Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 3 Algebra Ex 3.1
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 3 Algebra Ex 3.1
Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 3 Algebra Ex 3.1
Question 1.
Fill in the blanks.
- The exponential form 149 should be read as ______
- The expanded form of p3 q2 is ______
- When base is 12 and exponent is 17, its e×ponential form is _____
- The value of (14 × 21)0 is _____
Answers:
- 14 Power 9
- p × p × p × q × q
- 1217
- 1
Question 2.
Say True or False.
- 23 × 32 = 65
- 29 × 32 = (2 × 3)9×2
- 34 × 37= 311
- 20 × 10000
- 23 < 32
Answers:
- False
- False
- True
- True
- True
Question 3.
Find the value of the following.
- 26
- 112
- 54
- 93
Solution:
- 26 = 2 × 2 × 2 × 2 × 2 × 2 = 64
- 112 = 11 × 11 = 121
- 54 = 5 × 5 × 5 × 5 = 625
- 93 = 9 × 9 × 9 = 729
Question 4.
Express the following in e×ponential form.
- 6 × 6 × 6 × 6
- t × t
- 5 × 5 × 7 × 7 × 7
- 2 × 2 × a × a
Solution:
- 6 × 6 × 6 × 6 = 61+1+1+1 = 64 [Since am × an = am+n]
- t × t = t1+1 = t2
- 5 × 5 × 7 × 7 × 7 = 51+1 × 71+1+1 = 52 × 73
- 2 × 2 × a × a = 21+1 × a1+1 = 22 × a2 = (2a)2
Question 5.
E×press each of the following numbers using e×ponential form,
(i) 512
(ii) 343
(iii) 729
(iv) 3125
Solution:
(i) 512
512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 29 [Using product rule]
(ii) 343
343 = 7 × 7 × 7 = 71+1+1
= 73 [Using product rule]
(iii) 729
729 = 3 × 3 × 3 × 3 × 3 × 3
= 36 [Using product rule]
(iv) 3125
3125 = 5 × 5 × 5 × 5 × 5
= 55 [Using product rule]
Question 6.
Identify the greater number in each of the following.
(i) 63 or 36
(ii) 53 or 35
(iii) 28 or 82
Solution:
(i) 63 or 36
63 = 6 × 6 × 6 = 36 × 6 = 216
36 = 3 × 3 × 3 × 3 × 3 × 3 = 729
729 > 216 gives 36 > 63
∴ 36 is greater.
(ii) 53 or 35
53 = 5 × 5 × 5 = 125
35 = 3 × 3 × 3 × 3 × 3 = 243
243 > 125 gives 35 > 53
∴ 35 is greater.
(iii) 28 or 82
28 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256
82 = 8 × 8 = 64
256 > 64 gives 28 > 82
∴ 28 is greater.
Question 7.
Simplify the following
(i) 72 × 34
(ii) 32 × 24
(iii) 52 × 104
Solution:
(i) 72 × 34 = (7 × 7) × (3 × 3 × 3 × 3)
= 49 × 81 = 3969
(ii) 32 × 24 = (3 × 3) × (2 × 2 × 2 × 2)
= 9 × 16 = 144
(iii) 52 × 104 = (5 × 5) × (10 × 10 × 10 × 10)
= 25 × 10000 = 2,50,000
Question 8.
Find the value of the following.
(i) (-4)2
(ii) (-3) × (-2)3
(iii) (-2)3 × (-10)3
Solution:
(i) (-4)2 = (-1)2 × (4)2 [since am × bm = (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 × 23 × 103 [Grouping the terms of same base]
= (-1)6 × (2 × 10)3
[∵ am × bm = (a × b)m]
= 1 × 203 [since (-1)n = 1 if n is even]
= 8000
Question 9.
Simplify using laws of exponents.
(i) 35 × 38
(ii) a4 × a10
(iii) 7x × 72
(iv) 25 ÷ 23
(v) 188 ÷ 184
(vi) (64)3
(vii) (xm)0
(viii) 95 × 35
(ix) 3y × 12y
(x) 256 × 56
Solution:
Question 10.
If a = 3 and b = 2, then find the value of the following.
(i) ab + ba
(ii) aa – bb
(iii) (a + b)b
(iv) (a – b)a
Solution:
(i) ab + ba
a = 3 and b = 2
we get 32 + 23 = (3 × 3) + (2 × 2 × 2) = 9 + 8 = 17
(ii) (aa – bb)
Substituting a = 3 and b = 2
we get 32 – 22 = (3 × 3 × 3) – (2 × 2) = 27 – 4 = 23
(iii) (a + b)b
Substituting a = 3 and b = 2
we get (3 + 2)2 = 52 = 5 × 5 = 25
(iv) (a – b)a
Substituting a = 3 and b = 2
we get (3 – 2)3 = 13 = 1 × 1 × 1 = 1
Objective Type Questions
Question 12.
a × a × a × a × a equal to
(i) a5
(ii) 5 a
(iii) 5a
(iv) a + 5
Answer:
(i) a5
Question 13.
The exponential form of 72 is
(i) 72
(ii) 27
(iii) 22 × 33
(iv) 23 × 32
Answer:
(iv) 23 × 32
Question 14.
The value of x in the equation a13 = x3 × a10 is
(i) a
(ii) 13
(iii) 3
(iv) 10
Answer:
(i) a
Question 15.
How many zeros are there in 10010 ?
(i) 2
(ii) 3
(iii) 100
(iv) 20
Answer:
(iv) 20
Question 16.
240 + 240 is equal to
(i) 440
(ii) 280
(iii) 241
(iv) 480
Answer:
(iii) 241
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