Standard Form

Write big and small numbers in standard form, change them back, and calculate with them.

In the exam for Cambridge IGCSE Maths 0580, Edexcel IGCSE Maths 4MA1, Cambridge O Level Maths 4024 · 115 real past-paper questions below

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What Standard Form Is

Standard form is a short way to write very big or very small numbers. It is a number from 1 up to 10, times a power of 10.

Formula

A \times 1 0^{n}

A is the front number. It must be at least 1 and less than 10. n is a whole number that says how many places the point moves.

Worked example

1
$7.2 \times 1 0^{4}$ — The front number 7.2 is at least 1 and less than 10, so this is standard form.
2
$= 7.2 \times 10000$ — $1 0^{4}$ means 10 × 10 × 10 × 10, which is 10 000.
3
$= 72000$ — Multiplying by 10 000 moves the point 4 places to the right: 7.2 becomes 72 000.

Rule

A positive power gives a big number. A negative power gives a number less than 1.

Key idea

The Sun is about $1.5 \times 1 0^{8}$ km away, and light travels about $3 \times 1 0^{5}$ km every second. So sunlight takes about $\frac{1.5 \times 1 0 ^{8}}{3 \times 1 0 ^{5}} = 500$ seconds to reach you: just over 8 minutes.

Writing a Number in Standard Form

The 1-mark question: write a number in standard form. Put the point after the first digit that is not 0, then count how far it moved.

Worked example

1
$9400000$ — The question. It is a big number, so the power will be positive.
2
$9.4$ — Put the point after the first digit, 9. This makes the front number, which must be from 1 up to 10.
3
$9.4 \to 9400000$ — Count the jumps to get back to the question: the point jumps 6 places right.
4
$9.4 \times 1 0^{6}$ — 6 jumps of × 10 is $\times 1 0^{6}$ . Big number, so the power is positive.

Worked example

1
$0.000 48$ — The question. It is less than 1, so the power will be negative.
2
$4.8$ — Put the point after the first digit that is not 0, which is 4. This makes the front number. Zeros at the front do not count.
3
$4.8 \to 0.000 48$ — Count the jumps to get back to the question: the point jumps 4 places left.
4
$4.8 \times 1 0^{- 4}$ — 4 jumps of ÷ 10 is $\times 1 0^{- 4}$ . Less than 1, so the power is negative.

Some questions give a number like $360 \times 1 0^{5}$ . It has a power of 10, but the front number is too big. Fix the front number first.

Worked example

1
$360 \times 1 0^{5}$ — 360 is not less than 10, so this is not standard form yet.
2
$360 = 3.6 \times 1 0^{2}$ — Write 360 in standard form on its own: the point jumps 2 places.
3
$3.6 \times 1 0^{2} \times 1 0^{5}$ — Put that in place of 360. The value has not changed.
4
$3.6 \times 1 0^{7}$ — Multiplying adds the powers: 2 + 5 = 7. Now 3.6 is from 1 up to 10.

Common mistake

Writing $48 \times 1 0^{- 5}$ or $0.48 \times 1 0^{- 3}$ . They have the right value, but the front number must be from 1 up to 10.

Back to an Ordinary Number

The 1-mark question: write it as an ordinary number. The power tells you how many places to move the point. Positive means right, negative means left.

Worked example

1
$3.07 \times 1 0^{5}$ — The power is +5, so move the point 5 places right. A positive power means a big number.
2
$3.07 \to 307$ — The first 2 jumps use the digits that are already there, 0 and 7.
3
$307 \to 307000$ — 3 more jumps, with no digits left: put a 0 in each empty place. That is 5 jumps in all.

Worked example

1
$6.1 \times 1 0^{- 3}$ — The power is −3, so move the point 3 places left. A negative power means a number less than 1.
2
$6.1 \to 0.61$ — Jump 1: the point moves to the front of the 6.
3
$0.61 \to 0.0061$ — Jumps 2 and 3: put a 0 in each empty place in front of the 6.

Common mistake

Counting zeros, not places. $3.07 \times 1 0^{5}$ is not 3.07 with five zeros stuck on. It is 307 000.

