WBBSE Class 7 Maths Solutions For Algebra Chapter 3 Concept of Index

Algebra Chapter 3 Concept of Index

Question 1. Choose the correct Answer

1. If we express the number 1234500000 in an index. form as power of 10 then we get

1. 1234.5 x 10⁶
2. 12345 x 10⁶
3. 123.45 x 10⁶
4. 12.345 x 10⁶

Solution:

Given number 1234500000

Now Count the Zero one in the number is ‘5’

we must be present in a form as the power of 10.

∴ 12345 x 10⁵ = 1234.5x 10⁶ (Power should not be Odd)

Option (1) 1234.5 x 10⁶ is the correct answer.

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2. The value of \(\frac{3^4 \times 3^{-5}}{3^{-6}}\) is

1. 243
2. 343
3. 81
4. 1/9

⇒ \(\frac{3^4 \times 3^{-5}}{3^{-6}}\) is

Solution:

∵ a^m x aⁿ = a^m+n

Here a = 3, m = 4, n =-5

⇒ \(\frac{3^{4+(-5)}}{3^{-6}}\)

⇒ \(\frac{3^{-1}}{3^{-6}}\)

⇒ \(3^{-1)-(-6)}\) ∵ \(\frac{a^m}{a^n}=a^{m-n}\)

Here, a = 3, m = -1, n = -6

⇒  \(3^{-1+6}\)

⇒  \(3^5\)

⇒ 3 x 3 x 3 x 3 x 3

⇒ 243

∴ \(\frac{3^4 \times 3^{-5}}{3^{-6}}\) = 243

Option 1 is 243 is the correct answer

Class 7 Algebra Problems With Solutions

3. The value of (a³ x b)² x (a² x b^-3)³ is

1. \(\frac{a^6}{b^7}\)
2. \(a^6 \cdot b^7\)
3. \(a^7 \cdot b^6\)
4. \(\frac{a^7}{b^6}\)

Solution:

(a⁰ x b)² x (a² x b^-3)³  ∵ a = 1

⇒ (1x b)² x (a² x b^-3)³  ∵  (a x b)^m = a^m x b^m

⇒ (1)² x (b)² x (a²)³ x (b^-3)³

In first bracket (1 x b)²

a = 1, b = b, m = 2

In the second bracket (a² x b^-3)³

∵ a = a², b = b^-3, m = 3

⇒ 1 x b² x a^2×3 x b^-3×3

⇒ b² x a⁶ x b^-9

⇒ a⁶ x b² x b^-9

⇒ a⁶ × b^2-9

⇒ a⁶ x b^-7

Here, a = b,

m = 2

n = -9

⇒ a⁶ x \(\frac{1}{b^7}\)

⇒ \(\frac{a^6}{b^7}\)

Option 1 is correct

Question 2. Write true or false.

1. 2 x 10⁵ + 3 × 10⁴ +4 × 10³ + 5 × 10² + 6 × 10 + 7 = 765433
2. The expression of 6x6x6x6x6 in terms of the index of a prime number is (2⁵ x 3⁵)
3. The value of \(\frac{2^6 \times 8 \times 16}{2^{11} \times 2^2}\) is 1

Solution:

1. 2 x 10⁵ + 3 × 10⁴ + 4 x 10³ + 5 × 10² +6 × 10 + 7= 765433

⇒ 200000+30000+4000 + 500+60 +7 = 765433

⇒ 234,567 = 765433

∴ 2 x 10⁵ + 3 × 10⁴ + 4 × 10³ + 5 x 10² + 6 x 10 + 7 = 76.5433; False

2. 6 × 6 x 6 x 6 x 6 = (2⁵ х 3⁵)

⇒ 2×3×2×3×2×3×2×3×2×3 = 2⁵ × 3⁵

⇒ 2×2×2×2×2×3×3×3×3×3 = 2⁵ x 3⁵

⇒ 2⁵ x 3⁵ = 2⁵ x 3⁵

∴ 6x6x6x6x6 =(2⁵ x 3⁵) → True

WBBSE Class 7 Algebra Chapter 3 

3. The value of \(\frac{2^6 \times 8 \times 16}{2^{11} \times 2^2}\) is 1

⇒ \(\frac{2^8 \times 8 \times 16}{\frac{2^{14}}{2^5} \times 2^2}\)

⇒ \(\frac{128}{32 \times 4}\)

⇒ \(\frac{128}{128}=1\)

∴ \(\frac{2^6 \times 8 \times 16}{2^{11} \times 2^2}=1\); True

Question 3. Fill in the blanks:

1. (-5)² x (6)² = ( )⁵
2. 0.53= 0·053 x 10
3. The value of \(\frac{3 \times 7^2 \times 2^5}{14 \times 42}\) is

Solution:

1. (-5)² x(6)² = ( )²

⇒ 2.5×36

⇒ 900

⇒ (30)²

∴ (-5)² x (6)² = (30)²

2. 0.53 = 0.053 × 10¹

⇒ 0.053×10¹

⇒ 0.53

∴ 0.53= 0.053×10¹

3. The value of \(\frac{3 \times 7^2 \times 2^5}{14 \times 42}\)

⇒ \(\frac{3 \times 7^2 \times 2^5}{14 \times 42}\)

