Operations with numbers in standard form

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Mathematics: Applications and Interpretation · Number and algebra (SL) · SL · Guide 2021

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Operations with numbers in standard form

Standard form writes any positive number as $a \times 10^k$ , where $1 \le a < 10$ and $k$ is an integer.

The value of $a$ carries the digits; the power $k$ carries the size.

So 52 300 000 is $5.23 \times 10^{7}$ and 0.000 000 408 is $4.08 \times 10^{-7}$ .

Writing a number in standard form

Move the decimal point until exactly one non-zero digit sits in front of it.

Count the places moved: that count is the size of $k$ .

A large number (point moved left) gives a positive $k$ ; a number between 0 and 1 (point moved right) gives a negative $k$ .

Numbers from 1 up to, but not including, 10 have $k = 0$ .

The condition $1 \le a < 10$ is what makes the form unique.

The number 4800 is $48 \times 10^{2}$ and $0.48 \times 10^{4}$ , and both statements are true, but only $4.8 \times 10^{3}$ is standard form.

The value $a = 10$ is excluded, because $10 \times 10^{3}$ is really $1 \times 10^{4}$ .

Two limiting cases.

Zero has no standard form, since $a \ge 1$ makes $a \times 10^k$ positive for every $k$ .

A negative quantity takes a minus sign in front, as in $-3.2 \times 10^{4}$ .

For a number $x \ge 1$ , the power $k$ is the whole-number part of $\log_{10} x$ .

For example, $\log_{10} 52\,300\,000 = 7.718\ldots$ , so $k = 7$ .

This power of ten is the order of magnitude of the quantity, and it lets you compare sizes at a glance.

Significant figures live in a

Every significant figure of the number is a digit of $a$ .

To give an answer to 3 significant figures, round $a$ to 3 significant figures and leave $k$ alone.

One case changes $k$ : $9.996 \times 10^{5}$ rounds to $10.0 \times 10^{5}$ , which you must rewrite as $1.00 \times 10^{6}$ .

The zeros in $1.00$ are significant, so keep them.

Multiplying and dividing

Multiply the $a$ values and add the powers; divide the $a$ values and subtract the powers.

This is the exponent law $10^{m} \times 10^{n} = 10^{m+n}$ , with $10^{m} \div 10^{n} = 10^{m-n}$ .

The result often lands outside $1 \le a < 10$ , so finish by adjusting: $24 \times 10^{-4} = 2.4 \times 10^{1} \times 10^{-4} = 2.4 \times 10^{-3}$ .

The adjustment is a balance.

Divide $a$ by 10 and $k$ goes up by 1; multiply $a$ by 10 and $k$ goes down by 1.

The value of the number never changes.

Adding and subtracting

You cannot add digits with different place values, so first rewrite one number to share the other's power of ten.

Then add or subtract the $a$ values and adjust.

For example, $3.6 \times 10^{6} + 4.5 \times 10^{5} = 3.6 \times 10^{6} + 0.45 \times 10^{6} = 4.05 \times 10^{6}$ .

When the powers differ by a lot, the smaller number barely moves the larger one, which is a useful check.

Using the GDC, and what to write

The GDC is allowed on every paper, so the arithmetic is rarely the problem.

Enter a number such as $3.5 \times 10^{-3}$ with the calculator's power-of-ten key, and put brackets round each number in a quotient.

Without brackets, $6 \div 2 \times 10^{3}$ is read as $(6 \div 2) \times 10^{3}$ , not as $6 \div (2 \times 10^{3})$ .

The GDC shows very large and very small answers as, for example, 7.19812278E8.

That is a display convention, not mathematical notation, and the IB says it is not an acceptable answer.

Neither is $7.20^{8}$ , nor the digits 7.20 on their own.

Write the calculation you set up, copy enough digits to show the value, then write $7.20 \times 10^{8}$ .

Carry the full GDC value into the next step.

