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# NCERT Exemplar for Class 7 Maths Solutions Chapter 11 Exponents & Powers

Last updated date: 12th Aug 2024
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## Class 7 Maths NCERT Exemplar Solutions Chapter 11 Exponents & Powers

The chapter Exponents and Powers help students in learning how to write large numbers in a shorter form so that it becomes easy to read, understand and compare. A number that is placed in the superscript position to the right of another number or the variable indicates the repeated multiplication concept. NCERT Exemplar Class 7 Math Solutions Chapter 11 Exponents & Powers help students in understanding the important concepts of this chapter. There are a total of 97 questions that will help you extensively practice this chapter.

Some of the basic concepts Class 7 students learn from this chapter are:

• Definition of Exponents

• Exponent Properties

• Exponential formulas.

• Scientific notation.

• Orders of Power and magnitude

• Laws of Exponents

• Multiplying Powers with the same base

• Division of Powers with the same base

• Multiplication of Powers with the same Exponents

• Division of the Powers with the same Exponents

• and other related fundamentals of the chapter.

Once students understand all the concepts, they can very easily solve all the questions of the topics mentioned in the NCERT books. Even if they get stuck at a step they can easily refer to NCERT Exemplar Class 7 Math Solutions Chapter 11 Exponents & Powers.

## Benefits of NCERT Exemplar Class 7 Math Solutions Chapter 11

Access NCERT Exemplar Solutions for Class 7 Mathematics

Chapter 11 - Exponents and Power

(Examples, Easy Methods and Step by Step Solutions)

### Solved Examples

In Examples 1 to 3, there are four options, out of which one is correct. Write the correct one.

Example 1: Out of the following, the number which is not equal to $\dfrac{{ - {\mathbf{8}}}}{{{\mathbf{27}}}}$ is

(a) $- {\left( {\dfrac{{\mathbf{2}}}{{\mathbf{3}}}} \right)^{\mathbf{3}}}$

(b) ${\left( {\dfrac{{ - {\mathbf{2}}}}{{\mathbf{3}}}} \right)^{\mathbf{3}}}$

(c) $- {\left( {\dfrac{{ - {\mathbf{2}}}}{{\mathbf{3}}}} \right)^{\mathbf{3}}}$

(d) $\left( {\dfrac{{ - {\mathbf{2}}}}{{\mathbf{3}}}} \right) \times \left( {\dfrac{{ - {\mathbf{2}}}}{{\mathbf{3}}}} \right) \times \left( {\dfrac{{ - {\mathbf{2}}}}{{\mathbf{3}}}} \right)$

Ans: Option (c) is the answer.

$\because - {\left( {\dfrac{{ - 2}}{3}} \right)^3} = - \left( {\dfrac{{ - 8}}{{27}}} \right) = \dfrac{8}{{27}}$

Example 2: ${\left( { - {\mathbf{7}}} \right)^{\mathbf{5}}} \times {\left( { - {\mathbf{7}}} \right)^{\mathbf{3}}}$is equal to

(a) ${\left( { - {\mathbf{7}}} \right)^{\mathbf{8}}}$

(b) $- {\left( {\mathbf{7}} \right)^{\mathbf{8}}}$

(c) ${\left( { - {\mathbf{7}}} \right)^{{\mathbf{15}}}}$

(d) ${\left( { - {\mathbf{7}}} \right)^{\mathbf{2}}}$

Ans: Option (a) is the answer.

By fundamental of exponents, we know $\Rightarrow {a^n} \times {a^m} = {a^{n+m}}$

$\therefore $${\left( { - {\mathbf{7}}} \right)^{\mathbf{5}}} \times {\left( { - {\mathbf{7}}} \right)^{\mathbf{3}}}$$= {\left( { - 7} \right)^{5 + 3}} = {\left( { - 7} \right)^8}$

Example 3: For any two non-zero integers x any y, ${{\mathbf{x}}^{\mathbf{3}}} \div {{\mathbf{y}}^{\mathbf{3}}}$ is equal to

(a) $\dfrac{{{{\mathbf{x}}^{\mathbf{0}}}}}{{\mathbf{y}}}$

(b) ${\left( {\dfrac{{\mathbf{x}}}{{\mathbf{y}}}} \right)^{\mathbf{3}}}$

(c) $\dfrac{{{{\mathbf{x}}^{\mathbf{6}}}}}{{\mathbf{y}}}$

(d) $\dfrac{{{{\mathbf{x}}^{\mathbf{9}}}}}{{\mathbf{y}}}$

Ans: Option (b) is the answer.

Using law of exponents,

$\dfrac{\mathrm{a}^{\mathrm{m}}}{\mathrm{b}^{\mathrm{m}}}=\left(\dfrac{\mathrm{a}}{\mathrm{b}}\right)^{\mathrm{m}}=(\mathrm{a} \div \mathrm{b})^{\mathrm{m}}$

$\therefore {x^3} \div {y^3} = {\left( {\dfrac{x}{y}} \right)^3}$

In Examples 4 and 5, fill in the blanks to make the statements true.

Example 4: ${\left( {{{\mathbf{5}}^{\mathbf{7}}} \div {{\mathbf{5}}^{\mathbf{6}}}} \right)^{\mathbf{2}}}$= ________

Ans: Solving the given expression:

$\Rightarrow {\left( {{5^7} \div {5^6}} \right)^2}$

$\Rightarrow {\left( {\dfrac{{{5^6} \times {5^1}}}{{{5^6}}}} \right)^2} = {\left( 5 \right)^2} = 25$

Example 5: $\dfrac{{{{\mathbf{a}}^{\mathbf{7}}}{{\mathbf{b}}^{\mathbf{3}}}}}{{{{\mathbf{a}}^{\mathbf{5}}}{\mathbf{b}}}}$= __________

Ans: Solving the given expression:

$\Rightarrow \dfrac{{{a^7}{b^3}}}{{{a^5}b}}$

$\Rightarrow \dfrac{{\left( {{a^5} \times {a^2}} \right)\left( {{b^2} \times b} \right)}}{{{a^5}b}} = {a^2}{b^2} = {\left( {ab} \right)^2}$

In Examples 6 to 8, state whether the statements are True or False:

Example 6: In the number ${{\mathbf{7}}^{\mathbf{5}}},{\mathbf{5}}$ is the base and ${\mathbf{7}}$ is the exponent.

Ans: False, because 7 is the base and 5 is in power so 5 is the exponent.

Example 7: $\dfrac{{{{\mathbf{a}}^{\mathbf{4}}}}}{{{{\mathbf{b}}^{\mathbf{3}}}}} = \dfrac{{{\mathbf{a}} + {\mathbf{a}} + {\mathbf{a}} + {\mathbf{a}}}}{{{\mathbf{b}} + {\mathbf{b}} + {\mathbf{b}}}}$

Ans: False, because ${a^n} = n$times $a$. Therefore,

$\dfrac{{{a^4}}}{{{b^3}}} = \dfrac{{a \times a \times a \times a}}{{b \times b \times b}}$

Example 8: ${{\mathbf{a}}^{\mathbf{b}}} > {{\mathbf{b}}^{\mathbf{a}}}$is true, if a = 3 and b = 4; but false, if a = 2 and b = 3.

Ans: True, because if $a = 2{\text{and}}b{\text{ = }}{\text{3}}$ then

$\Rightarrow {a^b} = {\left( 2 \right)^3} = 8,{b^a} = {\left( 3 \right)^2} = 9$

$\therefore {a^b} < {b^a}$

Example 9: By what number should we multiply ${{\mathbf{3}}^{\mathbf{3}}}$ so that the product may be equal to ${{\mathbf{3}}^{\mathbf{7}}}$?

Ans: Let the number be x then according to question,

By the property of exponents $\Rightarrow {a^m} \div {a^n} = {a^{m - n}}$

$\Rightarrow {3^3} \times x = {3^7}$

$\Rightarrow x = \dfrac{{{3^7}}}{{{3^3}}} = {\left( 3 \right)^{7 - 3}}$

$\Rightarrow x = {\left( 3 \right)^4} = 81$

Therefore, $x = 81$ or ${3^4}$.

Example 10: Find $x$ so that ${\left( {\dfrac{5}{3}} \right)^5} \times {\left( {\dfrac{5}{3}} \right)^{11}} = {\left( {\dfrac{5}{3}} \right)^{8x}}$

Ans: Given: ${\left( {\dfrac{5}{3}} \right)^5} \times {\left( {\dfrac{5}{3}} \right)^{11}} = {\left( {\dfrac{5}{3}} \right)^{8x}}$

Solving the given expression by the property of exponents $\Rightarrow {a^n} \times {a^m} = {a^{n + m}}$

${\left( {\dfrac{5}{3}} \right)^5} \times {\left( {\dfrac{5}{3}} \right)^{11}} = {\left( {\dfrac{5}{3}} \right)^{8x}}$

Equating the powers because base is same so,

$\Rightarrow 5 + 11 = 8x$

$\Rightarrow 16 = 8x$

$\Rightarrow x = \dfrac{{16}}{8} = 2$

$\therefore x = 2$

Example 11: Express 648 in exponential notation.

Ans: Taking factors of 648:

$\underline {2|648}$

$\underline {2|324}$

$\underline {2|162}$

$\underline {3|81}$

$\underline {3|27}$

$\underline {3|3}$

$\underline {|1}$

Therefore,

$\Rightarrow 648 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 = {2^3} \times {3^4}$.

Example 12: Express 2,36,00,000 in standard form.

Ans: Standard form of $2,36,00,000$= $\dfrac{{23600000}}{{10000000}} \times 10000000 = 2.36 \times {10^7}$.

Example 13: Which of the two is larger: ${{\mathbf{3}}^{{\mathbf{12}}}}$or ${{\mathbf{6}}^{\mathbf{6}}}$ ?

Ans: On comparing ${3^{12}} = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 531441$

And ${6^6} = 6 \times 6 \times 6 \times 6 \times 6 \times 6 = 46656$ So, ${3^{12}} > {6^6}$ .

Example 14: Find $x$ such that ${\dfrac{1}{5}^5} \times {\dfrac{1}{5}^{19}} = {\dfrac{1}{5}^{8x}}$

Ans: Understand and Explore the Problem:

• If you are trying to find the value of x for the given equation.

Plan a Strategy

• Apply laws of exponents in the given equation to find the value of $x$.

Solve

• Given, ${\dfrac{1}{5}^5} \times {\dfrac{1}{5}^{19}} = {\dfrac{1}{5}^{8x}}$

Solving the given expression by the property of exponents $\Rightarrow {a^n} \times {a^m} = {a^{n + m}}$comparing the powers because base is same so,

$\Rightarrow 5 + 19 = 8x$

$\Rightarrow 8x = 24$

$\Rightarrow x = \dfrac{{24}}{8} = 3$

$\therefore x = 3$

Revise

• Put the value of $x$ in the equation and check if it satisfies the equation and LHS = RHS

$\Rightarrow 5 + 19 = 8x$

$\Rightarrow 24 = 8\left( 3 \right)$

$\Rightarrow 24 = 24$

$\therefore {\text{LHS}}{\text{ = }}{\text{RHS}}$

Hence, the verified value of $x$ is correct.

Exercise

In questions 1 to 22, there are four options, out of which one is correct. Write the correct one.

1. ${\left[ {{{\left( {-{\mathbf{3}}} \right)}^{\mathbf{2}}}} \right]^{\mathbf{3}}}$is equal to

(a) ${\left( {-{\mathbf{3}}} \right)^{\mathbf{8}}}$

(b) ${\left( {-{\mathbf{3}}} \right)^{\mathbf{6}}}$

(c) ${\left( {-{\mathbf{3}}} \right)^{\mathbf{5}}}$

(d) ${\left( {-{\mathbf{3}}} \right)^{{\mathbf{23}}}}$

Ans: Option (b) is the answer.

Given: ${\left[ {{{\left( {-{\mathbf{3}}} \right)}^{\mathbf{2}}}} \right]^{\mathbf{3}}}$

Solving the given expression by the property of exponents $\Rightarrow {{a^n}^m} = {a^{nm}}$

$\Rightarrow {\left[ {{{\left( {-{\mathbf{3}}} \right)}^{\mathbf{2}}}} \right]^{\mathbf{3}}}$

$\Rightarrow {\left( { - 3} \right)^{2 \times 3}} = {\left( { - 3} \right)^6}$

2. For a non-zero rational number x, ${{\mathbf{x}}^{\mathbf{8}}} \div {{\mathbf{x}}^{\mathbf{2}}}$is equal to

(a) ${{\mathbf{x}}^4}$

(b) ${{\mathbf{x}}^{\mathbf{6}}}$

(c) ${{\mathbf{x}}^{{\mathbf{10}}}}$

(d) ${{\mathbf{x}}^{{\mathbf{16}}}}$

Ans: Option (b) is the answer.

Given: ${x^{\mathbf{8}}} \div {x^{\mathbf{2}}}$

By the property of exponents $\Rightarrow {a^m} \div {a^n} = {a^{m - n}}$

$\Rightarrow {x^8} \div {x^2}$

$\Rightarrow \left( {\dfrac{{{x^8}}}{{{x^2}}}} \right) = {x^{8 - 2}} = {x^6}$

3. x is a non-zero rational number. Product of the square of x with the cube of x is equal to the

(a) second power of x

(b) third power of x

(c) fifth power of x

(d) sixth power of x

Ans: Option (c) is the answer.

Solving the given expression by the property of exponents $\Rightarrow {a^n} \times {a^m} = {a^{n + m}}$

$\Rightarrow {x^2} \times {x^3} = {\left( x \right)^{2 + 3}} = {x^5}$.

4. For any two non-zero rational numbers x and y,{x^5} \div {y^5} is equals to

(a) ${\left( {{\mathbf{x}} \div {\mathbf{y}}} \right)^1}$

(b) ${\left( {{\mathbf{x}} \div {\mathbf{y}}} \right)^0}$

(c) ${\left( {{\mathbf{x}} \div {\mathbf{y}}} \right)^5}$

(d) ${\left( {{\mathbf{x}} \div {\mathbf{y}}} \right)^{{\mathbf{10}}}}$

Ans: By the property of exponents $\Rightarrow {a^n} \div {b^n} = {\left( {a \div b} \right)^n}$

$\Rightarrow {x^5} \div {y^5} = {\left( {x \div y} \right)^5}$

Option (c) is correct.

