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GCSE Maths · EdexcelChapter 6 · Standard form, indices and surds
Chapter 6 · Foundation tier

GCSE Maths Standard Form, Indices and Surds (Foundation): worked examples and questions with answers

By the end of this chapter you will be able to

  • Use index notation and the laws of indices to multiply, divide and raise powers to powers, with numbers and with letters
  • Work with zero and negative indices
  • Calculate with roots and (Higher) with fractional indices
  • Write very large and very small numbers in standard form and change them back
  • Compare, order and calculate with numbers in standard form, with and without a calculator
  • Give exact answers in terms of π
  • (Higher) Simplify, add, multiply and expand with surds, and rationalise denominators

You should already know

  • Multiplying and dividing whole numbers and decimals (Chapter 1)
  • Square numbers, cube numbers and roots (Chapter 2)
  • Fractions and reciprocals (Chapter 3)
  • Rounding to significant figures (Chapter 4)
  • Using letters in algebra

Where this leads

  • Expanding brackets and factorising (Chapter 7)
  • Area and circumference of circles (Chapter 18)
  • Pythagoras and exact trigonometry (Chapter 20)
  • Growth and decay (Chapters 5 and 25)

Edexcel GCSE (9–1) Mathematics 1MA1, Foundation tier. Content references: N7, N8, N9, A4.

Printable worksheets. Every lesson in this chapter has a free worksheet and an answer sheet to download and print.

Exercise P  Do you remember?

Non-calculator
1
FWork out each of these.
(a)24
(b)3.7 × 1000
(c)6.4 ÷ 100
(d)√81 and 3√27
(e)Round 0.043756 to 3 significant figures.
6.1

Index notation and the laws of indices

Edexcel 1MA1: N6, N7, A4
Grade 2-5

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Index notation and the laws of indicesComing soon

Fold a sheet of paper in half and it has 2 layers. Fold it again and there are 4 layers. Fold again: 8. Every fold doubles the number of layers, so after 3 folds there are 2 × 2 × 2 layers. Mathematicians write this in a short way: 23, said “2 to the power 3”. This is called index notation.

34baseindex (power)3 × 3 × 3 × 3= 814 lots of 3 multiplied
In 34 the base is 3 and the index is 4: four lots of 3 are multiplied together.

The big number is the base. The small raised number is the index (plural: indices; it is also called the power or exponent). The index tells you how many lots of the base are multiplied together. You already know two special cases: “squared” means index 2 and “cubed” means index 3. Be careful: 23 is 2 × 2 × 2 = 8. It is not 2 × 3 = 6.

The index counts lots, and counting lots is easy. That is why there are three simple laws of indices. Each one comes straight from writing the powers out in full.

333333343234+2 = 64 lots + 2 lots = 6 lots
Law 1: 34 × 32. Four lots of 3 and then two more lots of 3 makes 4 + 2 = 6 lots of 3.

Law 1 (multiplying). When you multiply powers of the same base, add the indices. 34 × 32 = 34+2 = 36.

Law 2 (dividing). When you divide powers of the same base, subtract the indices. 35 ÷ 32 = (3 × 3 × 3 × 3 × 3) ÷ (3 × 3). Two of the 3s cancel on the top and bottom, leaving three of them: 35−2 = 33.

Law 3 (power of a power). (32)3 means three lots of 32, that is 32 × 32 × 32 = 32+2+2 = 36. Three lots of 2 is 3 × 2, so we multiply the indices.

These laws only work when the bases are the same. They also work with letters, because a letter stands for a number. For example x2 × x5 = x7. When there are numbers in front of the letters (the coefficients), deal with the numbers and the letters separately: 3x2 × 4x5 = (3 × 4) × x2+5 = 12x7. A letter on its own has index 1, so x = x1.

Key ideas

  • am × an = am+n   (add the indices)
  • am ÷ an = am−n   (subtract the indices)
  • (am)n = am×n   (multiply the indices)
  • The bases must be the same for the first two laws.
  • With coefficients: multiply or divide the numbers, then use the laws on each letter. (2a)3 = 23a3 = 8a3.
  • a1 = a.
Worked example 6.1.1
Write 2 × 2 × 2 × 2 × 2 as a power of 2, and work out its value.

Key words. In 34, the base is 3 and the index (also called the power or exponent) is 4. The index tells you how many lots of the base are multiplied together: 34 = 3 × 3 × 3 × 3 = 81.

Method. Count how many 2s are multiplied to get the index, then multiply them out.

  1. Count the 2s. The index of a power is the number of equal numbers being multiplied.
    There are 5 lots of 2, so the index is 5.
  2. Write it as a power. Base 2, index 5.
    2 × 2 × 2 × 2 × 2 = 25
  3. Multiply step by step. Do one multiplication at a time.
    2 × 2 = 4
    4 × 2 = 8
    8 × 2 = 16
    16 × 2 = 32
  4. Write the answer.
    25 = 32

Check. 25 = 22 × 23 = 4 × 8 = 32. ✓

Watch out. 25 is not 2 × 5 = 10. The index counts lots of 2 multiplied together.

Worked example 6.1.2
Write 34 × 32 as a single power of 3.

Key words. In 34, the base is 3 and the index (also called the power or exponent) is 4. The index tells you how many lots of the base are multiplied together: 34 = 3 × 3 × 3 × 3 = 81.

Method. When you multiply powers with the same base, you add the indices: am × an = am+n.

  1. Check that the bases are the same. These laws only work when the base (the big number) is the same in both powers.
    Both bases are 3.
  2. Write each power out in full. This shows what is really happening: we are counting how many lots of the base are multiplied.
    34 = 3 × 3 × 3 × 3
    32 = 3 × 3
    Together: 4 lots of 3 and 2 lots of 3 make 6 lots of 3.
  3. Add the indices. Joining the lots means adding the number of lots.
    4 + 2 = 6
  4. Write the answer as a single power. Keep the same base and use the new index.
    34 × 32 = 36

Check. Work out the numbers: 34 = 81 and 32 = 9. 81 × 9 = 729, and 36 = 729. ✓

Watch out. Add the indices when you multiply. Do not multiply the indices, and do not multiply the bases (32 × 34 is not 96).

Worked example 6.1.3
Write (57 × 52) ÷ 56 as a single power of 5.

Key words. In 34, the base is 3 and the index (also called the power or exponent) is 4. The index tells you how many lots of the base are multiplied together: 34 = 3 × 3 × 3 × 3 = 81.

Method. Use the laws of indices in order: add indices to multiply, subtract indices to divide.

  1. Check that every base is the same. The laws of indices only work when the bases match.
    Every base is 5.
  2. Work out the top first: multiply, so add the indices. am × an = am+n.
    7 + 2 = 9, so the top is 59
  3. Divide: subtract the indices. am ÷ an = am−n.
    5956 = 59 ÷ 56 = 53, because 9 − 6 = 3

Check. The total of the indices on top is 9 and on the bottom is 6; 9 − 6 = 3. ✓

Watch out. Deal with the multiplication in the top (and bottom) first, then divide. Only subtract once the top and bottom are each a single power.

Worked example 6.1.4
Work out (23)4 as a power of 2, and then as a number.

Key words. In 34, the base is 3 and the index (also called the power or exponent) is 4. The index tells you how many lots of the base are multiplied together: 34 = 3 × 3 × 3 × 3 = 81.

Method. To raise a power to a power, multiply the indices: (am)n = am×n.

  1. Write the bracket out as repeated multiplication. The outer index 4 means 4 lots of 23 multiplied together.
    (23)4 = 23 × 23 × 23 × 23
    = 23 + 3 + 3 + 3
    That is 4 lots of 3, which is 4 × 3 = 12.
  2. Multiply the indices. Each of the outer lots holds m of the base, so the total is m × n.
    3 × 4 = 12
  3. Write the answer as a single power. Keep the same base and use the new index.
    (23)4 = 212
  4. Work out the value. 212 means 12 lots of 2 multiplied together.
    212 = 2 × 2 × … × 2 (12 lots of 2) = 4096

Check. Work out the numbers: 23 = 8, and 84 = 4096. Also 212 = 4096. ✓

Watch out. A power of a power: multiply the indices. Adding them would be for 32 × 34, which is a different question.

Worked example 6.1.5
Simplify 3x2 × 4x5.

Key words. In 34, the base is 3 and the index (also called the power or exponent) is 4. The index tells you how many lots of the base are multiplied together: 34 = 3 × 3 × 3 × 3 = 81. A coefficient is the number in front of a letter, so in 7x3 the coefficient is 7. A letter on its own, like x, really means x1.

Method. Multiply the numbers, then add the indices of each letter: am × an = am+n.

  1. Multiply the numbers in front (the coefficients). The numbers can be dealt with separately from the letters.
    3 × 4 = 12
  2. Deal with the letter x: add the indices. The letter is the same in both terms, so the index law applies. (x on its own means x1.)
    x2 × x5 = x2 + 5 = x7
  3. Put the number and the letters together. Write the coefficient first, then the letters in alphabetical order.
    12x7

Check. Choose numbers: x = 2. The original expression has the value 1536, and the answer 12x7 also has the value 1536. ✓

Watch out. Add the indices of the same letter only. You cannot combine different letters: a2 × b3 stays as it is.

Worked example 6.1.6
Simplify (2a3)4.

Key words. In 34, the base is 3 and the index (also called the power or exponent) is 4. The index tells you how many lots of the base are multiplied together: 34 = 3 × 3 × 3 × 3 = 81. A coefficient is the number in front of a letter, so in 7x3 the coefficient is 7. A letter on its own, like x, really means x1.

Method. Raise the number and every letter to the power. Multiply the indices: (am)n = am×n.

  1. Raise the number in front to the power. Everything inside the bracket is raised to the power, and that includes the number.
    (2)4 = 2 × 2 × 2 × 2 = 16
  2. Raise the letter a to the power: multiply the indices. (am)n = am×n.
    (a3)4 = a3 × 4 = a12
  3. Put it all together.
    16a12

Check. Choose numbers: a = 2. The original expression has the value 65,536, and the answer 16a12 also has the value 65,536. ✓

Watch out. The number in front is raised to the power too: (2a)3 = 8a3, not 2a3.

Watch out

Do not multiply the bases and do not multiply the indices when you are multiplying powers. 32 × 34 is not 96. The base stays 3 and the indices add: 36. And you cannot combine powers of different letters or different numbers: x3 + x2 does not equal x5, and 23 × 32 is not 65.