⇒ \(\frac{3 \times 49 \times 32}{588}\)

⇒ \(\frac{4.704}{588}\)

= 8

∴ \(\frac{3 \times 7^2 \times 2^5}{14 \times 42}=8\)

Class 7 Maths Algebra Solutions WBBSE

Question 4. Express the following number in terms of the index

1. 4672
2. 12045
3. 400608

Solution:

1. 4672

Number in terms of index 4 ×10³+6×10²+7×10¹ +2

⇒ 4000+600+ 70+2

⇒ 4672

2. 12045

Number in terms of index 19 x 10⁴ +2 ×10³+ 4 ×10 + 5

⇒ 10000+2.000+40+5

⇒ 12,045

3. 4000608

Number in terms of index

⇒ 4×10⁶+6×10² +8

⇒ 4000000 +600+8

⇒ 4000608.

WBBSE Class 7 Maths Chapter 3 Answers

Question 5. Express in terms of the index of a prime number.

1. 1800
2. 882
3. 80

Solution:

1. 1800

Taking LCM of 1800

∴ 2 x 2 x 2 x -3 x 3 x 5 x 5

∴ 2³×3² x 5²

∴ 1800 = 2³ x 3² x 5²

Index of prime number.

2. 882

Taking LCM of 882

∴ 2 x3 × 3 ×7 x 7

∴2 x 3² x 7²

∴ 882=2 x 3² x 7² index of prime number.

3. 80

Taking LCM of 80.

∴ 2 x 2 x 2 x 2 x 5

∴ 2⁴ x 5

∴ 80 = 2⁴ x 5 index of prime number.

Class 7 Maths Chapter 3 Solved Exercises

Question 6. Simplify

1. \(\frac{x^{-2} \times x^7}{x^8 \times x^{-4}}\)
2. \(\frac{(18)^3 \times(2.5)^2}{(30)^4}\)

Solution:

1. \frac{x^{-2} \times x^7}{x^8 \times x^{-4}} \quad a^m \times d^n=a^{m-n}[/latex]

Here a = x, m = -2, n = 7 → Numerator

a = x, m = 8, n = -4 → Denominatore

⇒ \(\frac{x^{-2+7}}{x^{8+(-4)}}\)

⇒ \(\frac{x^5 x}{x^4}\)

∴ x

∴ \(\frac{x^{-2} \times x^7}{x^8 \times x^{-4}}=x\)

2. \(\frac{(18)^3 \times(2.5)^2}{(30)^4}\)

⇒ \(\frac{18 \times 18 \times 18 \times 25 \times 25}{30 \times 30 \times 30 \times 30}\)

⇒ \(\frac{58,32 \times 625}{810,000}\)

⇒ \(\frac{3.64,5,000}{810,000} \Rightarrow \frac{364.5}{810}=4.5\)

∴ \(\frac{(18)^3 \times(2.5)^2}{(30)^4}=4.5\)

Question 7. Find the value of

1. \(\frac{7^3 \times 2^6 \times 10}{3.5 \times 224}\)
2. \(\frac{3^4 \times 6^5}{72}\)

Solution:

1. \(\frac{7^3 \times 2^6 \times 10}{35 \times 224}\)

⇒ \(\frac{343 \times 64 \times 10}{35 \times 224}\)

⇒ \(\frac{219,520}{7840}\)

= 28

∴ \(\frac{7^3 \times 2^6 \times 10}{35 \times 224}=28\)

2. \(\frac{3^4 \times 6^5}{72}\)

⇒ \(\frac{81 \times 776}{72}\)

⇒ \(\frac{629,856}{72}=8748\)

⇒ \(\frac{3^4 \times 6^5}{72}=8748\)

WBBSE Maths Study Material Class 7

Question 8. Express in terms of index.

1. \(\frac{(a b)^6 \times\left(a^2 c\right)^2}{\left(a^4\right)^2 \times(b c)^5}\)
2. \(9 \times 27 \times 81 \times 243\) (index of 3)

Solution:

1. \(\frac{a^6 \times b^6 \times a^2 \times a^2 \times c^2}{a^4 \times a^4 \times b^5 \times c^5 c^3}\)

⇒ \(\frac{a^2 \times b}{c^3}\)

⇒ \(a^2 \times b \times c^{-3}\)

∴ \(\frac{(a b)^6 \times\left(a^2 c\right)^2}{\left(a^4\right)^2 \times(b c)^2}=a^2 \times b \times c^{-3}\)

2. \(9 \times 27 \times 81 \times 243\) (index of 3)

⇒ 3 x 3 x 9 x 3 x 9 x 9 x 3 x 81

⇒ 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 9 x 9

⇒ 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3

⇒ 3¹⁴

∴ 9 x 27 x 81 x 243 = 3¹⁴