Key idea

Asked how many zeros there are? Count jumps, not zeros. In $2.35 \times 1 0^{50}$ , the digits 3 and 5 use 2 of the 50 jumps, so 48 zeros follow the 5.

Multiplying and Dividing

The 2-mark question: work it out and give the answer in standard form. Deal with the front numbers and the powers of 10 separately.

Worked example

1
$(3 \times 1 0^{5}) \times (6 \times 1 0^{4})$ — The question. You can multiply in any order, so the front numbers and the powers can be done separately.
2
$3 \times 6 = 18$ — Multiply the front numbers as normal.
3
$1 0^{5} \times 1 0^{4} = 1 0^{9}$ — Multiplying powers of 10 adds the powers: 5 + 4 = 9.
4
$18 \times 1 0^{9}$ — Put the two parts together. Not finished: 18 is not less than 10.
5
$1.8 \times 1 0^{10}$ — 18 is 1.8 × 10. That extra 10 joins the power, so 9 + 1 = 10.

Worked example

1
$(1.2 \times 1 0^{9}) \div (4 \times 1 0^{3})$ — The question. Dividing works the same way: front numbers on their own, powers on their own.
2
$1.2 \div 4 = 0.3$ — Divide the front numbers as normal.
3
$1 0^{9} \div 1 0^{3} = 1 0^{6}$ — Dividing powers of 10 subtracts the powers: 9 − 3 = 6.
4
$0.3 \times 1 0^{6}$ — Put the two parts together. Not finished: 0.3 is less than 1.
5
$3 \times 1 0^{5}$ — 0.3 is 3 ÷ 10. Dividing by 10 takes 1 off the power, so 6 − 1 = 5.

Common mistake

Stopping at $18 \times 1 0^{9}$ . It is the right value but not standard form, so you lose the mark.

Key idea

Write the unfinished line, like $18 \times 1 0^{9}$ , before you fix it. In a 2-mark question that line can still earn 1 mark if the last step slips.

Negative Powers and Squares

Exams often put a minus in the power. The method is the same, but take care when you subtract a negative.

Worked example

1
$(3 \times 1 0^{5}) \div (6 \times 1 0^{- 3})$ — The question. The second power is negative.
2
$3 \div 6 = 0.5$ — Divide the front numbers as normal.
3
$1 0^{5} \div 1 0^{- 3} = 1 0^{5 - (- 3)}$ — Dividing subtracts the powers. Keep the −3 in brackets so its minus is not lost.
4
$5 - (- 3) = 5 + 3 = 8$ — Taking away a negative is the same as adding: the two minuses make a plus.
5
$0.5 \times 1 0^{8}$ — Put the two parts together. Not finished: 0.5 is less than 1.
6
$5 \times 1 0^{7}$ — 0.5 is 5 ÷ 10, so take 1 off the power: 8 − 1 = 7.

To square a number in standard form, square both parts.

Worked example

1
$(3 \times 1 0^{- 4})^{2}$ — Squared means times itself: $(3 \times 1 0^{- 4}) \times (3 \times 1 0^{- 4})$ .
2
$3^{2} = 9$ — Square the front number: 3 × 3 = 9.
3
$(1 0^{- 4})^{2} = 1 0^{- 8}$ — Squaring a power doubles it: −4 × 2 = −8. It is the same as −4 + (−4).
4
$9 \times 1 0^{- 8}$ — 9 is from 1 up to 10, so this is already standard form.

Key idea

Cubing works the same way: cube the front number and multiply the power by 3. So $(2 \times 1 0^{5})^{3} = 8 \times 1 0^{15}$ .

Common mistake

Writing $1 0^{5} \div 1 0^{- 3} = 1 0^{2}$ . Dividing by $1 0^{- 3}$ means taking away −3, which adds 3: the answer is $1 0^{8}$ .

Adding and Subtracting

The 2-mark question: a total or a difference. You cannot just add the powers here. The safe way is to write both as ordinary numbers first.

