Module 9 of 47Lesson 1 of 8 · AI 1.1 — Standard form

Course map · Topic 1: Number and Algebra

1.1 Standard form

SL & HL11 worked questions

Standard form is a short way to write very big and very small numbers, such as the distance to the Sun or the mass of an atom. You must be able to read it, write it, calculate with it and give answers in it.

✏️ Start with the questions. Try each one on paper, then open the solution and compare.

Stuck? Open the notes ↓

✏️ Practice questions

Try each question on paper first. Then open the solution and compare every step.

Question 1●○○ Warm-up✍️ By hand · Paper 1[3 marks]

Write the following numbers in standard form.

(a) $4\,560\,000$ [1]
(b) $0.000\,037\,2$ [1]
(c) $23.5\times10^{4}$ [1]

Show the full step-by-step solution

Step 1🧠 ThinkWhat are we looking for?

We need a number between 1 and 10 (the front number), times a power of 10. We move the decimal point until there is exactly one non-zero digit in front of it, and count how many places we moved it.

Part (a)

Step 1✍️ By handBig number → positive power

4\,560\,000

is really

4\,560\,000.0

. Move the point left until it sits after the 4:

4.56

. We moved it 6 places.

4\,560\,000 = 4.56\times1\,000\,000

= 4.56\times10^{6}

Part (b)

Step 1✍️ By handSmall number → negative power

Move the point right until it sits after the first non-zero digit, the 3:

3.72

. We moved it 5 places. The number is smaller than 1, so the power is negative.

0.000\,037\,2 = 3.72\times10^{-5}

Part (c)

Step 1✍️ By handFix the front number

23.5

is bigger than 10, so this is not standard form yet. Write

23.5

itself in standard form, then add the powers.

23.5 = 2.35\times10^{1}

23.5\times10^{4} = 2.35\times10^{1}\times10^{4}

= 2.35\times10^{5}

✅ Check it Turn each answer back:

4.56\times10^{6}

— move the point 6 places right — gives

4\,560\,000

. ✔

Final answer (a)

4.56\times10^{6}

(b)

3.72\times10^{-5}

(c)

2.35\times10^{5}

⚠️ Common mistake In (c), writing

23.5\times10^{4}

unchanged or

2.35\times10^{3}

. Making the front number smaller means the power must get bigger.

Question 2●○○ Warm-up✍️ By hand · Paper 1[2 marks]

Write the following as ordinary numbers.

(a) $6.08\times10^{5}$ [1]
(b) $9.1\times10^{-4}$ [1]

Show the full step-by-step solution

Step 1🧠 ThinkUndo standard form

A positive power means a big number: move the point to the right. A negative power means a small number: move the point to the left. Fill empty places with zeros.

Part (a)

Step 1✍️ By handMove 5 places right

6.08\times10^{5} = 6.08\times100\,000

= 608\,000

Part (b)

Step 1✍️ By handMove 4 places left

9.1\times10^{-4} = 9.1\div10\,000

= 0.000\,91

✅ Check it In (b) there are 3 zeros after the decimal point before the 9 — one fewer than the power 4. This is a quick way to check small numbers. ✔

Final answer (a)

608\,000

(b)

0.000\,91

Question 3●○○ Warm-up✍️ + 🖩 · Paper 1[2 marks]

Calculate

(3.2\times10^{5})\times(4.5\times10^{-2})

. Give your answer in the form

a\times10^{k}

, where

1\le a\lt 10

and

k\in\mathbb{Z}

Show the full step-by-step solution

Step 1🧠 ThinkFront numbers together, powers together

When we multiply, we can swap the order. So we multiply

3.2\times4.5

, and separately

10^{5}\times10^{-2}

. For powers of 10, multiplying means adding the powers:

5 + (-2) = 3

Step 2✍️ By handBy hand

3.2\times4.5 = 14.4

10^{5}\times10^{-2} = 10^{3}

14.4\times10^{3}

— but 14.4 is bigger than 10!

14.4 = 1.44\times10^{1}

1.44\times10^{1}\times10^{3} = 1.44\times10^{4}

Step 3🖩 On the GDCOn the GDC — use the EE key

3.2›EE💻 click EE on keypad›5›×💻 *›4.5›EE💻 click EE on keypad›(−)💻 type -›2›enter💻 Enter

In the normal setting the calculator shows the ordinary number 14400. If you switch to Scientific format (press doc → Settings → Document Settings → Exponential Format: Scientific), the same calculation shows 1.44E4.

