Tag Archives: Maths

Working out values from a pie chart

Working out values from a pie chart

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This is a typical question from a Dulwich College 11+ Maths paper that asks you to work out various quantities from a pie chart.

To answer questions like this, you have to be comfortable working with fractions and know that there are 360 degrees in a circle.

So how should you start?

The first question asks for the fraction of the school children who liked tennis.

To work this out, you just need to take the following steps:

  1. Put the number of degrees showing the tennis segment over 360 to create a fraction.
  2. Simplify the fraction.

The number of degrees is 45, so the fraction is 45/360.

The first step to simplifying fractions is to see if the numerator goes into the denominator, which it does in this case: 45/45 = 1 and 360/45 = 8, so the fraction is 1/8 in its lowest terms.

(By the way, for a complete guide to simplifying fractions, just read Working with Fractions.)

The second question asks how many of the children preferred cricket.

To answer this, you should be able to learn a bit from the first question.

To work this out, you just need to take the following steps:

  1. Put the number of degrees showing the cricket segment over 360 to create a fraction.
  2. Multiply that fraction by the number of school children in the survey, which is 240.

As with the first question, you need to work out the fraction of the children in the survey you’re dealing with.

In this case, it’s 60/360 or 1/6.

To find out the number of children, you just have to multiply by 240, which is 1/6 x 240 = 40.

The final question asks you to estimate (or guess) how many of the children would say their favourite sport was football out of the whole school of 1200 pupils.

To work this out, you just need to take the following steps:

  1. Work out the number of degrees taken up by the football segment of the pie chart.
  2. Put the number of degrees over 360 to create a fraction.
  3. Multiply that fraction by the number of children in the school, which is 1200.

To work out the number of degrees, it’s easier if you spot that the first half of the pie chart is composed of just football and tennis.

There are 180 degrees in total for that half, so taking away 45 degrees for the tennis-lovers gives you 135 degrees.

This works out at a fraction of 135/360 or 3/8.

Now, we only have data for the 240 children who’ve been surveyed, but that’s why we’re being asked to estimate the answer.

We have to assume that the other kids at school share the same preferences as the ones in the survey.

If we do that, all we need to do is multiply 3/8 by 1200 to get 3/8 x 1200 = 450.

And that’s it…!

Reflecting shapes in a mirror line

Reflecting shapes in a mirror line

 

 

 

 

 

 

 

 

 

This is a typical question from a Dulwich College 11+ Maths paper, and it asks you to draw a reflection of the triangle in the mirror line shown on the chart.

Dulwich papers tend to be a bit tricky, and this is not the easiest version of this kind of reflective symmetry question.

For a start, the mirror line is drawn at 45 degrees rather than being horizontal or vertical, and it doesn’t help that the diagram is a bit ‘squashed’, which means the mirror line is actually at around 40 degrees rather than 45!

So how should you do it?

The first thing to do is to imagine that you were looking at yourself in the mirror from, say, 30cm away.

Your reflection will appear ‘in’ the mirror, but it won’t be on the surface of the mirror, will it?

It’ll actually seem to be 30cm ‘behind’ the mirror – which is exactly the same distance as you are in front of it.

That’s important, and you’ll have to use that fact when you do the question.

The basic steps are these:

  1. Plot the ‘vertices’ (or corners) of the reflected shape one by one by drawing a small cross in pencil.
  2. Join them up using a ruler and pencil.

In order to plot each corner, you need to imagine that the corner is your face and that the mirror line is the mirror.

To see your reflection, you have to be standing right in front of the mirror – looking at an angle of 90 degrees to the mirror – so to ‘see’ the reflection of a corner, you have to do the same, looking at an angle of 90 degrees to the mirror line.

The distance from your face to the mirror is the same as the distance to the spot ‘behind’ the mirror where you see your reflection.

In the same way, the distance from the corner to the mirror line is the same as the distance to the spot ‘behind’ the mirror line where the reflected point should go.

If you use the diagram at the top of this article to help you, you should be able to see that the top of the triangle is one-and-a-half diagonal squares away from the mirror line.

That means you need to go another one-and-a-half diagonal squares the other side of the mirror line (continuing in the same direction) in order to plot the reflected point.

Now repeat this for the other corners of the triangle, which are four-and-a-half and three diagonal squares away from the mirror line.

Once you’ve done that, you can join up all three points using a ruler and pencil to make the reflected triangle.

Once you get the hang of it, you may not even need to plot all the corners: if it’s a simple shape like a square or a rectangle, then you might be able to draw it from scratch.

Just make sure you label the shape if the question asks you to.

And that’s it…!

How do I know if my child will get a place?

Parents at school gates with children

Parents at school gates with children

 

 

This is the question I get asked the most as a tutor. And even if parents don’t ask it directly, I know that it’s always lurking in the background somewhere…!

School entrance exams are very stressful for pupils and parents alike, and it would be nice to be able to reassure them by giving them all the pass marks for their target schools. Unfortunately, it’s much more complicated than that.

Schools adjust the marks from Common Entrance exams at 11+ and 13+ to allow for the different ages of the children. Some will have a birthday late in the school year, which means they’ll be ‘young for their year’, and it’s generally agreed that it would be unfair to penalise those children by asking them to compete directly against other pupils who might be up to 12 months older than they are.

That’s a big difference at such a young age, so schools ‘standardise’ marks using a formula that adjusts for the relative age of each pupil. That formula also includes adjustments for various other factors, so it’s impossible to know in advance what your child’s standardised score will be.

On top of that, schools don’t often publish their pass marks, so what are pupils and their parents to do?

Well, if you can get hold of your child’s standardised score – and that’s a big if! – then you can at least check whether that score has been good enough in the past to guarantee a place at certain schools. There’s a website called elevenplusexams.co.uk that has posted what they call ‘Entry Allocation Scores & Collated Cutoffs’ for a few schools in Essex. You can find the 2019 figures here, and you can also find out the results and offers for Bishop Vesey’s Grammar School, The Schools of King Edward VI in Birmingham and Sutton Coldfield Grammar School for Girls here. If your chosen schools are not on those sites, feel free to search for them online.

I’m sorry I can’t be of more help, but at least that’s a start.

Good luck…!

 

SOHCAHTOA

SOHCAHTOA (pronounced ‘soccer-toe-uh’) is a useful ‘mnemonic’ to remember the definitions of sines, cosines and tangents. Amazingly, I was never taught this at school, so I just had to look up all the funny numbers in a big book of tables without understanding what they meant. As a result, I was always a bit confused by trigonometry until I started teaching Maths and came across SOHCAHTOA quite by accident!

The reason it’s called SOHCAHTOA is because the letters of all three equations make up that word – if you ignore the equals signs…

First of all, let’s define our terms:

  • S stands for sine (or sin)
  • O stands for the opposite side of a right-angled triangle
  • H stands for the hypotenuse of a right-angled triangle
  • C stands for cosine (or cos)
  • A stands for the adjacent side of a right-angled triangle
  • T stands for tangent (or tan)
  • O stands for the opposite side of a right-angled triangle (again)
  • A stands for the adjacent side of a right-angled triangle (again)

Sines, cosines and tangents are just the numbers you get when you divide one particular side of a right-angled triangle by another. For a given angle, they never change – however big the triangle is.

Sine = Opposite ÷ Hypotenuse

Cosine = Adjacent ÷ Hypotenuse

Tangent = Opposite ÷ Adjacent

All these ratios were discovered by Indian and Arabic mathematicians some time before the 9th Century, but you can still use them today to help you work out the length of the sides of a right-angled triangle or one of the angles.

Each of these formulas can be rearranged to make two other formulas. (If it helps, you can put the three values in a number triangle with the one in the middle at the top). Let’s take the sine formula first:

Sine = Opposite ÷ Hypotenuse means:

  • Hypotenuse = Opposite ÷ Sine
  • Opposite = Hypotenuse x Sine

As long as you know the angle and the length of the opposite side or the hypotenuse, you can work out the length of the other one of those two sides.