Round only the final answer, and round $a$ , not $k$ .

If a question asks for "the form $a \times 10^{k}$ , where $1 \le a < 10$ and $k \in \mathbb{Z}$ ", that form carries the final mark.

Standard form in context

Contexts with very large or very small quantities come from astronomy, atoms and cells, data storage, energy, and national finances.

Two skills matter there.

Units. Convert before you combine. A cubic metre is $10^{3}$ litres, a micrometre is $10^{-6}$ m and a tonne is $10^{3}$ kg. Multiplying by $10^{3}$ adds 3 to $k$ .

Comparisons. "How many times larger" is a quotient, not a difference. Two quantities whose powers differ by 2 differ by a factor of roughly 100.

Standard form also turns up inside other topics.

A sequence total, a loan over many years or a small probability can produce a number that the GDC shows in E-notation, and the same rules apply to the answer line.

✓

EXAM FOCUS

How the IB examines this

Paper 1: a short question in two or three parts: write down a number in standard form, then find a product or quotient in a context (a density, a time, a number of particles), often with a unit conversion. The final A1 usually needs "the form $a \times 10^{k}$ ". Marks are lost for E-notation, for writing only the digits of $a$ , for $a$ left outside $1 \le a < 10$ , and for a conversion factor applied the wrong way.

Paper 1, accuracy: a question that mixes a stated accuracy with standard form ("to 3 significant figures, in the form ..."). The marks reward the set-up written first and the full GDC value seen. A 2 significant figure answer, or a rounding that should have changed $k$ but did not, loses the A1.

Paper 2: standard form inside a longer context question (energy, emissions, populations of particles, data), often linked to a sequence or a percentage. Parts chain: a "show that" for a value in standard form, then a "hence" that uses it, then a decision or comparison in context. Marks are lost by premature rounding between parts, by mixing units, and by a decision stated with no supporting value.

◆

Key terms

Standard form: a number written as $a \times 10^{k}$ with $1 \le a < 10$ and $k \in \mathbb{Z}$ .

Order of magnitude: the power of ten, $k$ , that gives the approximate size of a quantity.

Significant figures: the digits that carry the precision of a value; in standard form they are the digits of $a$ .

Calculator notation (E-notation): a display such as 2.5E-7, meaning $2.5 \times 10^{-7}$ ; never accepted as a final answer.

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Worked example 1: a quotient by hand (Paper 1, show that)

(a) Show that $\dfrac{(6.3 \times 10^{5}) \times (4 \times 10^{-2})}{1.8 \times 10^{-3}} = 1.4 \times 10^{7}$ .

(b) Find $4.6 \times 10^{-3} - 7 \times 10^{-4}$ , giving your answer in the form $a \times 10^{k}$ , where $1 \le a < 10$ and $k \in \mathbb{Z}$ .

1. (a) Collect the $a$ values: $\dfrac{6.3 \times 4}{1.8} = \dfrac{25.2}{1.8} = 14$ . M1

2. Collect the powers: $10^{5 + (-2) - (-3)} = 10^{6}$ , so the value is $14 \times 10^{6}$ . A1

3. Adjust: $14 \times 10^{6} = 1.4 \times 10^{1} \times 10^{6} = 1.4 \times 10^{7}$ . AG [2 marks]

4. (b) Match the powers: $4.6 \times 10^{-3} - 0.7 \times 10^{-3}$ . M1

5. $= 3.9 \times 10^{-3}$ . A1 [2 marks]

Answer: (a) shown, the line $14 \times 10^{6}$ comes before the given result; (b) $3.9 \times 10^{-3}$ .

A "show that" must reach the result without using it.

Typing the expression into the GDC and writing $1.4 \times 10^{7}$ shows nothing, so it earns nothing.

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Worked example 2: energy from a dam (Paper 1, GDC)

A hydroelectric dam produces $2.27 \times 10^{16}$ joules of energy in a year.