Normal format: ordinary number

Scientific format: 1.44E4 = 1.44 × 10⁴

💡 From now on In the rest of this page the GDC is set to Scientific format, so answers appear as standard form straight away.

Final answer

1.44\times10^{4}

⚠️ Common mistake Writing

14.4\times10^{3}

. It has the right value but it is not in standard form, so it loses the mark.

Question 4●●○ Standard✍️ + 🖩 · Paper 1[2 marks]

Calculate

\dfrac{6.3\times10^{8}}{1.8\times10^{-3}}

. Give your answer in standard form.

Show the full step-by-step solution

Step 1🧠 ThinkDividing: divide the front, subtract the powers

Divide

6.3

1.8

. For the powers, dividing means subtracting:

8 - (-3)

. Taking away a negative is the same as adding, so

8 - (-3) = 11

Step 2✍️ By handBy hand

\dfrac{6.3}{1.8} = 3.5

\dfrac{10^{8}}{10^{-3}} = 10^{8-(-3)} = 10^{11}

3.5\times10^{11}

Step 3🖩 On the GDCCheck with the GDC

6.3›EE💻 click EE on keypad›8›÷💻 /›1.8›EE💻 click EE on keypad›(−)💻 type -›3›enter💻 Enter

You can type the division with the ÷ key — no brackets needed, because each EE number is one single number to the calculator.

3.5E11 means 3.5 × 10¹¹

Final answer

3.5\times10^{11}

⚠️ Common mistake Doing

8 - 3 = 5

and getting

3.5\times10^{5}

. Watch the minus sign on the

-3

Question 5●●○ Standard🖩 GDC · Paper 1[4 marks]

Light travels at

3.00\times10^{8}

metres per second. The distance from the Sun to the Earth is about

1.496\times10^{11}

metres.

(a) Find the time, in seconds, that light takes to travel from the Sun to the Earth. Give your answer in the form $a\times10^{k}$ , where $1\le a\lt 10$ , $k\in\mathbb{Z}$ .[3]
(b) Convert your answer to minutes.[1]

Show the full step-by-step solution

Step 1🧠 ThinkWhich formula?

Time = distance ÷ speed. Distance is in metres and speed is in metres per second, so the time comes out in seconds.

Part (a)

Step 1✍️ By handSet up the division

t = \dfrac{1.496\times10^{11}}{3.00\times10^{8}}

Step 2🖩 On the GDCCalculate on the GDC

1.496›EE💻 click EE on keypad›11›÷💻 /›3›EE💻 click EE on keypad›8›enter💻 Enter

4.98667E2 = 498.667 seconds

Step 3✍️ By handRound to 3 significant figures

t = 4.98667\ldots\times10^{2}

t \approx 4.99\times10^{2}

secondsA1

Part (b)

Step 1✍️ By hand60 seconds in a minute

\dfrac{498.667}{60} = 8.31

minutes (3 s.f.)A1

💡 Remember Use the unrounded value 498.667 in (b), not 499. On the GDC just press ÷ 60 straight after — the calculator keeps the full answer.

Final answer (a)

4.99\times10^{2}

seconds (b)

8.31

minutes

Question 6●●○ Standard✍️ By hand · Paper 1[4 marks]

Without using a calculator, find the following. Give each answer in standard form.

(a) $4.2\times10^{6} + 3.5\times10^{5}$ [2]
(b) $7.1\times10^{-3} - 8\times10^{-4}$ [2]

Show the full step-by-step solution

Step 1🧠 ThinkWhy can't we just add 4.2 and 3.5?

4.2\times10^{6}

is millions, but

3.5\times10^{5}

is hundreds of thousands. They are different sizes, like adding metres to centimetres. We first write both with the same power of 10.

Part (a)

Step 1✍️ By handMake both powers 10⁶

3.5\times10^{5}

is the same as

0.35\times10^{6}

(the front number gets 10 times smaller, the power one bigger).

4.2\times10^{6} + 0.35\times10^{6}

= (4.2+0.35)\times10^{6}

= 4.55\times10^{6}

Part (b)

Step 1✍️ By handOr use ordinary numbers

0.0071 - 0.0008

= 0.0063

= 6.3\times10^{-3}

✅ Check it In (a),

4\,200\,000 + 350\,000 = 4\,550\,000 = 4.55\times10^{6}

. ✔

Final answer (a)

4.55\times10^{6}

(b)

6.3\times10^{-3}

⚠️ Common mistake Writing

7.7\times10^{6}

in (a) by adding 4.2 and 3.5 — the powers were different.

