1.1

Definition & Normalization of Standard Form

  • Official IB Syllabus Definition: A real number expressed in the form: $$a \times 10^k \quad \text{where } 1 \le a < 10 \text{ and } k \in \mathbb{Z}$$ The coefficient $a$ is called the mantissa or significand, and $k$ is the integer exponent (order of magnitude).
  • The Strict Coefficient Bounds ($1 \le a < 10$):
    • $a$ can equal $1$, but $a$ can never equal or exceed $10$.
    • Expressions like $45 \times 10^3$ or $0.35 \times 10^{-4}$ are not in standard form.
  • Decimal Point Shift Rules:
    • Moving decimal point left: increases $k$ by the number of steps ($45000 = 4.5 \times 10^4$).
    • Moving decimal point right: decreases $k$ by the number of steps ($0.00038 = 3.8 \times 10^{-4}$).
  • Normalization Protocol: Whenever a computation yields $a \ge 10$ or $a < 1$, adjust the coefficient and exponent immediately: $$48.6 \times 10^5 = (4.86 \times 10^1) \times 10^5 = 4.86 \times 10^6$$ $$0.027 \times 10^{-3} = (2.7 \times 10^{-2}) \times 10^{-3} = 2.7 \times 10^{-5}$$

Example 1

Converting & Normalizing into Standard Form

Convert the following into standard form $a \times 10^k$ where $1 \le a < 10$ and $k \in \mathbb{Z}$:
(a) $0.0000728$
(b) $842 \times 10^5$
(c) $0.059 \times 10^{-6}$

  1. Part (a): $0.0000728$
    Move decimal point $5$ places to the right to place it between $7$ and $2$: $$0.0000728 = 7.28 \times 10^{-5}$$
  2. Part (b): $842 \times 10^5$
    Convert coefficient $842 = 8.42 \times 10^2$: $$842 \times 10^5 = (8.42 \times 10^2) \times 10^5 = 8.42 \times 10^{2+5} = 8.42 \times 10^7$$
  3. Part (c): $0.059 \times 10^{-6}$
    Convert coefficient $0.059 = 5.9 \times 10^{-2}$: $$0.059 \times 10^{-6} = (5.9 \times 10^{-2}) \times 10^{-6} = 5.9 \times 10^{-2 + (-6)} = 5.9 \times 10^{-8}$$

Answer: (a) $7.28 \times 10^{-5}$; (b) $8.42 \times 10^7$; (c) $5.9 \times 10^{-8}$

1.1

Non-Calculator Operations in Standard Form (Paper 1)

  • Multiplication Law: Multiply coefficients and add powers of $10$: $$(a \times 10^p) \times (b \times 10^q) = (a \times b) \times 10^{p+q}$$ Always normalize the resulting product if $a \times b \ge 10$.
  • Division Law: Divide coefficients and subtract powers of $10$: $$\frac{a \times 10^p}{b \times 10^q} = \left(\frac{a}{b}\right) \times 10^{p-q}$$ Always normalize if $\frac{a}{b} < 1$.
  • Addition and Subtraction (Exponent Alignment): You cannot add or subtract coefficients directly unless both numbers share the same power of $10$!
    1. Identify the higher exponent and convert the smaller number to match it.
    2. Add or subtract the adjusted coefficients: $(a + b') \times 10^{p}$.
    3. Normalize the answer if necessary.

Example 2

Evaluating Non-Calculator Arithmetic in Standard Form

Without using a calculator, evaluate the following, expressing your answers in the form $a \times 10^k$ where $1 \le a < 10$ and $k \in \mathbb{Z}$:
(a) $\frac{(4.5 \times 10^7) \times (8.0 \times 10^{-3})}{1.5 \times 10^2}$
(b) $(6.2 \times 10^6) + (7.9 \times 10^5)$

  1. Part (a): Evaluate numerator product first: $$(4.5 \times 8.0) \times 10^{7 + (-3)} = 36.0 \times 10^4$$ Divide by denominator $1.5 \times 10^2$: $$\frac{36.0 \times 10^4}{1.5 \times 10^2} = \left(\frac{36.0}{1.5}\right) \times 10^{4 - 2} = 24.0 \times 10^2$$ Normalize: $24.0 \times 10^2 = 2.4 \times 10^1 \times 10^2 = 2.4 \times 10^3$.
  2. Part (b): Match exponents to $10^6$: $$7.9 \times 10^5 = 0.79 \times 10^6$$ Add the coefficients: $$(6.2 + 0.79) \times 10^6 = 6.99 \times 10^6$$ Since $1 \le 6.99 < 10$, the expression is already normalized.

