Chapter 6 – Exponents of RD Sharma Solutions for Class 7 is provided here. Students can get the PDF of the chapter from the given links. The syllabus given by the CBSE for all the classes is completely based on RD Sharma Solutions. Class 7 is a stage where several new topics are introduced to students. BYJU’S RD Sharma Solutions Class 7 Solutions for Maths are designed to help students solve the CBSE Class 7 Maths problems with ease.
Chapter 6 – Exponents, contains 4 exercises. RD Sharma Solutions for Class 7 are given here and contain the answers to all the questions present in these exercises. Let us have a look at some of the concepts that are being discussed in this chapter.
- Definition and meaning of exponents
- Finding the value of a number given in the exponential form
- Expressing the numbers in exponential form
- Laws of exponents
- Use of exponents in expressing large numbers in standard form
- Expanded exponential form
RD Sharma Solutions for Class 7 Maths Chapter 6 Exponents
Access answers to Maths RD Sharma Solutions for Class 7 Chapter 6 – Exponents
Exercise 6.1 Page No: 6.12
1. Find the values of each of the following:
(i) 132
(ii) 73
(iii) 34
Solution:
(i) Given 132
132 = 13 × 13 =169
(ii) Given 73
73 = 7 × 7 × 7 = 343
(iii) Given 34
34 = 3 × 3 × 3 × 3
= 81
2. Find the value of each of the following:
(i) (-7)2
(ii) (-3)4
(iii) (-5)5
Solution:
(i) Given (-7)2
We know that (-a) even number= positive number
(-a) odd number = negative number
We have, (-7)2 = (-7) × (-7)
= 49
(ii) Given (-3)4
We know that (-a) even number= positive number
(-a) odd number = negative number
We have, (-3)4 = (-3) × (-3) × (-3) × (-3)
= 81
(iii) Given (-5)5
We know that (-a) even number= positive number
(-a) odd number = negative number
We have, (-5)5 = (-5) × (-5) × (-5) × (-5) × (-5)
= -3125
3. Simplify:
(i) 3 × 102
(ii) 22 × 53
(iii) 33 × 52
Solution:
(i) Given 3 × 102
3 × 102 = 3 × 10 × 10
= 3 × 100
= 300
(ii) Given 22 × 53
22 × 53 = 2 × 2 × 5 × 5 × 5
= 4 × 125
= 500
(iii) Given 33 × 52
33 × 52 = 3 × 3 × 3 × 5 × 5
= 27 × 25
= 675
4. Simply:
(i) 32 × 104
(ii) 24 × 32
(iii) 52 × 34
Solution:
(i) Given 32 × 104
32 × 104 = 3 × 3 × 10 × 10 × 10 × 10
= 9 × 10000
= 90000
(ii) Given24 × 32
24 × 32 = 2 × 2 × 2 × 2 × 3 × 3
= 16 × 9
= 144
(iii) Given 52 × 34
52 × 34 = 5 × 5 × 3 × 3 × 3 × 3
= 25 × 81
= 2025
5. Simplify:
(i) (-2) × (-3)3
(ii) (-3)2 × (-5)3
(iii) (-2)5 × (-10)2
Solution:
(i) Given (-2) × (-3)3
(-2) × (-3)3 = (-2) × (-3) × (-3) × (-3)
= (-2) × (-27)
= 54
(ii) Given (-3)2 × (-5)3
(-3)2 × (-5)3 = (-3) × (-3) × (-5) × (-5) × (-5)
= 9 × (-125)
= -1125
(iii) Given (-2)5 × (-10)2
(-2)5 × (-10)2 = (-2) × (-2) × (-2) × (-2) × (-2) × (-10) × (-10)
= (-32) × 100
= -3200
6. Simplify:
(i) (3/4)2
(ii) (-2/3)4
(iii) (-4/5)5
Solution:
(i) Given (3/4)2
(3/4)2 = (3/4) × (3/4)
= (9/16)
(ii) Given (-2/3)4
(-2/3)4 = (-2/3) × (-2/3) × (-2/3) × (-2/3)
= (16/81)
(iii) Given (-4/5)5
(-4/5)5 = (-4/5) × (-4/5) × (-4/5) × (-4/5) × (-4/5)
= (-1024/3125)
7. Identify the greater number in each of the following:
(i) 25 or 52
(ii) 34 or 43
(iii) 35 or 53
Solution:
(i) Given 25 or 52
25 = 2 × 2 × 2 × 2 × 2
= 32
52 = 5 × 5
= 25
Therefore, 25 > 52
(ii) Given 34 or 43
34 = 3 × 3 × 3 × 3
= 81
43 = 4 × 4 × 4