A household uses $1.3 \times 10^{10}$ joules in a year.

(a) Find the number of households the dam could supply for a year. Give your answer in the form $a \times 10^{k}$ , where $1 \le a < 10$ and $k \in \mathbb{Z}$ , to 3 significant figures.

(b) Take a year to be 365 days. Find the mean energy the dam produces each second.

1. (a) Write the set-up: $\dfrac{2.27 \times 10^{16}}{1.3 \times 10^{10}}$ . Enter each number in brackets with the power-of-ten key.

2. The GDC shows 1746153.846, so the value is $1\,746\,153.8\ldots$

3. Round $a$ to 3 significant figures: $1.75 \times 10^{6}$ households.

4. (b) Seconds in a year: $365 \times 24 \times 3600 = 31\,536\,000 = 3.1536 \times 10^{7}$ .

5. Write the set-up: $\dfrac{2.27 \times 10^{16}}{3.1536 \times 10^{7}}$ . The GDC shows 719812278, so the value is $7.198\ldots \times 10^{8}$ .

6. To 3 significant figures, $a = 7.20$ . The final zero is significant, so it stays.

Answer: (a) $1.75 \times 10^{6}$ households; (b) $7.20 \times 10^{8}$ joules per second.

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Worked example 3: is the headline true? (Paper 1, interpretation)

The public debt of an invented country is $1.87 \times 10^{12}$ euros.

Its population is $4.62 \times 10^{7}$ .

A headline says the debt is "more than 40 000 euros for every resident".

(a) Determine whether the headline is correct.

(b) The mean annual income in the country is $3.15 \times 10^{4}$ euros. Interpret the debt per resident in terms of this income.

1. (a) Debt per resident $= \dfrac{1.87 \times 10^{12}}{4.62 \times 10^{7}}$ (M1)

2. $= 40\,476.19\ldots = 4.05 \times 10^{4}$ euros (3 s.f.) A1

3. $40\,476 > 40\,000$ , so the headline is correct. R1

4. (b) $\dfrac{40\,476.19\ldots}{3.15 \times 10^{4}} = 1.28$ (3 s.f.) (A1)

5. The debt per resident is about 1.28 times the mean annual income: a resident on that income would need nearly 1.3 years of their whole income to repay their share. R1

Answer: (a) correct, about $4.05 \times 10^{4}$ euros per resident; (b) about 1.28 years of mean income.

The decision in (a) earns its mark only after the value is written.

The R1 in (b) needs the meaning in the context (years of income), not just the number 1.28.

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Worked example 4: microplastics in a lake (Paper 2)

A lake holds $3.8 \times 10^{9}$ m³ of water.

Tests suggest a mean of 2.6 microplastic particles per litre.

Use $1 \text{ m}^{3} = 10^{3}$ litres.

(a) Write down the volume of the lake in litres, in the form $a \times 10^{k}$ , where $1 \le a < 10$ and $k \in \mathbb{Z}$ . [1]

(b) Show that the lake contains approximately $9.9 \times 10^{12}$ particles. [2]

(c) The mean mass of a particle is $4.5 \times 10^{-8}$ kg. Find the total mass of the particles in the lake, in kg. [2]

(d) A filtration plant removes $1.2 \times 10^{9}$ particles each day. Assume no new particles enter the lake. Find the number of days needed to remove 10% of the particles. [3]

In fact, $3.5 \times 10^{8}$ new particles enter the lake each day.