  • Unknown: hypotenuse
    Known: opposite and angle
    • If one of the angles of a right-angled triangle is 45° and the opposite side is 5cm, the formula for the length of the hypotenuse must be opposite ÷ sin(45°). The sine of 45° is 0.707 (to three decimal places), so hypotenuse = 5 ÷ 0.707 = 7cm (to the nearest cm).
  • Unknown: opposite
    Known: hypotenuse and angle
    • If one of the angles of a right-angled triangle is 45° and the hypotenuse is 5cm, the formula for the length of the opposite side must be hypotenuse x sin(45°). The sine of 45° is 0.707 (to three decimal places), so opposite = 5 x 0.707 = 4cm (to the nearest cm).

Equally, as long as you know the the hypotenuse and opposite side lengths, you can work out the angle by using the ‘arcsine’ or ‘inverse sine’ function on your calculator, which works out the matching angle for a given sine and is written as sin-1, eg sin(45°) = 0.707, which means sin-1(0.707) = 45°.

  • Unknown: angle
  • Known: opposite and hypotenuse
    • If the opposite side of a right-angled triangle is 4cm and the hypotenuse is 5cm, the formula for the angle must be sin-1(4÷5), or the inverse sine of 0.8. The sine of 53° (to the nearest degree) is 0.8, so the angle must be 53°.

We can do the same kind of thing with the cosine formula, except this time we’re dealing with the adjacent rather than the opposite side.

Cosine = Adjacent ÷ Hypotenuse means:

  • Hypotenuse = Adjacent ÷ Cosine
  • Adjacent = Hypotenuse x Cosine

As long as you know the angle and the length of the adjacent side or the hypotenuse, you can work out the length of the other one of those two sides.

  • Unknown: hypotenuse
    Known: adjacent and angle
    • If one of the angles of a right-angled triangle is 45° and the adjacent side is 5cm, the formula for the length of the hypotenuse must be adjacent ÷ cos(45°). The cosine of 45° is 0.707 (to three decimal places), so hypotenuse = 5 ÷ 0.707 = 7cm (to the nearest cm).
  • Unknown: adjacent
    Known: hypotenuse and angle
    • If one of the angles of a right-angled triangle is 45° and the hypotenuse is 5cm, the formula for the length of the adjacent side must be hypotenuse x cos(45°). The sine of 45° is 0.707 (to three decimal places), so adjacent = 5 x 0.707 = 4cm (to the nearest cm).

Equally, as long as you know the the hypotenuse and adjacent side lengths, you can work out the angle by using the ‘arccosine’ or ‘inverse cosine’ function on your calculator, which works out the matching angle for a given cosine and is written as cos-1, eg cos(45°) = 0.707, which means cos-1(0.707) = 45°.

  • Unknown: angle
  • Known: adjacent and hypotenuse
    • If the adjacent side of a right-angled triangle is 4cm and the hypotenuse is 5cm, the formula for the angle must be cos-1(4÷5), or the inverse cosine of 0.8. The sine of 37° (to the nearest degree) is 0.8, so the angle must be 37°.

Finally, we can do the same kind of thing with the tangent formula, except this time we’re dealing with the opposite and adjacent sides.

Tangent = Opposite ÷ Adjacent means:

  • Adjacent = Opposite ÷ Tangent
  • Opposite = Adjacent x Tangent

As long as you know the angle and the length of the opposite or adjacent side, you can work out the length of the other one of those two sides.

  • Unknown: adjacent
    Known: opposite and angle
    • If one of the angles of a right-angled triangle is 45° and the opposite side is 5cm, the formula for the length of the adjacent must be opposite ÷ tan(45°). The tangent of 45° is 1, so adjacent = 5 ÷ 1 = 5cm.
  • Unknown: opposite
    Known: adjacent and angle
    • If one of the angles of a right-angled triangle is 45° and the adjacent side is 5cm, the formula for the length of the opposite side must be adjacent x tan(45°). The tangent of 45° is 1, so opposite = 5 x 1 = 5cm.

Equally, as long as you know the the opposite and adjacent side lengths, you can work out the angle by using the ‘arctangent’ or ‘inverse tangent’ function on your calculator, which works out the matching angle for a given tangent and is written as tan-1, eg tan(45°) = 0.707, which means tan-1(0.707) = 45°.

  • Unknown: angle
  • Known: adjacent and hypotenuse
    • If the adjacent side of a right-angled triangle is 5cm and the hypotenuse is 5cm, the formula for the angle must be tan-1(5÷5), or the inverse tangent of 1. The tangent of 45° is 1, so the angle must be 45°.

Long multiplication

You can use short multiplication if you’re multiplying one number by another that’s in your times tables (up to 12). However, if you want to multiply by a higher number, you need to use long multiplication.

  • Write down the numbers one on top of the other with the smaller number on the bottom and a times sign on the left (just as you would normally), then draw three lines underneath to hold three rows of numbers.
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  • Multiply the top number by the last digit of the bottom number as you would normally.
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  • Write a zero at the end of the next answer line (to show that you’re multiplying by tens now rather than units).
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  • Multiply the top number by the next digit of the bottom number, starting to the left of the zero you’ve just added.
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  • Add the two answer lines together to get the final answer.
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Notes:

  • Some people write the tens they’ve carried right at the top of the sum, but that can get very confusing with three lines of answers!
  • Don’t forget to add the zero to the second line of your answer. If it helps, you can try writing it down as soon as you set out the sum (and before you’ve even worked anything out).
  • At 11+ level, long multiplication will generally be a three-digit number multiplied by a two-digit number, but the method will work for any two numbers, so don’t worry. If you have to multiply two three-digit numbers, say, you’ll just have to add another line to your answer.

Sample questions:

Have a go at these questions. Make sure you show your working – just as you’d have to do in an exam.

  1. 216 x 43
  2. 17 x 423
  3. 23 x 648
  4. 782 x 28
  5. 127 x 92

How to add, subtract, multiply and divide

The most important things you need to do in Maths are to add, subtract, divide and multiply. If you’re doing an entrance exam, and there’s more than one mark for a question, it generally means that you have to show your working. Even if it’s easy enough to do in your head, you still have to write down the sum on paper. That way, the examiner knows that you didn’t just guess!

Here are the basic operations:

Addition

The standard way to add numbers is the ‘column method’.

  • Write down the numbers one on top of the other (however many there are), with two lines under them and a plus sign on the left.
  • Add the first column of numbers on the right and put the answer between the lines.
    • If the total is more than 9, ‘carry’ the tens by putting that number in small handwriting under the next space on the answer line.
  • Add the next column of numbers working from the right and put the answer between the lines, adding any numbers below the line that have been carried.
  • If you get to the final column of numbers and the total is more than 9, you can write both digits on the answer line.
  • If you have more than two columns of numbers and the total is more than 9, you’ll have to ‘carry’ any tens again by putting that number in small handwriting under the next space on the answer line.
  • You can then finish off as normal.

Notes:

  • You don’t need the second line if you don’t want to use it.
  • You can also choose to put the carried numbers above the top line of the sum, but that gets a bit messy if you’re doing long multiplication, so it’s best to get into the habit of using this method.

Sample questions:

Have a go at these questions. Don’t just do them in your head. That’s too easy! Make sure you show your working – just as you’d have to do in an exam.

  1. 8 + 5
  2. 17 + 12
  3. 23 + 19
  4. 77 + 88
  5. 127 + 899

Subtraction

The standard way to subtract one number from another is again the ‘column method’, but this time it’s slightly different. For a start, you can only use this method with two numbers (not three or more), and you can’t use it for negative numbers.

  • Write down the two numbers one on top of the other, with the bigger one on top, the usual two lines under them and a minus sign on the left.
  • Working from the right, take away the first digit in the second number from the first digit in the first and write the answer on the answer line.
    • If you can’t do it because the digit on the top row is too small, you’ll have to ‘borrow’ a 10 from the digit in the next column.
      • Place a 1 above and to the left of the top right-hand digit to make a new number, in this case 12.
      • Cross out the digit you’re borrowing from, subtract 1 and write the new digit above and to the left of the old one.
      • You can now subtract as normal, so 12 – 7 = 5 in this case.
  • Working from the right, subtract the next digit in the bottom number from the next digit in the top number and put the answer between the lines.
  • Repeat this step until you’ve finished the sum.
    • Note that in this case you have to borrow 1 from the 2, leaving 1, and then borrow 1 from the 4, writing it next to the 1 so it makes 11. It may look like you’re borrowing 11, but you’re not. You’ve just had to write the two 1s next to each other.