Answer: (a) $2.4 \times 10^3$; (b) $6.99 \times 10^6$

1.1

Significant Figures & The IB 3 s.f. Examination Rule

  • Official IB Accuracy Convention: Unless otherwise specified in a question, all non-exact final numerical answers must be given to exactly 3 significant figures (or exact fractions/surds where appropriate).
  • Rules for Identifying Significant Figures in Standard Form: In $a \times 10^k$, all digits written in the coefficient $a$ are significant:
    • $4.5 \times 10^3 \implies 2\text{ s.f.}$
    • $4.50 \times 10^3 \implies 3\text{ s.f.}$ (the trailing zero is significant and required!).
    • $4.500 \times 10^3 \implies 4\text{ s.f.}$
  • Rounding Protocol: Inspect the 4th significant digit:
    • If $\ge 5$, round the 3rd digit up: $6.487 \times 10^4 \to 6.49 \times 10^4$.
    • If $< 5$, keep the 3rd digit: $2.143 \times 10^{-2} \to 2.14 \times 10^{-2}$.
  • Intermediate Calculation Rule: Always retain at least 4 to 5 significant figures (or store values in calculator memory) during multi-step working. Rounding prematurely causes catastrophic accuracy loss that forfeits final accuracy marks ($A0$).

Example 3

Multi-Step Calculation with 3 s.f. Rounding

Calculate $\sqrt{\frac{5.84 \times 10^8}{3.19 \times 10^3}}$, giving your answer in standard form correct to $3$ significant figures.

  1. Evaluate the radicand (inside fraction) with full precision: $$\frac{5.84 \times 10^8}{3.19 \times 10^3} = \left(\frac{5.84}{3.19}\right) \times 10^{8 - 3} = 1.830721 \times 10^5$$
  2. Take the square root: $$\sqrt{1.830721 \times 10^5} = \sqrt{183072.1} \approx 427.869$$
  3. Convert to standard form: $$427.869 = 4.27869 \times 10^2$$
  4. Round to 3 significant figures: The 4th significant digit is $8 \ge 5$, so round $7$ up to $8$: $$4.28 \times 10^2$$

Answer: $4.28 \times 10^2$

1.1

Orders of Magnitude & Scientific Modeling

  • Order of Magnitude Comparison: The order of magnitude is the power of $10$ closest to the number. The ratio of two scales $10^p$ and $10^q$ is $10^{p - q}$. For instance, a planetary distance of $10^{11} \text{ m}$ is $10^{11 - (-10)} = 10^{21}$ times larger than an atomic radius of $10^{-10} \text{ m}$.
  • Calculator $E$ Notation Warning: Calculators often display $3.45\text{E}8$ or $3.45\text{E}-5$. Never write $E$ on an IB examination script! You must write the full mathematical form: $3.45 \times 10^8$ or $3.45 \times 10^{-5}$.
  • Unit Conversions with Standard Form: Convert base units before calculating: $$\text{kilometres to metres: } \times 10^3, \quad \text{millimetres to metres: } \times 10^{-3}, \quad \text{hours to seconds: } \times 3600$$
Standard Form Architecture and Orders of Magnitude Diagram showing standard form decomposition into a normalized mantissa between 1 and 10 and integer power of ten along a logarithmic scale. a × 10 k Coefficient: 1 ≤ a < 10 Exponent: k ∈ ℤ (Integer) 10⁻¹⁰ m Atom 10⁻⁶ m Cell 10⁰ m Human 10⁶ m Earth 10¹¹ m Orbit
Figure 1.1: Structure of standard form $a \times 10^k$, where $1 \le a < 10$ and $k \in \mathbb{Z}$ compresses physical scales over 20+ orders of magnitude.