5. ${{\mathbf{a}}^{\mathbf{m}}} \times {{\mathbf{a}}^{\mathbf{n}}}$ is equals to

(a) ${\left( {{{\mathbf{a}}^{\mathbf{2}}}} \right)^{{\mathbf{mn}}}}$

(b) ${{\mathbf{a}}^{{\mathbf{m}} - {\mathbf{n}}}}$

(c) ${{\mathbf{a}}^{{\mathbf{m}} + {\mathbf{n}}}}$

(d) ${{\mathbf{a}}^{{\mathbf{mn}}}}$

Ans: Option (c) is the answer.

Solving the given expression by the property of exponents $\Rightarrow {a^m} \times {a^n} = {a^{m + n}}$

6. $\left( {{{\mathbf{1}}^{\mathbf{0}}} + {{\mathbf{2}}^{\mathbf{0}}} + {{\mathbf{3}}^{\mathbf{0}}}} \right)$ is equal to

(a) 0

(b) 1

(c) 3

(d) 6

Ans: Option (c) is the answer.

Given: $\left( {{1^0} + {2^0} + {3^0}} \right)$

By the property of exponents and powers $\Rightarrow {a^0} = 1$

$\therefore \left( {{1^0} + {2^0} + {3^0}} \right) = \left( {1 + 1 + 1} \right) = 3$.

7. Value of  $\dfrac{{{\mathbf{1}}{{\mathbf{0}}^{{\mathbf{22}}}} + {\mathbf{1}}{{\mathbf{0}}^{{\mathbf{20}}}}}}{{{\mathbf{1}}{{\mathbf{0}}^{{\mathbf{20}}}}}}$ is

(a) ${\mathbf{10}}$

(b) ${\mathbf{1}}{{\mathbf{0}}^{{\mathbf{42}}}}$

(c) ${\mathbf{101}}$

(d) ${\mathbf{1}}{{\mathbf{0}}^{{\mathbf{22}}}}$

Ans: Option (c) is the answer.

Given: $\dfrac{{{{10}^{22}} + {{10}^{20}}}}{{{{10}^{20}}}}$

Solving the given expression by the property of exponents $\Rightarrow {a^n} \times {a^m} = {a^{n + m}}$

$\Rightarrow $$\dfrac{{{{10}^{22}} + {{10}^{20}}}}{{{{10}^{20}}}} \Rightarrow \dfrac{{\left( {{{10}^{20}} \times {{10}^2}} \right) + {{10}^{20}}}}{{{{10}^{20}}}} = \dfrac{{{{10}^{20}} \times \left( {{{10}^2} + 1} \right)}}{{{{10}^{20}}}} \Rightarrow \left( {100 + 1} \right) = 101 8. The standard form of the number 12345 is (a) $1234.5 \times {10^1}$ (b) $123.45 \times {10^2}$ (c) $12.345 \times {10^3}$ (d) $1.2345 \times {10^4}$ Ans: Option (d) is the answer. Standard form of 12345 is = 1.2345 \times 10000 = 1.2345 \times {10^4} 9. If ${2^1998}-{2^1997}-{2^1996}+{2^1995}=K \times {2^1995}$, then the value of K Is: (a) 1 (b) 2 (c) 3 (d) 4 Ans: Option (c) is the answer. Solving the given expression by the property of exponents \Rightarrow {a^n} \times {a^m} = {a^{n + m}} and \Rightarrow {a^m} \div {a^n} = {a^{m - n}} Therefore, \Rightarrow {2^{1998}} - {2^{1997}} - {2^{1996}} + {2^{1995}} = {\text{K}} \times {{\text{2}}^{1995}} \Rightarrow {2^{1995}}\left( {{2^0} - {2^2} - {2^1} + {2^3}} \right) = {\text{K}} \times {{\text{2}}^{1995}} \Rightarrow \left( {1 - 4 - 2 + 8} \right) = {\text{K}} \Rightarrow \left( {1 - 6 + 8} \right) = {\text{K}} \Rightarrow \left( { - 5 + 8} \right) = {\text{K}} \Rightarrow {\text{K}} = 3 10. Which of the following is equal to 1? (a) {{\mathbf{2}}^{\mathbf{0}}} + {{\mathbf{3}}^0} + {{\mathbf{4}}^{\mathbf{0}}} (b) {{\mathbf{2}}^{\mathbf{0}}} \times {{\mathbf{3}}^0} \times {{\mathbf{4}}^{\mathbf{0}}} (c) ({{\mathbf{3}}^0} - {{\mathbf{2}}^{\mathbf{0}}}) \times {{\mathbf{4}}^{\mathbf{0}}} (d) \left( {{{\mathbf{3}}^0} - {{\mathbf{2}}^{\mathbf{0}}}} \right) \times \left( {{{\mathbf{3}}^0} + {{\mathbf{2}}^{\mathbf{0}}}} \right) Ans: Option (b) is the answer. By the property of exponents \Rightarrow {a^0} = 1 \Rightarrow \left( {{2^0} \times {3^0} \times {4^0}} \right) \Rightarrow \left( {1 \times 1 \times 1} \right) = 1 11. In standard form, the number 72105.4 is written as 7.21054 × ${\mathbf{1}}{{\mathbf{0}}^{\mathbf{n}}}$ where n is equal to (a) 2 (b) 3 (c) 4 (d) 5 Ans: Option (c) is the answer. Standard form of 72105.4 = 7.21054 \times {10^{\mathbf{n}}} Comparing the exponents: \Rightarrow 7.21054 \times {10^4} = 7.21054 \times {10^{\text{n}}} \Rightarrow {10^4} = {10^{\text{n}}} \therefore {\text{n}} = 4 12. Square of \left( {\dfrac{{ - {\mathbf{2}}}}{{\mathbf{3}}}} \right) is (a) \dfrac{{ - {\mathbf{2}}}}{{\mathbf{3}}} (b) \dfrac{{\mathbf{2}}}{{\mathbf{3}}} (c) \dfrac{{ - {\mathbf{4}}}}{{\mathbf{9}}} (d) \dfrac{{\mathbf{4}}}{{\mathbf{9}}} Ans: Option (d) is the answer. As per the given condition square of \left( {\dfrac{{ - 2}}{3}} \right): \Rightarrow {\left( {\dfrac{{ - 2}}{3}} \right)^2} = \dfrac{4}{9} 13. Cube of \left( {\dfrac{{ - {\mathbf{1}}}}{{\mathbf{4}}}} \right)is (a) \left( {\dfrac{{ - {\mathbf{1}}}}{{{\mathbf{12}}}}} \right) (b) \dfrac{{\mathbf{1}}}{{{\mathbf{16}}}} (c) \dfrac{{ - {\mathbf{1}}}}{{{\mathbf{64}}}} (d) \dfrac{{\mathbf{1}}}{{{\mathbf{64}}}} Ans: Option (d) is the answer. As per the given condition cube of \left( {\dfrac{{ - 1}}{4}} \right)is: \Rightarrow {\left( {\dfrac{{ - 1}}{4}} \right)^3} \Rightarrow \left( {\dfrac{{{{\left( { - 1} \right)}^3}}}{{{{\left( 4 \right)}^3}}}} \right) = \dfrac{{ - 1}}{{64}} 14. Which of the following is not equal to{\left( {\dfrac{{ - 5}}{4}} \right)^4}? (a) \dfrac{{{{\left( { - {\mathbf{5}}} \right)}^{\mathbf{4}}}}}{{{{\mathbf{4}}^{\mathbf{4}}}}} (b) \dfrac{{{{\mathbf{5}}^{\mathbf{4}}}}}{{{{\left( { - {\mathbf{4}}} \right)}^{\mathbf{4}}}}} (c) - \dfrac{{{{\mathbf{5}}^{\mathbf{4}}}}}{{{{\mathbf{4}}^{\mathbf{4}}}}} (d) \left( { - \dfrac{5}{4}} \right) \times \left( { - \dfrac{5}{4}} \right) \times \left( { - \dfrac{5}{4}} \right) \times \left( { - \dfrac{5}{4}} \right) Ans: Option (c) is the answer. By the property of fundamental of ‘Exponents and Powers’ {\left( {\dfrac{a}{b}} \right)^n} = \dfrac{{{{\left( a \right)}^n}}}{{{{\left( b \right)}^n}}} Therefore, \Rightarrow {\left( {\dfrac{{ - 5}}{4}} \right)^4} = \dfrac{{{{\left( { - 5} \right)}^4}}}{{{{\left( 4 \right)}^4}}} = {\dfrac{5^4}{4^4}} For option (c) -\dfrac{{{{\mathbf{5}}^{\mathbf{4}}}}}{{{{\mathbf{4}}^{\mathbf{4}}}}} value is not equal to \dfrac{ {5^4}}{{4^4}}. 15. Which of the following is not equal to 1? (a) \dfrac{{{{\mathbf{2}}^{\mathbf{3}}} \times {{\mathbf{3}}^{\mathbf{2}}}}}{{{\mathbf{4}} \times {\mathbf{18}}}} (b) \left[ {{{\left( { - {\mathbf{2}}} \right)}^{\mathbf{3}}} \times {{\left( { - {\mathbf{2}}} \right)}^{\mathbf{4}}}} \right] \div {\left( { - {\mathbf{2}}} \right)^{\mathbf{7}}} (c) \dfrac{{{{\mathbf{3}}^{\mathbf{0}}} \times {{\mathbf{5}}^{\mathbf{3}}}}}{{{\mathbf{5}} \times {\mathbf{25}}}} (d) \dfrac{{{{\mathbf{2}}^{\mathbf{2}}}}}{{{{\left( {{{\mathbf{7}}^{\mathbf{0}}} + {{\mathbf{3}}^{\mathbf{0}}}} \right)}^{\mathbf{3}}}}} Ans: Option (d) is the answer. By the property of fundamental of ‘Exponents and Powers’ {\left( {\dfrac{a}{b}} \right)^n} = \dfrac{{{{\left( a \right)}^n}}}{{{{\left( b \right)}^n}}}, {a^n} \times {a^m} = {a^{n + m}},{a^n} \div {a^m} = {a^{n - m}}, \left( {{a^0}} \right) = 1and {\left( {{a^n}} \right)^m} = {a^{n \times m}} Solving for (d) option: \Rightarrow \dfrac{{{2^2}}}{{{{\left( {{7^0} + {3^0}} \right)}^3}}} \Rightarrow \dfrac{{{2^2}}}{{{{\left( {1 + 1} \right)}^3}}} = \dfrac{4}{{{{\left( 2 \right)}^3}}} \Rightarrow \dfrac{4}{8} = \dfrac{1}{2} \therefore \dfrac{1}{2} \ne 1 16. ${\left( {\dfrac{{\mathbf{2}}}{{\mathbf{3}}}} \right)^{\mathbf{3}}} \times {\left( {\dfrac{{\mathbf{5}}}{{\mathbf{7}}}} \right)^{\mathbf{3}}}$is equal to (a) {\left( {\dfrac{{\mathbf{2}}}{{\mathbf{3}}} \times \dfrac{{\mathbf{5}}}{{\mathbf{7}}}} \right)^{\mathbf{9}}} (b) {\left( {\dfrac{{\mathbf{2}}}{{\mathbf{3}}} \times \dfrac{{\mathbf{5}}}{{\mathbf{7}}}} \right)^{\mathbf{6}}} (c) {\left( {\dfrac{{\mathbf{2}}}{{\mathbf{3}}} \times \dfrac{{\mathbf{5}}}{{\mathbf{7}}}} \right)^{\mathbf{3}}} (d) {\left( {\dfrac{{\mathbf{2}}}{{\mathbf{3}}} \times \dfrac{{\mathbf{5}}}{{\mathbf{7}}}} \right)^{\mathbf{0}}} Ans: Option (c) is the answer. By the property of fundamental of ‘Exponents and Powers’ \Rightarrow {a^n} \times {b^n} = {\left( {ab} \right)^n} Solving: \Rightarrow {\left( {\dfrac{2}{3}} \right)^3} \times {\left( {\dfrac{5}{7}} \right)^3} = {\left( {\dfrac{2}{3} \times \dfrac{5}{7}} \right)^3}Hence, (c) is the answer. 17. In standard form, the number 829030000 is written as K $\times {\text{ }}{\mathbf{1}}{{\mathbf{0}}^{\mathbf{8}}}$where K is equal to (a) 82903 (b) 829.03 (c) 82.903 (d) 8.2903 Ans: Option (d) is the answer. Standard form of 829030000 = 8.29030000 \times {10^8} = 8.2903 \times {10^8} Comparing to find the value of K \Rightarrow {\text{K}} \times {\text{1}}{{\text{0}}^8} = 8.2903 \times {10^8} \therefore {\text{K}} = {\text{8}}{\text{.2903}} 18. Which of the following has the largest value? (a) 0.0001 (b) \dfrac{{\mathbf{1}}}{{{\mathbf{10000}}}} (c) \dfrac{{\mathbf{1}}}{{{\mathbf{1}}{{\mathbf{0}}^{\mathbf{6}}}}} (d) \dfrac{{\mathbf{1}}}{{{\mathbf{1}}{{\mathbf{0}}^{\mathbf{6}}}}} \div {\mathbf{0}}.{\mathbf{1}} Ans: Option (a) and (b) is the answer. As 0.0001 = \dfrac{1}{{10000}} and \dfrac{1}{{10000}}are equal. Also, option c \Rightarrow \dfrac{1}{{{{10}^6}}} = \dfrac{1}{{1000000}} And option (d) \Rightarrow \dfrac{1}{{{{10}^6}}} \div 0.1 = \dfrac{1}{{{{10}^6}}} \times 10 = \dfrac{1}{{100000}} Therefore, option (c) and (d) have smaller values. 19. In standard form 72 crore is written as (a) 72 × ${\mathbf{1}}{{\mathbf{0}}^7}$ (b) 72 × ${\mathbf{1}}{{\mathbf{0}}^{\mathbf{8}}}$ (c) 7.2 × ${\mathbf{1}}{{\mathbf{0}}^{\mathbf{8}}}$ (d) 7.2 × ${\mathbf{1}}{{\mathbf{0}}^7}$ Ans: Option (c) is the answer. 72 crores = 720000000 Standard form of 720000000 = 7.2 \times {10^8} 20. For non-zero numbers a and b, {\left( {\dfrac{{\mathbf{a}}}{{\mathbf{b}}}} \right)^{\mathbf{m}}} \div {\left( {\dfrac{{\mathbf{a}}}{{\mathbf{b}}}} \right)^{\mathbf{n}}} , where m > n, is equal to (a) {\left( {\dfrac{{\mathbf{a}}}{{\mathbf{b}}}} \right)^{{\mathbf{mn}}}} (b) {\left( {\dfrac{{\mathbf{a}}}{{\mathbf{b}}}} \right)^{{\mathbf{m}} + {\mathbf{n}}}} (c) {\left( {\dfrac{{\mathbf{a}}}{{\mathbf{b}}}} \right)^{{\mathbf{m}} - {\mathbf{n}}}} (d) {\left( {{{\left( {\dfrac{{\mathbf{a}}}{{\mathbf{b}}}} \right)}^{\mathbf{m}}}} \right)^{\mathbf{n}}} Ans: Option (c) is the answer. By the property of fundamental of ‘Exponents and Powers’ {a^m} \div {a^n} = {a^{m - n}} Given: {\left( {\dfrac{a}{b}} \right)^m} \div {\left( {\dfrac{a}{b}} \right)^n} = {\left( {\dfrac{a}{b}} \right)^{m - n}} 21. Which of the following is not true? (a) ${{\mathbf{3}}^{\mathbf{2}}} > {{\mathbf{2}}^{\mathbf{3}}}$ (b) {{\mathbf{4}}^{\mathbf{3}}} = {{\mathbf{2}}^{\mathbf{6}}} (c) {{\mathbf{3}}^{\mathbf{3}}} = {\mathbf{9}} (d) {{\mathbf{2}}^{\mathbf{5}}} > {{\mathbf{5}}^{\mathbf{2}}} Ans: Option (c) is the answer. Given, {3^3} \ne 9 {3^3} = 27\therefore 27 \ne 9{\text{ but 27}}{\text{ > }}{\text{9}} 22. Which power of 8 is equal to {{\mathbf{2}}^{\mathbf{6}}}? (a) 3 (b) 2 (c) 1 (d) 4 Ans: Option (b) is the answer. \Rightarrow {2^6} = 64 {\text{and }}{\left( 8 \right)^2} = 64 OR \Rightarrow {\left( 8 \right)^x} = {2^6} \Rightarrow {\left( {{2^3}} \right)^x} = {2^6} \Rightarrow {2^{3x}} = {2^6} \Rightarrow 3x = 6 \Rightarrow x = \dfrac{6}{3} = 2 In questions 23 to 39, fill in the blanks to make the statements true. 23. {\left( { - 2} \right)^{31}} \times {\left( { - 2} \right)^{13}} = {\left( { - 2} \right)^ - } Ans: Given: {\left( { - 2} \right)^{31}} \times {\left( { - 2} \right)^{13}} = {\left( { - 2} \right)^ - } By the property of fundamental of ‘Exponents and Powers’ \Rightarrow {a^n} \times {a^m} = {a^{n + m}} Let (---) = x \therefore$${\left( { - 2} \right)^{31}} \times {\left( { - 2} \right)^{13}} = {\left( { - 2} \right)^x}$