= 64
Therefore, 34 > 43
(iii) Given 35 or 53
35 = 3 × 3 × 3 × 3 × 3
= 243
53 = 5 × 5 × 5
= 125
Therefore, 35 > 53
8. Express each of the following in exponential form:
(i) (-5) × (-5) × (-5)
(ii) (-5/7) × (-5/7) × (-5/7) × (-5/7)
(iii) (4/3) × (4/3) × (4/3) × (4/3) × (4/3)
Solution:
(i) Given (-5) × (-5) × (-5)
Exponential form of (-5) × (-5) × (-5) = (-5)3
(ii) Given (-5/7) × (-5/7) × (-5/7) × (-5/7)
Exponential form of (-5/7) × (-5/7) × (-5/7) × (-5/7) = (-5/7)4
(iii) Given (4/3) × (4/3) × (4/3) × (4/3) × (4/3)
Exponential form of (4/3) × (4/3) × (4/3) × (4/3) × (4/3) = (4/3)5
9. Express each of the following in exponential form:
(i) x × x × x × x × a × a × b × b × b
(ii) (-2) × (-2) × (-2) × (-2) × a × a × a
(iii) (-2/3) × (-2/3) × x × x × x
Solution:
(i) Given x × x × x × x × a × a × b × b × b
Exponential form of x × x × x × x × a × a × b × b × b = x4a2b3
(ii) Given (-2) × (-2) × (-2) × (-2) × a × a × a
Exponential form of (-2) × (-2) × (-2) × (-2) × a × a × a = (-2)4a3
(iii) Given (-2/3) × (-2/3) × x × x × x
Exponential form of (-2/3) × (-2/3) × x × x × x = (-2/3)2 x3
10. Express each of the following numbers in exponential form:
(i) 512
(ii) 625
(iii) 729
Solution:
(i) Given 512
Prime factorisation of 512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
= 29
(ii) Given 625
Prime factorisation of 625 = 5 x 5 x 5 x 5
= 54
(iii) Given 729
Prime factorisation of 729 = 3 x 3 x 3 x 3 x 3 x 3
= 36
11. Express each of the following numbers as a product of powers of their prime factors:
(i) 36
(ii) 675
(iii) 392
Solution:
(i) Given 36
Prime factorisation of 36 = 2 x 2 x 3 x 3
= 22 x 32
(ii) Given 675
Prime factorisation of 675 = 3 x 3 x 3 x 5 x 5
= 33 x 52
(iii) Given 392
Prime factorisation of 392 = 2 x 2 x 2 x 7 x 7
= 23 x 72
12. Express each of the following numbers as a product of powers of their prime factors:
(i) 450
(ii) 2800
(iii) 24000
Solution:
(i) Given 450
Prime factorisation of 450 = 2 x 3 x 3 x 5 x 5
= 2 x 32 x 52
(ii) Given 2800
Prime factorisation of 2800 = 2 x 2 x 2 x 2 x 5 x 5 x 7
= 24 x 52 x 7
(iii) Given 24000
Prime factorisation of 24000 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 5 x 5 x 5
= 26 x 3 x 53
13. Express each of the following as a rational number of the form (p/q):
(i) (3/7)2
(ii) (7/9)3
(iii) (-2/3)4
Solution:
(i) Given (3/7)2
(3/7)2 = (3/7) x (3/7)
= (9/49)
(ii) Given (7/9)3
(7/9)3 = (7/9) x (7/9) x (7/9)
= (343/729)
(iii) Given (-2/3)4
(-2/3)4 = (-2/3) x (-2/3) x (-2/3) x (-2/3)
= ((16/81)
14. Express each of the following rational numbers in power notation:
(i) (49/64)
(ii) (- 64/125)
(iii) (-12/16)
Solution:
(i) Given (49/64)
We know that 72 = 49 and 82= 64
Therefore (49/64) = (7/8)2
(ii) Given (- 64/125)
We know that 43 = 64 and 53 = 125
Therefore (- 64/125) = (- 4/5)3
(iii) Given (-1/216)
We know that 13 = 1 and 63 = 216
Therefore -1/216) = – (1/6)3
15. Find the value of the following:
(i) (-1/2)2 × 23 × (3/4)2
(ii) (-3/5)4 × (4/9)4 × (-15/18)2
Solution:
(i) Given (-1/2)2 × 23 × (3/4)2
(-1/2)2 × 23 × (3/4)2 = 1/4 × 8 × 9/16
= 9/8
(ii) Given (-3/5)4 × (4/9)4 × (-15/18)2
(-3/5)4 × (4/9)4 × (-15/18)2 = (81/625) × (256/6561) × (225/324)
= (64/18225)
16. If a = 2 and b= 3, the find the values of each of the following:
