) ×

(2 × 2 × 2 × 2 × 2 × 2 × 3)

= 2^{8} × 3^{4}

Thus 108 × 192 = 2^{8} × 3^{4}

(ii) 270

We have 270 = 2 × 3 × 3 × 3 × 5

= 2^{1} × 3^{3} × 5^{1}

= 2 × 3^{3} × 5

So 270 = 2 × 3^{3} × 5

(iii) 729 × 64

729 = 3 × 3 × 3 × 3 × 3 × 3

64 = 2 × 2 × 2 × 2 × 2 × 2

729 × 64 = (3 × 3 × 3 × 3 × 3 × 3)

× (2 × 2 × 2 × 2 × 2 × 2)

= 3^{6} × 2^{6}

∴ 729 × 64 = 3^{6} × 2^{6}

(iv) 768

We have 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

= 2^{8} × 3^{1}

= 2^{8} × 3

Thus 768 = 2^{8} × 3

Question 3.

Identify the greater number, wherever possible in each of the following.

(i) 5^{3} or 3^{5}

(ii) 2^{8} or 8^{2}

(iii) 100^{2} or 2^{1000}

(iv) 2^{10} or 10^{2}

Solution:

(i) 5^{3} or 3^{5} 5^{3} = 5 × 5 × 5 = 125

3^{5} = 3 × 3 × 3 × 3 × 3 = 243

243 > 125 ∴ 3^{5} > 5^{3}

(ii) 2^{8} or 8^{2} 2^{8} = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256

8^{2} = 8 × 8 = 64

256 > 64 ∴ 2^{8} > 8^{2}

(iii) 100^{2} or 2^{1000}

We have 100^{2} = 100 × 100 = 10000

2^{100} = (2^{10})^{10} = (2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2)^{10}

= (1024)^{10} = [(1024)^{2}]^{5}

= (1024 × 1024)^{5} = (1048576)^{5}

Since 1048576 > 10000

(1048576)^{5} > 10000

i.e., (1048576) > 100^{2}

(2^{10})^{10} > 100^{2}

2^{100} > 100^{2}

(iv) 2^{10} or 10^{2}

We have 2^{10} = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024

10^{2} = 10 × 10 = 100

Since1 1024 > 100

2^{10} > 10^{2}

Exercise 3.2

Question 1.

Find the unit digit of the following exponential numbers.

(i) 255^{223}

(ii) 8111^{1000}

(iii) 4866^{431}

Solution:

(i) 255^{223}

Unit digit of base 255 is 5 and power is 223.

Thus the unit digit of 255^{223} is 5.

(ii) 8111^{1000}

Unit digit of base 8111 is 1 and power is 1000.

Thus the unit digit of 8111^{1000} is 1.

(iii) 4866^{431}

Unit digit of base 4866 is 6 and power is 431.

Thus the unit digit of 4866^{431} is 6.

Question 2.

Find the unit digit of the numbers

(i) 1844^{671}

(ii) 1564^{100}

Solution:

(i) 1844^{671}

Unit digit of base 1844 is 4 and the power is 671 (odd power)

Therefore unit digit of 1844^{671} is 4

(ii) 1564^{100}

Unit digit of base 1564 is 4 and the power is 100 (even power)

Therefore unit digit of 1564^{100} is 6.

Question 3.

Find the unit digit of the numbers

(i) 999^{222}

(ii) 1549^{777}

Solution:

(i) 999^{222}

Unit digit of base 999 is 9 and the power is 222 (even power).

Therefore, unit digit of 999^{222} is 1.

(ii) 1549^{777}

Unit digit of base 1549 is 9 and the power is 777 (odd power).

Therefore, unit digit of 1549^{777} is 9.

Question 4.

Find the unit digit of 1549^{101} + 6541^{20}

Solution:

1549^{101} + 6541^{20}

In 1549^{101}, the unit digit of base 1549 is 9 and power is 101 (odd power).

Therefore, unit digit of the 1549^{101} is 9.

In 6541^{20}, the unit digit of base 6541 is 1 and power is 20 (even power).

Therefore, unit digit of the 6541^{20} is 1.

∴ Unit digit of 1549^{101} + 6541^{20} is 9 + 1 = 10

∴ Unit digit of 1549^{101} + 6541^{20} is 0.

Exercise 3.3

Question 1.

Find the degree of the following polynomials.

