Samacheer Class 7 Maths - Algebra Intext Questions

1. 2 3 × 2 5 = 2 3+5 = 2 8 [since a m × a n = a m+n ]
2. p 2 × p 4 = p 2+4 = p 6 [since a m × a n = a m+n ]
3. x 6 × x 4 = x 6 + 4 = x 10 [since a m × a n = a m+n ]
4. 3 1 × 3 5 × 3 4 = 3 1+5 × 3 4 [since a m × a n = a m+n ]
= 3 6 × 3 4 [since a m × a n = a m+n ]
= 3 10
5. (-1) 2 × (-1) 3 × (-1) 5
= (-1) 2+3 × (-1) 5 [Since a m × a n = a m+n ]
= (-1) 5 × (-1) 5
= (-1) 5+5 [Since a m × a n = a m+n ]
= (-1) 10

Try These (Text book Page No. 48)

1. (3 2 ) 3 = 3 2×3 = 3 6 [since (a m ) n = a m×n ]
2. [(-5)] 2 = (-5) 3×2 = (-5) 6 [since (a m ) n = a m×n ]
3. (20 6 ) 2 = 20 6×2 = 20 12 [since (a m ) n = a m×n ]
4. (10 3 ) 5 = 10 3×5 = 10 15 [since (a m ) n = a m×n ]

(i) 5 2 × 3 2 = (5 × 3) 2 = 15 2 [since a m × b m = (a × b) m ]
(ii) x 3 × y 3 = (x × y) 3 = (x y) 3
(iii) 7 4 × 8 4 = (7 × 8) 4 = 56 4

(i) 5 3 ÷ 2 3 = (\(\frac { 5 }{ 2 } \)) 3 – [Since \(\frac{a^{m}}{b^{m}}\) = (\(\frac { a }{ b } \)) m ]
(ii) (-2) 4 ÷ 3 4 = (\(\frac { -2 }{ 3 } \)) 4
(iii) 8 6 ÷ 5 6 = (\(\frac { 8 }{ 6 } \)) 6
(iv) 6 3 ÷ (-7) 3 = (\(\frac { 6 }{ -7 } \)) 3

Exercise 3.2
Try These (Text book Page No. 54)

(i) 106 21 Unit digit of base 106 is 6 and the power is 21 and is positive.
Thus the unit digit of 106 21 is 6.
(ii) 25 8 Unit digit of base 25 is 5 and the power is 8 and is positive.
Thus the unit digit of 25 8 is 5.
(iii) 31 18 Unit digit of base 31 is 1 and the power 18 and is positive.
Thus the unit digit of 31 18 is 1.
(iv) 20 10 Unit digit of base 20 is 0 and the power 10 and is positive.
Thus the unit digit of 20 10 is 0.
Try These (Text book Page No. 55)

(i) 64 11 Unit digit of base 64 is 4 and the power is 11 (odd power).
∴ Unit digit of 64 11 is 4.
(ii) 29 18 Unit digit of base 29 is 9 and the power is 18 (even power).
Therefore, unit digit of 29 18 is 1.
(iii) 79 19 Unit digit of base 79 is 9 and the power is 19 (odd power).
Therefore, unit digit of 7919 is 9.
(iv) 104 32 Unit digit of base 104 is 4 and the power is 32 (even power).
Therefore, unit digit of 104 32 is 6.

Exercise 3.3
Try These (Text book Page No. 35)

(i) 512

512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 2 9 [Using product rule]
(ii) 343

343 = 7 × 7 × 7 = 7 1+1+1
= 7 3 [Using product rule]
(iii) 729

729 = 3 × 3 × 3 × 3 × 3 × 3
= 3 6 [Using product rule]
(iv) 3125

3125 = 5 × 5 × 5 × 5 × 5
= 5 5 [Using product rule]

(i) 6 3 or 3 6
6 3 = 6 × 6 × 6 = 36 × 6 = 216
3 6 = 3 × 3 × 3 × 3 × 3 × 3 = 729
729 > 216 gives 3 6 > 6 3
∴ 36 is greater.
(ii) 5 3 or 3 5
5 3 = 5 × 5 × 5 = 125
3 5 = 3 × 3 × 3 × 3 × 3 = 243
243 > 125 gives 3 5 > 5 3
∴ 3 5 is greater.
(iii) 2 8 or 8 2
2 8 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256
8 2 = 8 × 8 = 64
256 > 64 gives 2 8 > 8 2
∴ 2 8 is greater.

(i) 7 2 × 3 4 = (7 × 7) × (3 × 3 × 3 × 3)
= 49 × 81 = 3969
(ii) 3 2 × 2 4 = (3 × 3) × (2 × 2 × 2 × 2)
= 9 × 16 = 144
(iii) 5 2 × 10 4 = (5 × 5) × (10 × 10 × 10 × 10)
= 25 × 10000 = 2,50,000

(i) (-4) 2 = (-1) 2 × (4) 2 [since a m × b m = (a × b) m ]
= 1 × 16 = 16 [since (-1) n = 1 if n is even]
(ii) (-3) × (-2) 3 = (-1) × (-3) × (-1) 3 × (-2) 3
= (-1) 4 × 24 [Grouping the terms of same base]
= 24
(iii) (-2) 3 × (-10) 3 = (-1) 3 × (-2) 3 × (-1) 3 × (-10) 3
= (-1) 3+3 × 2 3 × 10 3 [Grouping the terms of same base]
= (-1) 6 × (2 × 10) 3
[∵ a m × b m = (a × b) m ]
= 1 × 20 3 [since (-1) n = 1 if n is even]
= 8000

