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MathematicsClass 810.1 Introduction

10.1 Introduction | Class 8 Mathematics Notes

By ConceptScroll Team · Published on 17 July 2026 · 3 min read

10.1 Introduction | Class 8 Mathematics Notes

10.1 Introduction – this guide gives you a concise, exam-ready overview of 10.1 Introduction from Class 8 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.

10.4 Use of Exponents to Express Small Numbers in Standard Form

This section introduces the use of exponents to express very small numbers in standard form, similar to how large numbers are expressed. Standard form uses powers of 10 to write numbers compactly, especially useful in science and engineering. The section lists real-world examples of very large and very small quantities, such as the distance from Earth to Sun (149,600,000,000 m) and the size of a red blood cell (0.000007 m). Very large numbers are expressed as a × 10ⁿ where n is positive, and very small numbers as a × 10⁻ⁿ where n is positive. For example, 0.000007 = 7 × 10⁻⁶, and thickness of paper 0.0016 cm = 1.6 × 10⁻³ cm. The section explains how to convert decimal numbers to standard form by counting decimal places moved and adjusting the exponent accordingly. This method simplifies arithmetic with very large or small numbers and aids in comparison. The section also includes a table to classify numbers as very large or very small and exercises to practice writing numbers in standard form.

📊 Diagram: Figures showing examples of very large and very small numbers; table classifying numbers as very large or very small; examples of decimal to standard form conversion.

🧪 Activity: Classify given numbers as very large or very small and write them in standard form.

🔗 Connection: Leads to the next subsection on comparing very large and very small numbers using standard form.

Frequently asked questions

1. Evaluate. (i) $ 3^{-2} $ (ii) $ (-4)^{-2} $ (iii) $ \left( \frac{1}{2} \right)^{-5} $

Solution: (i) $3^{-2} = \frac{1}{3^2} = \frac{1}{9}$

(ii) $(-4)^{-2} = \frac{1}{(-4)^2} = \frac{1}{16}$

(iii) $\left( \frac{1}{2} \right)^{-5} = 2^5 = 32$

2. Simplify and express the result in power notation with positive exponent. (i) $ (-4)^5 \div (-4)^8 $ (ii) $ \left( \frac{1}{2^3} \right)^2 $ (iii) $ (-3)^4 \times \left( \frac{5}{3} \right)^4 $ (iv) $ (3^{-7} \div 3^{-10}) \times 3^{-5} $ (v) $ 2^{-3} \times (-7)^{-3} $

Solution: (i) $(-4)^5 \div (-4)^8 = (-4)^{5-8} = (-4)^{-3} = \frac{1}{(-4)^3} = \frac{1}{-64} = -\frac{1}{64}$

(ii) $\left( \frac{1}{2^3} \right)^2 = \left( 2^{-3} \right)^2 = 2^{-6} = \frac{1}{2^6} = \frac{1}{64}$

(iii) $(-3)^4 \times \left( \frac{5}{3} \right)^4 = \left( -3 \times \frac{5}{3} \right)^4 = (-5)^4 = 5^4 = 625$

(iv) $(3^{-7} \div 3^{-10}) \times 3^{-5} = 3^{-7 - (-10)} \times 3^{-5} = 3^{3} \times 3^{-5} = 3^{3-5} = 3^{-2} = \frac{1}{9}$

(v) $2^{-3} \times (-7)^{-3} = \frac{1}

3. Find the value of. (i) $ (3^0 + 4^{-1}) \times 2^2 $ (ii) $ (2^{-1} \times 4^{-1}) \div 2^{-2} $ (iii) $ \left( \frac{1}{2} \right)^{-2} + \left( \frac{1}{3} \right)^{-2} + \left( \frac{1}{4} \right)^{-2} $ (iv) $ (3^{-1} + 4^{-1} + 5^{-1})^0 $ (v) $ \left\{ \left( \frac{-2}{3} \right)^{-2} \right\}^2 $

Solution: (i) $3^0 = 1$, $4^{-1} = \frac{1}{4}$, so $(1 + \frac{1}{4}) \times 2^2 = \frac{5}{4} \times 4 = 5$

(ii) $(2^{-1} \times 4^{-1}) \div 2^{-2} = \left( \frac{1}{2} \times \frac{1}{4} \right) \div \frac{1}{2^2} = \frac{1}{8} \div \frac{1}{4} = \frac{1}{8} \times 4 = \frac{1}{2}$

(iii) $\left( \frac{1}{2} \right)^{-2} = 2^2 = 4$, $\left( \frac{1}{3} \right)^{-2} = 3^2 = 9$, $\left( \frac{1}{4} \right)^{-2} = 4^2 = 16$, sum = $4 + 9 + 16 = 29$

(iv) Any number to the power 0 is 1, so $(3^

4. Evaluate (i) $\frac{8^{-1} \times 5^3}{2^{-4}}$ (ii) $(5^{-1} \times 2^{-1}) \times 6^{-1}$

Solution: (i) $\frac{8^{-1} \times 5^3}{2^{-4}} = \frac{\frac{1}{8} \times 125}{\frac{1}{2^{-4}}} = \frac{\frac{125}{8}}{2^{-4}} = \frac{125}{8} \times 2^{4} = \frac{125}{8} \times 16 = 125 \times 2 = 250$

(ii) $(5^{-1} \times 2^{-1}) \times 6^{-1} = \frac{1}{5} \times \frac{1}{2} \times \frac{1}{6} = \frac{1}{60}$

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