If you can’t borrow from a digit because it’s a zero, just cross it out, write 9 above and to the left and borrow from the next digit to the left. If that’s a zero, too, just do the same again until you reach one that’s not zero.

Notes:

  • You don’t need the second line if you don’t want to use it.
  • If the answer to the sum in the last column on the left is zero, you don’t need to write it down, so your answer should be 17, say, not 017.
  • You don’t need to put commas in numbers that are more than 1,000.
  • You could cross out the numbers from top left to bottom right instead, but that leaves less room to write any little numbers above and to the left (where they have to go), so it’s best to get into the habit of using this method.

Sample questions:

Have a go at these questions. Don’t just do them in your head. That’s too easy! Make sure you show your working – just as you’d have to do in an exam.

  1. 8 – 5
  2. 17 – 12
  3. 43 – 19
  4. 770 – 681
  5. 107 – 89

Multiplication (or short multiplication)

This is short multiplication, which is meant for multiplying one number by another that’s in our times tables (up to 12). If you want to multiply by a higher number, you need to use long multiplication.

  • Write down the numbers one on top of the other with the single-digit number on the bottom, two lines underneath and a times sign on the left.
  • Multiply the last digit of the top number by the bottom number and put the answer between the lines.
    • If the total is more than 9, ‘carry’ the tens by putting that number in small handwriting under the next space on the answer line.
  • Working from the right, multiply the next digit of the top number by the bottom number, adding any number below the answer line.
    • As with addition, if you get to the final column of numbers and the total is more than 9, you can write both digits on the answer line.

Notes:

  • You don’t need the second line if you don’t want to use it.
  • You can also choose to put the carried numbers above the top line of the sum, but that gets a bit messy if you’re doing long multiplication, so it’s best to get into the habit of using this method.

Sample questions:

Have a go at these questions. Don’t just do them in your head. That’s too easy! Make sure you show your working – just as you’d have to do in an exam.

  1. 21 x 3
  2. 17 x 4
  3. 23 x 6
  4. 77 x 8
  5. 127 x 9

Division (or short division, or the ‘bus stop’ method)

This is short division, which is meant for dividing one number by another that’s in your times tables (up to 12). If you want to divide by a higher number, you need to use long division (see my article here). It’s called the ‘bus stop’ method because the two lines look a bit like the area where a bus pulls in at a bus stop.

  • Write down the number you’re dividing (the ‘dividend’), draw the ‘bus stop’ shape around it so that all the digits are covered and then write the number you’re dividing by (the ‘divisor’) on the left.
  • Try to divide the first digit of the dividend by the divisor. If it goes in exactly, write the answer on the answer line above the first digit of the dividend.
  • If it goes in, but there’s a remainder, write the answer on the answer line above the first digit of the dividend and then write the remainder above and to the left of the next digit in the dividend.
  • If it doesn’t go, then make a number out of the first two digits of the dividend and divide that number by the divisor, adding any remainder above and to the left of the next digit.
  • Repeat this process for each of the remaining digits, using any remainders to make a new number with the next digit.
  • If you divide one number by another in the middle of the dividend and it doesn’t go, then just put a zero on the answer line and combine the digit with the next one.

Notes:

  • If you have a remainder at the end of the sum, you can either show it as a remainder or you can put a decimal point above and below the line, add a zero to the dividend and carry on until you have no remainder left.
    • If the remainder keeps going, it’s likely to repeat the same digits over and over again. This is called a ‘recurring decimal’. Once you spot the pattern, you can stop doing the sum. Just put a dot over the digit that’s repeating or – if there’s more than one – put a dot over the first and last digit in the pattern.
  • If your handwriting is a bit messy, make sure you make the numbers quite large with a bit of space between them so that you can fit everything in!

Sample questions:

Have a go at these questions. Don’t just do them in your head. That’s too easy! Make sure you show your working – just as you’d have to do in an exam.

  1. 36 ÷ 3
  2. 172 ÷ 4
  3. 222 ÷ 6
  4. 816 ÷ 8
  5. 126 ÷ 9

Past papers

 

 

Here is a selection of over 5,000 past papers sorted by age group, subject, school and year, together with around 1,000 mark schemes. You’ll find links to over 100 other sources of past papers at the foot of the page.

The first few KS1 SATs papers are available free of charge. To join nearly 2,000 other subscribers in gaining access to the rest, please sign up on the Subscribe page or click the button below. A 12-month plan costs just £12.99 with no automatic renewal.

“Hi, Nick,

Long-time user of your priceless past paper collection here. Just wanted to say thank you for putting this all together — it truly is a phenomenal resource for assessment. And as a tutor/teacher/human myself, I’ve used the past papers as a fall-back more times than I’d necessarily like to admit!

Keep up the brilliant work on all fronts,

Sam”

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If you have specific needs and want to know in advance if papers will be available from a certain school in a certain year for a certain subject, then feel free to leave a comment or email me at info@nickdale.me, and I’ll check for you.

Once you’ve subscribed, finding papers is easy:

  • If you’re looking for papers for a particular level, subject or school, just search for it by name (using Ctrl-F or Cmd-F), eg 11+, English or Latymer.
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Papers are available from the following schools, as well as other organisations such as the Independent Schools Examinations Board (ISEB):

  • Aldenham, Alleyn’s
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  • Caterham School, Chigwell School, Christ’s Hospital, City of London Freemen’s School, City of London School, City of London School for Girls, Cheadle Hulme School, Colfe’s School, The Crossley Heath School
  • Dame Alice Owen’s School, Dulwich College
  • Eltham College, Emanuel School, Epsom College, Eton College
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  • Godolphin & Latymer, Greenshaw High School
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  • Nonsuch High School for Girls, North Halifax Grammar School, North London Collegiate School
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Key Stage 1 SATs

KS1 English grammar, punctuation and spelling Paper 2- questions (2019)
KS1 English grammar, punctuation and spelling test mark schemes (2019)
KS1 English reading Paper 1- reading prompt and answer booklet (2019)
KS1 English reading Paper 2- reading answer booklet (2019)
KS1 English reading Paper 2- reading booklet (2019)
KS1 English reading test mark schemes (2019)
KS1 English grammar, punctuation and spelling Paper 1- spelling (2018)
KS1 English grammar, punctuation and spelling Paper 2- questions (2018)
KS1 English grammar, punctuation and spelling test mark schemes (2018)
KS1 English reading Paper 1- reading prompt and answer booklet (2018)
KS1 English reading Paper 2- reading answer booklet (2018)
KS1 English reading Paper 2- reading booklet (2018)
KS1 English reading test mark schemes Paper 1- reading prompt and answer booklet and Paper 2- reading answer booklet (2018)
KS1 English grammar, punctuation and spelling Paper 1- spelling (2017)
KS1 English grammar, punctuation and spelling Paper 2- questions (2017)
KS1 English grammar, punctuation and spelling test mark schemes Paper 1- spelling and Paper 2- questions (2017)
KS1 English reading Paper 1- reading prompt and answer booklet (2017)
KS1 English reading Paper 2- reading answer booklet (2017)
KS1 English reading Paper 2- reading booklet (2017)
KS1 English reading test mark schemes Paper 1- reading prompt and answer booklet and Paper 2- reading answer booklet (2017)
KS1 English Spelling Test – Playtime (QCA, 2003)
KS1 English Spelling Test – Making Soup (QCA, 2004)
KS1 English – Sunflowers (QCA, 2003)
KS1 English – Friends and Playtime (QCA, 2004)

KS1 Maths Paper 1- arithmetic (2019)
KS1 Maths Paper 2- reasoning (2019)
KS1 Maths test mark schemes (2019)
KS1 Maths Paper 1- arithmetic (2018)
KS1 Maths Paper 2- reasoning (2018)
KS1 Maths- mark schemes (2018)
KS1 Maths Paper 1- arithmetic (2017)
KS1 Maths Paper 2- reasoning (2017)
KS1 Maths Paper 1- arithmetic (2015)
KS1 Maths Paper 2- reasoning (2015)
KS1 Maths test mark schemes Paper 1- arithmetic and Paper 2- reasoning (2017)
KS1 Maths Test Paper B (QCA, 2004)
KS1 Maths Test Paper B (QCA, 2003)
KS1 Maths Test Paper A (QCA, 2004)
KS1 Maths Test Paper A (QCA, 2003)
KS1 Maths Test Paper (QCA, 2002)
KS1 Maths Test Paper (QCA, 2001)
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Circle theorems

This article explains circle theorems, including tangents, sectors, angles and proofs (with thanks to Revision Maths).