(i) (a + b)a
(ii) (a b)b
(iii) (b/a)b
(iv) ((a/b) + (b/a))a
Solution:
(i) Consider (a + b)a
Given a = 2 and b= 3
(a + b)a = (2 + 3)2
= (5)2
= 25
(ii) Given a = 2 and b = 3
Consider, (a b)b = (2 × 3)3
= (6)3
= 216
(iii) Given a =2 and b = 3
Consider, (b/a)b = (3/2)3
= 27/8
(iv) Given a = 2 and b = 3
Consider, ((a/b) + (b/a))a = ((2/3) + (3/2))2
= (4/9) + (9/4)
LCM of 9 and 6 is 36
= 169/36
Exercise 6.2 Page No: 6.28
1. Using laws of exponents, simplify and write the answer in exponential form
(i) 23 × 24 × 25
(ii) 512 ÷ 53
(iii) (72)3
(iv) (32)5 ÷ 34
(v) 37 × 27
(vi) (521 ÷ 513) × 57
Solution:
(i) Given 23 × 24 × 25
We know that first law of exponents states that am × an × ap = a(m+n+p)
Therefore above equation can be written as 23 x 24 x 25 = 2(3 + 4 + 5)
= 212
(ii) Given 512 ÷ 53
According to the law of exponents we have am ÷ an = am-n
Therefore given question can be written as 512 ÷ 53 = 512 – 3 = 59
(iii) Given (72)3
According to the law of exponents we have (am)n = amn
Therefore given question can be written as (72)3 = 76
(iv) Given (32)5 ÷ 34
According to the law of exponents we have (am)n = amn
Therefore (32)5 ÷ 34 = 3 10 ÷ 34
According to the law of exponents we have am ÷ an = am-n
3 10 ÷ 34 = 3(10 – 4) = 36
(v) Given 37 × 27
We know that law of exponents states that am x bm = (a x b)m
37 × 27 = (3 x 2)7 = 67
(vi) Given (521 ÷ 513) × 57
According to the law of exponents we have am ÷ an = am-n
= 5(21 -13) x 57
= 58 x 57
According to the law of exponents we have am x an = a(m +n)
= 5(8+7) = 515
2. Simplify and express each of the following in exponential form:
(i) {(23)4 × 28} ÷ 212
(ii) (82 × 84) ÷ 83
(iii) (57/52) × 53
(iv) (54× x10y5)/ (54 × x7y4)
Solution:
(i) Given {(23)4 × 28} ÷ 212
{(23)4 x 28} ÷ 212= {212 x 28} ÷ 212 [According to the law of exponents we have (am)n = amn]
= 2(12 + 8) ÷ 212[According to the law of exponents we have am x an = a(m +n)]
= 220 ÷ 212 [According to the law of exponents we have am ÷ an = am-n]
= 2 (20 – 12)
= 28
(ii) Given (82 × 84) ÷ 83
(82 × 84) ÷ 83 [According to the law of exponents we have am x an = a(m +n)]
= 8(2 + 4) ÷ 83
= 86 ÷ 83[According to the law of exponents we have am ÷ an = am-n]
= 8(6-3) = 83 = (23)3 = 29
(iii) Given (57/52) × 53
= 5(7-2) x 53[According to the law of exponents we have am ÷ an = am-n]
= 55 x 53[According to the law of exponents we have am x an = a(m +n)]
= 5(5 + 3) = 58
(iv) Given (54× x10y5)/ (54 × x7y4)
= (54-4× x10-7y5-4) [According to the law of exponents we have am ÷ an = am-n]
= 50x3y1 [since 50 = 1]
= 1x3y
3. Simplify and express each of the following in exponential form:
(i) {(32)3 × 26} × 56
(ii) (x/y)12 × y24 × (23)4
(iii)(5/2)6 × (5/2)2
(iv) (2/3)5× (3/5)5
Solution:
(i) Given {(32)3 × 26} × 56
= {36 × 26} × 56[According to the law of exponents we have (am)n = amn]
= 66 × 56 [since law of exponents states that am x bm = (a x b)m]
= 306
(ii) Given (x/y)12 × y24 × (23)4
= (x12/y12) × y24 × 212
= x12 × y24-12 × 212[According to the law of exponents we have am ÷ an = am-n]
= x12 × y12 × 212
= (2xy)12
(iii) Given (5/2)6 × (5/2)2
= (5/2)6+2[According to the law of exponents we have am x an = a(m +n)]
= (5/2)8
(iv) Given (2/3)5× (3/5)5
= (2/5)5[since law of exponents states that am x bm = (a x b)m]