Question 7●●○ Standard🖩 GDC · Paper 1[3 marks]

Botswana has a population of approximately

2.36\times10^{6}

people and an area of approximately

5.82\times10^{5}\ \text{km}^2

. Find the population density of Botswana in people per

\text{km}^2

. Give your answer correct to 3 significant figures.

Show the full step-by-step solution

Step 1🧠 ThinkWhat is population density?

Population density tells us how many people there are, on average, for each square kilometre. So we share the people out over the area: people ÷ area.

Step 2✍️ By handWrite the division

\text{density} = \dfrac{2.36\times10^{6}}{5.82\times10^{5}}

Step 3🖩 On the GDCCalculate

2.36›EE💻 click EE on keypad›6›÷💻 /›5.82›EE💻 click EE on keypad›5›enter💻 Enter

4.05498E0 = 4.05498 × 10⁰ = 4.05498

Step 4✍️ By handRound

4.05498\ldots

\approx 4.05

people per

\text{km}^2

💡 Remember

10^{0} = 1

, so E0 just means “times 1”. An answer between 1 and 10 does not need a power of 10.

Final answer

4.05

people per

\text{km}^2

Question 8●●● Exam-level🖩 GDC · Paper 1[5 marks]

The mass of one hydrogen atom is

1.67\times10^{-27}

kg.

(a) Find the mass of $6.02\times10^{23}$ hydrogen atoms. Give your answer in kg, in the form $a\times10^{k}$ , where $1\le a\lt 10$ , $k\in\mathbb{Z}$ .[2]
(b) Write down your answer to part (a) in grams.[1]
(c) Find the number of hydrogen atoms in 1 kg of hydrogen. Give your answer in standard form.[2]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkMany atoms: multiply

If one atom has mass

1.67\times10^{-27}

kg, then

6.02\times10^{23}

atoms have that many times as much mass.

Step 2🖩 On the GDCMultiply on the GDC

1.67›EE💻 click EE on keypad›(−)💻 type -›27›×💻 *›6.02›EE💻 click EE on keypad›23›enter💻 Enter

1.00534 × 10⁻³ kg

Step 3✍️ By handRound to 3 s.f.

1.67\times10^{-27}\times6.02\times10^{23}

= 1.01\times10^{-3}

kgA1

Part (b)

Step 1✍️ By hand1 kg = 1000 g

1.00534\times10^{-3}\times1000 = 1.01

gA1

Part (c)

Step 1🧠 ThinkHow many fit into 1 kg?

“How many of these fit into that?” is a division:

1 \div

(mass of one atom).

Step 2🖩 On the GDCDivide on the GDC

1›÷💻 /›1.67›EE💻 click EE on keypad›(−)💻 type -›27›enter💻 Enter

5.98802 × 10²⁶ atoms

Step 3✍️ By handAnswer

\dfrac{1}{1.67\times10^{-27}}

= 5.99\times10^{26}

atomsA1

Final answer (a)

1.01\times10^{-3}

kg (b)

1.01

g (c)

5.99\times10^{26}

atoms

Question 9●●● Exam-level🖩 GDC · Paper 1[6 marks]

Each year, about

1.1\times10^{10}\ \text{m}^3

of water flows into the Okavango Delta. Assume the water arrives at a constant rate and that a year has 365 days.

(a) Find the volume of water that flows into the delta each day. Give your answer in standard form, correct to 3 significant figures.[2]
(b) An Olympic swimming pool holds $2.5\times10^{3}\ \text{m}^3$ of water. Find how many Olympic pools could be filled with one day's water. Give your answer in standard form.[2]
(c) Find the volume of water that flows into the delta each second.[2]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkShare the year's water over 365 days

The same amount arrives every day, so one day's water = the year's water ÷ 365.

Step 2🖩 On the GDCDivide by 365

1.1›EE💻 click EE on keypad›10›÷💻 /›365›enter💻 Enter

Line 2 uses the unrounded answer (ans)

Step 3✍️ By handAnswer to 3 s.f.

\dfrac{1.1\times10^{10}}{365}

= 3.01\times10^{7}\ \text{m}^3

per dayA1

Part (b)

Step 1✍️ By handHow many pools fit into one day's water?