Example 4

Real-World Astronomical Modeling

The distance from Earth to the Sun is approximately $1.496 \times 10^8\text{ km}$. Light travels at a speed of $3.00 \times 10^8\text{ m s}^{-1}$.
(a) Convert the Earth-Sun distance into metres in standard form.
(b) Calculate the time in seconds it takes for light to travel from the Sun to Earth, giving your answer to $3$ significant figures.

  1. Part (a): Convert km to metres ($1\text{ km} = 10^3\text{ m}$): $$d = (1.496 \times 10^8\text{ km}) \times 10^3\text{ m/km} = 1.496 \times 10^{8+3}\text{ m} = 1.496 \times 10^{11}\text{ m}$$
  2. Part (b): Use time formula $t = \frac{d}{v}$: $$t = \frac{1.496 \times 10^{11}}{3.00 \times 10^8} = \left(\frac{1.496}{3.00}\right) \times 10^{11 - 8} = 0.498667 \times 10^3$$
  3. Normalize and round to 3 significant figures: $$0.498667 \times 10^3 = 4.98667 \times 10^2 \approx 4.99 \times 10^2\text{ seconds} \quad (\text{or } 499\text{ s})$$

Answer: (a) $1.496 \times 10^{11}\text{ m}$; (b) $4.99 \times 10^2\text{ s}$

Example 5

Measurement Estimation & Percentage Error Analysis

A rectangular plot of land has exact measured length $L = 148.6\text{ m}$ and width $W = 72.4\text{ m}$. A surveyor estimates the length as $150\text{ m}$ and width as $70\text{ m}$.
(a) Calculate the exact area $A_E$ and the surveyor's estimated area $A_A$. [3 marks]
(b) Calculate the percentage error in the surveyor's estimated area, giving your answer to $3$ significant figures. [3 marks]

  1. Part (a): Compute exact and estimated areas: $$A_E = 148.6 \times 72.4 = 10\,758.64\text{ m}^2 \quad [M1][A1]$$ $$A_A = 150 \times 70 = 10\,500\text{ m}^2 \quad [A1]$$
  2. Part (b): Apply the percentage error formula: $$\epsilon = \left| \frac{v_A - v_E}{v_E} \right| \times 100\% = \left| \frac{10\,500 - 10\,758.64}{10\,758.64} \right| \times 100\% \quad [M1]$$ $$\epsilon = \left| \frac{-258.64}{10\,758.64} \right| \times 100\% = \frac{258.64}{10\,758.64} \times 100\% \approx 2.40402\% \quad [A1]$$ Round to $3$ significant figures: $$\epsilon = 2.40\% \quad [A1]$$

Answers: (a) $A_E = 10\,758.64\text{ m}^2$, $A_A = 10\,500\text{ m}^2$; (b) $\epsilon = 2.40\%$

Examiner Commentary: Candidates must always divide by the EXACT value $v_E$ in the denominator. Dividing by the approximate value $v_A = 10\,500$ is penalized with a method mark loss. Also note that trailing zeros in $2.40\%$ are required to demonstrate 3 significant figures.

Tier 1: Core Fluency (Questions 1–4)
Exercise 1 Paper 1 Core Skill

Write the following numbers in standard form $a \times 10^k$, where $1 \le a < 10$ and $k \in \mathbb{Z}$:

(a) $48\,500\,000$

(b) $0.00000307$

View Complete Solution
  1. Part (a): Shift decimal point $7$ places left: $$48\,500\,000 = 4.85 \times 10^7$$
  2. Part (b): Shift decimal point $6$ places right: $$0.00000307 = 3.07 \times 10^{-6}$$

Final Answer: (a) $4.85 \times 10^7$; (b) $3.07 \times 10^{-6}$

Exercise 2 Paper 1 Non-Calculator Product

Without using a calculator, evaluate $(2.5 \times 10^4) \times (6.0 \times 10^7)$, giving your answer in standard form.

