Mathematics support for students at the University of Suffolk.

Work in number is important in many degrees. The following guides contain notes and videos on a variety of topics in number.

Units are what we are measuring. If we were telling someone how far away something was we wouldn't simply say something like '20'. We would also need to communicate a unit of measurement, for example '20 metres', '20 kilometres' or '20 miles'.

Scientific measurements make use of the metric system to measure quantities.

Quantity Measured | Basic Unit | Symbol |
---|---|---|

Time | Second | s |

Mass | Gram | g |

Distance | Metre | m |

Volume | Litre or Cubic Metre | L or m³ |

Force | Newton | N |

Energy | Joule | J |

Power | Watt | W |

Current | Ampere | A |

Potential Difference | Volt | V |

Resistance | Ohm | |

Pressure | Pascal | Pa |

Frequency | Hertz | Hz |

Number | Mole | mol |

When writing a large or small number of these basic units we often change the unit via a prefix, so that the number is more manageable. For example the prefix kilo (k) represents 1000 basic units. So instead of 40000 (forty thousand) metres (40000 m) we would write 40 kilometres (40 km).

Name | Size | Symbol |
---|---|---|

Peta | Trillion - 1000000000000 | P |

Terra | Billion - 1000000000 | T |

Mega | Million - 1000000 | M |

Kilo | Thousand - 1000 | k |

Hecto | Hundred - 100 | h |

Deca | Ten - 10 | da |

Unit | 1 | |

Deci | Tenth - 0.1 | d |

Centi | Hundredth - 0.01 | c |

Milli | Thousandth - 0.001 | m |

Micro | Millionth - 0.000001 | µ or mc |

Nano | Billionth - 0.000000001 | n |

It is necessary that you can convert from one metric unit to another. For example to turn 0.125 L into mL (litres into millilitres). The table below can help you to do this:

Power of 10 | 9 | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 | -1 | -2 | -3 | -4 | -5 | -6 | -7 | -8 | -9 |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|

Unit | T | M | k | h | da | unit |
d | c | m | µ | n |

The power of 10 represents the number of zeros from the metric unit table (M - Mega -> 6 -> 1000000).

To use this table to convert between metric units we multiply by powers of 10 as we move to the right and divide by powers of 10 as we move to the left.

Example:

Convert 0.125 L into mL.

L is the unit, so begin a zero. We need to move 3 to the right to get to milli (m). Multiply 0.125 by 10 and by 10 and by 10 again (10 to the power of 3). It is quicker to multiply the 0.125 by 1000.

0.125 x 1000 = 125

Therefore 0.125 L is 125 mL

Example:

Convert 250000000 µm into km.

Starting in the µ column and moving to the k column requires 9 jumps to the left. This means divide by 10 to the power of 9 (divide by 1 with 9 zeroes after, 1000000000).

250000000 ÷ 1000000000 = 0.25

Therefore 250000000 µm = 0.25 km

The link will take you to ThoughtCo and to a 10 question quiz on converting between metric units:

Percentages are measurements out of 100. This guide will look at

- Calculating percentages of an amount
- Calculating percentage increase and decreases
- Writing one number as a percentage of another

**Find 45% of 230.**

Method 1:

A percentage breaks values into a hundred parts. So 1% of 230 splits 230 into 100 parts:

230 ÷ 100 = 2.3

Now that we know that 1% is 2.3 we can multiply this by 45 to get 45%:

2.3 × 45 = 103.5

45% of 230 is 103.5

Method 2:

45% means 45 out of 100, which means 45 divided by 100.

45 ÷ 100 = 0.45

The word of means ×. Therefore:

0.45 × 230 = 103.5

**Find 8% of 7500**

7500 ÷ 100 = 75

75 × 8 = 600

8% of 7500 is 600.

OR:

0.08 × 7500 = 600

(using that 8% is 8 ÷ 100 = 0.08)

**A 750 ml bottle of wine is 13.5% alcohol. How many ml of alcohol are in the bottle of wine?**

We need to find 13.5% of 750 ml.

Find 1%:

750 ÷ 100 = 7.5 ml

Now find 13.5%:

13.5 × 7.5 = 101.25 ml

Or we can change 13.5% into a decimal by 13.5 ÷ 100 = 0.135 and multiplying:

0.135 × 750 = 101.25 ml

**Quiz - Follow this link for a quiz at tutorials point on finding percentages of an amount.**

**Increase 250 by 15%**

The question is asking to add 15% of 250 onto 250. We can do this in a couple of ways:

Method 1 - Find 15% of 250 and add this value on.

First we find 15% of 250 by one of the methods from the previous section:

250 ÷ 100 = 2.5 15 × 2.5 = 37.5 or 0.15 × 250 = 37.5

We then add this value onto the 250:

250 + 37.5 = 287.5

Method 2 - Add 15% onto 100% and find this percentage of 250.

