WBBSE Solutions For Class 7 Maths Algebra Chapter 3 Concept Of Index Exercise 3 Solved Problems

Class 7 Math Solution WBBSE Algebra Chapter 3 Concept Of Index Exercise 3 Solved Problems

Power or Index: The product obtained by multiplying a number several times by itself is called the power or Index of that number.

⇒ 3 x 3 x 3 x 3 =3⁴ (Power or Index of 3 is 4)

⇒ a x a x a x a x a =a⁵ (Power or Index of a is 5)

⇒ Some important formulas on Index
⇒ [a is non zero integer and also m and n are integers]

1. a^m. aⁿ = a^{m + n}

2. a^m ÷ aⁿ = a^{m – n}

3. (a^m)ⁿ = a^mn

4. \(a^m=\frac{1}{a^{-m}}\)

5. a⁰ = 1

6. (ab)^m = a^m.bⁿ

7. \(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\)

Wbbse Class 7 Maths Solutions

Question 1. Choose the correct answer 

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1. If we express the number 35400000 in index of 10 we get

1. 354 x 10⁴
2. 354 x 10³
3. 354 x 10⁵
4. 354 x 10⁶

Solution: 35400000
= 354 x 100000
= 354 × 105

So the correct answer is 3. 354 x 10⁵

2. If we express the number 16489 in index form as power of 10 we get

1. 16.489 x 10²
2. 164.89 x 10²
3. 1.6489 x 10²
4. 1648.9 x 10²

Solution: 16489 = \(\frac{16489}{100}\) x 100

= 164.89 × 10²

So the correct answer is 2. 164.89 x 10²

3. The value of 7 x 10⁶ + 3 x 10⁴+ 4 x 10³ + 6 × 10² + 8 x 10 + 9 is

1. 734689
2. 7346089
3. 7340689
4. 7034689

Solution: 7 x 10⁶ + 3 x 10⁴+ 4 x 10³ + 6 × 10² + 8 x 10 + 9
= 7000000 + 30000 + 4000+ 600 + 80 +9
= 7034689

So the correct answer is 4. 7034689

Wbbse Class 7 Maths Solutions

Question 2. Write ‘true’ or ‘false’

1. The simplest form of (a⁷×a^-5)+(a^-3×a⁶) is \(\frac{1}{a}\)

Solution: (a⁷×a^-5)+(a^-3×a⁶)

= \(\frac{a^7 \times a^{-5}}{a^{-3} \times a^6}=\frac{a^{7-5}}{a^{-3+6}}=\frac{a^2}{a^3}=a^{2-3}=a^{-1}=\frac{1}{a}\)

So the statement is true.

2. (-5)⁴× (3)⁴= -50625

Solution: (-5)⁴× (3)⁴

= (-5) × (-5) × (-5) x (-5) × 3 × 3 × 3 × 3
= (+25) × (+25) x 81
= + 625 x 81
= 50625

So the statement is false.

3. The value of \(\frac{3 \times 7^2 \times 2^4}{21 \times 112}\) is 1

Solution: \(\frac{3 \times 7^2 \times 2^4}{21 \times 112}\)

= \(\frac{3 \times 7^2 \times 2^4}{3 \times 7 \times 2^4 \times 7}\)

= 3^1-1 x 7^2-1-1 x 2^4-4
= 3⁰ x 7⁰ x 2⁰
= 1 x 1 x 1
= 1

So the statement is true.

Question 3. Fill in the blanks

1. 20 x 18 x  = [2²×3×5]²

Solution: 20 x 18 x = 2⁴ x 3² x 5²

⇒ 2² × 5 ×3² x 2 x = 2⁴ x 3² x 5²

⇒ 2²⁺¹ x 5 x 3² x = 2⁴ x 3² x 5²

⇒   = \(\frac{2^4 \times 3^2 \times 5^2}{2^3 \times 5 \times 3^2}\)

=2^4-3 x 3^2-2 x 5^2-1
= 2¹x 3⁰ x 5¹

= 2 x 1 x 5
= 10

Wbbse Class 7 Maths Solutions

2. \(\left(x^2\right)^3 \times\left(y^{-3}\right)^2=\left(\frac{x}{y}\right)\)

Solution: \(\left(x^2\right)^3 \times\left(y^{-3}\right)^2\)

= x⁶ x y^-6

= \(\frac{x^6}{y^6}=\left(\frac{x}{y}\right)^6\)

So the answer is 6

3. 