Isosceles Triangle

Two Radii and a chord make an isosceles triangle.

Perpendicular Chord Bisection

The perpendicular from the centre of a circle to a chord will always bisect the chord (split it into two equal lengths).

Angles Subtended on the Same Arc

Angles subtended on the same arc

Angles formed from two points on the circumference are equal to other angles, in the same arc, formed from those two points.

Angle in a Semi-Circle

angle in a semi-circle

Angles formed by drawing lines from the ends of the diameter of a circle to its circumference form a right angle. So c is a right angle.

Proof

We can split the triangle in two by drawing a line from the centre of the circle to the point on the circumference our triangle touches.

Divide the triangle in two

We know that each of the lines which is a radius of the circle (the green lines) are the same length. Therefore each of the two triangles is isosceles and has a pair of equal angles.

Two isosceles triangles

But all of these angles together must add up to 180°, since they are the angles of the original big triangle.

Therefore x + y + x + y = 180, in other words 2(x + y) = 180.
and so x + y = 90. But x + y is the size of the angle we wanted to find.

Tangents

A tangent to a circle is a straight line which touches the circle at only one point (so it does not cross the circle- it just touches it).

A tangent to a circle forms a right angle with the circle’s radius, at the point of contact of the tangent.

angle with a tangent

Also, if two tangents are drawn on a circle and they cross, the lengths of the two tangents (from the point where they touch the circle to the point where they cross) will be the same.

Tangents from an external point are equal in length

Angle at the Centre

Angle at the centre

The angle formed at the centre of the circle by lines originating from two points on the circle’s circumference is double the angle formed on the circumference of the circle by lines originating from the same points. i.e. a = 2b.

Proof

You might have to be able to prove this fact:

proof diagram 1

OA = OX since both of these are equal to the radius of the circle. The triangle AOX is therefore isosceles and so ∠OXA = a
Similarly, ∠OXB = b

proof diagram 2

Since the angles in a triangle add up to 180, we know that ∠XOA = 180 – 2a
Similarly, ∠BOX = 180 – 2b
Since the angles around a point add up to 360, we have that ∠AOB = 360 – ∠XOA – ∠BOX
= 360 – (180 – 2a) – (180 – 2b)
= 2a + 2b = 2(a + b) = 2 ∠AXB

Alternate Segment Theorem

Alternate segment theorem

This diagram shows the alternate segment theorem. In short, the red angles are equal to each other and the green angles are equal to each other.

Proof

You may have to be able to prove the alternate segment theorem:

proof of alternate segment theorem

We use facts about related angles

A tangent makes an angle of 90 degrees with the radius of a circle, so we know that ∠OAC + x = 90.
The angle in a semi-circle is 90, so ∠BCA = 90.
The angles in a triangle add up to 180, so ∠BCA + ∠OAC + y = 180
Therefore 90 + ∠OAC + y = 180 and so ∠OAC + y = 90
But OAC + x = 90, so ∠OAC + x = ∠OAC + y
Hence x = y

Cyclic Quadrilaterals

cyclic quadrilateral is a four-sided figure in a circle, with each vertex (corner) of the quadrilateral touching the circumference of the circle. The opposite angles of such a quadrilateral add up to 180 degrees.

Area of Sector and Arc Length

A sector

If the radius of the circle is r,
Area of sector = πr2 × A/360
Arc length = 2πr × A/360

In other words, area of sector = area of circle × A/360
arc length = circumference of circle × A/360

Long division

Long division is on the syllabus for both 11+ and 13+ exams, so it’s important to know when and how to do it.

The basic idea is that it’s tricky to do short division when the number you’re dividing by (the ‘divisor’) is outside your times tables, ie more than 12. Using long division makes it easier by including a way of calculating the remainder using a proper subtraction sum. It also makes it neater because you don’t have to try and squeeze two-digit remainders in between the digits underneath the answer line (the ‘dividend’).

So how does it work? Well, the only difference involves the remainder. In normal short division, you work it out in your head and put it above and to the left of the next digit in the dividend. In long division, you work out the multiple of the divisor, write it down under the dividend and subtract one from the other to get the remainder. You then pull down the next digit of the dividend and put it on the end of the remainder, repeating as necessary.

To take the example at the top of the page, what is 522 divided by 18?

  1. How many 18s in 5?
  2. It doesn’t go
  3. How many 18s in 52?
  4. Two (write 2 on the answer line, and write 36 under the dividend with a line beneath it)
  5. What’s 52 – 36?
  6. 16 (write it on the next line)
  7. Pull down the next digit from the dividend (write it after the 16)
  8. How many 18s in 162?
  9. Nine (write it on the answer line, giving 29 as the answer, or ‘quotient’)

That’s the basic method, but here are a couple of tips to help you out.

The first is that you can make life easier for yourself by guessing round numbers. Working with numbers outside your times tables is tricky, so you can use ‘trial and error’ to come up with the right multiple of the divisor by trying ‘easy’ ones like 5 or 10. If it’s too big or too small, you can simply try again with a smaller or bigger number.

The second is that you can often divide the divisor by two to force it back into your times tables. Why divide by 18 when you can simply divide by nine and halve the result? You just have to be careful that you only deal in even multiples, eg 52 ÷ 18 is tricky, but the nearest even multiple of 9 is 4 (as 5 is an odd number and 6 x 9 = 54, which is too much), so the answer must be 4.

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Maths trick

maths trick

Here’s a Maths trick a friend of mine saw on QI. Who knows? It might make addition and subtraction just a little bit more fun!

  1. Find a book and write down the ninth word on p108, eg ‘becoming’.
  2. Ask someone to write down a number with three different digits in it, eg 321.
  3. Ask him to reverse it and take the smaller number away from the larger, eg 321 – 123 = 198.
  4. Ask him to write down the answer and, again, reverse it, eg 891.
  5. Ask him to add the two new numbers together, eg 198 + 891 = 1089.
  6. Ask him to find the ninth word on p108 of the book, eg ‘becoming’.
  7. Reveal the same word you wrote down earlier!

The trick is that the answer is always going to be 1089, whatever the number you first think of, so it should work every time – unless there’s a problem with the Maths!

Simultaneous equations

Simultaneous equations help you work out two variables at once.

download

Why do we have simultaneous equations? Well, there are two ways of looking at it.

  1. The first is that it solves a problem that seems insoluble: how do you work out two variables at once? For example, if x + y = 10, what are x and y? That’s an impossible question because x and y could literally be anything. If x was 2, then y would be 8, but if x was 100, then y would be -90, but if x was 0.5, then y would be 9.5 and so on. Simultaneous equations help us solve that problem by providing more data. Yes, we still can’t solve each equation individually, but having both of them allows us to solve for one variable and then the other.
  2. The second way of looking at simultaneous equations is to imagine that they describe two lines that meet. The x and y values are obviously different as you move along both lines, but they are identical at the point where they meet, and that is the answer to the question.

The next question is obviously ‘How do we solve simultaneous equations?’ The answer is simple in theory: you just have to add both equations together to eliminate one of the variables, at which point you can work out the second one and then put it back into one of the original equations to work out the first variable. However, it gets more and more complicated as the numbers get less and less ‘convenient’, so let’s take three examples to illustrate the three different techniques you need to know.