First we add 15% onto 100%:

100% + 15% = 115%

We now find 115% of 250 using one of the methods from the previous section:

250 ÷ 100 = 2.5 115 × 2.5 = 287.5 or 1.15 × 250 = 287.5

**A person is taking 20 ml of a particular drug. It is decided, for good reason, to increase the dosage by 40%. What is the new dosage?**

This question is asking us to increase 20 ml by 40%.

40% of 20 ml is 8 ml (refer to previous section).

20 ml + 8 ml = 28 ml

The new dosage should be 28 ml.

OR

100% + 40 % = 140%

140% of 20 ml is 28 ml

**Decrease 6540 by 23%**

Here we need to take away 23% of 6540 from 6540

Method 1 - Find 23% of 6540 and take this off 6540

23% of 6540 = 1504.2

6540 - 1504.2 = 5035.8

Method 2 - Subtract 23% from 100% and find this percentage of 6540

100% - 23% = 77%

77% of 6540 = 5035.8

**A person of mass 80 kg reduces their mass by 6%. What is their new mass?**

We need to decrease 80 kg by 6%.

6% of 80 kg = 4.8 kg

80 kg - 4.8 kg = 75.2 kg

OR

100% - 6% = 94%

94% of 80 kg = 75.2 kg

The new mass of the person is 75.2 kg.

**Quiz - Follow this link for a quiz at ProProfs on calculating percentage increases and decreases**

This section will look at writing one number as a percentage out of another number and calculating the percentage change between two numbers.

**What is 48 out of 60 as a percentage?**

We need to change the 'out of 60' to 'out of 100'. We can achieve this by first dividing the 48 by 60 and then multiplying by 100.

48 ÷ 60 × 100 = 80

48 out of 60 is the same as 80 out of 100, so 48 out of 60 is 80%.

**25 ml of ethanol is diluted in 250 ml of water. What percentage of the new solution is ethanol?**

Here we have 25 ml of ethanol out of a total of 275 ml (25 + 250). We need to write 25 out of 275 as a percentage.

25 ÷ 275 × 100 = 9.1

The solution is 9.1% ethanol.

**A supermarket increases the price of a can of beans from 60p to 68p. What is the percentage increase in price?**

We can use two methods to solve this. Either find the increase, 8p, and find this as a percentage of 60p (the original price) or find 68p as a percentage of 60p and find the percentage increase from 100%.

8 out 60:

8 ÷ 60 × 100 = 13.3 - there is a 13.3% increase in price

OR

68 out of 60:

68 ÷ 60 × 100 = 113.3 - 113.3% is a 13.3% increase (113.3% - 100%)

**A car reduces in value from £24000 to £20000 over the course of a year. What is the cars depreciation as a percentage?**

The car has reduced by £4000 from its original value of £24000.

4000 out of 24000

4000 ÷ 24000 × 100 = 16.7 - the car has depreciated by 16.7%

OR

20000 out of 24000

20000 ÷ 24000 × 100 = 83.3 - this is a 16.7% drop from 100% (100% - 83.3%)

**Quiz - Follow the link for a quiz at transum on writing one number as a percentage of another.**

It starts at Level 1, please try higher levels too.

Both standard form and engineering form are a method to simplify the writing of very large or very small numbers. They both show numbers written as powers of 10.

**Writing large numbers in standard form**

The number 5400000000000000 can be written by considering it's highest place value, in this case it's represented by the 5 (5 quadrillion). We then place the decimal point after this value and round to a suitable degree of accuracy, so in this case we could use 5.4. To maintain the size of this large number we work out how many times we would need to multiply 5,4 by 10 in order for it to be equal to the original 5400000000000000. Every time we multiply 5.4 by 10 the digits move to the left (or some consider the decimal point moving to the right) with zeros filling gaps.

5.4 x 10 = 54

54 x 10 = 540

540 x 10 = 5400

Continuing this we will see that we will need to keep multiplying by 10 on 15 occasions to get to the number we require. You can get to this answer by counting how many 'jumps' from 5.4 it takes to get to 5400000000000000. Multiplying by 10 and then by 10 and so on for 15 occasions can be written mathematically as 10^{15}.

We can therefore say 5400000000000000 = 5.4 x 10^{15}

More Examples:

:- Write 24200 in standard form

Place the decimal point after the 2, so 2.42. We then need to move across by 4, so multiply by 10^{4}

24200 = 2.42 x 10^{4}

:- Write 1453632054325 in standard form.

The decimal point will go after the 1. We will round the number (it makes no sense to write in standard form if we don't). We could round to 3 significant figures (i.e. 1.45 - the next digit after the 5 is a 3, which rounds down). Counting the number of jumps we get

1453632054325 = 1.45 x 10^{12}

:- Write 60295341 in standard form.

If we use 3 significant figures we will get 6.03 ( he digit after the 602 is a 9 which rounds up). Counting the number of jumps we get 7.