Solution: (-7)² x 8²
= 49 x 8²
= 7² x 8²
= (7 x 8)²
= (56)²

⇔ (-7)² x 8²
= (-7 x 8)²= (-56)²

So the answer is 56 or -56.

Question 4. Express 49 x 49 x 49 x 49 as a power 7

Solution: 49 x 49 x 49 x 49
= 7 x 7 x 7 x 7 x 7 x 7 x 7 x 7
= 7⁸

Wbbse Class 7 Maths Solutions

Question 5. Express the following numbers in index form as the power of 10 (taking 1, 2, and 3 places of the decimal)

1. 4678
2. 526824

Solution: 1. 4678

= \(\frac{4678}{10}\) x 10 =467.8 x 10

⇒ 4678 = \(\frac{4678}{100}\) x 100 = 46.78 x 10²

⇒ 4678 = \(\frac{4678}{1000}\) x 1000 = 4.678 x 10³

2. 526824

⇒ 526824 = \(\frac{526824}{10}\) x 10 = 52682.4 x 10

⇒ 526824 = \(\frac{526824}{100}\) x 100 = 5268.24 x 10²

⇒ 526824= \(\frac{526824}{1000}\)  x 1000 = 526.824 x 10³

Question 6. Form the numbers from their expanded form

1. 9 x 10⁴ + 3 x 10² + 7
2. 3 x 10⁷ + 4 x 10⁶ + 2 x 10⁴ +7 x 10² + 5

Solution:

1. 1. 9 x 10⁴ + 3 x 10² + 7

= 90000 + 300 + 7
= 90307

2. 3 x 10⁷ + 4 x 10⁶ + 2 x 10⁴ +7 x 10² + 5

= 30000000+4000000+ 20000 + 700 + 5
= 34020705

Question 7. Simplify and express each of them in powder form

1. \(\frac{10^3 \times 10^4}{2^5 \times 5^4}\)

2. \(\frac{4^8 \times a^{10} b^4}{4^3 \times a^4 b^2}\) [ a ≠ 0, b ≠ 0]

Solution:

1. \(\frac{10^3 \times 10^4}{2^5 \times 5^4}\)

= \(\frac{(2 \times 5)^3 \times(2 \times 5)^4}{2^5 \times 5^4}\)

= 2^3+4-5 x 5^3+4-4

= 2²x5³
= 4 × 125
= 500

2. \(\frac{4^8 \times a^{10} b^4}{4^3 \times a^4 b^2}\)

=4^8-3 x a^10-4 x b^4-2

= 4⁵ x a⁶ x b²

Question 8. So that \(a^m=\frac{1}{a^{-m}}\)

Solution: a^m = a^0-(-m)

= \(\frac{a^0}{a^{-m}}=\frac{1}{a^{-m}}\)

 

Class 7 Math Solution WBBSE Concept Of Index

Concept Of Index Exercise 5.1

Let’s expand the following in the index of 

Question 1. 8275
Solution:

8275 = 8 × 10³ + 2 × 10² +7 ×10 + 5

Question 2. 90925
Solution:

90925 = 9 × 10⁴ + 0 ×10³+9 ×10² +2 ×10 + 5

Question 3. 12578
Solution:

12578 = 1 × 10⁴ +2 × 10³ + 5 × 10² +7 ×10 + 8

Question  4. 7858
Solution:

7858 = 7 ×  10³ + 8 × 10² + 5×10 + 8

Class Vii Math Solution WBBSE Concept Of Index Exercise 5.2

Question 1. 100 = 10 ^____
Solution:

100  = 10 × 10

= 10²

Question 2. 27 = 3^____
Solution:

27 = 3 × 3 × 3

= 3³

Question 3.125 = 5^____
Solution:

125 = 5 × 5 × 5

= 5³

Question 4. 32 = 2^___
Solution:

32 =  2 × 2 × 2 × 2 × 2

= 2⁵

Question 5. 343 = 7____
Solution:

343 =  7 × 7 × 7

= 7³

Question 6.121 = 11______
Solution:

121=11

121= 11 × 11

=11²

Question 7. 625 = 5______
Solution: =

625 = 5 × 5 × 5 × 5

= 5⁴

Question 8. 2³  = ^____× ^_____×
Solution:

2³  = ^____× ^_____× = 2 × 2 ×2

= 2³

Question 9. 3⁴ = ^_____ ×^_____ ×^_____ ×
Solution: =

3⁴ = ^_____ ×^_____ ×^_____ × = 3 × 3 × 3 × 3=

= 3⁴

= 81

Question 10. 729 = 9^_____
Solution:

729 =  9 × 9 × 9

= 9³

Question 11. 2 × 2× 2 × 2 × 2 = ^_____
Solution:

2 × 2× 2 × 2 × 2 = 2⁵

Question 12. (- 2) × (- 2) × (- 2) = (- 2)_________
Solution:

2 × 2× 2 × 2 × 2 = (-2)³

Question 13. (- 2) ×(- 2) × (- 2) × (- 2) = (- 2) ________________
Solution:

(- 2) ×(- 2) × (- 2) × (- 2) = (- 2)

= (-2)⁴

Class Vii Math Solution WBBSE Concept Of Index Exercise 5.3

Let’s express the following numbers in the product of power from of their prime factors.

Question 1. 24.
Solution:

24= 2 × 2 × 2 × 3 = 2³ × 3

Question 2. 56.
Solution:

56 = 2× 2 × 2 ×7 = 2³ ×7

Question 3. 63.
Solution:

63. = 3 × 3 × 7 = 3² × 7

Question 4. 72
Solution:

72 = 2 × 2 × 2× 3 ×3 = 2³ × 3²

Question 5. 200
Solution:

200 = 2  ×2 × 2× 5× 5 = 2³ × 5²

Class Vii Math Solution WBBSE Concept Of Index Exercise 5.4

Let’s put > or < signs in the respective blank squares

Question 1. 5³ ____________ 3⁵
Solution:

5³ = 5× 5× 5

= 125

3⁵ = 3×3 × 3× 3×3

= 243

∴ 5³ < 3⁵

Question 2. 6²______________2⁶
Solution:

6² = 6 ×6

= 36

2⁶ = 2 × 2 × 2 × 2 × 2 × 2

= 64

∴ 6² < 2⁶

Question 3. 2⁴________4²
Solution:

2⁴ = 2 × 2 × 2 × 2

= 16

4²= 4 × 4 = 16

∴ 2⁴= 4²

Question 4. 7²______2⁷
Solution:

7² = 7x 7

= 49

2⁷ = 2×2 × 2 × 2 × 2 × 2 ×2

= 128

∴ 7²< 2⁷

Question 5. 3⁴_________4³
Solution:

3⁴ = 3 × 3 × 3 × 3

= 81

4³ = 4 × 4 × 4 = 64

∴ 3⁴> 4³

Question 6. 3⁵_______5³
Solution: 

3⁵ = 3 × 3 × 3 × 3 × 3

= 243

5³=5 × 5 × 5

= 125

∴ 3⁵>5³

WB Class 7 Math Solution Concept Of Index Exercise 5.5

Question 1. 2⁵ × 2⁷
Solution:

2⁵ × 2⁷ = 2⁵⁺⁷

= 2¹²

Question 2. (-3)¹⁸× (-3)¹²
Solution:

(-3)¹⁸× (-3)¹²= (3)¹⁸⁺¹²

= (-30)³⁰

Question 3. 10⁸ x 10²
Solution:

10⁸ x 10² = (10)⁸⁺²

= (10)¹⁰

Question 4. 2¹⁵ ÷ 2¹³ 
Solution:

2¹⁵ ÷ 2¹³ 

= (2)^15-13

= 2²

Question 5. 9¹⁵-9¹⁴
Solution:

9¹⁵-9¹⁴ = 9^15-14

= 9

Question 6. 11⁶÷ 11⁴
Solution:

11⁶÷ 11⁴ = 11^6-4

= 11²

Concept Of Index Exercise 5.6

Let’s fill the blank squares given below:

Question 1. 9²÷ 9² = ________________
Solution: =

9²÷ 9² = 9^2-2

= 9⁰

= 1

Question 2. 7³_______0 = 1
Solution:

7³_______0 = 1

7³ ÷7³ = 1

= 7⁰

Question 3. 11⁰= ______________
Solution:

11⁰ = 1

Question 4. 1 =13__________
Solution:

1 =13___

∴ 1= 13⁰= 1

∴ 1-1 = 0

Question 5. 1 =(- 13) _______________
Solution:

1 =(- 13)

= (-13)⁰

WB Class 7 Math Solution Concept Of Index Exercise – 5.7

Question 1. 6⁵÷2⁵ = ________________
Solution:

6⁵÷2⁵

= \(\frac{6^5}{2^5}\)

= \(\left(\frac{6}{2}\right)^5\)

= (3)⁵

Question 2. _____ =  7² ÷ 2²
Solution: 

= \(\frac{7^2}{2^2}\)