Simple addition and subtraction

The first step in solving simultaneous equations is to try and eliminate one of the variables by adding or subtracting them, but you can only do that if the number of the variable is the same in both. In theory, you could choose the first or the second term, but I find the one in the middle is the easiest, eg

4x + 2y = 10

16x – 2y = 10

Here, the number of the variables in the middle of the equations is the same, so adding them together will make them disappear:

20x = 20

It’s then simple to divide both sides by 20 to work out x:

x = 1

Once you have one variable, you can simply plug it back into one of the original equations to work out the other one, eg

4x + 2y = 10

4 x 1 + 2y = 10

4 + 2y = 10

2y = 6

y = 3

Answer: x = 1, y = 3

Multiplying one equation

If the number of variables in the middle is not the same, but one is a factor of the other, try multiplying one equation by whatever number is needed to make the number of the variables match, eg

4x + 2y = 10

7x + y = 10

Multiplying the second equation by 2 means the number of the y’s is the same:

4x + 2y = 10

14x + 2y = 20

The rest of the procedure is exactly the same, only this time we have to subtract rather than add the equations to begin with:

10x = 10

x = 1

The next part is exactly the same as the first example as we simply plug in x to find y:

4x + 2y = 10

4 x 1 + 2y = 10

4 + 2y = 10

2y = 6

y = 3

Answer: x = 1, y = 3

Multiplying both equations

If the number of variables in the middle is not the same, but neither is a factor of the other, find the lowest common multiple and multiply the two equations by whatever numbers are needed to reach it, eg

4x + 2y = 10

x + 3y = 10

The lowest common multiple of 2 and 3 is 6, which means we need to multiply the first equation by 3:

12x + 6y = 30

…and the second by 2:

2x + 6y = 20

As the number of variables in the middle is now the same, we can carry on as before by subtracting one from the other in order to find x:

10x = 10

x = 1

Again, the final part of the technique is exactly the same as we plug x into the first of the original equations:

4x + 2y = 10

4 x 1 + 2y = 10

4 + 2y = 10

2y = 6

y = 3

Answer: x = 1, y = 3

Practice questions

Job done! Now, here are a few practice questions to help you learn the rules. Find x and y in the following pairs of simultaneous equations:

  1. 2x + 4y = 16
    4x – 4y = 8
  2. 3x + 2y = 12
    5x + 2y = 16
  3. 12x – 4y = 28
    3x – 2y = 5
  4. 2x – y = 12
    3x – 2y = 17
  5. 4x + 3y = 24
    5x – 2y = 7
  6. 4x + 3y = 31
    5x + 4y = 40
  7. x + 4y = 23
    5x – 2y = 5
  8. 4x + 3y = 37
    2x – 3y = -13
  9. 2x + 4y = 16
    3x – 5y = -9
  10. 2x + 4y = 20
    3x + 3y = 21

How to become a private tutor

Adrian-Beckett_09032013_035

I’ve talked to a few people who wanted to become private tutors, so I thought I’d write down a few tips for anyone who’s interested.

How did I start out?

I started as a private tutor quite by accident. It was 2009, and I was finding it hard to get work as a freelance management consultant when I happened to read an article in the Telegraph called 10 Ways to Beat the Recession. The author mentioned a few ways of earning some extra cash, including becoming an extra on film sets – which I was already doing – and working as a private tutor. I’d never done any proper teaching before, although I was a golf coach, and I’d coached skiing a few times in the Alps, but I thought I’d sign up with a couple of agencies and see what happened. Within a week, I had two clients, and I’ve never looked back since!

What qualifications do I need?

The first and most important thing to say is that you don’t need any teaching qualifications! Yes, that’s right. You don’t need a PGCE, and you don’t need to have done any training as a teacher. As a private tutor, you are just that – private – so you don’t have to jump through all the Government hoops that a teacher in a state school would have to do. Obviously, potential clients want the best person to teach their child, so you need to show some sort of academic record, but that can be as little as a degree in English – which is what I had when I started. Admittedly, I went to Oxford, which probably counts for a lot with Russian billionaires (!), but you don’t need to have an Oxbridge degree to become a tutor. Far from it. However, what you probably will need is a criminal records check. This is just a piece of paper that certifies you haven’t been convicted of a criminal offence and was often known as a ‘CRB check’, although it’s now officially called an Enhanced Certificate from the Disclosure and Barring Service, or ‘DBS check’. You can’t apply for an ‘enhanced certificate’ yourself, but your tuition agency can help you. In fact, they may require you to have one and even to renew it every year or two. It costs around £18 and can take up to three months to arrive, so it’s worth applying as early as possible. Some agencies may charge up to £80 to make the application on your behalf, so be careful! You can find further information here.

What subjects can I teach?

You can teach whatever you like! Agencies will just ask you which subjects you offer and at what level, so you have complete freedom to choose. I focus on English and Maths, which are the most popular subjects, but that’s mostly led by demand from clients. They are the main subjects at 11+ level, so that’s what most people are looking for help with.

What age children can I teach?

Again, the choice is yours. I’ve taught students from as young as five to as old as 75, but the peak demand is at 11+ level, when the children are around 10 years old. I make it a rule that I’ll only teach a subject to a level that I’ve reached myself, such as GCSE or A-level, but clients sometimes take you by surprise. When I turned up to teach what I thought was going to be English to two boys, the nanny suddenly asked me to do Latin instead. When I said I hadn’t done any Latin since I was 15, she just said, “Oh, you’ll be fine…!”

What preparation do I need to do?

  • Research. One of the big attractions of tutoring for me is that the work is very enjoyable. I like teaching, and I like spending time with children, so it’s the perfect combination! The reason I stopped work as a management consultant was the constant stress, the persistent worry that I wasn’t up to the job, but teaching 10-year-olds never makes me feel like that. Whether it’s English or Maths, I’m confident in my ability to teach and never worry about being asked an impossible question. However, that doesn’t mean you can walk into your first lesson without doing any preparation at all. In my case, I wanted to teach English, so I needed to find out what kind of questions cropped up in 11+ and 13+ entrance exams and come up with a good method of answering them. Once I’d done that, I was ready. Maths was a bit easier, but I still looked through a few papers to make sure there was no risk of being blind-sided by something I’d forgotten how to do or had never studied. Whatever the subject you’re offering, I suggest you do the same.
  • Past papers. The other thing I needed to do was to find past papers to give to my pupils. That was a bit tricky in the early days until a kind parent gave me a collection of photocopied exams. After that, I carried a couple around with me to take to lessons, but it wasn’t a great solution, so I decided to create a website – this one. Over time, I collected dozens of past papers and wrote various articles on how to do different kinds of question in Maths, English and French. Now, I don’t have to carry around anything with me or spend time dictating notes. I can simply ask my pupils to look it up online. Setting up a website is pretty easy using WordPress or something similar, but you should feel free to use the resources on my past papers tab if you don’t want to go to the trouble yourself, and all my articles are available for free if you need them. The main ones I use for English are about doing comprehensions and writing stories, but there are plenty more. The website proved unexpectedly popular, and I had over 28,000 visitors last year! The other advantage is that it generated enough business for me not to need agencies any more. That means I can charge what I like, I don’t have to pay any commission, and I can have a direct relationship with all my clients without anybody acting as an intermediary – and often just getting in the way!
  • Business cards. I know it sounds a bit old-fashioned, but having business cards is very useful. If you’re just starting out, nobody knows your name, so paying a few quid to market your services is one of the best investments you can make. You never know when people will tell you they’re looking for a tutor, and it’s the easiest thing in the world to give them a business card. Even if you don’t have a website, it will at least tell them how to reach you, and you should get a lot more clients out of it.

How can I find work?