4. Write 9 × 9 × 9 × 9 × 9 in exponential form with base 3.
Solution:
Given 9 × 9 × 9 × 9 × 9 = (9)5 = (32)5
= 310
5. Simplify and write each of the following in exponential form:
(i) (25)3 ÷ 53
(ii) (81)5 ÷ (32)5
(iii) 98 × (x2)5/ (27)4 × (x3)2
(iv) 32 × 78 × 136/ 212 × 913
Solution:
(i) Given (25)3 ÷ 53
= (52)3 ÷ 53[According to the law of exponents we have (am)n = amn]
= 56 ÷ 53 [According to the law of exponents we have am ÷ an = am-n]
= 56 – 3
= 53
(ii) Given (81)5 ÷ (32)5[According to the law of exponents we have (am)n = amn]
= (81)5 ÷ 310[81 = 34]
= (34)5 ÷ 310 [According to the law of exponents we have (am)n = amn]
= 320 ÷ 310
= 320-10 [According to the law of exponents we have am ÷ an = am-n]
= 310
(iii) Given 98 × (x2)5/ (27)4 × (x3)2
= (32)8 × (x2)5/ (33)4× (x3)2[According to the law of exponents we have (am)n = amn]
= 316 × x10/312 × x6
= 316-12 × x10-6[According to the law of exponents we have am ÷ an = am-n]
= 34 × x4
= (3x)4
(iv) Given (32 × 78 × 136)/ (212 × 913)
= (32 × 7276 × 136)/(212× 133 × 73)[According to the law of exponents we have (am)n = amn]
= (212 × 76 × 136)/(212× 133 × 73)
= (76 × 136)/(133 × 73)
= 916/913[According to the law of exponents we have am ÷ an = am-n]
= 916-3
= 913
6. Simplify:
(i) (35)11 × (315)4 – (35)18 × (35)5
(ii) (16 × 2n+1 – 4 × 2n)/(16 × 2n+2 – 2 × 2n+2)
(iii) (10 × 5n+1 + 25 × 5n)/(3 × 5n+2 + 10 × 5n+1)
(iv) (16)7 ×(25)5× (81)3/(15)7 ×(24)5× (80)3
Solution:
(i) Given (35)11 × (315)4 – (35)18 × (35)5
= (3)55 × (3)60 – (3)90 × (3)25[According to the law of exponents we have (am)n = amn]
= 3 55+60 – 390+25
= 3115 – 3115
= 0
(ii) Given (16 × 2n+1 – 4 × 2n)/(16 × 2n+2 – 2 × 2n+2)
= (24 × 2(n+1) -22 × 2n)/(24 × 2(n+2) -2n+1 × 22) [According to the law of exponents we have (am)n = amn]
= 22 × 2(n+3-2n)/)22× 2(n+4-2n+1)
= 2n × 23 – 2n/ 2n × 24 – 2n × 2
= 2n(23 – 1)/ 2n(24 – 1) [According to the law of exponents we have am ÷ an = am-n]
= 8 -1 /16 -2
= 7/14
= (1/2)
(iii) Given (10 × 5n+1 + 25 × 5n)/(3 × 5n+2 + 10 × 5n+1)
= (10 × 5n+1 + 52 × 5n)/(3 × 5n+2 + (2 × 5) × 5n+1)
= (10 × 5n+1 + 5 × 5n+1)/(3 × 5n+2 + (2 × 5) × 5n+1) [According to the law of exponents we have (am)n = amn]
= 5n+1 (10+5)/ 5n+1 (10+15)[According to the law of exponents we have am ÷ an = am-n]
= 15/25
= (3/5)
(iv) Given (16)7 ×(25)5× (81)3/(15)7 ×(24)5× (80)3
= (16)7 ×(52)5× (34)3/(3 × 5 )7 ×(3 × 8)5× (16 × 5)3
= (16)7 ×(52)5× (34)3/37 × 57 × 35 × 85× 163× 53
= (16)7/ 85 × 16 3
= (16)4/85
= (2 × 8)4/85
= 24/8
= (16/8)
= 2
7. Find the values of n in each of the following:
(i) 52n × 53 = 511
(ii) 9 x 3n = 37
(iii) 8 x 2n+2 = 32
(iv) 72n+1 ÷ 49 = 73
(v) (3/2)4 × (3/2)5 = (3/2)2n+1
(vi) (2/3)10× {(3/2)2}5 = (2/3)2n – 2
Solution:
(i) Given 52n x 53 = 511
= 52n+3 = 511
On equating the coefficients, we get
2n + 3 = 11
⇒2n = 11- 3
⇒2n = 8
⇒ n = (8/2)
⇒ n = 4
(ii) Given 9 x 3n = 37
= (3)2 x 3n = 37