Comparing powers as base is same,

$\Rightarrow 31 + 13 = x$

$\Rightarrow x = 44$

$\therefore x = 44$

24. ${\left( { - 3} \right)^8} \div {\left( { - 3} \right)^5} = {\left( { - 3} \right)^{ - - - }}$

Ans: Given: ${\left( { - 3} \right)^8} \div {\left( { - 3} \right)^5} = {\left( { - 3} \right)^{ - - - }}$

By the property of the fundamentals of ‘Exponents and Powers, ${a^n} \div {a^m} = {a^{n - m}}$.

Let (---) = $x$

$\therefore $${\left( { - 3} \right)^8} \div {\left( { - 3} \right)^5} = {\left( { - 3} \right)^{ - - - }} Comparing powers as base is same, \Rightarrow \dfrac{{{{\left( { - 3} \right)}^8}}}{{{{\left( { - 3} \right)}^5}}} = {\left( { - 3} \right)^x} \Rightarrow {\left( { - 3} \right)^{8 - 5}} = {\left( { - 3} \right)^x} \Rightarrow 8 - 5 = x \therefore x = 3 25. {\left( {\dfrac{{{\mathbf{11}}}}{{{\mathbf{15}}}}} \right)^{\mathbf{4}}} \times {\left( { - - - } \right)^{\mathbf{5}}} = {\left( {\dfrac{{{\mathbf{11}}}}{{{\mathbf{15}}}}} \right)^{\mathbf{9}}} Ans: Given: {\left( {\dfrac{{11}}{{15}}} \right)^4} \times {\left( { - - - } \right)^5} = {\left( {\dfrac{{11}}{{15}}} \right)^9} Let (---) = x \therefore By the property of fundamental of ‘Exponents and Powers’{a^n} \div {a^m} = {a^{n - m}} \Rightarrow$${\left( {\dfrac{{11}}{{15}}} \right)^4} \times {\left( x \right)^5} = {\left( {\dfrac{{11}}{{15}}} \right)^9}$

$\Rightarrow {\left( x \right)^5} = {\left( {\dfrac{{11}}{{15}}} \right)^9} \div {\left( {\dfrac{{11}}{{15}}} \right)^4}$

$\Rightarrow {\left( x \right)^5} = {\left( {\dfrac{{11}}{{15}}} \right)^{9 - 4}}$

$\Rightarrow {\left( x \right)^5} = {\left( {\dfrac{{11}}{{15}}} \right)^5}$

If power is same, then we compare base therefore,

$\Rightarrow x = \dfrac{{11}}{{15}}$

$\therefore x = \dfrac{{11}}{{15}}$

26. ${\left( {\dfrac{{ - {\mathbf{1}}}}{{\mathbf{4}}}} \right)^{\mathbf{3}}} \times {\left( {\dfrac{{ - {\mathbf{1}}}}{{\mathbf{4}}}} \right)^{ - - - }} = {\left( {\dfrac{{ - {\mathbf{1}}}}{{\mathbf{4}}}} \right)^{{\mathbf{11}}}}$

Ans: Given: ${\left( {\dfrac{{ - 1}}{4}} \right)^3} \times {\left( {\dfrac{{ - 1}}{4}} \right)^{ - - - }} = {\left( {\dfrac{{ - 1}}{4}} \right)^{11}}$

Let (---) = $x$

By the property of fundamental of ‘Exponents and Powers’${a^n} \div {a^m} = {a^{n - m}}$

$\Rightarrow $${\left( {\dfrac{{ - 1}}{4}} \right)^3} \times {\left( {\dfrac{{ - 1}}{4}} \right)^x} = {\left( {\dfrac{{ - 1}}{4}} \right)^{11}} \Rightarrow {\left( {\dfrac{{ - 1}}{4}} \right)^x} = {\left( {\dfrac{{ - 1}}{4}} \right)^{11}} \div {\left( {\dfrac{{ - 1}}{4}} \right)^3} \Rightarrow {\left( {\dfrac{{ - 1}}{4}} \right)^x} = {\left( {\dfrac{{ - 1}}{4}} \right)^{11 - 3}} \Rightarrow {\left( {\dfrac{{ - 1}}{4}} \right)^x} = {\left( {\dfrac{{ - 1}}{4}} \right)^8} Comparing powers as base is same, \Rightarrow x = 8 \therefore x = 8 27. {\left[ {{{\left( {\dfrac{{\mathbf{7}}}{{{\mathbf{11}}}}} \right)}^{\mathbf{3}}}} \right]^{\mathbf{4}}} = {\left( {\dfrac{{\mathbf{7}}}{{{\mathbf{11}}}}} \right)^{ - - - }} Ans: Given: {\left[ {{{\left( {\dfrac{7}{{11}}} \right)}^3}} \right]^4} = {\left({\dfrac{7}{{11}}} \right)^{ - - - }} Let (---) = x By the property of fundamental of ‘Exponents and Powers’ {\left( {{a^n}} \right)^m} = {a^{n \times m}} \therefore {\left[ {{{\left( {\dfrac{7}{{11}}} \right)}^3}} \right]^4} = {\left( {\dfrac{7}{{11}}} \right)^{ - - - }} \Rightarrow {\left[ {{{\left( {\dfrac{7}{{11}}} \right)}^3}} \right]^4} = {\left( {\dfrac{7}{{11}}} \right)^x} \Rightarrow {\left[ {\dfrac{7}{{11}}} \right]^{3 \times 4}} = {\left( {\dfrac{7}{{11}}} \right)^x} \Rightarrow {\left( {\dfrac{7}{{11}}} \right)^{12}} = {\left( {\dfrac{7}{{11}}} \right)^x} Comparing powers as base is same, \therefore$$x = 12$

28. ${\left( {\dfrac{6}{{13}}} \right)^{10}} \div {\left[ {{{\left( {\dfrac{6}{{13}}} \right)}^5}} \right]^2} = {\left( {\dfrac{6}{{13}}} \right)^{ - - - }}$

Ans: Given: ${\left( {\dfrac{6}{{13}}} \right)^{10}} \div {\left[ {{{\left( {\dfrac{6}{{13}}} \right)}^5}} \right]^2} = {\left( {\dfrac{6}{{13}}} \right)^{ - - - }}$

Let (---) = $x$

By the property of fundamental of ‘Exponents and Powers’${a^n} \div {a^m} = {a^{n - m}}$,${\left( {{a^n}} \right)^m} = {a^{n \times m}}$

$\therefore $${\left( {\dfrac{6}{{13}}} \right)^{10}} \div {\left[ {{{\left( {\dfrac{6}{{13}}} \right)}^5}} \right]^2} = {\left( {\dfrac{6}{{13}}} \right)^x} \Rightarrow {\left( {\dfrac{6}{{13}}} \right)^{10}} \div {\left[ {\dfrac{6}{{13}}} \right]^{5 \times 2}} = {\left( {\dfrac{6}{{13}}} \right)^x} \Rightarrow {\left( {\dfrac{6}{{13}}} \right)^{10}} \div {\left( {\dfrac{6}{{13}}} \right)^{10}} = {\left( {\dfrac{6}{{13}}} \right)^x} \Rightarrow {\left( {\dfrac{6}{{13}}} \right)^{10 - 10}} = {\left( {\dfrac{6}{{13}}} \right)^x} \Rightarrow {\left( {\dfrac{6}{{13}}} \right)^0} = {\left( {\dfrac{6}{{13}}} \right)^x} \Rightarrow x = 0 29. {\left[ {{{\left( {\dfrac{{ - 1}}{4}} \right)}^{16}}} \right]^2} = {\left( {\dfrac{{ - 1}}{4}} \right)^{ - - - }} Ans: Given: {\left[ {{{\left( {\dfrac{{ - 1}}{4}} \right)}^{16}}} \right]^2} = {\left( {\dfrac{{ - 1}}{4}} \right)^{ - - - }} Let (---) = x By the property of fundamental of ‘Exponents and Powers’, and {\left( {{a^n}} \right)^m} = {a^{n \times m}} \therefore$${\left[ {{{\left( {\dfrac{{ - 1}}{4}} \right)}^{16}}} \right]^2} = {\left( {\dfrac{{ - 1}}{4}} \right)^x}$

$\Rightarrow {\left( {\dfrac{{ - 1}}{4}} \right)^{16 \times 2}} = {\left( {\dfrac{{ - 1}}{4}} \right)^x}$

$\Rightarrow {\left( {\dfrac{{ - 1}}{4}} \right)^{32}} = {\left( {\dfrac{{ - 1}}{4}} \right)^x}$

Comparing powers as base is same:

$\therefore x = 32$.

30. ${\left( {\dfrac{{13}}{{14}}} \right)^5} \div {\left( {\_\_\_} \right)^2} = {\left( {\dfrac{{13}}{{14}}} \right)^3}$

Ans: Given: ${\left( {\dfrac{{13}}{{14}}} \right)^5} \div {\left( {\_\_\_} \right)^2} = {\left( {\dfrac{{13}}{{14}}} \right)^3}$

Let (---) = $x$

By the property of fundamental of ‘Exponents and Powers’${a^n} \div {a^m} = {a^{n - m}}$

$\therefore $${\left( {\dfrac{{13}}{{14}}} \right)^5} \div {\left( x \right)^2} = {\left( {\dfrac{{13}}{{14}}} \right)^3} \Rightarrow {\left( {\dfrac{{13}}{{14}}} \right)^5} \div {\left( x \right)^2} = {\left( {\dfrac{{13}}{{14}}} \right)^3} \therefore {\left( {\dfrac{{13}}{{14}}} \right)^5} = {\left( {\dfrac{{13}}{{14}}} \right)^3} \times {\left( x \right)^2} \Rightarrow {\left( {\dfrac{{13}}{{14}}} \right)^5} \div {\left( {\dfrac{{13}}{{14}}} \right)^3} = {\left( x \right)^2} \Rightarrow {\left( {\dfrac{{13}}{{14}}} \right)^{5 - 3}} = {\left( x \right)^2} \Rightarrow {\left( {\dfrac{{13}}{{14}}} \right)^2} = {\left( x \right)^2} If powers are same then we compare base: \therefore$$x = \dfrac{{13}}{{14}}$

31. ${a^5} \times {a^5} \times {a^0} = {a^{ - - - }}$

Ans: Given: ${a^5} \times {a^5} \times {a^0} = {a^{ - - - }}$

Let (---) = $x$

Then,

$\Rightarrow {a^5} \times {a^5} \times {a^0} = {a^{ - - - }}$

$\Rightarrow {a^5} \times {a^5} \times {a^0} = {a^x}$

By the property of fundamental of ‘Exponents and Powers’$\Rightarrow {a^n} \times {a^m} = {a^{n + m}}$

$\Rightarrow {a^{5 + 5 + 0}} = {a^x}$

$\Rightarrow {a^{10}} = {a^x}$

$\therefore x = 10$

(Comparing powers as base is same)