(i) x^{5} – x^{4} + 3

(ii) 2 – y^{5} – y^{3} + 2y^{8}

(iii) 2

(iv) 5x^{3} + 4x^{2} + 7x

(v) 4xy + 7x^{2}y + 3xy^{3}

Solution:

(i) x^{5} – x^{4} + 3

The terms of the given expression are x^{5}, -x^{4}, 3.

Degree of each of the terms : 5, 4, 0

Term with highest degree: x^{5}

Therefore degree of the expression in 5.

(ii) 2 – y^{5} – y^{3} + 2y^{8}

The terms of the given expression are 2, -y^{5} ,-y^{3}, 2y^{8}.

Degree of each of the terms : 0, 2, 3, 8.

Term with highest degree: 2y^{8}

Therefore degree of the expression in 8.

(iii) 2

Degree of the constant term is 0.

∴ Degree of 2 is 0.

(iv) 5x^{3}+ 4x^{2} + 7x

The terms of the given expression are 5x^{3}, 4x^{2}, 7x

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

Term with highest degree: 5x^{3}

Therefore degree of the expression in 3.

(v) 4xy + 7x^{2}y + 3xy^{3}

The terms of the given expression are 4xy , 7x^{2}y, 3xy^{3}

Degree of each of the terms : 2, 3, 4

Term with highest degree: 3xy^{3}

Therefore degree of the expression in 4.

Question 2.

State whether a given pair of terms in like or unlike terms.

(i) 1,100

(ii) -7x,\(\frac { 5 }{ 2 } \)x

(iii) 4m^{2}p, 4mp^{2}

(iv) 12xz, 12x^{2}z^{2}

Solution:

(i) 1, 100 is a pair of like terms. [∵ 1 = x^{0} and 100 = 100 x^{0}]

(ii) -7x, \(\frac { 5 }{ 2 } \)x is a pair of like terms.

(iii) 4m^{2}p, Amp is a pair of unlike terms.

(iv) 12xz, 12x^{2}z^{2} is a pair of unlike terms.

Question 3.

Subtract 5a^{2} – 7ab + 5b^{2} from 3ab – 2a^{2} – 2b^{2} and find the degree of the expression

Solution:

(i) We have (3ab – 2a^{2} – 2b^{2}) – (5a^{2} – 7ab + 5b^{2})

= 3ab – 2a^{2} – 2b^{2} – 5a^{2} + 7ab – 5b^{2}

= (3ab + 7ab) + (- 2 – 5)a^{2} + (- 2 – 5)b^{2}

= 10 ab + (-7)a^{2} + (-7)b^{2}

= 10ab – 7a^{2} – 7b^{2}

Degree of the expression is 2.

Question 4.

Add x^{2} – y^{2} -1,y^{2} – 1 – x^{2}, 1 – x^{2} – y^{2} and find the degree of the expression.

Solution:

We have (x^{2} – y^{2} – 1) + (y^{2} – 1 – x^{2}) + (1 – x^{2} – y^{2})

= x^{2} – y^{2} – 1 + y^{2} – 1 – x^{2} + 1 – x^{2} – y^{2}

= (x^{2} – x^{2} – x^{2}) + (-y^{2} + y^{2} – y^{2}) + (- 1 – 1 + 1)

= (1 – 1 – 1)x^{2} + (- 1 + 1 – 1)y^{2} + ( – 2 + 1)

= (- 1) x^{2} + (- 1)y^{2} + (-1) = – x^{2} – y^{2} – 1

Degree of the expression is 2.

Question 5.

Find the degree of the terms

(i) x^{2}

(ii) 4xyz

(iii) \(\frac{7 x^{2} y^{4}}{x y}\)

(iv) \(\frac{x^{2} \times y^{2}}{x \times y^{2}}\)

Solution:

We have

(i) x^{2}

The exponent in x^{2} is 2. ∴ Degree of the term is 2.

(ii) 4xyz

In 4xyz the sum of the powers of x, y and z as 3.

(iii) \(\frac{7 x^{2} y^{4}}{x y}\)

We have \(\frac{7 x^{2} y^{4}}{x y}\) = 7x^{2-1} y^{4-1} = 7x^{1}y^{3} [Since \(\frac{a^{m}}{a^{n}}\) = am-n]

In 7 x^{1 }y^{3} the sum of the poweres of x and y is 4(1 + 3 = 4)

Thus degree of the expression is 4.

(iv) \(\frac{x^{2} \times y^{2}}{x \times y^{2}}\)

We have \(\frac{x^{2} \times y^{2}}{x \times y^{2}}\) = x^{2-1} y^{2-2} = x^{1} y^{0} = x^{1} [∵ y^{0} = 1]

The exponent of the expression is 1.