Tuition agencies are the best place to start, but there are different kinds. Some are online and simply require you to fill out a form for them to check and vet, but others ask you to go through an interview, either over the phone or in person. Either way, you need to put together a tailored CV that shows off your academic achievements and highlights any teaching experience you’ve had. This may not be very much at the beginning, but you simply need to show enough potential to get you through the door. Once you’ve shown enough aptitude and commitment to get accepted by a few agencies, you’ll rapidly build up your experience on the job.

Here is a list of the tuition agencies I’ve been in touch with, together with contact details where available. I’m based in London, so there is obviously a geographical bias there, but some of the agencies such as Fleet Tutors offer national coverage, and you can always search online for others in your local area.

Name

Email

Telephone

Website

A-Star Tuitionastartuition@btinternet.com01772 814739astartuition.com
Approved Tutorsapprovedtutors.co.uk
Athena Tuitionathenatuition.co.uk/
Beacon Tutorsinfo@beacontutors.co.uk020 8983 2158beacontutors.co.uk/
Bespoke Tuitionemma@bespoketuition.com07732 371880bespoketuition.com
Bigfoot Tutorstutors@bigfoottutors.com020 7729 9004bigfoottutors.com
Bright Young Things07702 019194brightyoungthingstuition.co.uk
Dulwich Tutorsinfo@dulwichtutors.com020 8653 3502dulwichtutors.com
Enjoy Educationkate@enjoyeducation.co.ukenjoyeducation.co.uk
Exam Confidence
First Tutorsfirsttutors.co.uk
Fleet Tutors0845 644 5452fleet-tutors.co.uk
Gabbitasgabbitas.com
Greater London Tutors020 7727 5599greaterlondontutors.com
Harrison Allenharrisonallen.co.uk
Holland Park Tuitionrecruitment@hollandparktuition.com020 7034 0800hollandparktuition.com
IPS Tutorsinfo@ipstutors.co.uk01509 265623ipstutors.co.uk
Ivy Educationivyeducation.co.uk
Kensington & Chelsea Tutorstutors@kctutors.co.uk020 7584 7987kctutors.co.uk
Keystone Tutorsenquiries@keystonetutors.com020 7351 5908keystonetutors.com
Kings Tutorsemily@kingstutors.co.ukkingstutors.co.uk
Knightsbridge Tutors07890 521390knightsbridgetutors.co.uk
Laidlaw Educationlaidlaweducation.co.uk
Mentor & Sonsandrei@mentorandsons.com07861 680377mentorandsons.com
Osborne Cawkwellenquiries@oc-ec.com020 7584 5355oc-ec.com
Personal Tutorsadmin@personal-tutor.co.ukpersonal-tutors.co.uk
Russell Education Groupjoe@russelleducationgroup.comn/a
Search Tutorssearchtutors.co.uk
Select My Tutorinfo@selectmytutor.co.ukselectmytutor.co.uk
SGA Educations@sga-education.comsga-education.com
Simply Learning Tuitionsimplylearningtuition.co.uk
The Tutor Pagesthetutorpages.com
Top Tutors020 8349 2148toptutors.co.uk
Tutor Houseinfo@tutorhouse.co.uk020 7381 6253tutorhouse.co.uk/
Tutor Hunttutorhunt.com
Tutorfairtutorfair.com
Tutors Internationaltutors-international.net
UK Tutorsuktutors.com
Westminster Tutorsexams@westminstertutors.co.uk020 7584 1288westminstertutors.co.uk
William Clarence Educationsteve@williamclarence.com020 7412 8988williamclarence.com
Winterwood winterwoodtutors.co.uk

That’s obviously a long list, but, to give you an idea, I earned the most from Adrian Beckett (teacher training), Bespoke Tuition, Bonas MacFarlane, Harrison Allen, Keystone Tutors, Mentor & Sons, Personal Tutors and Shawcross Bligh.

Once you’ve been accepted by and started working for a few agencies, you’ll soon see the differences. Some offer higher rates, some the option to set your own rates, some provide a lot of work, some offer the best prospects of jobs abroad. It all depends what you’re looking for.

Where will the lessons take place?

When I first started tutoring, I had to cycle to all my clients. I put a limit of half an hour on my travel time, but it still took a lot of time and effort to get to my pupils. Fortunately, I’m now able to teach at my home, either in person or online using Skype and an electronic whiteboard, which means my effective hourly rate has gone up enormously. Travel is still a little bit of a problem for most tutors, though, and I certainly couldn’t have reached my pupils without having a bicycle. I didn’t have a car, and public transport wasn’t really an option in most cases. You just have to decide how far you’re prepared to go: the further it is, the more business you’ll get, but the longer it’ll take to get there and therefore the lower your effective hourly rate.

The other possibility, of course, is teaching abroad. I’ve been lucky enough to go on teaching assignments in Belarus, Greece, Hong Kong, Kenya, Russia, Switzerland and Turkey, and it’s a great way to see the world. The clients can sometimes be a little bit difficult, and the children can sometimes behave like spoiled brats (!), but staying with a great client in a sunny getaway overseas can be a wonderful experience. The only reason I don’t apply for more foreign postings is that I don’t want to let down my existing clients – going away for three weeks just before the 11+ exams in January would NOT go down well!

When will the lessons take place?

If you’re teaching children, lessons will usually be in the after-school slot between 1600 and 2000 or at weekends. That does limit the amount of hours you can teach, but it’s up to you how much you want to work. I used to work seven days a week, but I eventually gave myself a day off and then another, so I now work Sundays to Thursdays with Friday and Saturday off. During the holidays, you lose a lot of regular clients when they disappear to the Maldives or somewhere for six weeks (!), but others might ask for an intensive sequence of lessons to take advantage of the extra time available, and there’s obviously a greater chance of a foreign assignment. All that means that the work is very seasonal, so you should expect your earnings to go up and down a bit and plan your finances accordingly.

What should I do during the lesson?

I generally teach from past papers, so I ask pupils to do a past paper for their homework and then mark it during the following lesson. ‘Marking’ means marking the questions, obviously, but it also means ‘filling in the gaps’ in the pupil’s knowledge. If he or she is obviously struggling with something, it’s worth spending a few minutes explaining the topic and asking a few practice questions. I’ve written a few articles on common problem areas in English and Maths, such as commas and negative numbers, so I often go through one of those and ask the pupil’s parents to print it out and put it in a binder. After a few weeks, that collection of notes gradually turns into a ready-made revision guide for the exams.

If the parents want you to work on specific topics, that’s also possible. For example, one mother wanted to help her son with ratios, so she printed out dozens of past papers and circled the ratio questions for him to do. He soon got the knack!

I approach English in a slightly different way to begin with. There are two types of question in the 11+, comprehensions and creative writing, so I generally spend the first lesson teaching pupils how to do one of those. I go through my article on the subject online and then ask them to answer a practice question by following the procedure I’ve outlined. They usually finish it off for their homework. After a few weeks of stories or comprehensions, I’ll switch to the other topic and do the same with that. I also ask pupils to write down any new words or words they get wrong in a vocabulary book because building vocabulary is very important for any type of English exam (and also for Verbal Reasoning). I ask them to fold the pages over in the middle so that they can put the words on the left and the meanings on the right (if necessary). Every few weeks, I can then give them a spelling test. If they can spell the words correctly and tell me what they mean, they can tick them off in their vocab book. Once they’ve ticked off a whole page of words, they can tick that off, too! I usually try to reinforce the learning of words by asking pupils to tell me a story using as many words as possible from their spelling test. It can be a familiar fairy story or something they make up, but it just helps to move the words from the ‘passive’ memory to the ‘active memory’, meaning that they actually know how to use them themselves rather than just understand them when they see them on the page.

What homework should I set?

Most children who have private lessons have pretty busy schedules, so I don’t want to overburden them. I generally set one exercise that takes around 30-45 minutes. That might be a Maths paper or an English comprehension or story, but it obviously depends on the subject and the level. Just make sure that the student writes down what needs to be done – a lot of them forget! You should also make a note in your diary yourself, just so that you can check at the start of the next lesson if the work has been done.

What feedback should I give the parents?