Worked example

1
$5.2 \times 1 0^{6} + 3.4 \times 1 0^{5}$ — Find the total. The powers are different (6 and 5), so the front numbers cannot simply be added.
2
$5.2 \times 1 0^{6} = 5200000$ — Write the first as an ordinary number: the point moves 6 places right.
3
$3.4 \times 1 0^{5} = 340000$ — Write the second as an ordinary number: the point moves 5 places right.
4
$5200000 + 340000 = 5540000$ — Add as normal, with the digits lined up by place value.
5
$= 5.54 \times 1 0^{6}$ — Back to standard form: point after the 5, and it jumps 6 places.

Worked example

1
$7 \times 1 0^{4} - 8 \times 1 0^{3}$ — Find the difference. Again the powers are different.
2
$7 \times 1 0^{4} = 70000$ — Write the first as an ordinary number: the point moves 4 places right.
3
$8 \times 1 0^{3} = 8000$ — Write the second as an ordinary number: the point moves 3 places right.
4
$70000 - 8000 = 62000$ — Subtract as normal.
5
$= 6.2 \times 1 0^{4}$ — Back to standard form: point after the 6, and it jumps 4 places.

Common mistake

Adding the front numbers when the powers are different. $5.2 + 3.4$ gives 8.6, which is wrong because the two numbers have different sizes.

Adding With Huge Powers

Some questions use powers like $1 0^{120}$ . That is too many zeros to write out. Instead, make the powers the same, then add the front numbers.

Formula

A \times 1 0^{n} + B \times 1 0^{n} = (A + B) \times 1 0^{n}

A and B are the front numbers. n is the power they share. It works only when the two powers are the same. The same rule works for −.

Worked example

1
$4.6 \times 1 0^{120} + 7 \times 1 0^{119}$ — Find the total. The powers are 120 and 119, so make them the same first.
2
$7 \times 1 0^{119} = 0.7 \times 1 0^{120}$ — Rewrite the smaller one. Dividing the front number by 10 and adding 1 to the power keeps the value the same.
3
$4.6 \times 1 0^{120} + 0.7 \times 1 0^{120}$ — Now both have $1 0^{120}$ .
4
$(4.6 + 0.7) \times 1 0^{120}$ — Same power, so add the front numbers. The $1 0^{120}$ stays as it is.
5
$= 5.3 \times 1 0^{120}$ — 4.6 + 0.7 = 5.3. It is from 1 up to 10, so this is the answer.

Worked example

1
$3 \times 1 0^{80} - 5 \times 1 0^{78}$ — Find the difference. The powers are 2 apart.
2
$5 \times 1 0^{78} = 0.05 \times 1 0^{80}$ — Move the point 2 places left in 5 and add 2 to the power. The value stays the same.
3
$(3 - 0.05) \times 1 0^{80}$ — Same power, so subtract the front numbers.
4
$= 2.95 \times 1 0^{80}$ — 3 − 0.05 = 2.95. It is from 1 up to 10, so this is the answer.

Key idea

Extended/Higher: letters in the power work the same. $3 \times 1 0^{k} + 6 \times 1 0^{k - 1} = 3 \times 1 0^{k} + 0.6 \times 1 0^{k} = 3.6 \times 1 0^{k}$ .

Common mistake

Adding the powers: $1 0^{120} + 1 0^{119}$ is not $1 0^{239}$ . Powers are added only when you multiply.

Times Bigger and Missing Numbers

“How many times bigger?” and “find n” questions are both divisions. Divide, then use the method from Multiplying and Dividing.

A city has $4.8 \times 1 0^{6}$ people and a town has $1.2 \times 1 0^{4}$ . How many times as many people live in the city?

Worked example

1
$\frac{4.8 \times 1 0 ^{6}}{1.2 \times 1 0 ^{4}}$ — “How many times” means divide the bigger number by the smaller one.
2
$4.8 \div 1.2 = 4$ — Divide the front numbers.
3
$1 0^{6} \div 1 0^{4} = 1 0^{2}$ — Dividing subtracts the powers: 6 − 4 = 2.
4
$4 \times 1 0^{2} = 400$ — Put the parts together. The city has 400 times as many people.

Find n when $2.5 \times 1 0^{3} \times n = 9 \times 1 0^{8}$ . Give n in standard form.