Exam tip

In an exam, show the indices adding: write 34+2 = 36. If the question says “work out the value” you must also give the final number.

Exercise 6A  Powers and the laws of indices with numbers

Non-calculator
1
FWrite each product as a power.
(a)5 × 5 × 5 × 5
(b)2 × 2 × 2 × 2 × 2 × 2
(c)10 × 10 × 10
(d)7 × 7
2
FWork out the value of each power.
(a)25
(b)34
(c)103
(d)53
(e)43
(f)62
3
FComplete the list of powers of 2: write each as a number.
(a)21
(b)22
(c)23
(d)24
(e)25
4
FWrite each as a single power.
(a)23 × 24
(b)52 × 56
(c)75 × 71
(d)103 × 103
5
FWrite each as a single power.
(a)68 ÷ 63
(b)49 ÷ 45
(c)37 ÷ 32
(d)106 ÷ 104
6
FWrite each as a single power.
(a)(32)4
(b)(53)2
(c)(25)3
(d)(102)3
7
FWork out the value of each. Write it as a single power first.
(a)23 × 22
(b)35 ÷ 33
(c)(22)3
(d)102 × 103
(e)54 ÷ 52
8
FWork out the value of each product.
(a)23 × 52
(b)22 × 33
(c)103 × 42
(d)32 × 24
9
FFind the missing index in each equation.
(a)34 × 3? = 39
(b)7? × 72 = 75
(c)58 ÷ 5? = 53
(d)2? ÷ 23 = 24
(e)(4?)3 = 412
10
FWrite each as a single power of the base.
(a)57 × 5256
(b)23 × 2524
(c)3932 × 33
(d)74 × 7372 × 71
11
RZoe says that 23 × 24 is the same as 47. Is she right? Explain your answer.
12
RDan writes 56 ÷ 52 = 53. Explain his mistake and write down the correct answer.
13
PSA piece of paper is folded in half again and again. Each fold doubles the number of layers.
(a)How many layers are there after 6 folds? Write the answer as a power of 2 and then as a number.
(b)What is the smallest number of folds that gives more than 1000 layers?
14
PSA message is sent in rounds. In round 1, one person sends it to 3 people. In every later round, each person who has just received it sends it to 3 new people. In round n, 3n people receive the message.
(a)How many people receive the message in round 5?
(b)Show that the number of people in round 3 multiplied by the number in round 2 is the number in round 5.
15
PSA cube has edges of length 102 cm. The volume of a cube is (edge)3. Work out the volume as a number of cm3.
16
RShow that 8 × 16 = 27.
17
Ra and b are positive whole numbers. 2a × 2b = 28 and 2a ÷ 2b = 22. Find a and b.

Exercise 6B  The laws of indices with letters

Non-calculator
1
FSimplify each expression.
(a)x3 × x4
(b)y5 × y2
(c)a × a6
(d)p3 × p3
2
FSimplify each expression.
(a)x9 ÷ x4
(b)y7 ÷ y2
(c)m8 ÷ m
(d)t5 ÷ t4
3
FSimplify each expression.
(a)(x3)2
(b)(y4)3
(c)(a2)5
(d)(p6)2
4
FSimplify each expression.
(a)3x2 × 4x5
(b)2y3 × 5y
(c)6a4 × 2a2
(d)7p × 3p3
5
FSimplify each expression.
(a)12y8 ÷ 4y3
(b)20a7 ÷ 5a2
(c)18x5 ÷ 6x
(d)8m6 ÷ 2m6
6
FSimplify each expression.
(a)(2a3)4
(b)(5x2)3
(c)(3y4)2
(d)(10p3)2
7
FSimplify each expression.
(a)3a2b × 5ab3
(b)4x2y × 2xy2
(c)6p3q2 ÷ 3pq
(d)10m5n4 ÷ 5m2n
8
RMia writes 2x3 × 3x4 = 6x12. Is she right? Give a reason.
9
RIs it true that x3 + x2 = x5? Use x = 2 to explain.
10
PSA rectangle has length 4x3 cm and width 3x2 cm. Write an expression for its area in the simplest form.
11
PSThe area of a rectangle is 24x7 cm2. Its length is 6x3 cm. Find its width.
12
PSThe three edges of a cuboid are 2a, 3a2 and 5a3. Find an expression for its volume.
6.2

Zero and negative indices

Edexcel 1MA1: N7, A4
Grade 4-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Zero and negative indicesComing soon

So far every index has been a positive whole number. But what do 20 and 2−1 mean? We cannot multiply “zero lots of 2”, so instead we follow the pattern.

23= 8÷222= 4÷221= 2÷220= 1÷22−1= 1/2÷22−2= 1/4÷22−3= 1/8Each step to the right divides by 2
Going one step to the right the index goes down by 1 and the value is divided by 2. Keep the pattern going and 20 = 1, 2−1 = 1/2, 2−2 = 1/4.

Look at the boxes from left to right: 8, 4, 2, then the next number must be 2 ÷ 2 = 1. So 20 = 1. The same thing happens with any base, so any non-zero number to the power 0 equals 1. You can also see it from the division law: a3 ÷ a3 = a3−3 = a0, but anything divided by itself is 1.

Continue the pattern past 1: the next value is 1 ÷ 2 = 1/2, so 2−1 = 1/2. Then 2−2 = 1/4 and 2−3 = 1/8. A negative index means “one over” (the reciprocal; see Chapter 3):

a−n = 1an

A negative index does not make the answer negative. It makes the answer small: a fraction. For a fraction base, the reciprocal turns the fraction upside down, so 23−2 = 322 = 94.

Two more things to know. 10−1 = 0.1, 10−2 = 0.01 and 10−3 = 0.001 (you will need these for standard form in section 6.4). And all the laws of section 6.1 still work with zero and negative indices. For example 52 × 5−4 = 52+(−4) = 5−2 = 125, and x3 ÷ x5 = x−2 = 1x2.

Key ideas

  • a0 = 1 for any number a that is not 0. (50 = 1, 10000 = 1.)
  • a1 = a.
  • a−n = 1 ÷ an. A negative index means “one over”, not a negative number.
  • (a/b)−n = (b/a)n: flip the fraction and make the index positive.
  • 10−1 = 0.1, 10−2 = 0.01, 10−3 = 0.001.
  • The laws of indices work for all whole-number indices, including zero and negative ones.
Worked example 6.2.1
Work out 50 + 2 × 30.

Key words. In 34, the base is 3 and the index (also called the power or exponent) is 4. The index tells you how many lots of the base are multiplied together: 34 = 3 × 3 × 3 × 3 = 81.

Method. A number (or bracket) to the power zero is 1.

  1. Remember the rule for a zero index. Any non-zero number raised to the power 0 equals 1. This is because an ÷ an = 1, and by the division law it is also an−n = a0.
    a0 = 1
  2. Replace every power with index 0 by 1, and work out the rest. Powers are worked out before multiplying and adding (BIDMAS): each zero power becomes 1.
    50 = 1 and 30 = 1
    50 + 2 × 30 = 1 + 2 × 1
    = 1 + 2 = 3
  3. Write the answer.
    3

Check. Look again: each 0 index was changed to 1 and nothing else was changed. ✓

Watch out. 50 = 1, not 0 and not 5. And in 5x0 only the x has the index 0, so it is 5 × 1 = 5.

Worked example 6.2.2
Work out 2−3.

Key words. In 34, the base is 3 and the index (also called the power or exponent) is 4. The index tells you how many lots of the base are multiplied together: 34 = 3 × 3 × 3 × 3 = 81.

Method. A negative index means "one over": a−n = 1 ÷ an. For a fraction, flip it upside down and make the index positive.

  1. Spot the negative index. A negative index does not make the answer negative. It means "one over".
    2−3: the index is −3
  2. Turn it into a reciprocal with a positive index. The rule is a−n = 1an. (Check with a pattern: 22 = 4, 21 = 2, 20 = 1, and each step is ÷ 2, so 2−1 = 12.)
    2−3 = 123
  3. Work out the bottom. 23 means 3 lots of 2 multiplied together.
    23 = 2 × 2 × 2 = 8
  4. Write the answer as a fraction.
    2−3 = 18

Check. Multiply the answer by 23: 18 × 8 = 1, and 2−3 × 23 = 20 = 1. ✓

Watch out. A negative index does not give a negative answer: 2−3 = 18, not a negative number.

Worked example 6.2.3
Work out (23)−2.

Key words. In 34, the base is 3 and the index (also called the power or exponent) is 4. The index tells you how many lots of the base are multiplied together: 34 = 3 × 3 × 3 × 3 = 81.

Method. A negative index means "one over": a−n = 1 ÷ an. For a fraction, flip it upside down and make the index positive.

  1. Spot the negative index. A negative index means "reciprocal". The reciprocal of a fraction is the fraction turned upside down.
    (23)−2: the index is −2
  2. Flip the fraction and change the index to positive. Negative index = reciprocal. The reciprocal of 23 is 32.
    (23)−2 = (32)2
  3. Raise the top and the bottom to the power. The index 2 applies to the numerator and to the denominator.
    3222 = 94
  4. Write the answer (simplify if possible).
    94

Check. Multiply the answer by (23)2 = 49: the product is 1, as (23)−2 × (23)2 = (23)0 = 1. ✓

Watch out. A negative index does not give a negative answer: (23)−2 = 94, not a negative number.

Worked example 6.2.4
Write 125 as a power of 5.

Key words. In 34, the base is 3 and the index (also called the power or exponent) is 4. The index tells you how many lots of the base are multiplied together: 34 = 3 × 3 × 3 × 3 = 81.

Method. Write the bottom of the fraction as a power, then use a−n = 1/an backwards.

  1. Look at the fraction. It has 1 on the top and a number on the bottom, so it is "one over" something. A negative index means "one over".
    125
  2. Write the denominator as a power of the base. 25 = 5 × 5 = 52
    125 = 152
  3. Use the rule for a negative index. "One over a power" is the same as the power with a negative index: 1an = a−n.
    152 = 5−2

Check. Convert back: 5−2 = 152 = 125. ✓

Watch out. The number on the bottom becomes the base. The index gets the minus sign.

Worked example 6.2.5
Work out 52 × 5−4, giving your answer as a fraction.