= (3)2+n = 37
On equating the coefficients, we get
2 + n = 7
⇒ n = 7 – 2 = 5
(iii) Given 8 x 2n+2 = 32
= (2)3 x 2n+2 = (2)5 [since 23 = 8 and 25 = 32]
= (2)3+n+2 = (2)5
On equating the coefficients, we get
3 + n + 2 = 5
⇒ n + 5 = 5
⇒ n = 5 -5
⇒ n = 0
(iv) Given 72n+1 ÷ 49 = 73
= 72n+1 ÷ 72 = 73 [since 49 = 72]
= 72n+1-2 = 73
= 72n-1=73
On equating the coefficients, we get
2n – 1 = 3
⇒ 2n = 3 + 1
⇒ 2n = 4
⇒ n =4/2 =2
(v) Given (3/2)4 × (3/2)5 = (3/2)2n+1
= (3/2)4+5 = (3/2)2n+1
= (3/2)9 = (3/2)2n+1
On equating the coefficients, we get
2n + 1 = 9
⇒ 2n = 9 – 1
⇒ 2n = 8
⇒ n =8/2 =4
(vi) Given (2/3)10× {(3/2)2}5 = (2/3)2n – 2
= (2/3)10 × (3/2)10 = (2/3)2n – 2
= 2 10 × 310/310 × 210 = (2/3)2n – 2
= 1 = (2/3)2n – 2
= (2/3)0 = (2/3)2n – 2
On equating the coefficients, we get
0 =2n -2
2n -2 =0
2n =2
n = 1
8. If (9n × 32 × 3n – (27)n)/ (33)5 × 23 = (1/27), find the value of n.
Solution:
Given (9n × 32 × 3n – (27)n)/ (33)5 × 23 = (1/27)
= (32)n × 33 × 3n – (33)n/ (315 × 23) = (1/27)
= 3(2n+2+n) – (33)n/ (315 × 23) = (1/27)
= 3(3n+2)– (33)n/ (315 × 23) = (1/27)
= 33n × 32 – 33n/ (315 × 23) = (1/27)
= 33n × (32 – 1)/ (315 × 23) = (1/27)
= 33n × (9 – 1)/ (315 × 23) = (1/27)
= 33n × (8)/ (315 × 23) = (1/27)
= 33n × 23/ (315 × 23) = (1/27)
= 33n/315 = (1/27)
= 33n-15 = (1/27)
= 33n-15 = (1/33)
= 33n-15 = 3-3
On equating the coefficients, we get
3n -15 = -3
⇒ 3n = -3 + 15
⇒ 3n = 12
⇒ n = 12/3 = 4
Exercise 6.3 Page No: 6.30
Express the following numbers in the standard form:
(i) 3908.78
(ii) 5,00,00,000
(iii) 3,18,65,00,000
(iv) 846 × 107
(v)723 × 109
Solution:
(i) Given 3908.78
3908.78 = 3.90878 x 103 [since the decimal point is moved 3 places to the left]
(ii) Given 5,00,00,000
5,00,00,000 = 5,00,00,000.00 = 5 x 107 [since the decimal point is moved 7 places to the left]
(iii) Given 3,18,65,00,000
3,18,65,00,000 = 3,18,65,00,000.00
= 3.1865 x 109 [since the decimal point is moved 9 places to the left]
(iv) Given846 × 107
846 × 107 = 8.46 x 102 x 10 [since the decimal point is moved 2 places to the left]
= 8.46 x 109 [since am x an = am+n]
(v) Given 723 × 109
723 × 109 = 7.23 x 102 x 109 [since the decimal point is moved 2 places to the left]
= 7.23 x 1011 [ since am x an = am+n]
2. Write the following numbers in the usual form:
(i) 4.83 × 107
(ii) 3.21 × 105
(iii) 3.5 × 103
Solution:
(i) Given 4.83 × 107
4.83 × 107 = 483 × 107-2 [since the decimal point is moved two places to the right]
= 483 × 105
= 4, 83, 00,000
(ii) Given 3.21 × 105
3.21 × 105 = 321 x 105-2 [since the decimal point is moved two places to the right]
= 321 x 103
= 3, 21,000
(iii) Given 3.5 × 103
3.5 × 103 = 35 x 103-1 [since the decimal point is moved one place to the right]
= 35 x 102
= 3,500
3. Express the numbers appearing in the following statements in the standard form:
(i) The distance between the Earth and the Moon is 384,000,000 meters.