32. 1 lakh = ${\mathbf{1}}{{\mathbf{0}}^{ - - - }}$

Ans: Given: 1 lakh = ${10^{ - - - }}$

1 lakh = 100000 = $10 \times 10 \times 10 \times 10 \times 10 = {10^5}$

Let (---) = $x$

Then,

$\Rightarrow 100000 \Rightarrow {10^5} = {10^x}$

Comparing powers as base is same,

$\therefore x = 5$

33. 1 million = ${\mathbf{1}}{{\mathbf{0}}^{ - - - }}$

Ans: Given: 1 million = ${10^{ - - - }}$

Let (---) = $x$

1 million = 10 lakh = 1 lakh x 10 = $10 \times 10 \times 10 \times 10 \times 10 \times 10$$=$${10^6}$

Then,

$\Rightarrow 1000000 \Rightarrow {10^6} = {10^x}$

$\Rightarrow {10^6} = {10^x}$

$\therefore x = 6$

(Comparing powers as base is same)

34. 729 = ${{\mathbf{3}}^{ - - - }}$

Ans: Given: 729 = ${3^{ - - - }}$

Let (---) = $x$

$\therefore 729 = {3^x}$

$\Rightarrow 729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = {3^6}$

$\Rightarrow {3^6} = {3^x}$

$\therefore x = 6$

(Comparing powers as base is same)

35. 432 = ${{\mathbf{2}}^{\mathbf{4}}}$ × ${{\mathbf{3}}^{ - - - }}$

Ans: Given: $432 = {2^4} \times {3^{ - - - }}$

Let (---) = $x$

$\Rightarrow 432 = {2^4} \times {3^x}$

$\Rightarrow 432 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 = {2^4} \times {3^3}$

$\therefore {2^4} \times {3^3} = {2^4} \times {3^x}$

$\Rightarrow {3^x} = {3^3}$

$\therefore x = 3$

(Comparing powers as base is same)

36. 53700000 = ––– × ${\mathbf{1}}{{\mathbf{0}}^{\mathbf{7}}}$

Ans: Given: 53700000 = ------$\times {10^7}$

Let (---) = $x$

$53700000 = 537 \times {10^5}$

${\text{Also, 5}}{\text{.37}}{\text{ = }}{\text{5}}{\text{.37}} \times {\text{1}}{{\text{0}}^2}$

$\therefore 5.3700000 \times {10^7} = x \times {10^7}$

$\therefore x = 5.37$

37. 88880000000 = ––– × ${\mathbf{1}}{{\mathbf{0}}^{{\mathbf{10}}}}$

Ans: Given: 8888000000 = -----$\times {10^{10}}$

Let (---) = $x$

$\Rightarrow 88880000000 = x \times {10^{10}}$

$88880000000 = 8888 \times {10^7} = 8.888 \times {10^7} \times {10^3}$

$\Rightarrow 8.888 \times {10^{10}} = x \times {10^{10}}$

$\therefore x = 8.888$

38. 27500000 = 2.75 × ${\mathbf{1}}{{\mathbf{0}}^{ - - - }}$

Ans: Given: 27500000 = 2.75 $\times {10^{ - - - }}$

Let (---) = $x$

By the property of fundamental of ‘Exponents and Powers’$\Rightarrow {a^n} \times {a^m} = {a^{n + m}}$

$\Rightarrow 27500000 = 2.75 \times {10^x}$

$27500000 = 275 \times {10^5} = 2.75 \times {10^5} \times {10^2}$

$\Rightarrow 2.75 \times {10^7} = 2.75 \times {10^x}$

$\therefore {10^7} = {10^x}$

$\therefore x = 7$

(Comparing powers as base is same)

39. 340900000 = 3.409 × ${\mathbf{1}}{{\mathbf{0}}^{ - - - }}$

Ans: Given: 340900000 = 3.409 $\times {10^{ - - - }}$

Let (---) = $x$

By the property of fundamental of ‘Exponents and Powers’$\Rightarrow {a^n} \times {a^m} = {a^{n + m}}$

$\Rightarrow 340900000 = 3.409 \times {10^x}$

$340900000 = 3.409 \times {10^5} \times {10^3}$

$\Rightarrow 3.409 \times {10^8} = 3.409 \times {10^x}$

$\therefore {10^8} = {10^x}$

$\therefore x = 8$

(Comparing powers as base is same)

40. Fill in the blanks with <, > or = sign.

(a) ${{\mathbf{3}}^{\mathbf{2}}}$ ______15

Ans: ${3^2}\_\_\_15$

$\because {3^2} = 3 \times 3 = 9$

$\therefore 9 < 15$

So, ${3^2} < 15$

(b) ${{\mathbf{2}}^{\mathbf{3}}}$ ______ ${{\mathbf{3}}^{\mathbf{2}}}$

Ans: ${2^3}\_\_\_{3^2}$

$\therefore {2^3} = 2 \times 2 \times 2 = 8$

${3^2} = 3 \times 3 = 9$

${\text{and }}{\text{8}}{\text{ < }}{\text{9}}$

$\therefore {2^3} < {3^2}$

(c) ${{\mathbf{7}}^{\mathbf{4}}}$ ______${{\mathbf{5}}^{\mathbf{4}}}$

Ans: ${7^{4}}\_\_\_\_{5^4}$

$\therefore {7^4} = 7 \times 7 \times 7 \times 7 = 2401$

${5^4} = 5 \times 5 \times 5 \times 5 = 625$

${\text{and }}2401 > 625$

$\therefore {7^4} > {5^4}$

(d) 10,000 ______ ${\mathbf{1}}{{\mathbf{0}}^{\mathbf{5}}}$

Ans: $10000\_\_\_{10^5}$

$\because 10000 = 10 \times 10 \times 10 \times 10 = {10^4}$

${\text{and 1}}{{\text{0}}^4} < {10^5}$

$\therefore {10^4} < {10^5}$

(e) ${{\mathbf{6}}^{\mathbf{3}}}$ _____${{\mathbf{4}}^{\mathbf{4}}}$

Ans: ${6^3}\_\_\_{4^4}$

$\therefore {6^3} = 6 \times 6 \times 6 = 216$

${4^4} = 4 \times 4 \times 4 \times 4 = 256$

${\text{and }}216 < 256$

$\therefore {6^3} < {4^4}$

In questions 41 to 65, state whether the given statements are True or False.

41. One million = ${\mathbf{1}}{{\mathbf{0}}^{\mathbf{7}}}$

Ans: False.

Because, 1 million = 10 lakh = 1 lakh x 10 = $10 \times 10 \times 10 \times 10 \times 10 \times 10$$=$${10^6}$

42. One hour = ${\mathbf{6}}{{\mathbf{0}}^{\mathbf{2}}}$ seconds

Ans: True,

Because, one hour = 60 minutes

And, 1 minute = 60 seconds

Therefore, one hour = $60 \times 60 = 3600 = {60^2}$seconds.

43. ${{\mathbf{1}}^{\mathbf{0}}} \times {\mathbf{0}} = {\mathbf{1}}$

Ans: False.

$\because {a^0} = 1$

$\Rightarrow {1^0} \times 0 = 1$

$\Rightarrow 1 \times 0 = 1$

$\Rightarrow 0 \ne 1$

44. ${\left( {-{\mathbf{3}}} \right)^{\mathbf{4}}} = -{\mathbf{12}}$

Ans: False.

$\because {\left( { - 3} \right)^4} = 81$

${\text{and, 81}} \ne - {\text{12}}$

45. ${{\mathbf{3}}^{\mathbf{4}}} > {{\mathbf{4}}^{\mathbf{3}}}$

Ans: True.

$\because {\left( 3 \right)^4} = 81$

${\left( 4 \right)^3} = 64$

${\text{and, 81}}{\text{ > }}{\text{64}}$

46. ${\left( {\dfrac{{ - {\mathbf{3}}}}{{\mathbf{5}}}} \right)^{{\mathbf{100}}}} = \dfrac{{ - {{\mathbf{3}}^{{\mathbf{100}}}}}}{{ - {{\mathbf{5}}^{{\mathbf{100}}}}}}$

Ans: True.

$\because {\left( {\dfrac{{ - 3}}{5}} \right)^{100}} = \dfrac{{{{\left( { - 3} \right)}^{100}}}}{{{{\left( 5 \right)}^{100}}}} = \dfrac{{{3^{100}}}}{{{5^{100}}}}$

$\Rightarrow \dfrac{{{3^{100}}}}{{{5^{100}}}} = \dfrac{{ - {3^{100}}}}{{ - {5^{100}}}}$

47. ${\left( {{\mathbf{10}} + {\mathbf{10}}} \right)^{{\mathbf{10}}}} = {\mathbf{1}}{{\mathbf{0}}^{{\mathbf{10}}}} + {\text{ }}{\mathbf{1}}{{\mathbf{0}}^{{\mathbf{10}}}}$

Ans: False.

$\because {\left( {10 + 10} \right)^{10}} = {\left( {20} \right)^{10}}$

$\Rightarrow {\left( {20} \right)^{10}} = {10^{10}} + {10^{10}}$

$\Rightarrow {\left( {20} \right)^{10}} = 2{\left( {10} \right)^{10}}$

$\therefore {20^{10}} \ne 2{\left( {10} \right)^{10}}$

(${a^n} \times {b^n} = {\left( {ab} \right)^n}$)

48. ${{\mathbf{x}}^{\mathbf{0}}} \times {{\mathbf{x}}^{\mathbf{0}}} = {{\mathbf{x}}^{\mathbf{0}}} \div {{\mathbf{x}}^{\mathbf{0}}}$ is true for all non-zero values of x.

Ans: True.

For example; x = 1

Then,

${1^0} \times {1^0} = {1^0} \div {1^0}\left( {{a^0} = 1} \right)$

$\therefore 1 = 1$

49. In the standard form, a large number can be expressed as a decimal number between 0 and 1, multiplied by a power of 10.

Ans: False, because $a \times {10^n}$ is the standard form of a number and $1 \leqslant a \leqslant 10$.

50. ${{\mathbf{4}}^{\mathbf{2}}}$ is greater than ${{\mathbf{2}}^{\mathbf{4}}}$.

Ans: False.

$\because {4^2} = 16$

${2^4} = 16$

${\text{so, }}{{\text{4}}^2} = {2^4}$

51.${{\mathbf{x}}^{\mathbf{m}}} + {{\mathbf{x}}^{\mathbf{m}}} = {{\mathbf{x}}^{{\mathbf{2m}}}}$, where x is a non-zero rational number and m is  a positive integer.

Ans:  False.

$\because {x^m} + {x^m} = 2{x^m}$

${x^m} + {x^m} \ne {x^{2m}}$

52.${{\mathbf{x}}^{\mathbf{m}}} \times {{\mathbf{y}}^{\mathbf{m}}} = {\left( {{\mathbf{x}} + {\mathbf{y}}} \right)^{{\mathbf{2m}}}}$, where x and y are non-zero rational numbers and m is a positive integer.

Ans: False.

$\because {x^m} \times {y^m} = {\left( {xy} \right)^m}$

${x^m} \times {y^m} \ne {\left( {x + y} \right)^{2m}}$

($\because {a^n} \times {b^n} = a{b^n}$)

53. ${{\mathbf{x}}^{\mathbf{m}}} \div {{\mathbf{y}}^{\mathbf{m}}} = {\left( {{\mathbf{x}} \div {\mathbf{y}}} \right)^{\mathbf{m}}}$, where x and y are non-zero rational numbers and m is a positive integer.

Ans: True.

By the property of fundamental of ‘Exponents and Powers’$\Rightarrow {a^m} \div {b^m} = {\left( {a \div b} \right)^m}$

54. ${{\mathbf{x}}^{\mathbf{m}}} \times {{\mathbf{x}}^{\mathbf{n}}} = {\left( {\mathbf{x}} \right)^{{\mathbf{m}} + {\mathbf{n}}}}$, where x is a non-zero rational number and m, n are positive integers.

Ans: True.

By the property of fundamental of ‘Exponents and Powers’$\Rightarrow {a^n} \times {a^m} = {a^{n + m}}$

55. ${{\mathbf{4}}^{\mathbf{9}}}$ is greater than ${\mathbf{1}}{{\mathbf{6}}^{\mathbf{3}}}$.

Ans: True.

$\because {16^3} = {\left( {{4^2}} \right)^3} = {4^6}$

$\therefore {4^9} > {16^3}$

56. ${\left( {\dfrac{{\mathbf{2}}}{{\mathbf{5}}}} \right)^{\mathbf{3}}} \div {\left( {\dfrac{{\mathbf{5}}}{{\mathbf{2}}}} \right)^{\mathbf{3}}} = {\mathbf{1}}$

Ans: False.

$\because {\left( {\dfrac{2}{5}} \right)^3} \div {\left( {\dfrac{5}{2}} \right)^3} = {\left( {\dfrac{2}{5}} \right)^3} \times {\left( {\dfrac{2}{5}} \right)^3}$

$\Rightarrow {\left( {\dfrac{2}{5}} \right)^{3 + 3}} = {\left( {\dfrac{2}{5}} \right)^6}$

$\therefore {\left( {\dfrac{2}{5}} \right)^6} \ne 1$

($\Rightarrow {a^n} \times {a^m} = {a^{n + m}}$ )

57. ${\left( {\dfrac{4}{3}} \right)^5} \times {\left( {\dfrac{5}{7}} \right)^5} = {\left( {\dfrac{4}{3} + \dfrac{5}{7}} \right)^5}$

Ans: False.

$\Rightarrow {\left( {\dfrac{4}{3}} \right)^5} \times {\left( {\dfrac{5}{7}} \right)^5} = {\left( {\dfrac{4}{3} + \dfrac{5}{7}} \right)^5}$

$\Rightarrow {\left( {\dfrac{4}{3} \times \dfrac{5}{7}} \right)^5} \ne {\left( {\dfrac{4}{3} + \dfrac{5}{7}} \right)^5}$

58. ${\left( {\dfrac{5}{8}} \right)^9} \div {\left( {\dfrac{5}{8}} \right)^4} = {\left( {\dfrac{5}{8}} \right)^4}$

Ans: False.

$\because {\left( {\dfrac{5}{8}} \right)^9} \div {\left( {\dfrac{5}{8}} \right)^4} = {\left( {\dfrac{5}{8}} \right)^4}$

$\Rightarrow {\left( {\dfrac{5}{8}} \right)^{9 - 4}} = {\left( {\dfrac{5}{8}} \right)^4}$

$\Rightarrow {\left( {\dfrac{5}{8}} \right)^5} \ne {\left( {\dfrac{5}{8}} \right)^4}$

(${a^n} \div {a^m} = {a^{n - m}}$)

59. ${\left( {\dfrac{7}{3}} \right)^2} \times {\left( {\dfrac{7}{3}} \right)^5} = {\left( {\dfrac{7}{3}} \right)^{10}}$

Ans: False.

