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

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

Exercise 3.1

Question 1.
Simplify the following and write the answer in Exponential form.
(i) 32 × 34 × 38
(ii) 615 ÷ 610
(iii) a3 × a2
(iv) 7x × 72
(v) (52)3 ÷ 53
Solution:
(i) 32 × 34 × 32 = 32+4+8 = 314 [∵ am × an = am+n]
So 32 × 34 × 32 = 314
(ii) 615 ÷ 610 = 615-10 = 65 [∵ am × an = am-n]
(iii) a3 × a2 = a3+2 = a5
(iv) 7x × 72 = 7x+2
(v) (52)3 ÷ 53 = 52×3 ÷ 53 = 56 ÷ 53 [∵ (am)n = am×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)
= 28 × 34
Thus 108 × 192 = 28 × 34

(ii) 270
We have 270 = 2 × 3 × 3 × 3 × 5

= 21 × 33 × 51
= 2 × 33 × 5
So 270 = 2 × 33 × 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)
= 36 × 26
∴ 729 × 64 = 36 × 26

(iv) 768
We have 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

= 28 × 31
= 28 × 3
Thus 768 = 28 × 3

Question 3.
Identify the greater number, wherever possible in each of the following.
(i) 53 or 35
(ii) 28 or 82
(iii) 1002 or 21000
(iv) 210 or 102
Solution:
(i) 53 or 35 53 = 5 × 5 × 5 = 125
35 = 3 × 3 × 3 × 3 × 3 = 243
243 > 125 ∴ 35 > 53

(ii) 28 or 82 28 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256
82 = 8 × 8 = 64
256 > 64 ∴ 28 > 82

(iii) 1002 or 21000
We have 1002 = 100 × 100 = 10000
2100 = (210)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) > 1002
(210)10 > 1002
2100 > 1002

(iv) 210 or 102
We have 210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024
102 = 10 × 10 = 100
Since1 1024 > 100
210 > 102

Exercise 3.2

Question 1.
Find the unit digit of the following exponential numbers.
(i) 255223
(ii) 81111000
(iii) 4866431
Solution:
(i) 255223
Unit digit of base 255 is 5 and power is 223.
Thus the unit digit of 255223 is 5.

(ii) 81111000
Unit digit of base 8111 is 1 and power is 1000.
Thus the unit digit of 81111000 is 1.

(iii) 4866431
Unit digit of base 4866 is 6 and power is 431.
Thus the unit digit of 4866431 is 6.

Question 2.
Find the unit digit of the numbers
(i) 1844671
(ii) 1564100
Solution:
(i) 1844671
Unit digit of base 1844 is 4 and the power is 671 (odd power)
Therefore unit digit of 1844671 is 4

(ii) 1564100
Unit digit of base 1564 is 4 and the power is 100 (even power)
Therefore unit digit of 1564100 is 6.

Question 3.
Find the unit digit of the numbers
(i) 999222
(ii) 1549777
Solution:
(i) 999222
Unit digit of base 999 is 9 and the power is 222 (even power).
Therefore, unit digit of 999222 is 1.

(ii) 1549777
Unit digit of base 1549 is 9 and the power is 777 (odd power).
Therefore, unit digit of 1549777 is 9.

Question 4.
Find the unit digit of 1549101 + 654120
Solution:
1549101 + 654120
In 1549101, the unit digit of base 1549 is 9 and power is 101 (odd power).
Therefore, unit digit of the 1549101 is 9.
In 654120, the unit digit of base 6541 is 1 and power is 20 (even power).
Therefore, unit digit of the 654120 is 1.
∴ Unit digit of 1549101 + 654120 is 9 + 1 = 10
∴ Unit digit of 1549101 + 654120 is 0.

Exercise 3.3

Question 1.
Find the degree of the following polynomials.
(i) x5 – x4 + 3
(ii) 2 – y5 – y3 + 2y8
(iii) 2
(iv) 5x3 + 4x2 + 7x
(v) 4xy + 7x2y + 3xy3
Solution:
(i) x5 – x4 + 3
The terms of the given expression are x5, -x4, 3.
Degree of each of the terms : 5, 4, 0
Term with highest degree: x5
Therefore degree of the expression in 5.

(ii) 2 – y5 – y3 + 2y8
The terms of the given expression are 2, -y5 ,-y3, 2y8.
Degree of each of the terms : 0, 2, 3, 8.
Term with highest degree: 2y8
Therefore degree of the expression in 8.

(iii) 2
Degree of the constant term is 0.
∴ Degree of 2 is 0.

(iv) 5x3+ 4x2 + 7x
The terms of the given expression are 5x3, 4x2, 7x
Degree of each of the terms : 3, 2, 1
Term with highest degree: 5x3
Therefore degree of the expression in 3.

(v) 4xy + 7x2y + 3xy3
The terms of the given expression are 4xy , 7x2y, 3xy3
Degree of each of the terms : 2, 3, 4
Term with highest degree: 3xy3
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) 4m2p, 4mp2
(iv) 12xz, 12x2z2
Solution:
(i) 1, 100 is a pair of like terms. [∵ 1 = x0 and 100 = 100 x0]
(ii) -7x, $$\frac { 5 }{ 2 }$$x is a pair of like terms.
(iii) 4m2p, Amp is a pair of unlike terms.
(iv) 12xz, 12x2z2 is a pair of unlike terms.

Question 3.
Subtract 5a2 – 7ab + 5b2 from 3ab – 2a2 – 2b2 and find the degree of the expression
Solution:
(i) We have (3ab – 2a2 – 2b2) – (5a2 – 7ab + 5b2)
= 3ab – 2a2 – 2b2 – 5a2 + 7ab – 5b2
= (3ab + 7ab) + (- 2 – 5)a2 + (- 2 – 5)b2
= 10 ab + (-7)a2 + (-7)b2
= 10ab – 7a2 – 7b2
Degree of the expression is 2.

Question 4.
Add x2 – y2 -1,y2 – 1 – x2, 1 – x2 – y2 and find the degree of the expression.
Solution:
We have (x2 – y2 – 1) + (y2 – 1 – x2) + (1 – x2 – y2)
= x2 – y2 – 1 + y2 – 1 – x2 + 1 – x2 – y2
= (x2 – x2 – x2) + (-y2 + y2 – y2) + (- 1 – 1 + 1)
= (1 – 1 – 1)x2 + (- 1 + 1 – 1)y2 + ( – 2 + 1)
= (- 1) x2 + (- 1)y2 + (-1) = – x2 – y2 – 1
Degree of the expression is 2.

Question 5.
Find the degree of the terms
(i) x2
(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) x2
The exponent in x2 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 $$\frac{7 x^{2} y^{4}}{x y}$$ = 7x2-1 y4-1 = 7x1y3 [Since $$\frac{a^{m}}{a^{n}}$$ = am-n]
In 7 x1 y3 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 $$\frac{x^{2} \times y^{2}}{x \times y^{2}}$$ = x2-1 y2-2 = x1 y0 = x1 [∵ y0 = 1]
The exponent of the expression is 1.