Key words. In 34, the base is 3 and the index (also called the power or exponent) is 4. The index tells you how many lots of the base are multiplied together: 34 = 3 × 3 × 3 × 3 = 81.

Method. When you multiply powers with the same base, you add the indices: am × an = am+n.

  1. Check that the bases are the same. These laws only work when the base (the big number) is the same in both powers.
    Both bases are 5.
  2. Say what the rule means. The index counts the lots of the base. Multiplying powers joins the lots together.
    52 has 2 lots of 5; 5−4 has −4 lots of 5.
  3. Add the indices. Joining the lots means adding the number of lots. The rule works for negative indices too, so take care with the signs.
    2 + (−4) = −2
  4. Write the answer as a single power. Keep the same base and use the new index.
    52 × 5−4 = 5−2
  5. Work out the value. A negative index means "one over" the positive power (see section 6.2).
    5−2 = 152 = 125

Watch out. Add the indices when you multiply. Do not multiply the indices, and do not multiply the bases (32 × 34 is not 96).

Worked example 6.2.6
Simplify x3 ÷ x5, and write your answer with a positive index.

Key words. In 34, the base is 3 and the index (also called the power or exponent) is 4. The index tells you how many lots of the base are multiplied together: 34 = 3 × 3 × 3 × 3 = 81. A coefficient is the number in front of a letter, so in 7x3 the coefficient is 7. A letter on its own, like x, really means x1.

Method. Divide the numbers, then subtract the indices of each letter: am ÷ an = am−n.

  1. Deal with the letter x: subtract the indices. The letter is the same in both terms, so the index law applies. (x on its own means x1.)
    x3 ÷ x5 = x3 − 5 = x−2 = 1x2
  2. Put the number and the letters together. Write the coefficient first, then the letters in alphabetical order.
    x−2
  3. Write with positive indices. A negative index means "one over": a−n = 1/an.
    x−2 = 1x2

Check. Choose numbers: x = 2. The original expression has the value 14, and the answer x−2 also has the value 14. ✓

Watch out. Subtract the indices (top minus bottom), and divide the numbers. Do not subtract the numbers.

Watch out

A negative index does not make a number negative. 2−3 is 18, not −8 and not −6. Also be careful with zero: 50 = 1, not 0. And in 5x0 the index 0 belongs to the x only, so it is 5 × 1 = 5, while (5x)0 = 1.

Exercise 6C  Zero and negative indices

Non-calculator
1
FWork out the value of each.
(a)70
(b)1000
(c)(−3)0
(d)20 + 30
(e)5 × 40
2
FWrite each as a fraction.
(a)2−1
(b)3−2
(c)10−3
(d)5−2
(e)4−1
(f)2−4
3
FWrite each fraction as a power with a negative index.
(a)18
(b)19
(c)1100
(d)1125
(e)132
4
FWork out each. Give your answers as whole numbers or fractions in their simplest form.
(a)(12)−3
(b)(23)−2
(c)(15)−2
(d)(34)−2
(e)(25)−1
5
FWrite each of these as a decimal.
(a)10−1
(b)10−2
(c)10−3
(d)10−4
(e)10−5
6
FComplete the pattern: 103 = 1000, 102 = 100, 101 = 10. Work out the value of the next three powers.
(a)100
(b)10−1
(c)10−2
7
FWork out the value of each. Give a whole number or a fraction.
(a)52 × 5−4
(b)23 ÷ 25
(c)(32)−1
(d)10−3 × 105
(e)7−2 × 7−1
8
FSimplify each. Write your answer with positive indices.
(a)x2 ÷ x6
(b)y3 ÷ y8
(c)a2 × a−5
(d)3x−2 × 4x5
9
RJack says that 3−2 = −9. Explain his mistake and work out the correct value.
10
RUse the division law to show that 50 = 1.
11
PSA virus is about 10−7 m long. A bacterial cell is about 10−5 m long. How many times longer is the cell than the virus?
12
PSWhich is larger, 2−3 or 3−2? Show your working.
13
PSA radioactive sample has a mass of 80 g. After n days its mass is 80 × 2−n grams. Work out the mass after 3 days and after 5 days.
14
RShow that (2−3 × 25) ÷ 2−2 = 16.

Exercise 6D  Index equations and mixed practice

Non-calculator
1
FFind the value of n in each equation.
(a)2n = 32
(b)3n = 81
(c)5n = 125
(d)10n = 1000
(e)2n = 128
2
FFind the value of n in each equation. (The answers are negative whole numbers.)
(a)2n = 18
(b)3n = 19
(c)10n = 1100
(d)5n = 15
(e)4n = 116
3
FFind the value of x.
(a)2x + 3 = 28
(b)3x − 1 = 34
(c)5x + 2 = 56
(d)10x − 2 = 103
4
FWork out each. Give your answer as a fraction.
(a)2−1 + 2−2
(b)3−1 + 3−2
(c)50 + 5−1
5
PSWrite these numbers in order, smallest first: 2−1, 20, 2−3, 22.
6
RWrite 18 as a power of 2, and write 8 × 132 as a power of 2.
(a)18
(b)8 × 132
7
RShow that (13)−2 × 3−4 = 19.
6.3

Roots and fractional indices

Edexcel 1MA1: N6, N7, A4
Grade 3-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Roots and fractional indicesComing soon

A root undoes a power. A square has area 25 cm2. Its side is the number that multiplies by itself to give 25. That number is 5, so the square root of 25 is 5. We write √25 = 5. A cube has volume 27 cm3. Its edge is the number that gives 27 when multiplied by itself three times, so the cube root of 27 is 3, written 3√27 = 3.

area 25side = 251/2= 5volume 27edge = 271/3= 3
Roots find the side of a square from its area and the edge of a cube from its volume.

Learn the perfect squares up to 152 = 225 and the cubes 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000 (see Chapter 2). Then roots of these are quick. You can also take roots of fractions and decimals: √(9/16) = 3/4 because (3/4)2 = 9/16, and √0.25 = 0.5 because 0.5 × 0.5 = 0.25.

Most numbers are not perfect squares, but we can say where their roots lie. √50 is between √49 = 7 and √64 = 8, so it is between 7 and 8. A calculator gives √50 = 7.07…. Use the √ key for square roots and the cube root key (often SHIFT then √) for cube roots.

Remember: (√a)2 = a. Equations such as x2 = 49 have two solutions, x = 7 and x = −7, because (−7)2 = 49 too. The √ sign itself means the positive root. You cannot take the square root of a negative number (no real number squares to −9), but a cube root of a negative number is fine: 3√(−8) = −2.

Key ideas

  • √a is the positive number that multiplies by itself to give a. 3√a multiplies three times.
  • (√a)2 = a, and x2 = k has two solutions, ±√k.
  • To place a root between two whole numbers use the perfect squares (or cubes) on each side.
  • Roots of fractions: take the root of the top and of the bottom.
Worked example 6.3.1
Work out √144.

Key words. The square root of a number undoes squaring: √49 = 7 because 72 = 49. The cube root undoes cubing: 3√27 = 3 because 33 = 27.

Method. Find the number that gives the original when raised to the power 2.

  1. Say what the question is asking. The square root of a number is the number that gives the original number when 2 lots of it are multiplied together. It undoes the power 2.
    √144: which number to the power 2 equals 144?
  2. Test the powers in order. List the powers until you reach the number.
    12 = 1, 22 = 4, 32 = 9, 42 = 16, 52 = 25, 62 = 36, 72 = 49, 82 = 64, 92 = 81, 102 = 100, 112 = 121, 122 = 144
  3. Write the answer. 122 = 144.
    12

Check. Check: 122 = 12 × 12 = 144. ✓

Watch out. The square root of 49 is 7, not 24.5. Do not halve the number: a root undoes a power.

Worked example 6.3.2
Work out √916 and √0.25.

Key words. The square root of a number undoes squaring: √49 = 7 because 72 = 49. The cube root undoes cubing: 3√27 = 3 because 33 = 27.

Method. Find the number that gives the original when raised to the power 2.

  1. Spot that the number is a fraction. A fraction is a top number over a bottom number, and we can deal with each separately.
    916: top 9, bottom 16
  2. Take the root of the top and of the bottom. For a fraction, √(a ÷ b) = √a ÷ √b.
    √9 = 3 and √16 = 4
  3. Write the answer as a fraction.
    34

Check. Check: 34 to the power 2 = 916. ✓

Watch out. The square root of 49 is 7, not 24.5. Do not halve the number: a root undoes a power.

Worked example 6.3.3
Write down the cube root of 125.

Key words. The square root of a number undoes squaring: √49 = 7 because 72 = 49. The cube root undoes cubing: 3√27 = 3 because 33 = 27.

Method. Find the number that gives the original when raised to the power 3.

  1. Say what the question is asking. The cube root of a number is the number that gives the original number when 3 lots of it are multiplied together. It undoes the power 3.
    3√125: which number to the power 3 equals 125?
  2. Test the powers in order. List the powers until you reach the number.
    13 = 1, 23 = 8, 33 = 27, 43 = 64, 53 = 125
  3. Write the answer. 53 = 125.
    5

Check. Check: 53 = 5 × 5 × 5 = 125. ✓

Watch out. The square root of 49 is 7, not 24.5. Do not halve the number: a root undoes a power.

Worked example 6.3.4
Between which two whole numbers does √50 lie?

Key words. A perfect square is the square of a whole number (1, 4, 9, 16…). A perfect cube is the cube of a whole number (1, 8, 27, 64…).

Method. Find the two perfect squares that the number lies between, then take roots.

  1. Find the perfect powers on either side of the number. We know the square root of perfect squares exactly, so use them as markers.
    72 = 49   and   82 = 64
  2. Place the number between them. 50 lies between 49 and 64.
    49 < 50 < 64
  3. Take the root of each part. Roots keep the order (a bigger number has a bigger root).
    7 < √50 < 8

Check. A calculator gives √50 = 7.071, which is between 7 and 8. ✓

Watch out. Between two whole numbers means one is below and the next one is above.

Worked example 6.3.5
Work out √36 + 3√27 − 23.

Method. Work out each root or power separately, then add and subtract.