View Complete Solution
  1. Multiply coefficients and add exponents: $$(2.5 \times 6.0) \times 10^{4 + 7} = 15.0 \times 10^{11}$$
  2. Normalize the coefficient: $$15.0 \times 10^{11} = (1.5 \times 10^1) \times 10^{11} = 1.5 \times 10^{12}$$

Final Answer: $1.5 \times 10^{12}$

Exercise 3 Paper 1 Non-Calculator Division

Without using a calculator, evaluate $\frac{1.44 \times 10^3}{1.2 \times 10^{-4}}$, giving your answer in standard form.

View Complete Solution
  1. Divide coefficients and subtract exponents: $$\left(\frac{1.44}{1.2}\right) \times 10^{3 - (-4)} = 1.2 \times 10^{3 + 4} = 1.2 \times 10^7$$
  2. Check normalization: Since $1 \le 1.2 < 10$, no further adjustment is required.

Final Answer: $1.2 \times 10^7$

Exercise 4 Paper 1 Exponent Alignment Addition

Without using a calculator, evaluate $(5.2 \times 10^5) + (7.8 \times 10^4)$, giving your answer in standard form.

View Complete Solution
  1. Convert both terms to the common exponent $10^5$: $$7.8 \times 10^4 = 0.78 \times 10^5$$
  2. Add the coefficients: $$(5.2 + 0.78) \times 10^5 = 5.98 \times 10^5$$
  3. Confirm standard form: $1 \le 5.98 < 10$, so the result is fully normalized.

Final Answer: $5.98 \times 10^5$

Tier 2: Exam Modeling (Questions 5–7)
Exercise 5 Paper 2 Calculator Precision & 3 s.f.

Calculate the value of $\frac{(6.34 \times 10^{-4}) \times (8.91 \times 10^9)}{4.27 \times 10^2}$, giving your answer in standard form correct to $3$ significant figures.

View Complete Solution
  1. Multiply the numerator terms: $$(6.34 \times 8.91) \times 10^{-4 + 9} = 56.4894 \times 10^5$$
  2. Divide by the denominator: $$\frac{56.4894 \times 10^5}{4.27 \times 10^2} = \left(\frac{56.4894}{4.27}\right) \times 10^{5 - 2} = 13.22937 \times 10^3$$
  3. Normalize into standard form: $$13.22937 \times 10^3 = 1.322937 \times 10^4$$
  4. Round to 3 significant figures: The 4th digit is $2 < 5$, so keep $1.32$: $$1.32 \times 10^4$$

Final Answer: $1.32 \times 10^4$

Exercise 6 Paper 2 Biological Modeling

A sample of blood contains $5.2 \times 10^6$ red blood cells per cubic millimetre ($\text{mm}^3$). An adult has approximately $5.0\text{ litres}$ of blood.
Given that $1\text{ litre} = 10^6\text{ mm}^3$, calculate the total number of red blood cells in the adult's body, giving your answer in standard form.

View Complete Solution
  1. Convert volume to $\text{mm}^3$: $$\text{Volume} = 5.0\text{ L} \times 10^6\text{ mm}^3\text{/L} = 5.0 \times 10^6\text{ mm}^3$$
  2. Multiply concentration by total volume: $$\text{Total cells} = (5.2 \times 10^6\text{ cells/mm}^3) \times (5.0 \times 10^6\text{ mm}^3)$$ $$= (5.2 \times 5.0) \times 10^{6 + 6} = 26.0 \times 10^{12}$$
  3. Normalize into standard form: $$26.0 \times 10^{12} = 2.6 \times 10^{13}$$

Final Answer: $2.6 \times 10^{13}$ cells

Exercise 7 Paper 1 Order of Magnitude Ratio

The radius of a hydrogen atom is approximately $5.3 \times 10^{-11}\text{ m}$. The radius of the Sun is approximately $6.96 \times 10^8\text{ m}$.
Calculate how many times larger the radius of the Sun is than the radius of a hydrogen atom, giving your answer in standard form to $3$ significant figures.

