krit.club logo

Number - Standard form

Grade 11IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

β€’

Definition: Standard form (scientific notation) is written as AΓ—10nA \times 10^n, where 1≀A<101 \le A < 10 and nn is an integer.

β€’

Positive indices (n>0n > 0): Used to represent very large numbers. The index nn represents how many places the decimal point moves to the left.

β€’

Negative indices (n<0n < 0): Used to represent very small numbers (between 0 and 1). The index nn represents how many places the decimal point moves to the right.

β€’

Multiplication/Division: Use index laws for the powers of 10 and multiply/divide the decimal numbers separately.

β€’

Addition/Subtraction: Convert numbers to the same power of 10 or convert them to ordinary numbers before performing the operation.

πŸ“Formulae

AΓ—10nΒ (whereΒ 1≀A<10,n∈Z)A \times 10^n \text{ (where } 1 \le A < 10, n \in \mathbb{Z})

(aΓ—10n)Γ—(bΓ—10m)=(aΓ—b)Γ—10n+m(a \times 10^n) \times (b \times 10^m) = (a \times b) \times 10^{n+m}

(aΓ—10n)Γ·(bΓ—10m)=(aΓ·b)Γ—10nβˆ’m(a \times 10^n) \div (b \times 10^m) = (a \div b) \times 10^{n-m}

πŸ’‘Examples

Problem 1:

Write 0.0000452 in standard form.

Solution:

4.52Γ—10βˆ’54.52 \times 10^{-5}

Explanation:

To get a number between 1 and 10, the decimal point must move 5 places to the right. Since it is a small number, the index is negative.

Problem 2:

Calculate (3Γ—105)Γ—(4Γ—103)(3 \times 10^5) \times (4 \times 10^3), giving your answer in standard form.

Solution:

1.2Γ—1091.2 \times 10^9

Explanation:

Multiply the numbers: 3Γ—4=123 \times 4 = 12. Add the indices: 105+3=10810^{5+3} = 10^8. This gives 12Γ—10812 \times 10^8. Since 12>1012 > 10, we rewrite it as 1.2Γ—101Γ—108=1.2Γ—1091.2 \times 10^1 \times 10^8 = 1.2 \times 10^9.

Problem 3:

Calculate (6.4Γ—107)Γ·(8Γ—10βˆ’2)(6.4 \times 10^7) \div (8 \times 10^{-2}).

Solution:

8Γ—1088 \times 10^8

Explanation:

Divide the numbers: 6.4Γ·8=0.86.4 \div 8 = 0.8. Subtract the indices: 107βˆ’(βˆ’2)=10910^{7 - (-2)} = 10^9. This gives 0.8Γ—1090.8 \times 10^9. Since 0.8<10.8 < 1, we rewrite it as 8Γ—10βˆ’1Γ—109=8Γ—1088 \times 10^{-1} \times 10^9 = 8 \times 10^8.

Problem 4:

Find the value of (2.5Γ—104)+(3Γ—103)(2.5 \times 10^4) + (3 \times 10^3).

Solution:

2.8Γ—1042.8 \times 10^4

Explanation:

Align the powers of 10. 3Γ—1033 \times 10^3 is the same as 0.3Γ—1040.3 \times 10^4. Adding them gives (2.5+0.3)Γ—104=2.8Γ—104(2.5 + 0.3) \times 10^4 = 2.8 \times 10^4.