  1. Work out each root or power on its own. A root undoes a power; a power is repeated multiplication.
    √36 = 6, because 6 × 6 = 36
    3√27 = 3, because 3 × 3 × 3 = 27
    23 = 2 × 2 × 2 = 8
  2. Put the values into the calculation. Keep the signs the same.
    √36 + 3√27 − 23 = 6 + 3 − 8
  3. Work out the answer. Add and subtract from left to right.
    6 + 3 − 8 = 1

Check. Work through it again from the left: 6 + 3 − 8 = 1. ✓

Watch out. Do each root or power first (BIDMAS: powers and roots before adding and subtracting).

Watch out

The square root of 36 is 6, not 18. A root does not halve the number. And √(9 + 16) is not √9 + √16: √25 = 5 but 3 + 4 = 7. Add or subtract inside the root first.

Exercise 6E  Roots without a calculator

Non-calculator
1
FWork out each square root.
(a)√36
(b)√81
(c)√121
(d)√169
(e)√400
(f)√25
2
FWork out each cube root.
(a)3√8
(b)3√27
(c)3√64
(d)3√125
(e)3√1000
(f)3√216
3
FWork out each square root.
(a)√916
(b)√425
(c)√136
(d)√4964
4
FWork out each root. Give your answer as a decimal.
(a)√0.25
(b)√0.81
(c)√0.0144
(d)3√0.027
(e)3√0.008
5
FWork out each calculation.
(a)√49 + √81
(b)3√64 − √16
(c)52 − 3√125
(d)√100 + 33 − √4
6
FBetween which two whole numbers does each root lie?
(a)√20
(b)√70
(c)3√30
(d)3√100
(e)√200
7
FSolve each equation.
(a)x2 = 49
(b)x2 = 121
(c)x3 = 27
(d)x3 = 125
8
FWork out each value.
(a)(√7)2
(b)(√13)2
(c)(√2)2
(d)(√50)2
9
PSA square patio has an area of 81 m2. Work out the length of one side, and the length of the fence needed to go all the way around the patio.
10
PSA cube-shaped box has a volume of 216 cm3. Work out the total length of all 12 edges of the box.
11
PSA square has an area of 150 cm2. Between which two whole numbers is the length of its side?
12
RIs √(9 + 16) the same as √9 + √16? Show your working.
13
RShow that √50 is greater than 7 but less than 7.1.
14
RExplain why √(−9) cannot be worked out using real numbers.
15
RFind a number between 50 and 100 that is both a square number and a cube number.

Exercise 6F  Powers and roots with a calculator

Calculator allowed
1
FUse a calculator to work out each root. Give your answers to 3 significant figures.
(a)√7
(b)√130
(c)√2000
(d)3√50
(e)3√300
2
FUse a calculator to work out each power. Give your answers to 3 significant figures.
(a)2.54
(b)1.0510
(c)3.2−2
(d)0.95
(e)8−2
3
PSA square picture has an area of 90 cm2. Work out the length of one side to 1 decimal place.
4
PS£1500 is invested. After 5 years it is worth £1500 × 1.045. Work out its value, to the nearest penny.
5
PSWork out √(5.22 + 3.92).
6
FUse a calculator to work out each value. Give your answers to 3 significant figures.
(a)√11
(b)√59
(c)1.56
(d)0.5−3
7
RShow, using a calculator, that 3√100 is between 4.6 and 4.7.
6.4

Standard form: large and small numbers

Edexcel 1MA1: N9, N7
Grade 4-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Standard form: large and small numbersComing soon

The Sun is about 150,000,000,000 m from the Earth. A hydrogen atom is about 0.0000000001 m wide. These numbers are long and easy to get wrong. Standard form is a short, standard way to write very big and very small numbers. It uses powers of 10 (see section 6.2).

10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²10⁰10²10⁴10⁶10⁸10¹⁰hydrogen atom1 × 10⁻¹⁰ mwidth of a hair7 × 10⁻⁵ ma person1.7 × 10⁰ mMount Everest8.8 × 10³ mdiameter of Earth1.3 × 10⁷ mEarth to Sun1.5 × 10¹¹ mSizes in metres: each step along the line is a factor of 100
Real things on a line of powers of 10. Each step along the line is a factor of 100.

A number is in standard form when it is written as A × 10n where

Big numbers have a positive power; small numbers (less than 1) have a negative power. For example 150,000,000,000 = 1.5 × 1011 and 0.0000000001 = 1 × 10−10.

To write a number in standard form: (1) put the decimal point after the first non-zero digit to make A; (2) count how many places the decimal point moved; (3) that count is the power. Moving to the left gives a positive power, and moving to the right gives a negative power. For example 45,000,000: the point moves 7 places left, so 4.5 × 107. And 0.00034: the point moves 4 places right, so 3.4 × 10−4.

3.2 × 10⁵320000startendmoves 5 places rightThe digits stay in order. The gaps are filled with zeros.3.2 × 10⁵ = 320000
Going back: 3.2 × 105 means move the decimal point 5 places right, filling with zeros, to get 320,000.

To change back to an ordinary number do the opposite: a positive power moves the decimal point right, a negative power moves it left, one place for each power of 10. The number is not made negative by a negative power: 4 × 10−2 = 0.04.

Sometimes a number looks like standard form but the front number is not between 1 and 10, for example 45 × 103. Fix it by changing A and the power together: 45 = 4.5 × 10, so 45 × 103 = 4.5 × 10 × 103 = 4.5 × 104. On a calculator, standard form is often shown with an E: 3.6E8 means 3.6 × 108.

Key ideas

  • Standard form: A × 10n with 1 ≤ A < 10 and n a whole number.
  • Big number: positive power. Number between 0 and 1: negative power.
  • Count how many places the decimal point moves. Left means positive, right means negative.
  • To fix 45 × 103: write 45 as 4.5 × 10, then add 1 to the power.
  • A calculator shows 3.6 × 108 as 3.6E8.
Worked example 6.4.1
Write 45,000,000 in standard form.

Key words. Standard form is a way to write very big or very small numbers as A × 10n, where A is at least 1 and less than 10 (so exactly one non-zero digit before the decimal point) and n is a whole number (positive, negative or zero).

Method. Write the number as A × 10n: put the decimal point after the first non-zero digit, then count how many places it moved.

  1. Find the first non-zero digit. In standard form A is at least 1 and less than 10, so A has exactly one non-zero digit in front of the decimal point. That digit is the first non-zero digit of the number.
    The first non-zero digit of 45,000,000 is 4.
  2. Place the decimal point just after it to make A. The digits of A are the same as the digits of the number (without the zeros at the ends).
    A = 4.5
  3. Count how many places the decimal point has moved. Starting from 45000000 the decimal point has to move 7 places to the left to sit after the 4. Each place moved is a factor of 10.
    45000000 → 4.5: 7 places to the left
  4. Decide the sign of the power. The number is 10 or more (a big number). We made it smaller to get A, so we must multiply by 10 to put the size back.
    n is positive: n = 7
  5. Write the answer. Put A and the power of 10 together.
    4.5 × 107

Check. Convert back: 4.5 × 107 = 45000000 = 45,000,000. ✓

Watch out. A big number has a positive power and a small number (less than 1) has a negative power. Do not count the zeros, count the places the point moves.

Worked example 6.4.2
Write 0.00034 in standard form.

Key words. Standard form is a way to write very big or very small numbers as A × 10n, where A is at least 1 and less than 10 (so exactly one non-zero digit before the decimal point) and n is a whole number (positive, negative or zero).

Method. Write the number as A × 10n: put the decimal point after the first non-zero digit, then count how many places it moved.

  1. Find the first non-zero digit. In standard form A is at least 1 and less than 10, so A has exactly one non-zero digit in front of the decimal point. That digit is the first non-zero digit of the number.
    The first non-zero digit of 0.00034 is 3.
  2. Place the decimal point just after it to make A. The digits of A are the same as the digits of the number (without the zeros at the ends).
    A = 3.4
  3. Count how many places the decimal point has moved. Starting from 0.00034 the decimal point has to move 4 places to the right to sit after the 3. Each place moved is a factor of 10.
    0.00034 → 3.4: 4 places to the right
  4. Decide the sign of the power. The number is less than 1 (a small number). We made it bigger to get A, so we must divide by 10 to put the size back. Dividing by 10 is the same as multiplying by 10−1.
    n is negative: n = −4
  5. Write the answer. Put A and the power of 10 together.
    3.4 × 10−4

Check. Convert back: 3.4 × 10−4 = 0.00034 = 0.00034. ✓

Watch out. A big number has a positive power and a small number (less than 1) has a negative power. Do not count the zeros, count the places the point moves.

Worked example 6.4.3
Write 3.07 × 104 as an ordinary number.

Key words. Standard form is a way to write very big or very small numbers as A × 10n, where A is at least 1 and less than 10 (so exactly one non-zero digit before the decimal point) and n is a whole number (positive, negative or zero).

Method. A positive power of 10 makes the number bigger (move the decimal point right). A negative power makes it smaller (move the decimal point left).

  1. Look at the power of 10. The power is 4, which is positive, so the number is big. Multiplying by 10 moves every digit one place to the left, which is the same as moving the decimal point one place to the right.
    104 means multiply by 10, 4 times
  2. Move the decimal point 4 places to the right. Move it one place at a time. If there are not enough digits, fill the gaps with zeros.
    3.07 → 30.7 → 307 → 3070 → 30700
  3. Write the answer. The last number in the chain is the answer.
    30,700

Check. Convert back to standard form: 30700 has first non-zero digit 3, so A = 3.07 and n = 4. ✓

Watch out. A negative power makes a number smaller, but it does not make it negative.

Worked example 6.4.4
Write 4.1 × 10−5 as an ordinary number.

Key words. Standard form is a way to write very big or very small numbers as A × 10n, where A is at least 1 and less than 10 (so exactly one non-zero digit before the decimal point) and n is a whole number (positive, negative or zero).

Method. A positive power of 10 makes the number bigger (move the decimal point right). A negative power makes it smaller (move the decimal point left).

  1. Look at the power of 10. The power is −5, which is negative, so the number is small. Dividing by 10 moves every digit one place to the right, which is the same as moving the decimal point one place to the left.
    10−5 means divide by 10, 5 times
  2. Move the decimal point 5 places to the left. Move it one place at a time and fill the new places with zeros.
    4.1 → 0.41 → 0.041 → 0.0041 → 0.00041 → 0.000041
  3. Write the answer. The last number in the chain is the answer.
    0.000041

Check. Convert back to standard form: 0.000041 has first non-zero digit 4, so A = 4.1 and n = −5. ✓

Watch out. A negative power makes a number smaller, but it does not make it negative.