View Complete Solution
  1. Formulate the ratio: $$\text{Ratio} = \frac{6.96 \times 10^8}{5.3 \times 10^{-11}} = \left(\frac{6.96}{5.3}\right) \times 10^{8 - (-11)} = \left(\frac{6.96}{5.3}\right) \times 10^{19}$$
  2. Divide the coefficients: $$\frac{6.96}{5.3} \approx 1.313207...$$
  3. Round to 3 significant figures: $$1.31 \times 10^{19}$$

Final Answer: $1.31 \times 10^{19}$ times larger

Tier 3: Complex Applications & IB Exam Modeling (Questions 8–10)
Exercise 8 Paper 1 Exponent Algebraic Solving

Given that $(4.0 \times 10^k)^2 = 1.6 \times 10^{13}$, where $k \in \mathbb{Z}$, find the value of $k$.

View Complete Solution
  1. Square the left-hand side expression: $$(4.0 \times 10^k)^2 = (4.0)^2 \times (10^k)^2 = 16.0 \times 10^{2k}$$
  2. Normalize the left-hand side into standard form: $$16.0 \times 10^{2k} = 1.6 \times 10^1 \times 10^{2k} = 1.6 \times 10^{2k + 1}$$
  3. Equate with the right-hand side: $$1.6 \times 10^{2k + 1} = 1.6 \times 10^{13}$$
  4. Equate the exponents of $10$: $$2k + 1 = 13 \implies 2k = 12 \implies k = 6$$

Final Answer: $k = 6$

Exercise 9 Paper 2 Relative Difference & Percentages

Two celestial bodies have masses $M_1 = 5.97 \times 10^{24}\text{ kg}$ and $M_2 = 7.35 \times 10^{22}\text{ kg}$.
(a) Calculate $M_1 - M_2$ in standard form to $3$ significant figures.
(b) Express $M_2$ as a percentage of $M_1$, correct to $3$ significant figures.

View Complete Solution
  1. Part (a): Convert $M_2$ to exponent $10^{24}$: $$M_2 = 0.0735 \times 10^{24}\text{ kg}$$ $$M_1 - M_2 = (5.97 - 0.0735) \times 10^{24} = 5.8965 \times 10^{24}\text{ kg}$$ Round to 3 significant figures: $$5.90 \times 10^{24}\text{ kg}$$
  2. Part (b): Calculate the percentage: $$\text{Percentage} = \frac{7.35 \times 10^{22}}{5.97 \times 10^{24}} \times 100\% = \left(\frac{7.35}{5.97}\right) \times 10^{22 - 24 + 2}\%$$ $$= \left(\frac{7.35}{5.97}\right) \times 10^0\% \approx 1.23115\% \approx 1.23\%$$

Final Answer: (a) $5.90 \times 10^{24}\text{ kg}$; (b) $1.23\%$

Exercise 10 Paper 1 Multi-part Velocity & Unit Conversion

A space probe travels at a steady speed of $5.4 \times 10^4\text{ km h}^{-1}$.

(a) Express this speed in metres per second ($\text{m s}^{-1}$) in standard form.

(b) The probe needs to reach a destination located $8.1 \times 10^8\text{ km}$ away. Calculate the time in hours needed for the journey, giving your answer in standard form.

(c) Convert the travel time found in (b) into days, rounded to the nearest integer.

View Complete Solution
  1. Part (a): Convert km/h to m/s ($1\text{ km/h} = \frac{1000\text{ m}}{3600\text{ s}} = \frac{1}{3.6}\text{ m/s}$): $$\text{Speed} = \frac{5.4 \times 10^4 \times 10^3}{3600} = \frac{5.4 \times 10^7}{3.6 \times 10^3} = \left(\frac{5.4}{3.6}\right) \times 10^{7 - 3} = 1.5 \times 10^4\text{ m s}^{-1}$$
  2. Part (b): Calculate travel time in hours: $$t = \frac{\text{Distance}}{\text{Speed}} = \frac{8.1 \times 10^8\text{ km}}{5.4 \times 10^4\text{ km h}^{-1}} = \left(\frac{8.1}{5.4}\right) \times 10^{8 - 4} = 1.5 \times 10^4\text{ hours}$$
  3. Part (c): Convert hours to days: $$\text{Days} = \frac{1.5 \times 10^4}{24} = \frac{15000}{24} = 625\text{ days}$$