Divide one day's water by the size of one pool. On the GDC press ÷ straight away — the calculator types ans for you, so it uses the full value

3.0137\ldots\times10^{7}

\dfrac{3.0137\times10^{7}}{2.5\times10^{3}}

= 1.21\times10^{4}

poolsA1

Part (c)

Step 1🧠 ThinkHow many seconds are in a year?

1 day = 24 hours, 1 hour = 3600 seconds. So one year =

365\times24\times3600

seconds. Put the whole product in brackets when you divide.

Step 2🖩 On the GDCDivide by the number of seconds in a year

1.1›EE💻 click EE on keypad›10›÷💻 /›(365×24×3600)›enter💻 Enter

≈ 349 m³ every second

Step 3✍️ By handAnswer

\dfrac{1.1\times10^{10}}{365\times24\times3600}

= 3.49\times10^{2}\ \text{m}^3

per secondA1

Final answer (a)

3.01\times10^{7}\ \text{m}^3

(b)

1.21\times10^{4}

pools (c)

3.49\times10^{2}\ \text{m}^3

per second (about 349 m³)

⚠️ Common mistake In (c), typing

\dfrac{1.1E10}{365}

·24·3600. Without brackets the calculator divides by 365 and then multiplies by 24 and 3600.

Question 10●●● Exam-level✍️ + 🖩 · Paper 1[6 marks]

Let

x = 2.4\times10^{5}

and

y = 6\times10^{-3}

(a) Find $xy$ . Give your answer in the form $a\times10^{k}$ , where $1\le a\lt 10$ , $k\in\mathbb{Z}$ .[2]
(b) Find $\dfrac{x}{y}$ , giving your answer in standard form.[2]
(c) Find $x^{2} + \dfrac{1}{y}$ , giving your answer in standard form correct to 3 significant figures.[2]

Show the full step-by-step solution

Part (a)

Step 1✍️ By handMultiply: add the powers

xy = (2.4\times6)\times10^{5+(-3)}

= 14.4\times10^{2}

= 1.44\times10^{3}

Part (b)

Step 1✍️ By handDivide: subtract the powers

\dfrac{x}{y} = \dfrac{2.4}{6}\times10^{5-(-3)}

= 0.4\times10^{8}

= 4\times10^{7}

💡 Remember 0.4 is less than 1, so it is not allowed in front.

0.4 = 4\times10^{-1}

, which makes the power go down by one:

10^{8}\to10^{7}

Part (c)

Step 1🧠 ThinkStore x and y first

Typing long numbers again and again causes mistakes. Store them in the calculator: type the number, press ctrl var (the sto→ key), then the letter. Then just type x and y.

Step 2🖩 On the GDCStore, then calculate

2.4›EE💻 click EE on keypad›5›sto→💻 type :=›x›enter💻 Enter›6›EE💻 click EE on keypad›(−)💻 type -›3›sto→💻 type :=›y›enter💻 Enter›x^2+1/y›enter💻 Enter

x² +

\dfrac{1}{y}

= 5.76 × 10¹⁰

Step 3✍️ By handWhy did $\dfrac{1}{y}$ make no difference?

x^{2} = 5.76\times10^{10} = 57\,600\,000\,000

\dfrac{1}{y} = 166.67

57\,600\,000\,000 + 166.67 = 57\,600\,000\,166.67

\approx 5.76\times10^{10}

(3 s.f.)A1

✅ Check it 166.67 is tiny compared with 57.6 billion, so after rounding to 3 significant figures it disappears. ✔

Final answer (a)

1.44\times10^{3}

(b)

4\times10^{7}

(c)

5.76\times10^{10}

Question 11●●● Exam-level🖩 GDC · Paper 2[6 marks]

A computer in a Gaborone data centre can carry out

3.5\times10^{9}

operations per second.

(a) Find the number of operations it can carry out in one hour. Give your answer in standard form.[2]
(b) A weather simulation needs $8\times10^{15}$ operations. Find the time, in seconds, that the computer needs for the simulation. Give your answer in the form $a\times10^{k}$ , where $1\le a\lt 10$ , $k\in\mathbb{Z}$ .[2]
(c) Convert your answer to part (b) into hours, correct to the nearest hour.[2]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkSeconds in one hour

1 hour = 60 minutes =

60\times60 = 3600

seconds. The computer does

3.5\times10^{9}

operations in each of those seconds.