I generally have a quick chat with the mother or father (or nanny) after the lesson to report on what we did during the lesson, what problems the child had and what homework I’ve set. This is also a good time to make any changes to the schedule, for instance if the family goes on holiday. If that’s not possible, I’ll email the client with a ‘lesson report’. Some agencies such as Bonas MacFarlane make this a part of their timesheet system.

How much will I get paid?

When I first started, I had absolutely no idea how much I was worth, and I ended up charging only £10 an hour, which is not much more than I pay my cleaner! Fortunately, a horrified friend pointed out that it should be ‘at least’ £35 an hour, and I upped my rates immediately. I now charge £60 an hour for private lessons, whether online or in person. Unfortunately, some agencies such as Fleet Tutors don’t allow you to set your own rates, so that’s one thing to bear in mind when deciding which agencies to work with. However, they did provide me with quite a bit of work when I first started, so it’s swings and roundabouts. The pay scale often varies depending on the age of the student and the level taught, so you’ll probably earn more for teaching older students at GCSE level or above if the agency sets the prices. If you have any private clients, you can obviously set whatever rate you like, depending on where you live, the age of your pupils, whether lessons are online or in person and so on. Personally, I only have one rate (although I used to charge an extra £5 for teaching two pupils at the same time), and I raise it by £5 every year to allow for inflation and extra demand. Tutoring is more and more popular than ever these days, and I read somewhere that over half of pupils in London have private lessons over the course of their school careers, so don’t sell yourself short! You should be able to make around £25,000 a year, which is not bad going for a couple of hours’ work a day!

Foreign jobs are a little different, and there is a ‘standard’ rate of around £800 a week including expenses. That means your flights and accommodation are all covered, and you can even earn a bit more on the side if you decide to rent out your home on Airbnb while you’re away! When it comes to day-to-day expenses such as taxis and food and drink, it’s important to negotiate that with the agency before accepting the job. It’s no good complaining about having to live in the client’s house and buy your own lunches when you’re in Moscow or Bratislava! It can be a dream job, but just make sure you look at it from every angle:

  • What subjects will I be teaching?
  • How many hours will I have to teach?
  • How many days off will I get per week?
  • Where will the lessons take place?
  • How do I get to and from my accommodation?
  • How long is the assignment? (I refuse anything more than three months.)
  • Where will I be staying? (NEVER at the client’s house!)
  • How old are the children?
  • Will I have any other responsibilities (eg ferrying the children to and from school)?
  • Do I need a visa?
  • What is the weekly rate?
  • What expenses are included (eg flights, accommodation, taxis, food, drink)?

How do I get paid?

Most agencies ask for a timesheet and pay their tutors monthly via BACS payments directly into their bank accounts. That’s a bit annoying from a cash flow point of view, but there’s not much you can do about it – other than using a different agency. When it comes to private clients, I generally ask for cash after the lesson, but it’s even more convenient if they can pay via standing order – as long as you can trust them! I once let a client rack up over £600 in fees because he tended to pay in big lump sums every few weeks, but then his business folded, and I had to use a Government website to try and chase him up. Fortunately, his wife saw the email and paid my bill, but it took months to sort out. Normally, though, the worst that happens is that a client just doesn’t have the right change and promises to pay the following week, so you just need to keep track of who owes what.

I hope all this helps. Good luck!

 

You must be crazy!

2+2=5

Why? What’s wrong with that…?

Teachers and tutors ask pupils to check their work, but how can you do that in Maths without doing the whole sum all over again? Well, you can’t! So how are you supposed to check your work?

What you have to understand first of all is that checking everything is right is very different from checking nothing is obviously wrong. To check everything is right means doing the whole paper twice, but you obviously don’t have time to do that. Checking nothing is obviously wrong is much easier because it just means doing a ‘quick and dirty’ calculation in your head. It doesn’t guarantee that the answer is right, but it’s a good compromise. I call it ‘sanity checking’, which means making sure your answers are not crazy! Unfortunately, there isn’t one method that works for every question – it depends on what type of question it is – but here are a few examples:

Algebra

If you have to ‘solve for 𝑥’ and it’s a difficult question, try putting your answer back into the original equation and seeing if one side equals the other, eg if you think 𝑥 = 5, then that works for 2𝑥 + 6 = 16, but not for 3𝑥 + 2 = 5. That would be crazy!

Multiplication

Every multiplication sum starts with multiplying the last digit of each number together, so try doing that when you’ve got your answer and checking if the last digit of the result is equal to the last digit of the answer, eg 176 x 467 is going to end in a 2 because 6 x 7 = 42, which also ends with a 2. Your answer couldn’t end in any other number. That would be crazy!

Rounding

If you have any kind of sum that involves adding, subtracting, multiplication or division, an easy way to check it is to round the numbers to one or two significant figures (eg to the nearest hundred) and work out the answer in your head. If it’s close enough, then your answer is not obviously wrong. If it’s nowhere near, then you’ll have to do it again, eg 1.7 x 3.4 is close to 2 x 3, so the answer might be 5.78, but it wouldn’t be 57.8. That would be crazy!

Units

Most answers in Maths tests need some kind of unit, such as kg, m, cm or ml. Sometimes, the units are provided, but sometimes they’re not. If they’re not, you just need to make sure that you use the right ones, eg if the scale of a map is 1:100,000, the distance represented by 9.8cm is 9.8km, not 9.8m. That would be crazy!

 

Negative numbers

negative-numbers

Working with negative numbers can be confusing, but a few simple rules can help you add, subtract, multiply and divide successfully:

Adding and subtracting

If the operator and the sign are the same, then you must add the number, eg 4 – -3 = 4 + 3 = 7 or 4 + +3 = 7.

If the operator and the sign are different, then you must subtract the number, eg 4 + -3 = 4 – 3 = 1 or 4 – +3 = 1.

Here are a few practice questions:

  1. 14 – -3
  2. -4 + -7
  3. 21 – +8
  4. -8 + -7
  5. 6 – -4

Multiplication and division

If the signs of the two numbers are the same, then you must make the answer positive, eg -4 x -3 = 4 x 3 = 12 or +4 x +3 = 4 x 3 = 12.

If the signs of the two numbers are different, then you must make the answer negative, eg -4 x +3 = -12 or +4 x -3 = -12.

Here are a few practice questions:

  1. 1 × -3
  2. -8 ÷ -4
  3. 2 × -7
  4. -14 ÷ -7
  5. 6 × -4

Useful terms in Maths

maths

Maths is complicated, but a good first step on the road to understanding it is to get to know the most useful terms. There are lists in the front of the Bond books, but here’s my own contribution. I hope it helps!

Algebra: expressions using letters to represent unknown values, eg 2(x + 3) = 16.

Angles: there are three types of angle, depending on the number of degrees.

  • acute angles are between 0 and 90 degrees.
  • obtuse angles are between 90 and 180 degrees.
  • reflex angles are between 180 and 360 degrees.

Arc: part of the circumference of a circle.

Averages: there are three types of average, and they are all useful in different ways.

  • The mean is found by adding up all the values and dividing the total by how many there are, eg the mean of the numbers 1-10 is 5.5, as 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55, and 55 ÷ 10 = 5.5.
  • The mode is the most common value (or values), eg the mode of 1, 2, 2, 3, 4, 5 is 2.
  • The median of an odd number of values sorted by size is the one in the middle, eg the median of the numbers 1-5 is 3. The median of an even number of values is the mean of the two numbers in the middle, eg the median of the numbers 1-10 is 5.5, as 5 and 6 are the numbers in the middle, and 11 ÷ 2 = 5.5.

Chord: a straight line drawn between two points on the circumference of a circle.

Circumference: the distance all the way round the edge of a circle.

Congruent: triangles are congruent if they are the same shape and size, eg two right-angled triangles with sides of 3cm, 4cm and 5cm would be ‘congruent’, even if one is the mirror image of the other. You can prove that two triangles are congruent by using any of the following methods: SAS (Side-Angle-Side), SSS (Side-Side-Side), ASA (Angle-Side-Angle), AAS (Angle-Angle-Side) and RHS or HL (Right-angle-Hypotenuse-Side or Hypotenuse-Leg). If all three measurements of the angles and/or sides are equal, the triangles are congruent. You can only create a congruent copy of a triangle by translation, reflection or rotation. (Note: congruence is the same as similarity, except that the triangles have to be the same size.)