Final Answer: (a) $1.5 \times 10^4\text{ m s}^{-1}$; (b) $1.5 \times 10^4\text{ hours}$; (c) $625\text{ days}$

Formula Booklet SL 1.1

Standard Form Definition

  • Mathematical Definition: $$a \times 10^k$$
  • Coefficient Constraint: $$1 \le a < 10$$
  • Exponent Constraint: $$k \in \mathbb{Z} \quad (\text{Integer: } \dots, -2, -1, 0, 1, 2, \dots)$$
Must Memorize

Arithmetic Laws in Standard Form

  • Multiplication: $$(a \times 10^p) \times (b \times 10^q) = (ab) \times 10^{p+q}$$
  • Division: $$\frac{a \times 10^p}{b \times 10^q} = \left(\frac{a}{b}\right) \times 10^{p-q}$$
  • Powers: $$(a \times 10^p)^n = a^n \times 10^{pn}$$
  • Addition / Subtraction: $$a \times 10^p \pm b \times 10^q = (a \pm b \cdot 10^{q-p}) \times 10^p \quad (p \ge q)$$
IB Exam Rules

Precision & Notation Protocols

  • 3 Significant Figures: Unless told otherwise, all non-exact answers must be rounded to exactly 3 s.f.
  • Exact Numbers: Leave integers and fractions exact (e.g. $\frac{7}{8}$ rather than $0.875$ if not requested as decimal).
  • No Calculator Display Symbols: Never write `3.45E7` or `3.45 * 10^7`. Write $3.45 \times 10^7$.
  • Significant Trailing Zeros: $4.50 \times 10^3$ has 3 s.f.; dropping the zero ($4.5 \times 10^3$) forfeits precision marks.

Critical IB Examination Pitfalls in Standard Form

  • Failing to Normalize the Mantissa: Writing $25.4 \times 10^6$ instead of $2.54 \times 10^7$ or $0.78 \times 10^{-3}$ instead of $7.8 \times 10^{-4}$ loses accuracy marks ($A0$).
  • Writing Calculator $E$ Notation: The symbol $E$ or $EE$ is calculator shorthand. On exam scripts, the symbol $\times 10^k$ is mandatory.
  • Adding Numbers Without Matching Powers: $(3 \times 10^4) + (4 \times 10^3) \neq 7 \times 10^7$. You must convert to $(3 + 0.4) \times 10^4 = 3.4 \times 10^4$.
  • Truncating Without Rounding: Truncating $4.288 \times 10^3$ to $4.28 \times 10^3$ instead of rounding up to $4.29 \times 10^3$ incurs an accuracy penalty.
  • Premature Rounding: Rounding intermediate steps to 3 s.f. can propagate errors into the 3rd figure of the final answer. Store intermediate values in your calculator.

Section A: Diagnostic Check

4 core concept check questions. Immediate feedback and scoring.

Score: 0 / 4

Question 1

Which of the following is written correctly in standard form $a \times 10^k$?

Correct! Standard form requires $1 \le a < 10$. In option C, $a = 4.50$ satisfies $1 \le 4.50 < 10$, and $k = 6 \in \mathbb{Z}$.

Question 2

What is the result of $(8.0 \times 10^5) \times (5.0 \times 10^{-2})$ in standard form?

Correct! $(8.0 \times 5.0) \times 10^{5 + (-2)} = 40.0 \times 10^3 = (4.0 \times 10^1) \times 10^3 = 4.0 \times 10^4$.

Question 3

Evaluate $(3.4 \times 10^7) + (2.0 \times 10^6)$ without a calculator:

Correct! Align powers of ten: $2.0 \times 10^6 = 0.2 \times 10^7$. Adding gives $(3.4 + 0.2) \times 10^7 = 3.6 \times 10^7$.