(i) a b + b a
a = 3 and b = 2
we get 3 2 + 2 3 = (3 × 3) + (2 × 2 × 2) = 9 + 8 = 17
(ii) (a a – b b )
Substituting a = 3 and b = 2
we get 3 2 – 2 2 = (3 × 3 × 3) – (2 × 2) = 27 – 4 = 23
(iii) (a + b) b
Substituting a = 3 and b = 2
we get (3 + 2) 2 = 5 2 = 5 × 5 = 25
(iv) (a – b) a
Substituting a = 3 and b = 2
we get (3 – 2) 3 = 1 3 = 1 × 1 × 1 = 1

(i) 25 23
Unit digit of base 25 is 5 and power is 23. Thus the unit digit of 25 23 is 5.
(ii) 11 10
Unit digit of base 11 is 1 and power is 10. Thus the unit digit of 11 10 is 1.
(iii) 46 15
Unit digit of base 46 is 6 and power is 15. Thus the unit digit of 46 15 is 6.
(iv) 100 12
Unit digit of base 100 is 0 and power is 12. Thus the unit digit of 100 12 is 0.
(v) 29 21
Unit digit of base 29 is 9 and power is 21 (odd power).
Therefore, unit digit of 29 21 is 9.
(vi) 19 12
Unit digit of base 19 is 9 and power is 12 (even power).
Therefore, unit digit of 19 12 is 1.
(vii) 24 25
Unit digit of base 24 is 4 and power is 25 (odd power).
Therefore, unit digit of 24 25 is 4.
(viii) 34 16
Unit digit of base 34 is 4 and power is 16 (even power).
Therefore, unit digit of 34 16 is 6.

(i) 114 20 + 115 21 + 116 22
In 114 20 unit digit of base 114 is 4 and power is 20 (even power).
∴ Unit digit of 114 20 is 6.
In 115 21 unit digit of base 115 is 5 and power is 21 (Positive Integer).
∴ Unit digit of 115 21 is 5.
In 116 22 unit digit of base 116 is 6 and power is 22 (Positive Integer).
∴ Unit digit of 116 22 is 6.
∴ Unit digit of 114 20 + 115 21 + 116 22 can be obtained by adding 6 + 5 + 6 = 17.
Unit digit of 114 20 + 115 21 + 116 22 is 7.
(ii) 10000 10000 + 11111 11111
In 10000 10000 the unit digit of base 10000 is 0 and power is 10000.
Unit digit of 10000 10000 is 0.
In 11111 11111 the unit digit of base 11111 is 1 and power is 11111.
Unit digit of 11111 11111 is 1.
Unit digit of 10000 100000 + 11111 11111 is 0 + 1 = 1
Objective Type Question

(i) 5x 2
In 5x 2 , the exponent is 2. Thus the degree of the expression is 2.
(ii) -7ab
In -7ab, the sum of powers of a and b is 2. (That is 1 + 1 = 2).
Thus the degree of the expression is 2.
(iii) 12pq 2 r 2
In 12pq 2 r 2 , the sum of powers of p, q and r is 5. (That is 1 +2 + 2 = 5).
Thus the degree of the expression is 5.
(iv) -125
Here – 125 is the constant term. Degree of constant term is 0.
∴ Degree of -125 is 0.
(v) 3z
The exponent is 3z is 1.
Thus the degree of the expression is 1.

6 2 × 6 m = 6 5
6 2+m = 6 5 [Since a m × a n = a m+n ]
Equating the powers, we get
2 + m = 5
m = 5 – 2 = 3

In 124 128 , the unit digit of base 124 is 4 and the power is 128 (even power).
Therefore, unit digit of 124 128 is 4.
Also in 126 124 , the unit digit of base 126 is 6 and the. power is 124 (even power).
Therefore, unit digit of 126 124 is 6.
Product of the unit digits = 6 × 6 = 36
∴ Unit digit of the 124 128 × 126 124 is 6.

In 16 23 , the unit digit of base 16 is 6 and the power is 23 (odd power).
Therefore, unit digit of 16 23 is 6.
In 71 48 , the unit digit of base 71 is 1 and the power is 48 (even power).
Therefore, unit digit of 71 48 is 1.
Also in 59 61 , the unit digit of base 59 is 9 and the power is 61 (odd power).
Therefore, unit digit of 59 61 is 9.
Sum of the unit digits = 6 + 1 + 9 = 16
∴ Unit digit of the given expression is 6.

The terms of the given expression are 2a 3 bc, 3a 3 b + 3a 3 c – 2a 2 b 2 c 2
Degree of each of the terms: 5,4,4,6.
Terms with the highest degree: – 2a 2 b 2 c 2
Therefore degree of the expression is 6.

Members of the groups LEADERS = 256
Members is individual groups of the members of LEADERS = 256
Total members who receive the message

= 256 × 256 = 2 8 × 2 8
2 8+8 = 2 16
= 65536
Totally 65536 members receive the message.

Given x = 3; y = 4 and z = -2.
w = x 2 – y 2 + z 2 – xyz
w = 3 2 – 4 2 + (-2) 2 – (3)(3)(-2)
w = 9 – 16 + 4 + 24
w = 37 – 16
w = 21

Let the two adjacent sides of the rectangle as
l = 2x 2 – 5xy + 3z 2 and b = 4xy – x 2 y + 3z 2

Perimeter of the rectangle
= 2(l + b) = 2(2x 2 – 5xy + 3z 2 + 4xy – x 2 – z 2 )
= 4x 2 – 10xy + 6z 2 + 8xy – 2x 2 – 2z 2
= 4x 2 – 2x 2 – 10xy + 8xy + 6z 2 – 2z 2
= x 2 (4 – 2) + xy (-10 + 8) + z 2 (6 – 2z 2 )
Perimeter = 2x 2 – 2xy + 4z 2
Degree of the expression is 2.
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