(e) Find the first day on which the lake contains fewer than $9.5 \times 10^{12}$ particles. [3]

(f) State one assumption made in your answer to (e). [1]

1. (a) $3.8 \times 10^{9} \times 10^{3} =$ $3.8 \times 10^{12}$ litres A1 [1 mark]

2. (b) $3.8 \times 10^{12} \times 2.6$ M1

3. $= 9.88 \times 10^{12}$ A1

4. $\approx 9.9 \times 10^{12}$ AG [2 marks]

5. (c) $9.88 \times 10^{12} \times 4.5 \times 10^{-8}$ (M1)

6. $= 444\,600 =$ $4.45 \times 10^{5}$ kg A1 [2 marks]

7. (d) 10% of the particles: $9.88 \times 10^{11}$ (A1)

8. $\dfrac{9.88 \times 10^{11}}{1.2 \times 10^{9}} = 823.33\ldots$ M1

9. 824 days A1 [3 marks]

10. (e) net daily fall $= 1.2 \times 10^{9} - 3.5 \times 10^{8} = 8.5 \times 10^{8}$ (A1)

11. $9.88 \times 10^{12} - 8.5 \times 10^{8}n < 9.5 \times 10^{12}$ , so $n > 447.05\ldots$ M1

12. day 448 A1 [3 marks]

13. (f) the inflow and the removal stay at the same daily rates (or: the particles are spread evenly, so the sample concentration represents the whole lake) A1 [1 mark]

Answer: (a) $3.8 \times 10^{12}$ litres; (c) $4.45 \times 10^{5}$ kg; (d) 824 days; (e) day 448.

Where marks go: in (b) the line $9.88 \times 10^{12}$ must be seen, because the given value is rounded.

Using $9.9 \times 10^{12}$ in (c) gives 445 500 kg, which still rounds to $4.46 \times 10^{5}$ and loses the A1.

In (d) and (e), 823 and 447 are the classic slips: after 823 days the 10% is not yet removed, so the count rounds up.

The particle count in (e) falls by the same amount each day, an arithmetic sequence.

✦

Exam tip

Before you leave a standard form answer, check three things: $a$ has one non-zero digit before the point, $a$ has the accuracy asked for (3 significant figures by default, trailing zeros kept), and the power is written as $\times 10^{k}$ , never as E or as a power of $a$ .

Common mistakes

Copying 7.198E8 or $7.20^{8}$ from the screen, or writing only 7.20. Only $7.20 \times 10^{8}$ earns the final A1.

Adding the powers when adding numbers: $3.6 \times 10^{6} + 4.5 \times 10^{5}$ is not $8.1 \times 10^{11}$ . Match the powers first; add powers only when multiplying.

Leaving $a$ outside the interval, as in $14 \times 10^{6}$ . Adjust to $1.4 \times 10^{7}$ .

Rounding the next step from a rounded value: using $9.9 \times 10^{12}$ instead of $9.88 \times 10^{12}$ changes the third figure of the next answer.

Getting the sign of $k$ wrong for small numbers: 0.0003 is $3 \times 10^{-4}$ , not $3 \times 10^{4}$ .

Dividing by a power of ten typed without brackets, so the GDC multiplies by it instead.

Check your understanding

1. [Paper 1] In this question give each answer in the form $a \times 10^{k}$ , where $1 \le a < 10$ and $k \in \mathbb{Z}$ .
(a) Write down 0.000 072 6 in this form. [1]
(b) Write down 3 140 000 000 in this form. [1]
(c) Find $\dfrac{(2.5 \times 10^{-4}) \times (8.4 \times 10^{9})}{1.5 \times 10^{3}}$ . [2]

2. [Paper 1] Let $p = 8.75 \times 10^{6}$ and $q = 3.2 \times 10^{-4}$ .
(a) Find $pq$ , giving your answer in standard form. [2]
(b) Find $\dfrac{p}{q}$ , giving your answer in standard form to 3 significant figures. [2]
(c) A student writes the answer to (b) as $2.73^{10}$ . Explain why this does not earn the final mark. [1]

3. [Paper 1] A star is $5.1 \times 10^{14}$ km from Earth. Light travels at $3.00 \times 10^{5}$ km per second.
(a) Write down the distance to the star in metres, in standard form. [1]
(b) Find the time, in seconds, that light from the star takes to reach Earth. Give your answer in standard form. [2]
(c) Hence find this time in years, taking one year to be 365 days. [2]