Question 4

A calculator screen shows `2.7849E-4`. What is this number correctly written on an IB exam to $3$ significant figures?

Correct! The 4th digit is $4 < 5$, so round down to $2.78$. Replace calculator `E-4` with $\times 10^{-4}$, giving $2.78 \times 10^{-4}$.

Section B: Examination Practice

Multi-step questions with markscheme breakdowns.

Paper 1 (Non-Calculator) [6 marks]

Let $A = 4.5 \times 10^6$ and $B = 1.5 \times 10^{-2}$.

(a) Find the value of $A \times B$, giving your answer in the form $a \times 10^k$ where $1 \le a < 10$ and $k \in \mathbb{Z}$. [2 marks]
(b) Find the value of $\frac{A}{B}$, giving your answer in the form $a \times 10^k$ where $1 \le a < 10$ and $k \in \mathbb{Z}$. [2 marks]
(c) Given that $C = 5.0 \times 10^5$, evaluate $A - C$ in standard form. [2 marks]
View Markscheme
(a)
$A \times B = (4.5 \times 1.5) \times 10^{6 + (-2)}$ M1
$= 6.75 \times 10^4$
Since $1 \le 6.75 < 10$, $\mathbf{6.75 \times 10^4}$ A1

(b)
$\frac{A}{B} = \left(\frac{4.5}{1.5}\right) \times 10^{6 - (-2)}$ M1
$= 3.0 \times 10^{6 + 2} = \mathbf{3.0 \times 10^8}$ A1

(c)
Convert $C$ to exponent $10^6$: $C = 0.5 \times 10^6$ M1
$A - C = (4.5 - 0.5) \times 10^6 = \mathbf{4.0 \times 10^6}$ A1
Paper 2 (Calculator Allowed) [6 marks]

A rectangular silicon chip has length $l = 3.65 \times 10^{-3}\text{ m}$ and width $w = 2.40 \times 10^{-3}\text{ m}$.

(a) Calculate the area of the chip in square metres ($\text{m}^2$), giving your answer in standard form correct to $3$ significant figures. [2 marks]
(b) The chip contains $1.42 \times 10^7$ transistors evenly distributed over its surface. Calculate the transistor density in transistors per square metre ($\text{transistors m}^{-2}$), giving your answer in standard form to $3$ significant figures. [4 marks]
View Markscheme
(a)
$\text{Area} = l \times w = (3.65 \times 10^{-3}) \times (2.40 \times 10^{-3})$ M1
$= (3.65 \times 2.40) \times 10^{-6} = 8.76 \times 10^{-6}\text{ m}^2$
$\mathbf{8.76 \times 10^{-6}\text{ m}^2}$ A1

(b)
$\text{Density} = \frac{\text{Total Transistors}}{\text{Area}} = \frac{1.42 \times 10^7}{8.76 \times 10^{-6}}$ M1
$= \left(\frac{1.42}{8.76}\right) \times 10^{7 - (-6)} = 0.1621 \times 10^{13}$ A1
Normalize into standard form: $1.621 \times 10^{12}$ M1
Round to 3 s.f.: $\mathbf{1.62 \times 10^{12}\text{ transistors m}^{-2}}$ A1

Unit Overview

All Lessons in Number & Algebra (Topic 1)

1.1 Standard Form Scientific notation, powers-of-ten operations, and the IB 3 s.f. rule. 1.2 Arithmetic Sequences & Series Common difference, general $n$th term, and partial sum formulas. 1.3, 1.8 Geometric Sequences & Infinite Sums Common ratio, finite sums, and convergence to infinity when $|r| < 1$. 1.4 Financial Applications Compound interest, annual depreciation, and TVM solver modeling. 1.5, 1.7 Exponents & Logarithms Laws of exponents, logarithm rules, base changes, and natural logs. 1.6, 1.9 Proof & Binomial Theorem Deductive proof, Pascal's triangle, and binomial expansion terms.