Step 2🖩 On the GDCMultiply

3.5›EE💻 click EE on keypad›9›×💻 *›3600›enter💻 Enter

1.26 × 10¹³ operations

Step 3✍️ By handAnswer

3.5\times10^{9}\times3600

= 1.26\times10^{13}

operationsA1

Part (b)

Step 1🧠 ThinkTime = work ÷ speed

The computer does

3.5\times10^{9}

operations each second. To find how many seconds it needs, divide the total operations by the operations per second.

Step 2🖩 On the GDCDivide, then convert to hours

8›EE💻 click EE on keypad›15›÷💻 /›3.5›EE💻 click EE on keypad›9›enter💻 Enter›÷💻 /›3600›enter💻 Enter

2.29 × 10⁶ s, then 634.9 hours

Step 3✍️ By handAnswer to (b)

\dfrac{8\times10^{15}}{3.5\times10^{9}}

= 2.29\times10^{6}

secondsA1

Part (c)

Step 1✍️ By handDivide by 3600 and round

\dfrac{2.28571\ldots\times10^{6}}{3600}

= 634.92\ldots \approx 635

hoursA1

✅ Check it 635 hours is about

635\div24\approx 26

days. A huge simulation taking a few weeks is believable. ✔

Final answer (a)

1.26\times10^{13}

(b)

2.29\times10^{6}

seconds (c) 635 hours

🆘 Stuck on a question? Open the notes, GDC toolkit and common mistakes

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

🎯 What you must know

A number in standard form looks like $a\times10^{k}$ , where $1\le a\lt 10$ and $k$ is an integer (a whole number, which may be negative).
Big numbers (10 or more) have a positive power: $52\,000 = 5.2\times10^{4}$ .
Small numbers (between 0 and 1) have a negative power: $0.0068 = 6.8\times10^{-3}$ .
The power $k$ counts how many places the decimal point moves so that exactly one non-zero digit is in front of it.
Multiply: multiply the front numbers, add the powers. Divide: divide the front numbers, subtract the powers. Then fix $a$ so that $1\le a\lt 10$ .
Add or subtract: you cannot just add the front numbers. Change both numbers to the same power of 10 (or to ordinary numbers) first.
The GDC writes $3.2\times10^{5}$ as 3.2E5. The letter E is calculator language only — never write E in your answer.

Standard form

a\times10^{k},\quad 1\le a\lt 10,\ k\in\mathbb{Z}

🧠 Learn this one

🖩 Your GDC toolkit for this subtopic

Follow the steps on your own TI-Nspire CX II. The screens show what you should see.

Type a number in standard form, e.g. 3.2 × 10⁵

3.2›EE💻 click EE on keypad›5

Type 3.2.
Press EE (it types E, which means “× 10 to the power”).
Type 5.

Normal format: ordinary number

Worked example: Question 3 — On the GDC — use the EE key

Type a negative power, e.g. 4.5 × 10⁻²

4.5›EE💻 click EE on keypad›(−)💻 type -›2

Type 4.5.
Press EE (it types E, which means “× 10 to the power”).
Press (−) (the negative key, not the minus key).
Type 2.

Normal format: ordinary number

Worked example: Question 3 — On the GDC — use the EE key

Make every answer appear in standard form: doc → Settings → Document Settings → Exponential Format → Scientific → OK

Press doc.
Choose Settings, then Document Settings.
Press tab until Exponential Format is highlighted, press click and choose Scientific.
Press tab to OK and press enter.

Set Exponential Format to Scientific

⚠️ Where students lose marks

Leaving the front number outside the range, e.g. writing $14.4\times10^{3}$ . The front number must be at least 1 and less than 10: $1.44\times10^{4}$ .
Getting the sign of the power wrong: small numbers (less than 1) need a negative power.
Copying the calculator display such as 3.5E11 as your answer. Write $3.5\times10^{11}$ .
Typing $10^{5}$ as 10 EE 5. That means $10\times10^{5}=10^{6}$ . Use either EE 5 on its own or 10 ^ 5.
Forgetting to round the front number to 3 significant figures: $4.986\ldots\times10^{2}$ becomes $4.99\times10^{2}$ .

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL AI 1.1 — Standard form

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL

1.1 Standard form

✏️ Practice questions

🎯 What you must know

🖩 Your GDC toolkit for this subtopic

⚠️ Where students lose marks