4. [Paper 1] A coin has a mass of 7.5 g, and 92% of its mass is copper. One copper atom has a mass of $1.06 \times 10^{-25}$ kg.
(a) Find the mass of copper in the coin, in kg, in standard form. [2]
(b) Find the number of copper atoms in the coin. Give your answer in standard form to 3 significant figures. [2]
(c) Find how many times larger this number is than $7.5 \times 10^{18}$ . [2]

5. [Paper 1] Let $x = 6 \times 10^{n}$ and $y = 1.5 \times 10^{n+2}$ , where $n \in \mathbb{Z}$ .
(a) Show that $xy = 9 \times 10^{2n+2}$ . [2]
(b) Find $x + y$ , giving your answer in the form $a \times 10^{k}$ , where $1 \le a < 10$ and $k$ is an expression in $n$ . [3]
(c) Find $\dfrac{y}{x}$ , giving your answer in standard form. [2]

6. [Paper 1] The table gives the masses of four invented moons.

Moon

Mass (kg)

$9.48 \times 10^{19}$

$3670 \times 10^{18}$

$0.0812 \times 10^{23}$

$1.48 \times 10^{20}$

(a) Write the masses of moons B and C in standard form. [2]
(b) List the moons in order of mass, smallest first. [1]
(c) Find how many times heavier moon C is than moon A. [2]

7. [Paper 1] A red blood cell is modelled as a disc of diameter 8.2 micrometres, where 1 micrometre $= 10^{-6}$ m.
(a) Write down the diameter in metres, in standard form. [1]
(b) Find the area of the circular face of the disc, in m², in standard form to 3 significant figures. [2]
(c) Find the number of complete cells that fit side by side, touching, along a line 1 cm long. [2]

8. [Paper 1] A sheet of printer paper is $9.7 \times 10^{-5}$ m thick. A student claims that a stack of one million sheets would be taller than a building 100 m high.
(a) Write down one million in standard form. [1]
(b) Find the height of the stack in metres. [2]
(c) Determine whether the student's claim is correct. [1]

9. [Paper 1] Let $w = (3.9985 \times 10^{7}) \times (2.5 \times 10^{-3})$ .
(a) Find the exact value of $w$ . [1]
(b) Write $w$ in standard form to 3 significant figures. [2]
(c) A student gives the answer to (b) as $10.0 \times 10^{4}$ . State why this is not in standard form. [1]

10. [Paper 2] A container ship burns $2.25 \times 10^{5}$ kg of fuel each day at sea. Each kilogram of fuel burned produces 3.11 kg of carbon dioxide (CO₂).
(a) Find the mass of CO₂ produced each day at sea. Give your answer in the form $a \times 10^{k}$ , where $1 \le a < 10$ and $k \in \mathbb{Z}$ , to 3 significant figures. [2]
(b) A voyage lasts 34 days at sea. Show that one voyage produces approximately $2.38 \times 10^{7}$ kg of CO₂. [2]
(c) The shipping company's fleet completes $1.45 \times 10^{3}$ such voyages each year. Find the CO₂ the fleet produces in a year, in tonnes, where 1 tonne $= 10^{3}$ kg. [3]
(d) Ships worldwide produce about $1.06 \times 10^{9}$ tonnes of CO₂ each year. Find the percentage of this produced by the company's fleet. [2]

The government introduces a levy of 85 euros for each tonne of CO₂.

(e) Find the company's yearly levy, in euros, in standard form to 3 significant figures. [2]
(f) A cleaner fuel produces 18% less CO₂ per kilogram burned, but costs 0.21 euros more per kilogram. The ships burn the same mass of fuel. Determine whether switching the whole fleet to the cleaner fuel would reduce the company's yearly costs. Justify your answer. [4]
(g) State one assumption you made in answering part (c). [1]

Try the questions before checking your work.

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