60295341 = 6.03 x 10^{7}

The 2 minute youtube video by Mark Willis goes through an example of writing a large number in standard form.

Standard form is also useful for writing very small numbers. We can achieve this by multiplying by a negative power of 10. This results in the number becoming smaller (it's the same as dividing by powers of 10).

The number 0.0000000572. We first move the decimal point to after the 5 so that we have 5.72. We then need to move digits to the right (or decimal point to the left if you prefer) 8 places to get back to the original 0.0000000572. We write this number as 5.72 x 10^{-8}

More examples:

:- Write 0.024 in standard form

First we write the number as 2.4 and then count the jumps to get to 0.024, which will be 2.

0.024 = 2.4 x 10^{-2}

:- Write 0.00000007 in standard form.

First we write the number as 7 and then count the jumps, in this case 8.

0.00000007 = 7 x 10^{-8}

The 3 minute youtube video from maths3000 goes through examples of writing small numbers in standard form.

Engineering form is very similar to standard form. The power of 10 however must be a multiple of 3. This means that we will have either 1, 2 or 3 digits before the decimal point.

Consider the number 45000000. To convert this into engineering form we need to count back in threes. 45 __000__ __000__ we can do this twice, out power of 10 will therefore be 6 (2 lots of 3). The engineering form of the number is 45 x 10^{6}

More examples:

:- Convert 12400000000 into engineering form

Count back in threes 12 400 000 000 - begin with the number 12.4 and we need to jump 9 places (3 lots of 3).

12400000000 = 12.4 x 10^{9}

:- Convert 2315700000000000000 into engineering form

2 315 700 000 000 000 000 - begin with the number 2.3157 and do 6 lots of 3 jumps (18)

2315700000000000000 = 2.3157 x 10^{18}

There may be occasions where you want a number written in standard form or engineering notation back into an ordinary number. This can be achieved by multiplying by 10 (for positive powers) or dividing by 10 (for negative powers). It can be done by moving the decimal point the correct number of places.

Consider 4.87 x 10^{5}

To convert this number back into ordinary form we would multiply 4.87 by 10000 (10 to power of 5) or move the decimal 5 places to the right. So 4.87 x 10^{5 } = 487000.

When the power is negative we move the decimal point to the left.

Consider 1.6 x 10^{-7}

Now we would divide 1.6 by 10000000 or move the decimal point 7 places to the left. 0.00000016.

Most modern calculators have a button that looks like [x10* ^{x}*]. Older calculators may have a button that looks like [EXP]. Both of these buttons allow numbers to be entered into standard form. For the remainder of this section we will use [] to represent the button for your calculator.

To enter the number 8.554 x 10^{34} into the calculator we would type 8.554 [] 34 (note there is no need to type [x] or 10). On older calculators this number may be displayed as 8.554 E+34 or 8.554 +34 or 8.554 34.

This allows for calculations of large/small numbers to be quickly made on the calculator. For example:

What is 8.9 x 10^{48} / 4.65 x 10^{-12}?

Enter on calculator 8.9 [] 48 ÷ 4.65 [] (-) 12 (for the -12 your calculator will either have a negative button (minus in brackets (-)) or a ± button (to use this type 12 then press this button).

If done correctly you should get 1.91 x 10^{57}

More examples:

:- The mass of a neutron is 1.6725 x 10^{-24} and the mass of an electron is 9.109 x 10^{-28}. Find the difference in mass between the two and find the total mass of 3 neutrons and 2 electrons.

Using a calculator (where [] represents the standard form button)

1.6725 [] -24 - 9.109 [] -28 gives an answer of 1.67 x 10^{-24} g.

3 x 1.675 [] -24 + 2 x 9.109 [] -28 gives an answer of 5.03 x 10^{-24} g

Answers:

The difference is masses is 1.67 x 10^{-24} g correct to 3 significant figures

The total mass of 3 neutrons and 2 electrons is 5.03 x 10^{-24} g correct to 3 significant figures.

:- 32 g of sulphur contains 6.02 x 10^{23} atoms. What is the mass of 1 atom of sulphur?

Answer:

6.02 x 10^{23} atoms = 32 g

1 atom = 32/6.02 x 10^{23} = 5.32 x 10^{-23}

(calculator 32 ÷ 6.02 [] 23)

To enter numbers in standard form on a spreadsheet, such as Excel, you use E after the number.

3.4E+9, 1.65E-19 represent the number 3.4 x 10^{9} and 1.65 x 10^{-19} respectively.

Standard form (or scientific notation) and engineering notation are used to express very large and very small numbers. They are written as a number multiplied by a power of 10. For large numbers the power of 10 is positive and for small numbers the power of 10 is negative. The link below takes you to a 12 question quiz at proprofs.com

Notes by

Phillip Roberts

Mathematics and Statistics Adviser

University of Suffolk

2017

Email: p.roberts2@uos.ac.uk