$\because {\left( {\dfrac{7}{3}} \right)^2} \times {\left( {\dfrac{7}{3}} \right)^5} = {\left( {\dfrac{7}{3}} \right)^{10}}$

$\Rightarrow {\left( {\dfrac{7}{3}} \right)^{5 + 2}} = {\left( {\dfrac{7}{3}} \right)^{10}}$

$\Rightarrow \therefore {\left( {\dfrac{7}{3}} \right)^7} \ne {\left( {\dfrac{7}{3}} \right)^{10}}$

(By the property $\Rightarrow {a^n} \times {a^m} = {a^{n + m}}$ )

60. ${{\mathbf{5}}^{\mathbf{0}}} \times {\mathbf{2}}{{\mathbf{5}}^{\mathbf{0}}} \times {\mathbf{12}}{{\mathbf{5}}^{\mathbf{0}}} = {\left( {{{\mathbf{5}}^{\mathbf{0}}}} \right)^{\mathbf{6}}}$

Ans: True.

$\Rightarrow \because {a^0} = 1$

$\Rightarrow {5^0} \times {25^0} \times {125^0} = {\left( {{5^0}} \right)^6}$

$\Rightarrow 1 \times 1 \times 1 = {1^6}$

$\therefore 1 = 1$

61. ${\mathbf{876543}}{\text{ }} = {\text{ }}{\mathbf{8}} \times {\mathbf{1}}{{\mathbf{0}}^{\mathbf{5}}} + {\mathbf{7}} \times {\mathbf{1}}{{\mathbf{0}}^{\mathbf{4}}} + {\mathbf{6}} \times {\mathbf{1}}{{\mathbf{0}}^{\mathbf{3}}} + {\mathbf{5}} \times {\mathbf{1}}{{\mathbf{0}}^{\mathbf{2}}} + {\mathbf{4}} \times {\mathbf{1}}{{\mathbf{0}}^{\mathbf{1}}} + {\mathbf{3}} \times {\mathbf{1}}{{\mathbf{0}}^{\mathbf{0}}}$

Ans: True.

$8 \times {10^5} = 800000$

$7 \times {10^4} = 70000$

$6 \times {10^3} = 6000$

$5 \times {10^2} = 500$

$4 \times 10 = 40$

$3 \times {10^0} = 3 \times 1 = 3$

${\text{now, 800000 + 70000 + 6000 + 500 + 40 + 3 = 876543}}$

62. ${\mathbf{600060}} = {\mathbf{6}} \times {\mathbf{1}}{{\mathbf{0}}^{\mathbf{5}}} + {\mathbf{6}} \times {\mathbf{1}}{{\mathbf{0}}^{\mathbf{2}}}$

Ans: False.

$\because 6 \times {10^5} = 600000$

$6 \times {10^2} = 600$

$\therefore 600000 + 600 = 600600$

$\therefore 600060 \ne 600600$

63. ${\text{ }}{\mathbf{4}} \times {\mathbf{1}}{{\mathbf{0}}^{\mathbf{5}}} + {\mathbf{3}} \times {\mathbf{1}}{{\mathbf{0}}^{\mathbf{4}}} + {\mathbf{2}} \times {\mathbf{1}}{{\mathbf{0}}^{\mathbf{3}}} + {\mathbf{1}} \times {\mathbf{1}}{{\mathbf{0}}^{\mathbf{0}}} = {\mathbf{432010}}$

Ans: False.

$400000 + 30000 + 2000 + 1 = 432001$

$\therefore 432001 \ne 432010$

64. ${\mathbf{8}} \times {\mathbf{1}}{{\mathbf{0}}^{\mathbf{6}}} + {\mathbf{2}} \times {\mathbf{1}}{{\mathbf{0}}^{\mathbf{4}}} + {\mathbf{5}} \times {\mathbf{1}}{{\mathbf{0}}^{\mathbf{2}}} + {\mathbf{9}} \times {\mathbf{1}}{{\mathbf{0}}^{\mathbf{0}}} = {\mathbf{8020509}}$

Ans: True.

$\Rightarrow 8 \times {10^6} + 2 \times {10^4} + 5 \times {10^2} + 9 \times {10^0}$

$8000000 + 20000 + 500 + 9 = 8020509$

65. ${{\mathbf{4}}^{\mathbf{0}}} + {{\mathbf{5}}^{\mathbf{0}}} + {{\mathbf{6}}^{\mathbf{0}}} = {\left( {{\mathbf{4}} + {\mathbf{5}} + {\mathbf{6}}} \right)^{\mathbf{0}}}$

Ans: False.

$\because {a^0} = 1$

$\Rightarrow {4^0} + {5^0} + {6^0} = {\left( {4 + 5 + 6} \right)^0}$

$\Rightarrow 1 + 1 + 1 = {\left( {15} \right)^0}$

$\therefore 3 \ne 1$

66. Arrange in ascending order:

${{\mathbf{2}}^{\mathbf{5}}},{{\mathbf{3}}^{\mathbf{3}}},{{\mathbf{2}}^{\mathbf{3}}} \times {\mathbf{2}},{\left( {{{\mathbf{3}}^{\mathbf{3}}}} \right)^{\mathbf{2}}},{{\mathbf{3}}^{\mathbf{5}}},{{\mathbf{4}}^{\mathbf{0}}},{{\mathbf{2}}^{\mathbf{3}}} \times {{\mathbf{3}}^{\mathbf{1}}}$

Ans: ${2^5} = 2 \times 2 \times 2 \times 2 \times 2 = 32$

${3^3} = 3 \times 3 \times 3 = 27$

${2^3} \times 2 = 2 \times 2 \times 2 \times 2 = 16$

${\left( {{3^3}} \right)^2} = {3^6} = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 729\left( {\because {{\left( {{a^n}} \right)}^m} = {a^{nm}}} \right)$

${3^5} = 3 \times 3 \times 3 \times 3 \times 3 = 243$

${4^0} = 1\left( {\because {a^0} = 1} \right)$

${2^3} \times {3^1} = 2 \times 2 \times 2 \times 3 = 24$

$\therefore 1 < 16 < 24 < 27 < 32 < 243 < 729$

${4^0} < {2^3} \times 2 < {2^3} \times {3^1} < {3^3} < {2^5} < {3^5} < {\left( {{3^3}} \right)^2}$

67. Arrange in descending order:

${{\mathbf{2}}^{{\mathbf{2}} + {\mathbf{3}}}},{\text{ }}{\left( {{{\mathbf{2}}^{\mathbf{2}}}} \right)^{\mathbf{3}}},{\text{ }}{\mathbf{2}} \times {{\mathbf{2}}^{\mathbf{2}}},{\text{ }}\dfrac{{{{\mathbf{3}}^{\mathbf{5}}}}}{{{{\mathbf{3}}^{\mathbf{2}}}}},{\text{ }}{{\mathbf{3}}^{\mathbf{2}}} \times {{\mathbf{3}}^{\mathbf{0}}},{\text{ }}{{\mathbf{2}}^{\mathbf{3}}} \times {{\mathbf{5}}^{\mathbf{2}}}$

Ans: ${2^{2 + 3}} = {2^5} = 32$

${\left( {{2^2}} \right)^3} = {2^{2 \times 3}} = {2^6} = 64$

$2 \times {2^2} = {2^3} = 8$

$\dfrac{{{3^5}}}{{{3^2}}} = {3^3} = 27$

${3^2} \times {3^0} = {3^2} \times 1 = 9$

${2^3} \times {5^2} = 8 \times 25 = 200$

${\text{so, 200}}{\text{ > }}{\text{64 > 32}}{\text{ > }}{\text{27}}{\text{ > 9}}{\text{ > 8}}$

$\therefore {2^3} \times {5^2} > {\left( {{2^2}} \right)^3} > {2^{2 + 3}} > \dfrac{{{3^5}}}{{{3^2}}} > {3^2} \times {3^0} > 2 \times {2^2}$

68. By what number should ${\left( {-{\text{ }}{\mathbf{4}}} \right)^{\mathbf{5}}}$be divided so that the quotient may be equal to${\left( { - {\mathbf{4}}} \right)^{\mathbf{3}}}$?

Ans: Let the number be x

Then,

$\Rightarrow \dfrac{{{{\left( { - 4} \right)}^5}}}{x} = {\left( { - 4} \right)^3}$

$\Rightarrow \dfrac{{{{\left( { - 4} \right)}^5}}}{{{{\left( { - 4} \right)}^3}}} = x$

$\left( {\because {a^n} \div {a^m} = {a^{n - m}}} \right)$

$\therefore {\left( { - 4} \right)^{5 - 3}} = x$

$\Rightarrow x = {\left( { - 4} \right)^2}$

$\therefore x = 16$

69. Find m so that ${\left( {\dfrac{2}{9}} \right)^3} \times {\left( {\dfrac{2}{9}} \right)^6} = {\left( {\dfrac{2}{9}} \right)^{2m - 1}}$

Ans: Given: ${\left( {\dfrac{2}{9}} \right)^3} \times {\left( {\dfrac{2}{9}} \right)^6} = {\left( {\dfrac{2}{9}} \right)^{2m - 1}}$

$\Rightarrow \left( {\because {a^n} \times {a^m} = {a^{n + m}}} \right)$

$\Rightarrow {\left( {\dfrac{2}{9}} \right)^{3 + 6}} = {\left( {\dfrac{2}{9}} \right)^{2m - 1}}$

$\Rightarrow {\left( {\dfrac{2}{9}} \right)^9} = {\left( {\dfrac{2}{9}} \right)^{2m - 1}}$

Comparing power as base is same:

$\Rightarrow 2m - 1 = 9$

$\Rightarrow 2m = 9 + 1$

$\Rightarrow 2m = 10$

$\therefore m = 5$

70. If $\dfrac{p}{q} = {\left( {\dfrac{3}{2}} \right)^2} \div {\left( {\dfrac{9}{4}} \right)^0}$ find the value of ${\left( {\dfrac{{\mathbf{p}}}{{\mathbf{q}}}} \right)^{\mathbf{3}}}$

Ans: Given: $\dfrac{{\text{p}}}{{\text{q}}} = {\left( {\dfrac{3}{2}} \right)^2} \div {\left( {\dfrac{9}{4}} \right)^0}$

We know, $\left( {{a^0} = 1} \right)$

Therefore,

$\Rightarrow \dfrac{{\text{p}}}{{\text{q}}} = {\left( {\dfrac{3}{2}} \right)^2} \div {\left( {\dfrac{9}{4}} \right)^0}$

$\Rightarrow \dfrac{{\text{p}}}{{\text{q}}} = {\left( {\dfrac{3}{2}} \right)^2} \div 1$

$\Rightarrow \dfrac{{\text{p}}}{{\text{q}}} = {\left( {\dfrac{3}{2}} \right)^2} = \left( {\dfrac{9}{4}} \right)$

$\therefore \dfrac{{\text{p}}}{{\text{q}}} = \left( {\dfrac{9}{4}} \right)$

Now, ${\left( {\dfrac{{\text{p}}}{{\text{q}}}} \right)^3} = {\left( {\dfrac{9}{4}} \right)^3} = \dfrac{{9 \times 9 \times 9}}{{4 \times 4 \times 4}} = \dfrac{{729}}{{64}}$.

71. Find the reciprocal of the rational number ${\left( {\dfrac{1}{2}} \right)^2} \div {\left( {\dfrac{2}{3}} \right)^3}$.

Ans: Given: ${\left( {\dfrac{1}{2}} \right)^2} \div {\left( {\dfrac{2}{3}} \right)^3}$

We know, $\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \times \dfrac{d}{c}$ therefore,

$\Rightarrow {\left( {\dfrac{1}{2}} \right)^2} \div {\left( {\dfrac{2}{3}} \right)^3} = {\left( {\dfrac{1}{2}} \right)^2} \times {\left( {\dfrac{3}{2}} \right)^3}$

$\Rightarrow \left( {\dfrac{1}{4}} \right) \times \left( {\dfrac{{27}}{8}} \right)$

$\Rightarrow \dfrac{{27}}{{32}}$

72. Find the value of:

(a) ${{\mathbf{7}}^{\mathbf{0}}}$

Ans: $\because {a^0} = 1$

$\Rightarrow {7^0} = 1$

(b) ${{\mathbf{7}}^{\mathbf{7}}} \div {{\mathbf{7}}^{\mathbf{7}}}$

Ans: ${7^7} \div {7^7} = 1$

(c) ${\left( {-{\mathbf{7}}} \right)^{{\mathbf{2}} \times {\mathbf{7}}-{\mathbf{6}}-{\mathbf{8}}}}$

Ans: ${\left( {-{\mathbf{7}}} \right)^{{\mathbf{2}} \times {\mathbf{7}}-{\mathbf{6}}-{\mathbf{8}}}}$ = ${ - 7}^{14 - 14} = { - 7} ^0$

$\because {a^0} = 1$

$\Rightarrow {\left( { - 7} \right)^0} = 1$

(d) $\left( {{{\mathbf{2}}^{\mathbf{0}}} + {{\mathbf{3}}^{\mathbf{0}}} + {{\mathbf{4}}^{\mathbf{0}}}} \right)\left( {{{\mathbf{4}}^{\mathbf{0}}}-{{\mathbf{3}}^{\mathbf{0}}}-{{\mathbf{2}}^{\mathbf{0}}}} \right)$

Ans: By the property of exponents and power: ${a^0} = 1$

$\left( {{2^0} + {3^0} + {4^0}} \right)\left( {{4^0} - {3^0} - {2^0}} \right)$

$\Rightarrow \left( {1 + 1 + 1} \right)\left( {1 - 1 - 1} \right)$

$\Rightarrow \left( 3 \right)\left( { - 1} \right) = - 3$

(e) ${\mathbf{2}} \times {\mathbf{3}} \times {\mathbf{4}} \div {{\mathbf{2}}^{\mathbf{0}}} \times {{\mathbf{3}}^{\mathbf{0}}} \times {{\mathbf{4}}^{\mathbf{0}}}$