Worked example

1
$n = \frac{9 \times 1 0 ^{8}}{2.5 \times 1 0 ^{3}}$ — n is multiplied by $2.5 \times 1 0^{3}$ , so divide both sides by it. Division undoes multiplication.
2
$9 \div 2.5 = 3.6$ — Divide the front numbers.
3
$1 0^{8} \div 1 0^{3} = 1 0^{5}$ — Dividing subtracts the powers: 8 − 3 = 5.
4
$n = 3.6 \times 1 0^{5}$ — 3.6 is from 1 up to 10, so this is standard form. Check: 2.5 × 3.6 = 9 and 3 + 5 = 8.

Common mistake

Dividing the small number by the big one. That gives an answer less than 1, but “how many times bigger” must be more than 1.

Standard Form on a Calculator

On a calculator paper, let the calculator do the work. Type each power with the $\times 1 0^{x}$ key (on some calculators it is EXP), and put brackets round the bottom of a fraction.

Worked example

1
$\frac{5.3 \times 1 0 ^{4} + 8 \times 1 0 ^{3}}{2.4 \times 1 0 ^{- 2}}$ — The question. Give the answer in standard form, correct to 3 significant figures.
2
$5.3 \times 1 0^{4} + 8 \times 1 0^{3} = 61000$ — Work out the top first, and write it down. A correct line like this can earn a mark even if a later step slips.
3
$61000 \div (2.4 \times 1 0^{- 2}) = 2541 666.6 \dots$ — Now divide by the bottom. The brackets make the calculator divide by all of it.
4
$2.541 666 \dots \times 1 0^{6}$ — Point after the 2: it jumps 6 places, so the power is 6.
5
$2.54 \times 1 0^{6}$ — Round the front number to 3 significant figures: 2.541… becomes 2.54. The power does not change.

Key idea

If the display shows 2.541666667E6, the E6 means $\times 1 0^{6}$ . Write it as $2.54 \times 1 0^{6}$ , never as E6.

Common mistake

Typing the bottom with the × key and the power key, and no brackets. The calculator then divides by 2.4 only and multiplies by $1 0^{- 2}$ , giving 254.1…, which is wrong. Use the $\times 1 0^{x}$ key or brackets.

Which Method? Read the Question

The words tell you which kind it is.

You see	You do
Write 0.00048 in standard form	point after the first digit, count the jumps
Write as an ordinary number	move the point: right if the power is positive, left if negative
Multiply or divide	front numbers together, then add or subtract the powers
Total or difference	ordinary numbers first, then back to standard form
Powers far apart or huge, like 10¹²⁰	make the powers the same, then add or subtract the front numbers
Which is largest or smallest?	compare the powers first, then the front numbers
How many times bigger? Find n	divide, then fix the front number
Correct to 3 significant figures	use the ×10ˣ key, then round the front number

Put $6.2 \times 1 0^{- 3}, 0.5 \times 1 0^{- 2}, 4 \times 1 0^{1}, 1.9 \times 1 0^{- 2}$ in order, smallest first.

Worked example

1
$0.5 \times 1 0^{- 2} = 5 \times 1 0^{- 3}$ — First fix any number that is not in standard form. 0.5 is less than 1, so make it 5 and take 1 off the power.
2
$1 0^{- 3} < 1 0^{- 2} < 1 0^{1}$ — Compare the powers. −3 is smaller than −2, so the $1 0^{- 3}$ numbers are the smallest.
3
$5 \times 1 0^{- 3} < 6.2 \times 1 0^{- 3}$ — Two numbers share $1 0^{- 3}$ , so compare their front numbers: 5 is less than 6.2.
4
$0.5 \times 1 0^{- 2}, 6.2 \times 1 0^{- 3}, 1.9 \times 1 0^{- 2}, 4 \times 1 0^{1}$ — The order, smallest first. Write each number the way the question gave it.

Common mistake

Thinking $1 0^{- 3}$ is bigger than $1 0^{- 2}$ because 3 is bigger than 2. $1 0^{- 3} = 0.001$ is smaller than $1 0^{- 2} = 0.01$ .

Key idea

Before you write the answer, check the front number is at least 1 and less than 10.

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