Worked example 6.4.5
Write 45 × 103 in standard form.

Key words. Standard form is a way to write very big or very small numbers as A × 10n, where A is at least 1 and less than 10 (so exactly one non-zero digit before the decimal point) and n is a whole number (positive, negative or zero).

Method. Change the front number A so that it is between 1 and 10, and adjust the power of 10 to match.

  1. Check the rule for standard form. In A × 10n, A must be at least 1 and less than 10.
    A = 45, which is 10 or more, so this is not in standard form.
  2. Write 45 as a number between 1 and 10 times a power of 10. Moving the decimal point 1 place to the left makes the number 10 times smaller, so we must multiply by 101 to keep the value the same.
    45 = 4.5 × 101
  3. Replace A in the original and combine the powers of 10. Powers of 10 multiply by adding the indices: 10a × 10b = 10a+b.
    45 × 103 = 4.5 × 101 × 103
    = 4.5 × 101 + 3
  4. Write the answer.
    4.5 × 104

Check. Both forms are the same number, 45000: 45 × 1000 = 45000 and 4.5 × 10000 = 45000. ✓

Watch out. If you make A smaller by moving the point left, the power of 10 gets bigger by the same number of places (and the other way round).

Worked example 6.4.6
Write 299,792,458 in standard form correct to 3 significant figures.

Key words. Standard form is a way to write very big or very small numbers as A × 10n, where A is at least 1 and less than 10 (so exactly one non-zero digit before the decimal point) and n is a whole number (positive, negative or zero).

Method. Write the number as A × 10n: put the decimal point after the first non-zero digit, then count how many places it moved.

  1. Round to 3 significant figures first. Significant figures count from the first non-zero digit.
    299,792,458 = 300,000,000 to 3 s.f.
  2. Find the first non-zero digit. In standard form A is at least 1 and less than 10, so A has exactly one non-zero digit in front of the decimal point. That digit is the first non-zero digit of the number.
    The first non-zero digit of 300,000,000 is 3.
  3. Place the decimal point just after it to make A. The digits of A are the same as the digits of the number (without the zeros at the ends).
    A = 3.00
  4. Count how many places the decimal point has moved. Starting from 300000000 the decimal point has to move 8 places to the left to sit after the 3. Each place moved is a factor of 10.
    300000000 → 3.00: 8 places to the left
  5. Decide the sign of the power. The number is 10 or more (a big number). We made it smaller to get A, so we must multiply by 10 to put the size back.
    n is positive: n = 8
  6. Write the answer. Put A and the power of 10 together.
    3.00 × 108

Check. Convert back: 3.00 × 108 = 300000000 (which is 300,000,000 to 3 s.f.). ✓

Watch out. A big number has a positive power and a small number (less than 1) has a negative power. Do not count the zeros, count the places the point moves.

Watch out

Count how far the decimal point moves, not how many zeros you can see. And check the sign: a number less than 1 must have a negative power. 0.0072 = 7.2 × 10−3, not 7.2 × 103.

Exam tip

A is always at least 1 and less than 10. If you get 12 × 105 or 0.7 × 104 you are not finished: it is not standard form yet.

Exercise 6I  Writing numbers in standard form

Non-calculator
1
FIs each number in standard form? Give a reason.
(a)3.4 × 105
(b)34 × 104
(c)0.5 × 103
(d)9.99 × 10−2
(e)10 × 102
(f)6 × 100
2
FWrite each number in standard form.
(a)6000
(b)45,000
(c)380,000
(d)5,600,000
(e)91,000,000
(f)720
3
FWrite each number in standard form.
(a)0.7
(b)0.06
(c)0.0035
(d)0.00072
(e)0.000081
(f)0.0000046
4
FWrite each as an ordinary number.
(a)5 × 103
(b)2.4 × 104
(c)7.08 × 105
(d)1.5 × 106
(e)9.2 × 107
5
FWrite each as an ordinary number.
(a)4 × 10−1
(b)6.2 × 10−3
(c)3.07 × 10−4
(d)8.5 × 10−5
(e)1.1 × 10−6
6
FThese numbers are not in standard form. Write each in standard form.
(a)45 × 103
(b)0.6 × 105
(c)120 × 10−4
(d)0.08 × 107
(e)250 × 102
7
FA calculator display shows each of these. Write the value in standard form and as an ordinary number.
(a)3.6E8
(b)2.4E-3
(c)7.1E5
(d)5E-2
8
FWrite each number in standard form, correct to the number of significant figures shown.
(a)4,523,000 (2 s.f.)
(b)0.0003846 (2 s.f.)
(c)299,792,458 (3 s.f.)
(d)6,371,000 (3 s.f.)
9
PSWrite each of these facts in standard form.
(a)The world population is about 8,100,000,000.
(b)The Moon is about 384,400 km from the Earth.
(c)A sheet of paper is about 0.0001 m thick.
(d)A hydrogen atom has a mass of 0.00000000000000000000000000167 kg.
10
PSWrite each number in standard form.
(a)25 million
(b)half a billion (a billion is 1,000,000,000)
(c)three thousandths
(d)1 hundred-thousandth
11
RExplain why 10 is written as 1 × 101 in standard form and why 0.1 is written as 1 × 10−1.
12
RAli writes 0.0072 = 7.2 × 103. Explain his mistake and write the correct standard form.
13
RN = 4.5 × 106. Write 10N and N ÷ 100 in standard form.
(a)10N
(b)N ÷ 100
14
RWhich of these is not equal to 3.5 × 104? A: 35,000 B: 350 × 102 C: 0.35 × 105 D: 3.5 × 10−4
15
RThe number 7.2 × 10n is greater than 500 and less than 50,000. List all the possible whole-number values of n.

Exercise 6J  Standard form in context

Non-calculator
1
PSWrite each measurement in standard form.
(a)The diameter of the Sun is 1,392,000 km.
(b)The wavelength of red light is 0.0000007 m.
(c)The population of the UK is about 67,000,000.
(d)A human hair is about 0.00007 m wide.
(e)The nearest star is about 40,000,000,000,000 km away.
2
PSWrite each of these as an ordinary number.
(a)The Earth is 1.5 × 108 km from the Sun.
(b)A virus is 1.2 × 10−7 m long.
(c)A computer can do 3 × 109 operations per second.
(d)A grain of pollen has a mass of 2.5 × 10−9 kg.
3
FWrite each of these numbers in standard form.
(a)85,000
(b)0.0092
(c)630,000,000
(d)0.00000058
(e)1,200,000,000
4
FWrite each number as an ordinary number.
(a)6.3 × 104
(b)9 × 10−3
(c)1.07 × 106
(d)4.4 × 10−2
5
FWrite each number in standard form.
(a)32 × 106
(b)0.45 × 103
(c)560 × 10−5
(d)0.09 × 104
6
RZoe says: “50 × 104 is bigger than 5 × 104, so it must be in standard form.” Is she right? Explain.
7
PSA scientist measures a bacterium as 0.0000025 m long. She writes 25 × 10−7 m. Write this correctly in standard form.
8
RShow that 3 × 105 is 100 times bigger than 3 × 103.
6.5

Comparing and calculating with standard form

Edexcel 1MA1: N9
Grade 5-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Comparing and calculating with standard formComing soon

Standard form makes big numbers easy to compare. Look at the powers of 10 first. A bigger power means a bigger number, because the front numbers are all between 1 and 10 and cannot make up the difference. So 3.2 × 105 is bigger than 9.9 × 104. If the powers are the same, compare the front numbers. Be careful with negative powers: −3 is bigger than −5, so 10−3 = 0.001 is bigger than 10−5 = 0.00001.

SunEarth1.5 × 10¹¹ mlight travels at 3 × 10⁸ m/s
Standard form makes it easy to calculate with huge numbers, such as the time light takes to reach us from the Sun.

Multiplying. Multiply the front numbers, add the powers of 10 (law 1 of indices), then check that A is between 1 and 10. (2 × 103) × (3 × 104) = 6 × 107. If the front numbers multiply to 10 or more, adjust: (4 × 105) × (3 × 102) = 12 × 107 = 1.2 × 108.

Dividing. Divide the front numbers and subtract the powers. If A comes out less than 1, adjust: (3 × 105) ÷ (5 × 10−2) = 0.6 × 107 = 6 × 106.

Adding and subtracting. The powers of 10 must be the same first (like fractions needing a common denominator). 3.2 × 104 + 8.5 × 103 = 3.2 × 104 + 0.85 × 104 = 4.05 × 104.

With a calculator use the ×10x key (sometimes EXP), then write the answer in standard form to the accuracy asked for. Always estimate first by rounding each A to a whole number. In real problems decide which operation is needed, for example time = distance ÷ speed.

Higher tier: you may also need to find a power or a square root of a number in standard form: (2 × 103)3 = 23 × 109 = 8 × 109. Raise A to the power and multiply the index of 10 by the power.

Key ideas

  • Compare the powers of 10 first, then the front numbers.
  • Multiply: multiply the A values and add the powers. Divide: divide the A values and subtract the powers.
  • Add or subtract: make the powers the same first.
  • After any calculation check that A is at least 1 and less than 10; if not, adjust A and the power together.
  • Calculator: use the ×10x key and check with an estimate.
Worked example 6.5.1
Which is larger, 3.2 × 105 or 9.9 × 104?

Key words. Standard form is a way to write very big or very small numbers as A × 10n, where A is at least 1 and less than 10 (so exactly one non-zero digit before the decimal point) and n is a whole number (positive, negative or zero).

Method. Compare the powers of 10 first. If the powers are the same, compare the front numbers.

  1. Check both numbers are in standard form. The front numbers must be between 1 and 10 for this comparison to work.
    3.2 and 9.9 are both at least 1 and less than 10.
  2. Compare the powers of 10 first. A bigger power of 10 means a bigger number, because both front numbers are between 1 and 10 and so cannot make up the difference.
    105 and 104: 5 > 4
  3. Decide. The powers are different, so the powers decide it. We do not need to look at the front numbers.
    3.2 × 105 > 9.9 × 104

Check. As ordinary numbers: 320000 and 99000, and 320000 is bigger than 99000. ✓

Watch out. Do not compare the front numbers first. 9.9 × 102 is much smaller than 1.1 × 105.

Worked example 6.5.2
Write these in order, smallest first: 2.5 × 104, 7.1 × 10−3, 2.1 × 104, 9 × 103.