Ans: ${\mathbf{2}} \times {\mathbf{3}} \times {\mathbf{4}} \div {{\mathbf{2}}^{\mathbf{0}}} \times {{\mathbf{3}}^{\mathbf{0}}} \times {{\mathbf{4}}^{\mathbf{0}}}$

By the property of exponents and power: ${a^0} = 1$

$24 \div 1 = 24$

(f) $\left( {{{\mathbf{8}}^{\mathbf{0}}}-{{\mathbf{2}}^{\mathbf{0}}}} \right) \times \left( {{{\mathbf{8}}^{\mathbf{0}}} + {{\mathbf{2}}^{\mathbf{0}}}} \right)$

Ans: By the property of exponents and power: ${a^0} = 1$

$\left( {{8^0} - {2^0}} \right) \times \left( {{8^0} + {2^0}} \right)$

$\Rightarrow \left( {1 - 1} \right)\left( {1 + 1} \right)$

$\Rightarrow \left( 0 \right)\left( 2 \right) = 0$

73. Find the value of n, where n is an integer and ${2^{n - 5}} \times {6^{2n - 4}} = \dfrac{1}{{1{2^4} \times 2}}$

Ans: Given: ${2^{n - 5}} \times {6^{2n - 4}} = \dfrac{1}{{{{12}^4} \times 2}}$

Solving:

= ${2^{n - 5}} \times {6^{2n - 4}} = \dfrac{1}{{{{12}^4} \times 2}}$

by the property of exponents and power: ${a^n} \div {a^m} = {a^{n - m}}$

$\Rightarrow {2^{n - 5}} \times {6^{2n - 4}} = \dfrac{1}{{{{12}^4} \times 2}}$

$\Rightarrow \dfrac{{{2^n}}}{{{2^5}}} \times \dfrac{{{6^{2n}}}}{{{6^4}}} = \dfrac{1}{{{{\left( {2 \times 6} \right)}^4} \times 2}}$

$\Rightarrow {2^n}{.6^{2n}} = \dfrac{{{2^5}{{.6}^4}}}{{{2^4}{{.6}^4}.2}}$

$\Rightarrow {2^n}{.6^{2n}} = 1$

$\Rightarrow {2^n}.{\left( {{6^2}} \right)^n} = 1$

$\Rightarrow {2^n}{.36^n} = 1$

$\left( {\because {a^n}.{b^n} = a{b^n}} \right)$

$\Rightarrow {\left( {2 \times 36} \right)^n} = 1$

$\Rightarrow {72^n} = 1$

$\Rightarrow {72^n} = {72^0}$

$\therefore n = 0$

74. Express the following in usual form:

(a) 8.01 × ${\mathbf{1}}{{\mathbf{0}}^{\mathbf{7}}}$

Ans: $8.01 \times {10^7}$

$\Rightarrow \dfrac{{801}}{{100}} \times {10^7} = 801 \times {10^5}$

$\Rightarrow 80100000$

(b) 1.75 × ${\mathbf{1}}{{\mathbf{0}}^{ - {\mathbf{3}}}}$

Ans: $1.75 \times {10^{ - 3}}$

$\Rightarrow \dfrac{{175}}{{100}} \times {10^{ - 3}} = 175 \times {10^{ - 5}}$

$\Rightarrow \dfrac{{175}}{{{{10}^5}}} = \dfrac{{175}}{{100000}} = 0.00175$

75. Find the value of

(a) ${{\mathbf{2}}^{\mathbf{5}}}$

Ans: ${2^5}$

$= {2^5} = 2 \times 2 \times 2 \times 2 \times 2 = 32$

(b) $\left( { - {{\mathbf{3}}^{\mathbf{5}}}} \right)$

Ans: ${\left( { - 3} \right)^5}$

$\Rightarrow {\left( { - 3} \right)^5} = - 3 \times - 3 \times - 3 \times - 3 \times - 3 = - 243$

(c) $- {\left( { - {\mathbf{4}}} \right)^4}$

Ans: $- {\left( { - 4} \right)^4}$

$\Rightarrow - {\left( { - 4} \right)^4} = - \left( { - 4 \times - 4 \times - 4 \times - 4} \right) = -256$

76. Express the following in exponential form:

(a) 3 × 3 × 3 × a × a × a ×a

Ans: 3 × 3 × 3 × a × a × a ×a

$= {3^3} \times {a^4} = 27{a^4}$

(b) a × a × b × b × b × c × c × c × c

Ans: a × a × b × b × b × c × c × c × c

$= {a^2} \times {b^3} \times {c^4}$$= {a^2}{b^3}{c^4} (c) s × s × t × t × s × s × t Ans: s × s × t × t × s × s × t = {s^4} \times {t^3}$$= {s^4}{t^3}$

77. How many times of 30 must be added together to get a sum equal to ${\mathbf{3}}{{\mathbf{0}}^{\mathbf{7}}}$?

Ans: If x times 30 is added then,

$\Rightarrow 30 \times x = {30^7}$

$\Rightarrow x = {30^6}$

Therefore, 30 must be added ${30^6}$ times to get ${30^7}$.

78. Express each of the following numbers using exponential notations:

(a) 1024

Ans: 1024

$\Rightarrow 1024 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = {2^{10}}$

(b) 1029

Ans: 1029

$\Rightarrow 1029 = 3 \times 7 \times 7 \times 7 = {3^1} \times {7^3}$

(c) $\dfrac{{{\mathbf{144}}}}{{{\mathbf{875}}}}$

Ans: $\dfrac{{144}}{{875}}$

$\Rightarrow \dfrac{{144}}{{875}} = \dfrac{{2 \times 2 \times 6 \times 6}}{{5 \times 5 \times 5 \times 7}}$

$\Rightarrow \dfrac{{{2^2} \times {6^2}}}{{{5^3} \times {7^1}}}$

79. Identify the greater number, in each of the following:

(a) ${{\mathbf{2}}^{\mathbf{6}}}{\text{ }}{\mathbf{or}}{\text{ }}{{\mathbf{6}}^{\mathbf{2}}}$

Ans: ${2^6}{\text{or}}{{\text{6}}^2}$

${2^6} = 64$

${6^2} = 36$

$\Rightarrow 64 > 36$

$\therefore {2^6} > {6^2}$

(b) ${{\mathbf{2}}^{\mathbf{9}}}{\text{ }}{\mathbf{or}}{\text{ }}{{\mathbf{9}}^{\mathbf{2}}}$

Ans: ${2^9}{\text{or }}{{\text{9}}^2}$

${2^9} = 512$

${9^2} = 81$

$\Rightarrow 512 > 81$

$\therefore {2^9} > {9^2}$

(c) ${\mathbf{7}}.{\mathbf{9}} \times {\mathbf{1}}{{\mathbf{0}}^{\mathbf{4}}}{\text{ }}{\mathbf{or}}{\text{ }}{\mathbf{5}}.{\mathbf{28}} \times {\mathbf{1}}{{\mathbf{0}}^{\mathbf{5}}}$

Ans: $7.9 \times {10^4}{\text{or 5}}{\text{.28}} \times {\text{1}}{{\text{0}}^5}$

$\Rightarrow 7.9 \times {10^4} = 79 \times {10^3} = 79000$

$\Rightarrow 5.28 \times {10^5} = 528 \times {10^3} = 528000$

$\Rightarrow 528000 > 79000$

$\therefore 5.28 \times {10^5} > 7.9 \times {10^4}$

80. Express each of the following as a product of powers of their prime factors:

(a) 9000

Ans: 9000

Taking factors of 9000:

$\underline {2|9000}$

$\underline {2|4500}$

$\underline {2|2250}$

$\underline {3|1125}$

$\underline {3|375}$

$\underline {5|125}$

$\underline {5|25}$

$\underline {5|5}$

$\underline {|1}$

Therefore,

$9000 = 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5 \times 5 = {2^3}{3^2}{5^3}$

(b) 2025

Ans: 2025

Taking factors of 2025:

$\underline {3|2025}$

$\underline {3|675}$

$\underline {3|225}$

$\underline {3|75}$

$\underline {5|25}$

$\underline {5|5}$

$\underline {|1}$

Therefore,

$2025 = 3 \times 3 \times 3 \times 3 \times 5 \times 5 = {3^4}{5^2}$

(c) 800

Ans: 800

Taking factors of 800:

$\underline {2|800}$

$\underline {2|400}$

$\underline {2|200}$

$\underline {2|100}$

$\underline {2|50}$

$\underline {5|25}$

$\underline {5|5}$

$\underline {|1}$

Therefore,

$800 = 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5 = {2^5}{5^2}$

81. Express each of the following in single exponential form:

(a) ${{\mathbf{2}}^{\mathbf{3}}} \times {{\mathbf{3}}^{\mathbf{3}}}$

Ans: ${2^3} \times {3^3}$

$= {\left( {2 \times 3} \right)^3} = {6^3}$ $\left( {\because {a^n} \times {a^m} = {a^{n + m}}} \right)$

(b) ${{\mathbf{2}}^{\mathbf{4}}} \times {{\mathbf{4}}^{\mathbf{2}}}$

Ans: ${2^4} \times {4^2}$

$= {2^2} \times {\left( {{2^2}} \right)^2} = {2^2} \times {2^4} = {2^{2 + 4}} = {2^6}$                              $\left( {\because {a^n} \times {a^m} = {a^{n + m}}} \right)$

(c) ${{\mathbf{5}}^{\mathbf{2}}} \times {{\mathbf{7}}^{\mathbf{2}}}$

Ans: ${5^2} \times {7^2}$

$= {\left( {5 \times 7} \right)^2} = {\left( {35} \right)^2}$ $\left( {\because {a^m} \times {b^m} = a{b^m}} \right)$

(d) ${\left( { - {\mathbf{5}}} \right)^{\mathbf{5}}} \times \left( { - {\mathbf{5}}} \right)$

Ans: ${\left( { - 5} \right)^5} \times \left( { - 5} \right)$

$= {\left( { - 5} \right)^{5 + 1}} = {\left( { - 5} \right)^6}$ $\left( {\because {a^n} \times {a^m} = {a^{n + m}}} \right)$

(e) ${\left( {-{\mathbf{3}}} \right)^{\mathbf{3}}} \times {\left( {-{\mathbf{10}}} \right)^{\mathbf{3}}}$

Ans: ${\left( { - 3} \right)^3} \times {\left( { - 10} \right)^3}$

$= {\left( { - 3 \times - 10} \right)^3} = {\left( {30} \right)^3}$ $\left( {\because {a^m} \times {b^m} = a{b^m}} \right)$

(f) ${\left( {-{\mathbf{11}}} \right)^{\mathbf{2}}} \times {\left( {-{\mathbf{2}}} \right)^{\mathbf{2}}}$

Ans: ${\left( { - 11} \right)^2} \times {\left( { - 2} \right)^2}$

$= {\left( { - 11 \times - 2} \right)^2} = {\left( {22} \right)^2}$ $\left( {\because {a^m} \times {b^m} = a{b^m}} \right)$

82. Express the following numbers in standard form:

(a) 76,47,000

Ans: Standard form of 7647000

$\Rightarrow 7647000 = 7647 \times {10^3}$

$\Rightarrow 7.647 \times {10^3} \times {10^3} = 7.647 \times {10^6}$

(b) 8,19,00,000

Ans: Standard form of 81900000

$\Rightarrow 81900000 = 819 \times {10^5}$

$\Rightarrow 8.19 \times {10^2} \times {10^5} = 8.19 \times {10^7}$

(c) 5, 83,00,00,00,000

Ans: Standard form of 583000000000

$\Rightarrow 583000000000 = 583 \times {10^9}$

$\Rightarrow 5.83 \times {10^2} \times {10^9} = 5.83 \times {10^{11}}$

(d) 24 billion

Ans: Standard form of 24 billion:

24 billion = 24,00,00,00,000

$\Rightarrow 24000000000 = 24 \times {10^9}$

$\Rightarrow 2.4 \times 10 \times {10^9} = 2.4 \times {10^{10}}$

83. The speed of light in vacuum is 3 × ${\mathbf{1}}{{\mathbf{0}}^{\mathbf{8}}}$ m/s. Sunlight takes about 8 minutes to reach the earth. Express distance of Sun from Earth in standard form.