(ii) Diameter of the Earth is 1, 27, 56,000 meters.
(iii) Diameter of the Sun is 1,400,000,000 meters.
(iv) The universe is estimated to be about 12,000,000,000 years old.
Solution:
(i) Given the distance between the Earth and the Moon is 384,000,000 meters.
The distance between the Earth and the Moon is 3.84 x 108 meters.
[Since the decimal point is moved 8 places to the left.](ii) Given diameter of the Earth is 1, 27, 56,000 meters.
The diameter of the Earth is 1.2756 x 107 meters.
[Since the decimal point is moved 7 places to the left.](iii) Given diameter of the Sun is 1,400,000,000 meters.
The diameter of the Sun is 1.4 x 109 meters.
[Since the decimal point is moved 9 places to the left.](iv) Given the universe is estimated to be about 12,000,000,000 years old.
The universe is estimated to be about 1.2x 1010 years old.
[Since the decimal point is moved 10 places to the left.]Exercise 6.4 Page No: 6.31
1. Write the following numbers in the expanded exponential forms:
(i) 20068
(ii) 420719
(iii) 7805192
(iv) 5004132
(v) 927303
Solution:
(i) Given 20068
20068 = 2 x 104 + 0 x 103 + 0 x 102 + 6 x 101 + 8 x 100
(ii) Given 420719
420719 = 4 x 105 + 2 x 104 + 0 x 103 + 7 x 102 + 1 x 101 + 9 x 100
(iii) Given 7805192
7805192 = 7 x 106 + 8 x 105 + 0 x 104 + 5 x 103 + 1 x 102 + 9 x 101 + 2 x 100
(iv) Given 5004132
5004132 = 5 x 106 + 0 x 105 + 0 x 104 + 4 x 103 + 1 x 102 + 3 x 101 + 2 x 100
(v) Given 927303
927303 = 9 x 105 + 2 x 104 + 7 x 103 + 3 x 102 + 0 x 101 + 3 x 100
2. Find the number from each of the following expanded forms:
(i) 7 × 104 + 6 × 103 + 0 × 102 + 4 × 101 + 5 × 100
(ii) 5 × 105 + 4 × 104 + 2 × 103 + 3 × 100
(iii) 9 × 105 + 5 × 102 + 3 × 101
(iv) 3 × 104 + 4 × 102 + 5 × 100
Solution:
(i) Given 7 × 104 + 6 × 103 + 0 × 102 + 4 × 101 + 5 × 100
= 7 x 10000 + 6 x 1000 + 0 x 100 + 4 x 10 + 5 x 1
= 70000 + 6000 + 0 + 40 + 5
= 76045
(ii) Given 5 × 105 + 4 × 104 + 2 × 103 + 3 × 100
= 5 x 100000 + 4 x 10000 + 2 x 1000 + 3 x 1
= 500000 + 40000 + 2000 + 3
= 542003
(iii) Given 9 × 105 + 5 × 102 + 3 × 101
= 9 x 100000 + 5 x 100 + 3 x 10
= 900000 + 500 + 30
= 900530
(iv) Given 3 × 104 + 4 × 102 + 5 × 100
= 3 x 10000 + 4 x 100 + 5 x 1
= 30000 + 400 + 5
= 30405
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