Key words. Standard form is a way to write very big or very small numbers as A × 10n, where A is at least 1 and less than 10 (so exactly one non-zero digit before the decimal point) and n is a whole number (positive, negative or zero).

Method. Order by the power of 10 first; use the front number to break ties.

  1. List the power of 10 of each number. The power of 10 is the main thing that decides the size.
    2.5 × 104: power 4
    7.1 × 10−3: power −3
    2.1 × 104: power 4
    9 × 103: power 3
  2. Order by the powers of 10. We need smallest to largest. Remember that −7 is smaller than −3.
    power −3, power 3, power 4, power 4
  3. If two numbers have the same power, compare their front numbers. With equal powers, the bigger front number is the bigger number.
    power 4: 2.5 and 2.1 → smaller first is 2.1
  4. Write the numbers in order.
    7.1 × 10−3 ,  9 × 103 ,  2.1 × 104 ,  2.5 × 104

Check. As ordinary numbers: 0.0071, 9000, 21000, 25000 — these increase along the list. ✓

Watch out. A negative power is smaller than a positive power, and a larger negative power (like −8) is smaller than −2.

Worked example 6.5.3
Work out (2.5 × 106) × (4 × 10−2). Give your answer in standard form.

Key words. Standard form is a way to write very big or very small numbers as A × 10n, where A is at least 1 and less than 10 (so exactly one non-zero digit before the decimal point) and n is a whole number (positive, negative or zero).

Method. Multiply the front numbers, add the powers of 10, then make sure the front number is between 1 and 10.

  1. Write the multiplication with the number parts together and the powers of 10 together. Multiplication can be done in any order, so we can group the front numbers and group the powers of 10.
    (2.5 × 106) × (4 × 10−2) = (2.5 × 4) × (106 × 10−2)
  2. Multiply the front numbers. Do this like an ordinary decimal multiplication.
    Ignore the decimal points: 25 × 4 = 100
    There are 1 decimal place altogether in 2.5 and 4, so the answer has 1 decimal place: 2.5 × 4 = 10
  3. Multiply the powers of 10: add the indices. 10a × 10b = 10a+b.
    106 × 10−2 = 106 + (−2) = 104
  4. Put the two parts together.
    10 × 104
  5. Check: is the front number between 1 and 10? 10 is 10 or more, so this is not standard form yet. Write 10 = 1 × 10 and combine the powers of 10.
    10 × 104 = 1 × 10 × 104 = 1 × 104 + 1 = 1 × 105
  6. Write the answer.
    1 × 105

Check. As ordinary numbers: 2500000 × 0.04 = 100000, which is 1 × 105. ✓

Watch out. Add the powers when you multiply (do not multiply them). Always check the front number at the end: 12 × 107 is not in standard form.

Worked example 6.5.4
Work out (3 × 105) ÷ (5 × 10−2).

Key words. Standard form is a way to write very big or very small numbers as A × 10n, where A is at least 1 and less than 10 (so exactly one non-zero digit before the decimal point) and n is a whole number (positive, negative or zero).

Method. Divide the front numbers, subtract the powers of 10, then make sure the front number is between 1 and 10.

  1. Write the division with the number parts together and the powers of 10 together. We can split the division into the front numbers and the powers of 10.
    (3 × 105) ÷ (5 × 10−2) = (3 ÷ 5) × (105 ÷ 10−2)
  2. Divide the front numbers. Do this like an ordinary decimal division.
    3 ÷ 5 = 0.6
  3. Divide the powers of 10: subtract the indices. 10a ÷ 10b = 10a−b.
    105 ÷ 10−2 = 105 − (−2) = 107
  4. Put the two parts together.
    0.6 × 107
  5. Check: is the front number between 1 and 10? 0.6 is less than 1, so this is not standard form yet. Write 0.6 = 6 × 10−1 and combine the powers of 10.
    0.6 × 107 = 6 × 10−1 × 107 = 6 × 10−1 + 7 = 6 × 106
  6. Write the answer.
    6 × 106

Check. Multiply back: 6 × 5 = 30 and the powers give 106 × 10−2 = 104, so we get 30 × 104 = 300000, which is the number we started with. ✓

Watch out. Subtract the powers when you divide, and take care with negative powers: 4 − (−2) = 6. If the front number comes out less than 1, adjust it.

Worked example 6.5.5
Work out 3.2 × 104 + 8.5 × 103.

Key words. Standard form is a way to write very big or very small numbers as A × 10n, where A is at least 1 and less than 10 (so exactly one non-zero digit before the decimal point) and n is a whole number (positive, negative or zero).

Method. Add standard form numbers: give them the same power of 10, add the front numbers, then tidy the answer back into standard form.

  1. Make the powers of 10 the same. We can only add when the powers of 10 match, just as we can only add fractions with the same denominator. Change the number with the smaller power so that it uses 104. Each increase of 1 in the power means moving the decimal point of the front number one place to the left.
    8.5 × 103 = 0.85 × 104
  2. Add the front numbers. The power of 10 stays the same: it just tells us the size of the unit.
    3.2 + 0.85 = 4.05
    so the result is 4.05 × 104
  3. Write the answer.
    4.05 × 104

Check. As ordinary numbers: 32000 + 8500 = 40500, which is 4.05 × 104. ✓

Watch out. You cannot just add the front numbers when the powers are different: 3 × 104 + 2 × 103 is not 5 × 107.

Worked example 6.5.6
Light travels at 3 × 108 m/s. The Sun is 1.5 × 1011 m from the Earth. How long does light take to reach the Earth? Give your answer in seconds in standard form.

Key words. Standard form is a way to write very big or very small numbers as A × 10n, where A is at least 1 and less than 10 (so exactly one non-zero digit before the decimal point) and n is a whole number (positive, negative or zero).

Method. Divide the front numbers, subtract the powers of 10, then make sure the front number is between 1 and 10.

  1. Decide what to do. The Sun is a long way from the Earth. We want a time. Time = distance ÷ speed, so divide the distance by the speed of light.
    time = (1.5 × 1011) ÷ (3 × 108)
  2. Write the division with the number parts together and the powers of 10 together. We can split the division into the front numbers and the powers of 10.
    (1.5 × 1011) ÷ (3 × 108) = (1.5 ÷ 3) × (1011 ÷ 108)
  3. Divide the front numbers. Do this like an ordinary decimal division.
    Multiply both numbers by 10 to make them whole (the answer does not change): 1.5 ÷ 3 = 15 ÷ 30 = 0.5
  4. Divide the powers of 10: subtract the indices. 10a ÷ 10b = 10a−b.
    1011 ÷ 108 = 1011 − 8 = 103
  5. Put the two parts together.
    0.5 × 103
  6. Check: is the front number between 1 and 10? 0.5 is less than 1, so this is not standard form yet. Write 0.5 = 5 × 10−1 and combine the powers of 10.
    0.5 × 103 = 5 × 10−1 × 103 = 5 × 10−1 + 3 = 5 × 102
  7. Write the answer with its unit.
    5 × 102 s

Check. Multiply back: 5 × 3 = 15 and the powers give 102 × 108 = 1010, so we get 15 × 1010 = 150000000000, which is the number we started with. ✓

Watch out. Subtract the powers when you divide, and take care with negative powers: 4 − (−2) = 6. If the front number comes out less than 1, adjust it.

Worked example 6.5.7
Use a calculator to work out (4.5 × 107) × (2.4 × 10−3). Give your answer in standard form to 3 significant figures.

Key words. Standard form is a way to write very big or very small numbers as A × 10n, where A is at least 1 and less than 10 (so exactly one non-zero digit before the decimal point) and n is a whole number (positive, negative or zero).

Method. Use the ×10x key on a calculator, then write the display in standard form to the accuracy asked for.

  1. Enter the calculation using the ×10x key. The ×10x key (on some calculators it is EXP) types "× 10 to the power" in one go. For a negative power, press the (−) key before the power number. Brackets stop the calculator mixing the numbers up.
    Press: 4.5 [×10x] 7 [×] 2.4 [×10x] [(−)] 3 [=]
  2. Read the answer from the display. Your calculator may show the answer as an ordinary number or in its own standard form style, such as 2.4E−3. The digits are what matter.
    Display: 1.08 × 105 (about)
  3. Round to 3 significant figures and write in standard form. Round the front number A to 3 significant figures: look at the digit after the last one you keep, and round up if it is 5 or more. The power of 10 stays the same unless the rounding makes A reach 10.
    1.08 × 105

Check. Estimate by rounding each front number to the nearest whole number: (5 × 107) × (2 × 10−3) ≈ 1 × 105, which is close to our answer. ✓

Watch out. Use the ×10x key, not separate × 10 ^ keys, unless you use brackets. Always check the power is sensible with a quick estimate.

Watch out

Do not add the powers when you add numbers: 3 × 104 + 2 × 103 is not 5 × 107. Make the powers match first. And after multiplying, check A: 12 × 107 is not in standard form.

Exam tip

In calculator questions write down the calculation you typed, then the answer, then round. If a question says “in standard form” a plain decimal answer will lose marks.

Exercise 6K  Comparing and ordering standard form

Non-calculator
1
FWrite down the larger number in each pair.
(a)3.2 × 105 and 9.9 × 104
(b)4.5 × 10−3 and 4.5 × 10−5
(c)6.1 × 107 and 6.8 × 107
(d)2 × 10−6 and 9 × 10−7
(e)1.2 × 1010 and 9.9 × 109
2
FPut < or > between each pair.
(a)7.5 × 103   ?   2.1 × 106
(b)3.3 × 10−2   ?   1.9 × 10−4
(c)5.6 × 108   ?   5.1 × 108
(d)9 × 10−5   ?   1 × 10−4
3
FWrite each list in order, smallest first.
(a)2.5 × 104 , 7.1 × 10−3 , 2.1 × 104 , 9 × 103
(b)5.2 × 10−6 , 4.9 × 10−5 , 1.1 × 10−5 , 8 × 10−7
4
FWrite this list in order, largest first: 6 × 108, 6 × 10−8, 5.9 × 108, 1 × 109.
5
PSThe distances of four planets from the Sun, in km, are: Mercury 5.79 × 107, Venus 1.08 × 108, Earth 1.5 × 108, Mars 2.28 × 108. Write the planets in order, closest to the Sun first.
6
PSThe mass of an electron is 9.11 × 10−31 kg. The mass of a proton is 1.67 × 10−27 kg. Which has the smaller mass?
7
RExplain why 9.9 × 102 is less than 1.1 × 105, even though 9.9 is bigger than 1.1.
8
RWrite down a number in standard form that is bigger than 3 × 104 and smaller than 3 × 105.
9
RTrue or false: 5 × 10−3 > 5 × 10−2? Explain.