Ans: Given:  speed of light in vacuum is = $3 \times {10^8}{\text{m}}{{\text{s}}^{ - 1}}$

Sunlight reaches in 8 minutes = $8 \times 60 = 480{\text{s}}$

As, Distance = speed x time

$\therefore 3 \times {10^8} \times 480 = 1440 \times {10^8}{\text{m}}$

$\Rightarrow {\text{1}}{\text{.440}} \times {\text{1}}{{\text{0}}^3} \times {10^8}$

$\Rightarrow 1.440 \times {10^{11}}{\text{m}}$

$\left( {\because {a^n} \times {a^m} = {a^{n + m}}} \right)$

84. Simplify and express each of the following in exponential form:

(a) $\left[ {{{\left( {\dfrac{3}{7}} \right)}^4} \times {{\left( {\dfrac{3}{7}} \right)}^5}} \right] \div {\left( {\dfrac{3}{7}} \right)^7}$

Ans: $\left[ {{{\left( {\dfrac{3}{7}} \right)}^4} \times {{\left( {\dfrac{3}{7}} \right)}^5}} \right] \div {\left( {\dfrac{3}{7}} \right)^7}$

Solving:

$\Rightarrow \left[ {{{\left( {\dfrac{3}{7}} \right)}^4} \times {{\left( {\dfrac{3}{7}} \right)}^5}} \right] \div {\left( {\dfrac{3}{7}} \right)^7}$

$\Rightarrow \left[ {{{\left( {\dfrac{3}{7}} \right)}^{4 + 5}}} \right] \div {\left( {\dfrac{3}{7}} \right)^7}\left( {\because {a^n} \times {a^m} = {a^{n + m}}} \right)$

$\Rightarrow {\left( {\dfrac{3}{7}} \right)^9} \div {\left( {\dfrac{3}{7}} \right)^7}\left( {\because {a^n} \div {a^m} = {a^{n - m}}} \right)$

$\Rightarrow {\left( {\dfrac{3}{7}} \right)^2}$

(b) $\left[ {{{\left( {\dfrac{7}{{11}}} \right)}^5} \div {{\left( {\dfrac{7}{{11}}} \right)}^2}} \right] \times {\left( {\dfrac{7}{{11}}} \right)^2}$

Ans: $\left[ {{{\left( {\dfrac{7}{{11}}} \right)}^5} \div {{\left( {\dfrac{7}{{11}}} \right)}^2}} \right] \times {\left( {\dfrac{7}{{11}}} \right)^2}$

Solving: $\left[ {{{\left( {\dfrac{7}{{11}}} \right)}^5} \div {{\left( {\dfrac{7}{{11}}} \right)}^2}} \right] \times {\left( {\dfrac{7}{{11}}} \right)^2}$

$\Rightarrow \left[ {{{\left( {\dfrac{7}{{11}}} \right)}^5} \div {{\left( {\dfrac{7}{{11}}} \right)}^2}} \right] \times {\left( {\dfrac{7}{{11}}} \right)^2}$

$\Rightarrow \left[ {{{\left( {\dfrac{7}{{11}}} \right)}^{5 - 2}}} \right] \times {\left( {\dfrac{7}{{11}}} \right)^2}$

$\Rightarrow {\left( {\dfrac{7}{{11}}} \right)^3} \times {\left( {\dfrac{7}{{11}}} \right)^2}$

$\Rightarrow {\left( {\dfrac{7}{{11}}} \right)^{3 + 2}} = {\left( {\dfrac{7}{{11}}} \right)^5}$

$\left( {\because {a^n} \times {a^m} = {a^{n + m}}} \right)$, $\left( {\because {a^n} \div {a^m} = {a^{n - m}}} \right)$

(c) ${\left( {{{\mathbf{3}}^{\mathbf{7}}} \div {{\mathbf{3}}^5}} \right)^{\mathbf{4}}}$

Ans: ${\left( {{3^7} \div {3^5}} \right)^4}$

Solving: ${\left( {{3^7} \div {3^5}} \right)^4}$

$\Rightarrow {\left( {\dfrac{{{3^7}}}{{{3^5}}}} \right)^4} = {\left( {{3^{7 - 5}}} \right)^4}$

$\Rightarrow {\left( {{3^2}} \right)^4} = {3^8}$

$\left( {\because {a^n} \div {a^m} = {a^{n - m}}} \right)$, ${\left( {{a^n}} \right)^m} = {a^{nm}}$

(d) $\left( {\dfrac{{{{\mathbf{a}}^{\mathbf{6}}}}}{{{{\mathbf{a}}^{\mathbf{4}}}}}} \right) \times {{\mathbf{a}}^{\mathbf{5}}} \times {{\mathbf{a}}^{\mathbf{0}}}$

Ans: $\left( {\dfrac{{{a^6}}}{{{a^4}}}} \right) \times {a^5} \times {a^0}$

Solving: $\left( {\dfrac{{{a^6}}}{{{a^4}}}} \right) \times {a^5} \times {a^0}$

$\Rightarrow \left( {{a^{6 - 4}}} \right) \times {a^5} \times {a^0}$

$\Rightarrow {a^2} \times {a^5} \times {a^0}$

$\Rightarrow {a^{2 + 5 + 0}} = {a^7}$

$\left( {\because {a^n} \div {a^m} = {a^{n - m}}} \right)$,$\left( {\because {a^n} \times {a^m} = {a^{n + m}}} \right)$

(e) $\left[ {{{\left( {\dfrac{3}{5}} \right)}^3} \times {{\left( {\dfrac{3}{5}} \right)}^8}} \right] \div \left[ {{{\left( {\dfrac{3}{5}} \right)}^2} \times {{\left( {\dfrac{3}{4}} \right)}^4}} \right]$

Ans: $\left[ {{{\left( {\dfrac{3}{5}} \right)}^3} \times {{\left( {\dfrac{3}{5}} \right)}^8}} \right] \div \left[ {{{\left( {\dfrac{3}{5}} \right)}^2} \times {{\left( {\dfrac{3}{5}} \right)}^4}} \right]$

Solving: $\left[ {{{\left( {\dfrac{3}{5}} \right)}^3} \times {{\left( {\dfrac{3}{5}} \right)}^8}} \right] \div \left[ {{{\left( {\dfrac{3}{5}} \right)}^2} \times {{\left( {\dfrac{3}{5}} \right)}^4}} \right]$

$\Rightarrow {\left( {\dfrac{3}{5}} \right)^{3 + 8}} \div {\left( {\dfrac{3}{5}} \right)^{2 + 4}}$

$\Rightarrow {\left( {\dfrac{3}{5}} \right)^{11}} \div {\left( {\dfrac{3}{5}} \right)^6}$

$\Rightarrow {\left( {\dfrac{3}{5}} \right)^{11 - 6}} = {\left( {\dfrac{3}{5}} \right)^5}$

$\left( {\because {a^n} \times {a^m} = {a^{n + m}}} \right)$ , $\left( {\because {a^n} \div {a^m} = {a^{n - m}}} \right)$

(f) $\left( {{{\mathbf{5}}^{{\mathbf{15}}}} \div {{\mathbf{5}}^{{\mathbf{10}}}}} \right) \times {{\mathbf{5}}^{\mathbf{5}}}$

Ans: $\left( {{5^{15}} \div {5^{10}}} \right) \times {5^5}$

Solving:

$\Rightarrow \left( {{5^{15 - 10}}} \right) \times {5^5}$

$\Rightarrow {5^5} \times {5^5} = {5^{5 + 5}} = {5^{10}}$

$\left( {\because {a^n} \div {a^m} = {a^{n - m}}} \right)$,$\left( {\because {a^n} \times {a^m} = {a^{n + m}}} \right)$

85. Evaluate

(a) $\dfrac{{{7^8} \times {a^{10}}{b^7}{c^{12}}}}{{{7^6} \times {a^8}{b^4}{c^{12}}}}$

Ans: Given:  $\dfrac{{{7^8} \times {a^{10}}{b^7}{c^{12}}}}{{{7^6} \times {a^8}{b^4}{c^{12}}}}$

By the property of exponents and power: ${a^n} \times {a^m} = {a^{n + m}}$, ${a^n} \div {a^m} = {a^{n - m}}$

$= {7^2} \times {a^2}{b^3}{c^0}\left( {{a^0} = 1} \right)$

$\therefore {7^2} \times {a^2}{b^3}$

(b) $\dfrac{{{5^4} \times {7^4} \times {2^7}}}{{8 \times 49 \times {5^3}}}$

Ans: Given: $\dfrac{{{5^4} \times {7^4} \times {2^7}}}{{8 \times 49 \times {5^3}}}$

By the property of exponents and power: ${a^n} \times {a^m} = {a^{n + m}}$, ${a^n} \div {a^m} = {a^{n - m}}$

$= \dfrac{{{5^4} \times {7^4} \times {2^7}}}{{{2^3} \times {7^2} \times {5^3}}}$

$= {5^{4 - 3}} \times {7^{4 - 2}} \times {2^{7 - 3}}$

$= {5^1} \times {7^2} \times {2^4}$

$= 5 \times 49 \times 16 = 3920$

(c) $\dfrac{{125 \times {5^2} \times {a^7}}}{{1{0^3} \times {a^4}}}$

Ans: Given: $\dfrac{{125 \times {5^2} \times {a^7}}}{{{{10}^3} \times {a^4}}}$

By the property of exponents and power: ${a^n} \times {a^m} = {a^{n + m}}$, ${a^n} \div {a^m} = {a^{n - m}}$

$= \dfrac{{{5^3} \times {5^2} \times {a^7}}}{{{{\left( {5 \times 2} \right)}^3} \times {a^4}}}$

$= \dfrac{{{5^5} \times {a^7}}}{{{5^3} \times {2^3} \times {a^4}}}$

$= \dfrac{{{5^{5 - 3}} \times {a^{7 - 4}}}}{{{2^3}}} = \dfrac{{{5^2} \times {a^3}}}{{{2^3}}}$

$= \dfrac{{25{a^3}}}{8}$

(d) $\dfrac{{{3^4} \times 1{2^3} \times 36}}{{{2^5} \times {6^3}}}$

Ans: Given: $\dfrac{{{3^4} \times {{12}^3} \times 36}}{{{2^5} \times {6^3}}}$

By the property of exponents and power: ${a^n} \times {a^m} = {a^{n + m}}$, ${a^n} \div {a^m} = {a^{n - m}}$

$= \dfrac{{{3^4} \times {{\left( {3 \times 4} \right)}^3} \times {6^2}}}{{{2^5} \times {6^3}}}$

$= \dfrac{{{3^4} \times {3^3} \times {4^3} \times {6^2}}}{{{2^5} \times {6^3}}}$

$= \dfrac{{{3^{4 + 3}} \times {4^3} \times {6^2}}}{{{2^5} \times {6^3}}}$

$= \dfrac{{{3^7} \times {{\left( {{2^2}} \right)}^3} \times {{\left( {2 \times 3} \right)}^2}}}{{{2^5} \times {{\left( {2 \times 3} \right)}^3}}} = \dfrac{{{3^7} \times {2^6} \times {2^2} \times {3^2}}}{{{2^5} \times {2^3} \times {3^3}}}$

$= \dfrac{{{3^{7 + 2}} \times {2^{6 + 2}}}}{{{2^{5 + 3}} \times {3^3}}} = \dfrac{{{3^9} \times {2^8}}}{{{2^8} \times {3^3}}}$

$= \dfrac{{{3^9}}}{{{3^3}}} = {3^{9 - 3}} = {3^6}$

$= {3^6} = 729$

(e) ${\left( {\dfrac{{6 \times 10}}{{{2^2} \times {5^3}}}} \right)^2} \times \dfrac{{25}}{{27}}$

Ans: given: ${\left( {\dfrac{{6 \times 10}}{{{2^2} \times {5^3}}}} \right)^2} \times \dfrac{{25}}{{27}}$

By the property of exponents and power: ${a^n} \times {a^m} = {a^{n + m}}$, ${a^n} \div {a^m} = {a^{n - m}}$

$= {\left( {\dfrac{{2 \times 3 \times 5 \times 2}}{{{2^2} \times {5^3}}}} \right)^2} \times \dfrac{{25}}{{27}}$

$= {\left( {\dfrac{{{2^2} \times 3 \times 5}}{{{2^2} \times {5^3}}}} \right)^2} \times \dfrac{{{5^2}}}{{{3^3}}}$

$= {\left( {\dfrac{3}{{ {5^2}}}} \right)^2} \times \dfrac{{{5^2}}}{{{3^3}}}$

$= \left( {\dfrac{{{3^2}}}{{ {5^4}}}} \right) \times \dfrac{{{5^2}}}{{{3^3}}}$

$= \dfrac{{{3^2}}}{{ {5^2} \times {3^3}}} = \dfrac{1}{{{5^2} \times 3}}$

$= \dfrac{1}{{ 3 \times 25}} = \dfrac{1}{{75}}$

(f) $\dfrac{{1{5^4} \times 1{8^3}}}{{{3^3} \times {5^2} \times 1{2^2}}}$

Ans: Given: $\dfrac{{{{15}^4} \times {{18}^3}}}{{{3^3} \times {5^2} \times {{12}^2}}}$

By the property of exponents and power: ${a^n} \times {a^m} = {a^{n + m}}$, ${a^n} \div {a^m} = {a^{n - m}}$

$= \dfrac{{{{\left( {3 \times 5} \right)}^4} \times {{\left( {2 \times 3 \times 3} \right)}^3}}}{{{3^3} \times {5^2} \times {{\left( {2 \times 2 \times 3} \right)}^2}}}$

$= \dfrac{{{5^4} \times {3^4} \times {2^3} \times {3^3} \times {3^3}}}{{{3^3} \times {5^2} \times {2^2} \times {2^2} \times {3^2}}}$

$= \dfrac{{{5^4} \times {3^4} \times {2^3} \times {3^3} \times {3^3}}}{{{3^3} \times {5^2} \times {2^2} \times {2^2} \times {3^2}}}$

$= \dfrac{{{5^4} \times {3^{4 + 3 + 3}} \times {2^3}}}{{{3^{3 + 2}} \times {5^2} \times {2^{2 + 2}}}} = \dfrac{{{5^{4 - 2}} \times {3^{10}} \times {2^3}}}{{{3^5} \times {2^4}}}$

$= {5^2} \times {3^{10 - 5}} \times {2^{3 - 4}}$

$= \dfrac{{{5^2} \times {3^5}}}{2} = \dfrac{{25 \times 243}}{2}$

$= \dfrac{{6075}}{2}$

(g) $\dfrac{{{6^2} \times {9^2} \times 2{5^3}}}{{{3^2} \times {4^2} \times 1{5^6}}}$

Ans: Given: $\dfrac{{{6^4} \times {9^2} \times {{25}^3}}}{{{3^2} \times {4^2} \times {{15}^6}}}$

By the property of exponents and power: ${a^n} \times {a^m} = {a^{n + m}}$, ${a^n} \div {a^m} = {a^{n - m}}$

$= \dfrac{{{{\left( {2 \times 3} \right)}^4} \times {{\left( {{3^2}} \right)}^2} \times {{\left( {{5^2}} \right)}^3}}}{{{3^2} \times {{\left( {{2^2}} \right)}^2} \times {{\left( {3 \times 5} \right)}^6}}}$

$= \dfrac{{{2^4} \times {3^4} \times {3^4} \times {5^6}}}{{{3^2} \times {2^4} \times {3^6} \times {5^6}}}$

$= \dfrac{{{2^4} \times {3^8} \times {5^6}}}{{{3^8} \times {2^4} \times {5^6}}} = 1$

86. Express the given information in Scientific notation (standard form) and then arrange them in ascending order of their size.

 S.No. Deserts of the world Area (sq. Kilometers) 1. Kalahari, South Africa 932,400 2. Thar, India 199,430 3. Gibson, Australia 155,400 4. Great Victoria, Australia 647,500 5. Sahara, North Africa 8.598,800

Ans: Representing the given information in scientific notation:

Numbers written in scientific notation are compared with factor and power of ten.