= \(\left(\frac{7}{2}\right)^5\)

Question 3. 10² = ______ × ________ 
Solution:

10² = 10 × 10

Question 4. (4)²  × 6² = ________²
Solution:

(4)²  × 6²

(-4² × 6²)  = (-24)²

Question 5. (5)⁰ = ______________
Solution:

(5)⁰ =  1

6.  \(\left(\frac{2}{3}\right)^3\)
Solution:

⇒ \(\left(\frac{2}{3}\right)^3\)

= \(\frac{2^3}{3^3}=\frac{8}{27}\)

Concept Of Index Exercise – 5.8

Question 1. Let us express 8 × 8 × 8 as power of 2.
Solution :

Given

8 × 8 × 8

= 2³ × 2³ × 2³⁺³⁺³

= (2)⁹

Question 2. 25 × 25 × 25 × 25 to be expessed as power of 5.
Solution :

Given

25 × 25 × 25 × 25

= 5² × 5² × 5²× 5²

= (5)^2+2+2+2

= (5)⁸

Question 3. Let’s express 36 × 36 × 36 as the power of 6.
Solution :

Given

36 × 36 ×  36

= (6)² × (6)² × (6)²

= (6)^{2+2 +2}  = 6⁶

Question 4. Let’s express 81 x 81 as power of 3.
Solution :

Given

81 x 81

= 3⁴ × 3⁴ = (3)⁴⁺⁴

= 3⁸

Question 5. Let us find the values of the following

1. \(\frac{2^6 \times 3^5}{(6)^5}\)
Solution:

Given

⇒ \(\frac{2^6 \times 3^5}{(6)^5}\)

= \(\frac{2^6 \times 3^5}{(2 \times 3)^5}\)

= \(\frac{2^6 \times 3^5}{2^5 \times 3^5}\)

= 2¹

= 2

2. \(\frac{10^3 \times 10^4}{2^5 \times 5^4}\)
Solution:

Given

⇒ \(\frac{10^3 \times 10^4}{2^5 \times 5^4}\)

= \(\frac{(2 \times 5)^3 \times(2 \times 5)^4}{2^5 \times 5^4}=\frac{2^3 \times 5^3 \times 2^4 \times 5^4}{2^5 \times 5^4}\)

= \(\frac{(2)^3 \times(2)^4 \times 5^3}{2^5}=2^{3+4-5} \times 5^3\)

2² × 5³ = 4 × 125

= 500

3. \(\frac{5^9 \times 5^6}{5^7}\)
Solution:

Given

⇒ \(\frac{5^9 \times 5^6}{5^7}\)

= \(5^{9+6-7}\)

= 5⁸

4. \(\frac{6^4 \times 3^8}{3^{12}}\)
Solution:

Given

⇒ \(\frac{6^4 \times 3^8}{3^{12}}\)

= \(\frac{(3 \times 2)^4 \times 3^8}{3^{12}}\)

= \(\frac{(3)^{4+8} \times 2^4}{3^{12}}\)

= 2 × 2 × 2 × 2

= 16

5. \(\frac{25^2 \times 25^5}{5^{10}}\)

Given

⇒ \(\frac{25^2 \times 25^5}{5^{10}}\)

= \(\frac{\left(5^2\right)^2 \times\left(5^2\right)^5}{5^{10}}\)

= \(\frac{5^4 \times 5^{10}}{5^{10}}\)

= 5⁴

5 × 5 × 5 × 5 = 625

6. \(\frac{2^3 \times 3^9}{3^6 \times 6^3}\)

Given

⇒ \(\frac{2^3 \times 3^9}{3^6 \times 6^3}\)

⇒  \(\frac{2^3 \times 3^9}{3^6 \times(2 \times 3)^3}\)

= \(\frac{2^3 \times 3^9}{3^6 \times 2^3 \times 3^3}\)

=\(\frac{3^9}{3^9}\)=1

7. \(\left(\frac{a^7}{a^5}\right) \times \begin{gathered}
a^2 \\
(a \neq 0)
\end{gathered}\)
Solution:

⇒ \(\left(\frac{a^7}{a^5}\right) \times \begin{gathered}
a^2 \\
(a \neq 0)
\end{gathered}\)

= \(\left(\frac{a^7}{a^5}\right) \times a^2\)

= \(\left(a^{7-5}\right) \times a^2\)

a² × a² = a⁴

8.\(\frac{3 \times 7^2 \times 2^4}{21 \times 112}\)
Solution:

Given

⇒ \(\frac{3 \times 7^2 \times 2^4}{21 \times 112}\)

= \(\frac{3 \times 2^4 \times 7^2}{3 \times 7 \times 7 \times 16}\)

= \(\frac{7^2 \times 2^4}{7^2 \times 2^4}\)

= 1

WB Class 7 Math Solution Concept Of Index Exercise 5.9

Question 1. Let’s express the distances given below in index of 10 and try to get a better idea of the distances Distance of Mercury from Sun is 57900000 km. Distances of Mars and Jupiter from Sun are 227900000 km. and 778300000 km respectively.
Solution :

Given

Distances of Mars and Jupiter from Sun are 227900000 km. and 778300000 km respectively.

1. Distance of Mercury from Sun = 57900000 km.

= 579 × 10⁵ km.

2. Distance of Mars from Sun = 227900000 km

= 2279 ×10⁵ km.

3. Distance of Jupiter from Sun = 778300000 km.

= 7783 × 10⁵ km.

Question 2. Let’s fill in the gaps –

1. The distance between Earth and the Moon is 384,000,000 m. = 384 x____________ m.
Solution :

⇒ Distance between Earth and Moon

⇒ 384,000,000 m = 384 x 10 ⁶

∴  10⁶

2. The speed of light in a vacuum is 3,00,000,000 m/sec. = 3 x_m/sec.
Solution:

⇒ The speed of light in a vacuum

= 3,00,000,000 m/sec. = 3 ×10⁸ m/sec

∴  10⁸

3. Let’s express the following number in index form as power of 10 (taking 1,2 and 3 places of decimal)

1. 978 =
Solution:

= 97. 8 × 10

= 9. 78 × 10²

= 0.978  × 10³

2. 1592170 =
Solution:

= 159217 × 10

= 15921 .7 × 10²

= 1592 .17 × 10²³

Question 4. Let’s form the numbers from their expanded form given below –

1. 3 ×10³ +2 ×10² +7 × 10+ 2
Solution:

3 ×10³ +2 ×10² +7 × 10+ 2 = 3000 + 200 + 70 + 2

= 3272

2. 2  × 10³+3 ×10+ 5
Solution:

2  × 10³+3 ×10+ 5 = 2000 + 30 + 5

= 2035

3. 8 ×10⁴ + 2 × 10³ + 3 × 10² + 6
Solution:

8 ×10⁴ + 2 × 10³ + 3 × 10² + 6

= 80000 + 2000 + 300 + 6

= 82306

4. 9 ×10⁴ +5 ×10³ +6 × 10² + 6 ×10
Solution:

9 ×10⁴ +5 ×10³ +6 × 10² + 6 ×10

= 90000 + 5000 + 600 + 70

= 95670

Question 5. Let’s simplify and express each of them in powder form

1. \(\frac{2^3 \times 3^5 \times 16}{3 \times 32}\)
Solution:

= \(\frac{2^3 \times 3^5 \times 2^4}{3 \times 2^5}=\frac{2^{3+4} \times 3^5}{2^5 \times 3}\)

= \(2^{7-5} \times 3^{5-1}\)

= \(2^2 \times 3^4\)‘

= 2² × 9²

= (18)²

= 18²

= 324

2. \(\left[\left(6^2\right)^3 \times 6^4\right] \div 6^7\)
Solution:

\(\left[\left(6^2\right)^3 \times 6^4\right] \div 6^7\) = 6⁶ × 6⁴÷ 6⁷

= 6^6+4-7

= 6³

3. \(\frac{3 \times \cdot 7^2 \times 11^0}{21 \times 7}\)
Solution:

⇒ \(\frac{3 \times 7^2 \times 1}{3 \times 7 \times 7}\)

= \(\frac{7^2}{7^2}\)

= 1

4. \(\left(3^0+2^0\right) \times 5^0\)
Solution:

(1+1) ×1

= 2 ×1 = 2

5. \(\frac{4^5 \times a^8 b^3}{4^5 \times a^5 b^2}(b \neq 0)\)
Solution:

6. \(\frac{2^8 x^7}{\left(2^2\right)^3 \times x^3}=\frac{2^8 \times x^7}{2^6 \times x^3}\)

⇒ \(\frac{2^8 x^7}{\left(2^2\right)^3 \times x^3}=\frac{2^8 \times x^7}{2^6 \times x^3}\)

= \(2^{8-6} \times x^{7-3}=2^2 \times x^4\)

= \((2)^2 \times\left(x^2\right)^2=\left(2 x^2\right)^2\)