Exercise 6L  Calculating with standard form (no calculator)

Non-calculator
1
FWork out each multiplication. Give your answers in standard form.
(a)(2 × 103) × (3 × 104)
(b)(4 × 105) × (2 × 10−2)
(c)(1.5 × 106) × (4 × 103)
(d)(2.5 × 10−3) × (4 × 10−2)
(e)(3 × 108) × (3 × 10−5)
2
FWork out each division. Give your answers in standard form.
(a)(8 × 109) ÷ (2 × 104)
(b)(6 × 107) ÷ (3 × 10−2)
(c)(9 × 105) ÷ (4 × 102)
(d)(3 × 105) ÷ (5 × 10−2)
(e)(4.5 × 10−3) ÷ (1.5 × 102)
(f)(1.2 × 106) ÷ (4 × 10−2)
3
FWork out each. Give your answers in standard form.
(a)(3.2 × 104) + (8.5 × 103)
(b)(5.2 × 106) + (4.1 × 106)
(c)(6 × 105) − (2.5 × 104)
(d)(7.2 × 10−3) − (3.1 × 10−3)
(e)(9.5 × 107) + (7.5 × 106)
(f)(5.2 × 104) − (4.9 × 104)
4
PSA rectangular field is 3 × 104 m long and 2 × 103 m wide. Work out its area in standard form.
5
PSA printer prints 4 × 103 pages every hour. How many pages does it print in 5 × 102 hours?
6
PSA grain of sand has a mass of 2.5 × 10−5 kg. How many grains of sand are in 1 kg? Give your answer in standard form.
7
PSA country has a population of 6 × 107. Each person uses 1.5 × 102 litres of water a day. How many litres does the whole country use in a day?
8
PSA company earned £3.6 × 106 in 2020 and £5.2 × 106 in 2021. Work out the total it earned over the two years.
9
RSam writes: 3 × 104 + 2 × 103 = 5 × 107. Is he correct? Show the correct answer.
10
RShow that (2 × 103) × (5 × 104) is 1 × 108, and explain why you have to change the front number.

Exercise 6M  Calculator questions and context

Calculator allowed
1
FUse a calculator to work out each. Give your answers in standard form to 3 significant figures.
(a)(4.5 × 107) × (2.4 × 10−3)
(b)(5.97 × 1024) ÷ (7.35 × 1022)
(c)(3.2 × 105) × (4.7 × 10−2) ÷ (1.6 × 104)
(d)(6.84 × 108) + (3.9 × 109)
2
FWork out each. Give your answers in standard form to 3 significant figures.
(a)(2.7 × 104)2
(b)√(8.1 × 107)
(c)(9.2 × 10−3) − (3.7 × 10−4)
3
PSLight travels at 3 × 108 m/s. The Sun is 1.496 × 1011 m from the Earth. How many seconds does light take to reach the Earth? Give your answer to 3 significant figures in standard form.
4
PSThe mass of the Earth is 5.97 × 1024 kg and the mass of the Moon is 7.35 × 1022 kg. How many times heavier is the Earth than the Moon? Give your answer in standard form to 3 significant figures.
5
PSA region has an area of 2.4 × 105 km2 and a population of 6.7 × 107. Work out the population density (people per km2) to 3 significant figures in standard form.
6
PSA memory card holds 2 × 1012 bytes. One photo uses 3.5 × 106 bytes. How many photos fit on the card? Give your answer to 3 significant figures in standard form.
7
PSThere are 6.02 × 1023 atoms in one mole of a substance. How many atoms are there in 3.5 moles? Give your answer to 3 significant figures in standard form.
8
PSJupiter has a diameter of 1.43 × 105 km. The Earth has a diameter of 1.27 × 104 km. How much bigger is the diameter of Jupiter? Give your answer to 3 significant figures in standard form.
9
PSThe galaxy Andromeda is 2.5 × 106 light-years away. One light-year is 9.46 × 1012 km. How far away is Andromeda in km? Give your answer to 3 significant figures in standard form.
10
RShow, by estimating, that (3.9 × 105) × (2.1 × 10−2) is approximately 8 × 103.
6.6

Exact answers with π

Edexcel 1MA1: N8, G17
Grade 4-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Exact answers with πComing soon

The number π (pi) is the circumference of any circle divided by its diameter. Its decimal never ends and never repeats: 3.14159265…. Because it cannot be written exactly as a decimal or fraction, any answer found with a rounded value of π is only approximate. If you want an exact answer you leave π as the symbol π. Questions say “give your answer in terms of π” or “leave your answer as a multiple of π”.

radius 6 cmarea = π × r × r
The area of a circle is π × r × r. For radius 6 cm the exact area is π × 36 = 36π cm2.

Remember the formulae: circumference C = πd = 2πr and area A = πr2 (more on circles in Chapter 18). Square only the radius, then write the number first and π after it: 36π, not π36. Treat π like a letter in algebra. 3π + 5π = 8π (like 3x + 5x = 8x), 4 × 3π = 12π, and 12π ÷ 4 = 3π.

Parts of circles use fractions: a semicircle is half and a quarter circle is one quarter. Take the fraction of the exact answer: a quarter of 64π is 16π. For a perimeter, the curved part is a fraction of the circumference and the straight edges are ordinary lengths, so the answer has two terms, such as (5π + 10) cm. Compound shapes: find each area exactly and add. If you need a decimal at the end, use the π button on your calculator and round at the very end.

8 cm5 cm
A rectangle 8 cm by 5 cm with a semicircle of diameter 8 cm: area = 40 + 8π cm2.

Key ideas

  • Exact answer: leave π in the answer, with the number in front: 36π.
  • C = πd = 2πr and A = πr2. Use the radius for the area (half the diameter).
  • Treat π like a letter: 3π + 5π = 8π.
  • A fraction of a circle: take that fraction of the exact area (or circumference).
  • Only use the π button, and round, when you want a decimal.
Worked example 6.6.1
Work out the area of a circle with radius 6 cm. Give your answer in terms of π.

Key words. π (pi) is the number 3.14159… that you get when you divide the circumference of any circle by its diameter. It never ends and never repeats, so we cannot write it exactly as a decimal. An answer left in terms of π (such as 12π) is exact: nothing has been rounded.

Method. Put the numbers into the formula but leave π as the symbol π. Then the answer is exact.

  1. Write the formula for the area of a circle. The area of a circle is π times the radius squared.
    A = πr2
  2. Substitute the radius. Replace r with the radius. Leave π as the symbol.
    A = π × 62
  3. Work out the square first. In BIDMAS, powers come before multiplication. Square the radius, not the π.
    62 = 6 × 6 = 36
  4. Write the number in front of π. Write the exact answer with the number first and then π, and add the units.
    36π cm2

Check. Use the π button to get a decimal and compare: 36π = 113.097 to 3 decimal places, so about 113.1 cm2 to 1 d.p. This is similar in size to 113.04 found using 3.14. ✓

Watch out. Do not replace π by 3.14 or press the π key if the question says "in terms of π" or "give your answer exactly". Also square only the radius.

Worked example 6.6.2
Work out the circumference of a circle with diameter 10 cm, in terms of π.

Key words. π (pi) is the number 3.14159… that you get when you divide the circumference of any circle by its diameter. It never ends and never repeats, so we cannot write it exactly as a decimal. An answer left in terms of π (such as 12π) is exact: nothing has been rounded.

Method. Put the numbers into the formula but leave π as the symbol π. Then the answer is exact.

  1. Find the radius. The radius is half of the diameter.
    radius = 10 ÷ 2 = 5 cm
  2. Write the formula for the circumference. The circumference is the distance around the circle: π times the diameter, which is also 2 × π × radius.
    C = πd   or   C = 2πr
  3. Substitute. Use the diameter, which is double the radius. Leave π as the symbol.
    d = 2 × 5 = 10 cm
    C = π × 10
  4. Write the number in front of π. Write the exact answer with the number first and then π, and add the units.
    10π cm

Check. Use the π button to get a decimal and compare: 10π = 31.416 to 3 decimal places, so about 31.4 cm to 1 d.p. This is similar in size to 31.4 found using 3.14. ✓

Watch out. Do not replace π by 3.14 or press the π key if the question says "in terms of π" or "give your answer exactly". Also square only the radius.

Worked example 6.6.3
Work out 3π + 5π.

Key words. π (pi) is the number 3.14159… that you get when you divide the circumference of any circle by its diameter. It never ends and never repeats, so we cannot write it exactly as a decimal. An answer left in terms of π (such as 12π) is exact: nothing has been rounded.

Method. Treat π as if it were a letter. Work with the numbers in front of it and keep π in the answer.

  1. Treat π like a letter in algebra. It is a number, but we do not write it as a decimal. So 3π + 5π works just like 3x + 5x = 8x: 3 lots of π and 5 lots of π make 8 lots of π.
    3π + 5π
  2. Add the numbers in front of π. The number in front tells us how many lots of π there are.
    3 + 5 = 8
  3. Write the answer with π.
    8π

Check. Use 3.14 for π as a rough check: 3 lots of 3.14 and the answer 8 lots of 3.14 = 25.12, which has the right size. ✓

Watch out. π does not change when you add or subtract numbers in front of it. 3π + 5π = 8π, not 8π2.

Worked example 6.6.4
The area of a quarter circle of radius 8 cm. Leave π in your answer.

Key words. π (pi) is the number 3.14159… that you get when you divide the circumference of any circle by its diameter. It never ends and never repeats, so we cannot write it exactly as a decimal. An answer left in terms of π (such as 12π) is exact: nothing has been rounded.

Method. Find the area of the whole circle, then take the fraction of it that the shape is. Leave π in the answer.

  1. Write the area of the whole circle. A = πr2.
    A = π × 82 = 64π cm2
  2. Take the fraction of it. The shape is a quarter of the circle, so multiply by 14.
    14 × 64π = 16π
  3. Write the answer with units.
    16π cm2

Check. As a decimal this is about 50.3 cm2, which is less than the whole circle (201.1 cm2). ✓

Watch out. Use the radius. If you are given the diameter, halve it first.