Standard form = $a \times {10^k}$

Therefore,

1. given: Area of Kalahari, South Africa = 932,400

In standard form:

$932400 = 9324 \times {10^2}$

$= 9.324 \times {10^3} \times {10^2}$

$= 9.324 \times {10^5}$

2. given: Area of Thar, India = 199,430

In standard form:

$199430 = 19943 \times {10^1}$

$= 1.9943 \times {10^4} \times {10^1}$

$= 1.9943 \times {10^5}$

3. given: Area of Gibson, Australia = 155,400

In standard form:

$155400 = 1554 \times {10^2}$

$= 1.554 \times {10^3} \times {10^2}$

$= 1.554 \times {10^5}$

4. given: Area of Great Victoria, Australia = 647500

In standard form:

$647500 = 6475 \times {10^2}$

$= 6.475 \times {10^3} \times {10^2}$

$= 6.475 \times {10^5}$

5. given: Area of Sahara, North Africa = 8,598,800

In standard form:

$8598800 = 85988 \times {10^2}$

$= 8.5988 \times {10^4} \times {10^2}$

$= 8.5988 \times {10^6}$

Because power of 10 is same in all standard form so we will compare factors for arranging the data in ascending order:

Therefore,

Gibson, Australia < Thar, India < Great Victoria, Australia < Kalahari, South Africa < Sahara, North Africa.

87. Express the given information in Scientific notation and then arrange them in descending order of their size.

 S.No. Name of the planet Mass (in kg) 1. Mercury 330000000000000000000000 2. Venus 48700000000000000000000000000 3. Earth 59800000000000000000000000000 4. Mars 64200000000000000000000000 5. Jupiter 19000000000000000000000000000 6. Saturn 56900000000000000000000000000 7. Uranus 869000000000000000000000000000 8. Neptune 1020000000000000000000000000000 9. Pluto 131000000000000000000000000

Ans: Representing the given information in scientific notation:

 S.No. Name of the planet Mass (in kg) 1. Mercury $3.3 \times {10^{23}}$ 2. Venus $4.87 \times {10^{24}}$ 3. Earth $5.98 \times {10^{24}}$ 4. Mars $6.42 \times {10^{23}}$ 5. Jupiter $1.9 \times {10^{27}}$ 6. Saturn $5.69 \times {10^{26}}$ 7. Uranus $8.69 \times {10^{25}}$ 8. Neptune $1.02 \times {10^{26}}$ 9. Pluto $1.31 \times {10^{22}}$

Numbers written in scientific notation are compared with factor and power of ten.

Therefore,

$1.9 \times {10^{27}}$> $5.69 \times {10^{26}}$ > $1.02 \times {10^{26}}$ > $8.69 \times {10^{25}}$ > $5.98 \times {10^{24}}$ > $4.87 \times {10^{24}}$ > $6.42 \times {10^{23}}$ > $3.3 \times {10^{23}}$ >$1.31 \times {10^{22}}$

Hence, Descending order of size is:

Jupiter > Saturn > Neptune > Uranus > Earth > Venus > Mars > Mercury > Pluto.

88. Write the number of seconds in scientific notation.

 S.No. Unit Value in Seconds 1. 1 minute 60 2. 1 Hour 3,600 3. 1 Day 86,400 4. 1 Month 2,600,000 5. 1 Year 32,000,000 6. 10 Years 3,20,000,000

Ans: 1. 1 minute = 60 seconds

Therefore, in scientific notation = $6 \times {10^1}$s.

2. 1 hour = 60 minute and 1 minute = 60 seconds

So, 1 hour = 60 x 60 = 3600 s

Therefore, in scientific notation = $3.6 \times {10^3}$s.

3. 1 day = 24 hours

1 hour = 60 minutes

1 minute = 60 seconds

So, 1 day = 24 x 60 x 60 = 86,400 seconds

Therefore, in scientific notation = $864 \times {10^2} = 8.64 \times {10^2} \times {10^2} = 8.64 \times {10^4}$s.

4.  1 month = 26,00,000 seconds

Therefore, in scientific notation = $26 \times {10^5} = 2.6 \times 10 \times {10^5} = 2.6 \times {10^6}$s.

5. 1 year = 32,000,000 seconds

Therefore, in scientific notation = $32 \times {10^6} = 3.2 \times 10 \times {10^6} = 3.2 \times {10^7}$s.

6. 10 year = 1 year x 10 = 320,000,000 x 10 = 320,000,000 seconds

Therefore, in scientific notation = $320 \times {10^6} = 32 \times {0^7} = 3.2 \times {10^8}$s.

89. On our own planet Earth, 361,419,000 square kilometers of area is covered with water and 148,647,000 square kilometers of area is covered by land. Find the approximate ratio of area covered with water to area covered by land by converting these numbers into scientific notation.

Ans: Given:

Area covered by land = 148647000 ${\text{k}}{{\text{m}}^2}$

Area covered by water = 361419000 ${\text{k}}{{\text{m}}^2}$

In standard form:

Area covered by land = 148647000 ${\text{k}}{{\text{m}}^2}$

= $1.48647 \times {10^{8}}{\text{k}}{{\text{m}}^2}$

Area covered by water = 361419000 ${\text{k}}{{\text{m}}^2}$

=$3.61419 \times {10^8}{\text{k}}{{\text{m}}^2}$

Now, ratio of water to land = $\dfrac{{3.61419 \times {{10}^8}{\text{k}}{{\text{m}}^2}}}{{1.48647 \times {{10}^8}{\text{k}}{{\text{m}}^2}}}$

$= \dfrac{{3.6}}{{1.5}} = 12:5$

Therefore, ratio = 12:5

90. If, ${2^{n + 2}} - {2^{n + 1}} + {2^n} = c \times {2^n}$ , find the value of c.

Ans: Given:  ${2^{n + 2}} - {2^{n + 1}} + {2^n} = {\mathbf{c}} \times {2^n}$

$\Rightarrow \left( {{2^n} \times {2^2}} \right) - \left( {{2^n} \times {2^1}} \right) + {2^n} = c \times {2^n}$

$\Rightarrow {2^n}\left( {4 - 2 + 1} \right) = c \times {2^n}$

$\Rightarrow {2^n}\left( 3 \right) = c \times {2^n}$

$\therefore c = 3$

91. A light-year is a distance that light can travel in one year. 1 light-year = 9,460,000,000,000 km.

(a) Express one light-year in scientific notation.

Ans: Given:

1 light-year = 9,460,000,000,000 km.

Therefore, in standard form = $9.46 \times {10^{10}} \times 100{\text{km}}$

$= 9.46 \times {10^{12}}{\text{km}}$

(b) The average distance between Earth and Sun is 1.496 × ${\mathbf{1}}{{\mathbf{0}}^{\mathbf{8}}}$ km. Is the distance between Earth and the Sun greater than, less than or equal to one light-year.

Ans: Average distance between Earth and Sun is = 1.496 × ${10^8}$ km

Therefore, distance between Earth and Sun:

$= \dfrac{{1.496}}{{10000}} \times {10^8} \times {10^4}{\text{km}}$

$= 0.0001496 \times {10^{12}}{\text{km}}$

So, $9.46 > 0.0001496$.

Hence, the distance between Earth and Sun is less than one light year.

92. Geometry Application: The number of diagonals of an n-sided figure is $\dfrac{{\mathbf{1}}}{{\mathbf{2}}}\left( {{{\mathbf{n}}^{\mathbf{2}}} - {\mathbf{3n}}} \right)$.Use the formula to find the number of diagonals for a 6-sided figure (hexagon).

Ans: Given:

Number of diagonals = $\dfrac{1}{2}\left( {{n^2} - 3n} \right)$

Number of sides in hexagon = 6

Therefore, $n = 6$

Now, Number of diagonals = $\dfrac{1}{2}\left( {{n^2} - 3n} \right)$

$= \dfrac{1}{2}\left( {{6^2} - 3\left( 6 \right)} \right)$

$= \dfrac{1}{2}\left( {36 - 18} \right)$

$= \dfrac{{18}}{2} = 9$

Hence, number of diagonals = 9.

93. Life Science: Bacteria can divide in every 20 minutes. So, 1 bacterium can multiply to 2 in 20 minutes. 4 in 40 minutes, and so on. How many bacteria will there be in 6 hours? Write your answer using exponents, and then evaluate.

Ans:  1 h = 60 min

6 h = 60 x 6 min = 360 min

Also, given that: A Bacteria doubles itself in every 20 min.

So, the number of times it will double itself = $\dfrac{{360}}{{20}} = 18$min.

∴ In 6 h = $2{\text{ }} \times 2 \times \ldots \times {\text{ }}2$(18 times) = ${2^{18}}$.

94. Blubber makes up 27 percent of a blue whale’s body weight. Deepak found the average weight of blue whales and used it to calculate the average weight of their blubber. He wrote the amount as ${{\mathbf{2}}^{\mathbf{2}}} \times {{\mathbf{3}}^{\mathbf{2}}} \times {\mathbf{5}} \times {\mathbf{17}}$kg. Evaluate this amount.

Ans: Calculated weight by Deepak = ${2^2} \times {3^2} \times 5 \times 17$kg

$= 2 \times 2 \times 3 \times 3 \times 5 \times 17$ kg

$= 4 \times 9 \times 5 \times 17{\text{kg}}$

${\text{ = }}{\text{36}} \times {\text{5}} \times {\text{17}}{\text{kg}}$

${\text{ = }}{\text{180}} \times {\text{17}}{\text{kg}}$

${\text{ = 3060}}{\text{kg}}$

So, the weight calculated by Deepak = 3060 kg.

95. Life Science Application: The major components of human blood are red blood cells, white blood cells, platelets and plasma. A typical red blood cell has a diameter of approximately 7 × ${\mathbf{1}}{{\mathbf{0}}^{ - {\mathbf{6}}}}$ meters. A typical platelet has a diameter of approximately 2.33 × ${\mathbf{1}}{{\mathbf{0}}^{ - {\mathbf{6}}}}$ meter. Which has a greater diameter, a red blood cell or a platelet?

Ans: Given:

Diameter of platelet = 2.33 x ${10^{ - 6}}$ m

Diameter of red blood cell = 7 x ${10^{ - 6}}$ m

The number with the greater power of 10 is greater than the number with the lesser power of 10.

If the power of ten is equal, then the number with the greater factor is the larger number. Therefore, a red blood cell has a greater diameter than a platelet.

$\because 7 > 2.33$.

96. A googol is the number 1 followed by 100 zeroes.

(a) How is a googol written as a power?

Ans: As 1 googol = 100 times 10 $= {10^{100}}$

(b) How is a googol times a googol written as a power?

Ans: Googol time’s googol:

Because, 1 googol = 100 times 10

Therefore, googol times googol = ${10^{100}} \times {10^{100}}$.

97. What’s the error? A student said that $\dfrac{{{{\mathbf{3}}^{\mathbf{5}}}}}{{{{\mathbf{9}}^{\mathbf{5}}}}}$ is the same as $\dfrac{{\mathbf{1}}}{{\mathbf{3}}}$. What mistake has the student made?

Ans: Given:

$= \dfrac{{{3^5}}}{{{9^5}}}$ can be written as $\dfrac{{{3^5}}}{{{{\left( {3 \times 3} \right)}^5}}} = \dfrac{{{3^5}}}{{{{\left( {{3^2}} \right)}^5}}} = \dfrac{{{3^5}}}{{{3^{10}}}}$$\left( {\because {a^n} \times {a^m} = {a^{n + m}}} \right)$

$\Rightarrow {3^5} \times {3^{ - 10}} = {3^{5 + \left( { - 10} \right)}} = {3^{5 - 10}} = {3^{ - 5}}$

$\therefore {3^{ - 5}} = \dfrac{1}{{{3^5}}}$

Thus, multiplying the base by its exponent, is the error.

The concept clarity of this chapter is very important for Class 7th students as after understanding this chapter, students will be able to know how to use formulas, scientific notations, etc concepts to solve the problems. Besides that, the solutions are created to guide the students through the difficult problems and their steps. The solutions will also help you easily understand some essential identities that are used in Exponents. If we prefer the exam perspective, then the solutions of this chapter will help students to commit fewer mistakes. When solutions are there, students can also check whether they are going the right way while practicing and solving questions or not. The prime objective of these solutions is to help students understand the practicality of the chapter.

## FAQs on NCERT Exemplar for Class 7 Maths Solutions Chapter 11 Exponents & Powers

1. Is it mandatory to practice all the questions mentioned in NCERT Exemplar Class 7 Math Solutions Chapter 11 Exponents & Powers?

Every example and question mentioned in NCERT Exemplar Class 7 Exponents & Powers chapters will help the students in understanding the important concepts they have read in the chapter. Practicing all the questions will give them an in-depth understanding of the subject and every exercise belonging to the chapter. So even if a question arrives differently, students will be able to solve it because they have a clear concept clarity of the Exponents and Powers chapter.

2. What are the number of questions given in the Class 7th Exponents & Powers chapter?

The chapter Exponents & Powers of Class 7 NCERT Exemplar consists of a total of 97 questions. Every question of the chapter has a different concept attached to it that helps students in having the actual concept clarity.  Besides that, the questions also have their sub-parts for better practice of a single formula. Some of the questions are in exponential form, some are a little complex, and some involve both easy and in-depth calculations.

3. What are the different concepts students learn in the Class 7th Exponents & Powers chapter?

The different concepts students of Class 7 learn from this chapter in terms of Mathematics include learning about Exponents, finding which exponent is greater, prime factorisation, Power of 1 and -1, what are negative Exponents, and zero Exponents. In addition to these, students also learn the law of Exponents, writing numbers as Powers of 10 and the standard form of Exponents. All the exercises and question tests are based on these concepts of the chapter only.

4. How helpful is NCERT Exemplar Class 7 Math Solutions Chapter 11 Exponents & Powers?

When students are practicing questions from Exemplar at some stages of the question or the problem, they will face difficulties. Especially if they are doing self-studies, then this booklet of exemplar solutions will help them identify the steps they have missed or understand the way to solve the problem. The solutions can be very helpful for the parents as well so that even if they are assessing their children they can do the corrections with the help of this NCERT Exemplar solutions PDF.

5. How can I download NCERT Exemplar solutions for Math and science chapters?

Students can very easily download the latest exemplar problem solutions for Class 7th in the form of PDF from Vedantu’s website. In the solution PDF, you will find solved answers to every question mentioned in the chapters. Every solution will be explained in detail with a step-by-step solution to each question for a better and easy understanding. You can also download important NCERT Exemplar Class 6 Science Solutions Chapter 7 Getting to Know Plants