Worked example 6.6.5
Work out the perimeter of a semicircle of radius 5 cm. Leave your answer in terms of π.

Key words. π (pi) is the number 3.14159… that you get when you divide the circumference of any circle by its diameter. It never ends and never repeats, so we cannot write it exactly as a decimal. An answer left in terms of π (such as 12π) is exact: nothing has been rounded.

Method. Add the curved length (a fraction of the circumference) to the straight edges. Leave π in the answer.

  1. Split the perimeter into the curved part and the straight part(s). The perimeter is the distance all the way round the shape. A semicircle has one curved edge and straight edges.
    perimeter = curved edge + straight edge(s)
  2. Find the length of the curved edge. The curved edge is half of the circumference of the full circle.
    half of 2πr = πr = π × 5 = 5π cm
  3. Find the length of the straight edge(s).
    the diameter = 2 × 5 = 10 cm
  4. Add them. Leave π in the answer. A multiple of π and a whole number cannot be combined into one term, so the answer has two parts.
    (5π + 10) cm

Check. As a decimal: 15.71 + 10 = 25.71 cm (about). This is a sensible length for a shape with radius 5 cm. ✓

Watch out. Do not forget the straight edge(s). The perimeter of a semicircle is more than half the circumference of the circle.

Worked example 6.6.6
Work out the exact area of a rectangle 8 cm by 5 cm with a semicircle of diameter 8 cm on one end.

Key words. π (pi) is the number 3.14159… that you get when you divide the circumference of any circle by its diameter. It never ends and never repeats, so we cannot write it exactly as a decimal. An answer left in terms of π (such as 12π) is exact: nothing has been rounded.

Method. Split into a rectangle and a semicircle, find each area exactly, then add.

  1. Split the shape into a rectangle and a semicircle. It is easier to find the areas of simple shapes and add them.
    area = rectangle + semicircle
  2. Find the area of the rectangle. area = length × width
    8 × 5 = 40 cm2
  3. Find the radius of the semicircle. The diameter of the semicircle is the width of the rectangle. The radius is half the diameter.
    radius = 8 ÷ 2 = 4 cm
  4. Find the area of the semicircle. A semicircle is half a circle: half of πr2.
    12 × π × 42 = 12 × π × 16 = 8π cm2
  5. Add the two areas. Leave π in the answer. The whole numbers and the π part cannot be combined into one term.
    (40 + 8π) cm2

Check. As a decimal: 40 + 25.13 = 65.13 cm2 (about). The rectangle alone is 40 cm2, so this is a little bigger. ✓

Watch out. Use the radius (half the width) in the circle formula, not the whole width.

Watch out

Do not replace π by 3.14 in an exact answer, and do not square the π. Area is π × (radius)2: for radius 6 it is 36π, not 6π. And use the radius: if you are given the diameter, halve it first.

Exercise 6N  Exact answers with π

Non-calculator
1
FWork out the area of each circle. Give your answers in terms of π.
(a)radius 3 cm
(b)radius 5 cm
(c)radius 6 cm
(d)radius 10 cm
2
FWork out the circumference of each circle. Give your answers in terms of π.
(a)diameter 8 cm
(b)diameter 10 cm
(c)diameter 14 cm
(d)diameter 20 cm
3
FWork out the circumference of each circle. Give your answers in terms of π.
(a)radius 4 cm
(b)radius 7 cm
(c)radius 9 cm
4
FWork out each. Leave π in your answers.
(a)3π + 5π
(b)9π − 4π
(c)12π − π
(d)7π + 2π
5
FWork out each. Leave π in your answers.
(a)4π × 3
(b)34π × 8
(c)5π × 6
(d)6π ÷ 2
(e)10π ÷ 4
6
FUse a calculator to write each exact answer as a decimal.
(a)36π cm2 (to 1 decimal place)
(b)10π cm (to 1 decimal place)
(c)25π m2 (to 2 decimal places)
7
FWork out the area of each shape. Leave π in your answers.
(a)a quarter circle of radius 8 cm
(b)a quarter circle of radius 10 cm
(c)a semicircle of radius 6 cm
(d)a semicircle of radius 4 cm
8
PSWork out the perimeter of each shape. Leave π in your answers.
(a)a semicircle of radius 5 cm
(b)a semicircle of diameter 12 cm
(c)a quarter circle of radius 8 cm
9
PSWork out the exact area of each shape. It is a rectangle with a semicircle on one end (the diameter of the semicircle is the width of the rectangle).
(a)rectangle 8 cm by 5 cm
(b)rectangle 12 cm by 7 cm
10
PSA cylindrical can has radius 3 cm and height 10 cm. Work out its volume in terms of π.
11
PSFind the radius of each circle.
(a)area 49π cm2
(b)area 100π cm2
(c)circumference 14π cm
(d)circumference 30π cm
12
PSA circular running track has an outer radius of 10 m and an inner radius of 6 m. Work out the exact area of the track.
13
RIs the fraction 22/7 equal to π? Explain your answer.
14
RSam says the area of a circle with radius 6 cm is 12π cm2. Explain his mistake and give the correct exact area.
15
RShow that the area of a semicircle with diameter 10 cm is 12.5π cm2.
16
RCircle A has radius 2 cm and circle B has radius 6 cm. Show that the area of B is 9 times the area of A.

Chapter summary

SkillMethod
Index laws (same base)Multiply: add indices. Divide: subtract indices. Power of a power: multiply indices. a1 = a.
Zero and negative indicesa0 = 1. a−n = 1/an. For fractions, flip the fraction and make the index positive.
Indices with lettersDeal with the number and each letter separately. (2a3)4 = 16a12.
Roots√a: number that multiplies by itself to give a. Between two whole numbers: use the perfect squares or cubes on either side. x2 = k has two solutions.
Fractional indices (Higher)am/n = (n-th root of a)m. Do the root first. Negative: reciprocal first. Solve xp/q = k by raising to the power q/p.
Standard formA × 10n with 1 ≤ A < 10. Big number: positive n. Number less than 1: negative n. Count the places the decimal point moves.
Calculating in standard formMultiply: multiply the A values, add the powers. Divide: divide the A values, subtract the powers. Add or subtract: make the powers match. Then tidy so that 1 ≤ A < 10.
Comparing standard formCompare the powers of 10 first; if equal compare the front numbers.
Exact answers with πLeave π in the answer: area 36π. Treat π like a letter. Use the radius for area.
Simplifying surds (Higher)√a × √b = √(ab). Use the largest square factor: √12 = 2√3. Add or subtract only like surds.
Expanding with surds (Higher)Every term × every term. √a × √a = a. (a + √b)(a − √b) = a2 − b.
Rationalising (Higher)k/(c√n): multiply top and bottom by √n. k/(a + √b): multiply top and bottom by the conjugate (a − √b).

Can you…?

  • I can write numbers using index notation and use the three index laws with numbers and letters.
  • I know that a0 = 1 and that a negative index means “one over”.
  • I can work out roots, place a root between two whole numbers, and use a calculator for powers and roots.
  • I can write any large or small number in standard form and convert back.
  • I can compare, order and calculate with numbers in standard form, with and without a calculator.
  • I can give exact answers for circle questions in terms of π.
  • (Higher) I can work with fractional indices and solve index equations.
  • (Higher) I can simplify, add, multiply and expand surds.
  • (Higher) I can rationalise a denominator, including using the conjugate.

Key terms

Base
The number that is multiplied in a power. In 34 the base is 3.
Index (power, exponent)
The small raised number that says how many lots of the base are multiplied. Plural: indices.
Reciprocal
One divided by a number. The reciprocal of 5 is 1/5 and a−1 = 1/a.
Root
The number that gives the original when raised to a power. √25 = 5 and 3√8 = 2.
Perfect square / cube
The square or cube of a whole number: 1, 4, 9, 16… and 1, 8, 27, 64…
Standard form
A number written as A × 10n with 1 ≤ A < 10 and n a whole number.
Significant figures
The digits of a number counted from the first non-zero digit.
Exact answer
An answer that has not been rounded, such as 36π or 2√3.
Surd
A square root that is not a whole number or fraction, such as √2. It is irrational.
Irrational number
A number that cannot be written as a fraction. Its decimal never ends or repeats.
Like surds
Surds with the same number under the root, for example 3√2 and 5√2.
Rationalise the denominator
Rewrite a fraction so that its bottom is a whole number, with no surd.
Conjugate
For a + √b, the conjugate is a − √b (the sign in the middle changed). Their product is a2 − b.

Exercise 6Z  Check-up (non-calculator)

Non-calculator
1
Write 45,000 in standard form.
2
Write 3.2 × 10−3 as an ordinary number.
3
Write (25 × 23) ÷ 24 as a single power of 2.
4
Work out 5−2.
5
Simplify 3x2 × 4x3.
6
Work out (2 × 103) × (4 × 105). Give your answer in standard form.
7
Work out the area of a circle with radius 5 cm. Give your answer in terms of π.
8
Work out √49 + 3√8.
9
Which is larger, 3.2 × 105 or 9.9 × 104?
10
Write 10−2 as a decimal.

Total 20 marks. Allow about 20 minutes.

Exercise 6X  Exam-style questions

Non-calculator
1
FWrite 7,300,000 in standard form.[1 mark]
2
FWrite 4.5 × 10−4 as an ordinary number.[1 mark]
3
FWrite 23 × 24 as a single power of 2.[1 mark]
4
FWork out (3.2 × 104) × (2.5 × 103). Give your answer in standard form.[2 marks]
5
FSimplify 6a5 ÷ 2a2.[2 marks]
6
FWork out 4−2. Give your answer as a fraction.[2 marks]
7
FWrite these numbers in order of size, smallest first: 6.1 × 103, 8.2 × 10−2, 5.9 × 103, 1.7 × 104.[2 marks]
8
FA shape is made from a rectangle 8 cm by 3 cm and a semicircle of diameter 8 cm on one end. Work out the exact area of the shape.[3 marks]
9
FLight travels at 3 × 108 m/s. The Moon is 3.84 × 108 m from the Earth. Work out how many seconds light takes to travel from the Moon to the Earth. Give your answer to 3 significant figures in standard form.[3 marks]
10
FA cube has a volume of 343 cm3. Work out the total surface area of the cube.[3 marks]

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