Cube: the result of multiplying any number by itself twice, eg 8 is the cube of 2, as 2 x 2 x 2 = 8.

Cube root: the number that has to be multiplied by itself twice to make another number, eg 2 is the cube root of 8, as 2 x 2 x 2 = 8.

Cuboid: a solid with a rectangle for each of the six sides, eg a shoe box.

Denominator: the number on the bottom of a fraction, eg 2 is the denominator of ½.

Diameter: the length of a line drawn across a circle passing through the centre.

Equation: any line of numbers and operators with an equals sign in the middle, showing that the two sides balance, eg 4x + 12 = 34.

Factor: a number that goes into another number evenly, eg 6 is a factor or 12.

Fibonacci series: a sequence of numbers created by adding the previous two numbers together to get the next one, eg 1, 1, 2, 3, 5, 8, 13…

Formula: a way of calculating the answer to a common problem using letters or words, eg the formula for distance is speed x time (or D = S x T).

Highest common factor (or HCF): the highest number that goes into two other numbers evenly, eg the HCF of 12 and 18 is 6.

Improper fraction: a fraction that is greater than one (in other words, the numerator is greater than the denominator), eg 9/5.

Lowest common multiple (or LCM) / Lowest common denominator (or LCD): the lowest number that is divisible by two other numbers, eg the LCM of 6 and 8 is 24.

Multiple: a number that can be divided evenly by another number, eg 12 is a multiple of 6.

Numerator: the number on the top of a fraction, eg 3 is the numerator of ¾.

Order of operations: the sequence of doing basic mathematical sums when you have a mixture of, say, addition and multiplication. BIDMAS (or BODMAS) is a good way of remembering it, as it stands for:

  • Brackets
  • Indices/Order (in other words, squares, cubes and so on)
  • Division
  • Multiplication
  • Addition
  • Subtraction

Note that addition doesn’t come ‘before’ subtraction – these operations have to be done in the order in which they occur in the sum, and it makes a difference to the answer, eg 4 – 3 + 2 = 3 if you do the operations in order, which is correct, but you’d get the wrong answer of -1 if you did 3 + 2 first.

Operator: the sign telling you which mathematical operation to do. The most common ones are +, -, x and ÷.

Parallel: two lines are parallel if they will never meet, eg the rails on a railway line.

Perimeter: the distance all the way round the outside of a shape.

Perpendicular: at 90 degrees to each other.

Pi (or π): a constant used to work out the circumference and area of circles, often shown as 22/7 or 3.14 although it’s actually an ‘irrational’ number, which means it goes on for ever.

Prime factors: the lowest prime numbers that can be multiplied together to make a given number, eg the prime factors of 12 are 2² x 3.

Prime numbers: a number that can only be divided by itself and one, eg 2, 3, 5, 7, 11, 13…

Probability: the chance of something happening, calculated as the number of ways of getting what you want divided by the total number of possible outcomes, eg the chance of a coin toss being heads is ½ as there is one ‘heads’ side but two sides in total. To work out the probability of a sequence of events, you have to multiply the individual probabilities together, eg the chance of a coin toss being heads twice in a row is ½ x ½ = ¼

Product: the result of multiplying two numbers together, eg 35 is the product of 5 and 7.

Quadrilateral: a four-sided shape such as the following:

  • Kite: a quadrilateral with two pairs of equal sides next to each other (or ‘adjacent’ to each other).
  • Parallelogram: a quadrilateral with opposite sides parallel to each other.
  • Rectangle: a quadrilateral with two opposite pairs of equal sides and four right angles.
  • Rhombus: a quadrilateral with equal sides.
  • Square: a quadrilateral with equal sides and four right angles.
  • Trapezium: a quadrilateral with one pair of parallel sides. (Note: an isosceles trapezium is symmetrical.)

Radius: the distance from the centre of a circle to the circumference.

Range: the highest minus the lowest value in a list, eg the range of the numbers 1-10 is 9.

Regular: a shape is regular if all its sides and angles are equal, eg a 50p piece is a regular (-ish!) heptagon.

Right angle: an angle of 90 degrees.

Sector: a ‘slice’ of a circle in between two radii.

Segment: a part of a circle separated from the rest by a chord.

Shapes: the name of each shape depends on the number of sides. Here are the first 12.

  • Quadrilaterals have four sides.
  • Pentagons have five sides.
  • Hexagons have six sides.
  • Heptagons have seven sides.
  • Octagons have eight sides.
  • Nonagons have nine sides.
  • Decagons have 10 sides.
  • Hendecagons have 11 sides.
  • Dodecagons have 12 sides.

Similar: triangles are similar if they are the same shape, but not necessarily the same size, eg a right-angled triangle with sides of 3cm, 4cm and 5cm is ‘similar’ to a right-angled triangle with sides of 6cm, 8cm and 10cm. (Note: similarity is the same as congruence, except that the triangles don’t have to be the same size.)

Square number: the result of multiplying any number by itself, eg 49 is a square number, as 7 x 7 = 49.

Square root: the number that has to be multiplied by itself to make another number, eg 6 is the square root of 36, as 6 x 6 = 36.

Sum: the result of adding two numbers together, eg 17 is the sum of 8 and 9.

Tangent: either a straight line that touches the circumference of a circle OR the length of the opposite side of a triangle divided by the length of the adjacent side

Triangles: there are four main types, each with different properties.

  • equilateral triangles have all three sides the same length and all three angles the same.
  • isosceles triangles have two sides the same length and two angles the same.
  • scalene triangles have three sides of different lengths with three different angles.
  • right-angled triangles have one 90-degree angle.

Variable: an unknown in algebra, eg x or y.

Algebra

Nothing makes the heart of a reluctant mathematician sink like an algebra question.algebra

Algebra is supposed to make life easier. By learning a formula or an equation, you can solve any similar type of problem whatever the numbers involved. However, an awful lot of students find it difficult, because letters just don’t seem to ‘mean’ as much as numbers. Here, we’ll try to make life a bit easier…

Gathering terms

X’s and y’s look a bit meaningless, but that’s the point. They can stand for anything. The simplest form of question you’ll have to answer is one that involves gathering your terms. That just means counting how many variables or unknowns you have (like x and y). I like to think of them as pieces of fruit, so an expression like…

2x + 3y – x + y

…just means ‘take away one apple from two apples and add one banana to three more bananas’. That leaves you with one apple and four bananas, or…

x + 4y

Here are a few practice questions:

  1. 3x + 4y – 2x + y
  2. 2m + 3n – m + 3n
  3. p + 2q + 3p – 3q
  4. 2a – 4b + a + 4b
  5. x + y – 2x + 2y

Multiplying out brackets

This is one of the commonest types of question. All you need to do is write down the same expression without the brackets. To take a simple example:

2(x + 3)

In this case, all you need to do is multiply everything inside the brackets by the number outside, which is 2, but what do we do about the ‘+’ sign? We could just multiply 2 by x, write down ‘+’ and then multiply 2 by 3:

2x + 6

However, that gets us into trouble if we have to subtract one expression in brackets from another (see below for explanation) – so it’s better to think of the ‘+’ sign as belonging to the 3. In other words, you multiply 2 by x and then 2 by +3. Once you’ve done that, you just convert the ‘+’ sign back to an operator. It gives exactly the same result, but it will work ALL the time rather than just with simple sums!

Here are a few practice questions:

  1. 2(a + 5)
  2. 3(y + 2)
  3. 6(3 + b)
  4. 3(a – 3)
  5. 4(3 – p)

Solving for x

Another common type of question involves finding out what x stands for (or y or z or any other letter). The easiest way to look at this kind of equation is using fruit again. In the old days, scales in a grocery shop sometimes had a bowl on one side and a place to put weights on the other. To weigh fruit, you just needed to make sure that the weights and the fruit balanced and then add up all the weights. The point is that every equation always has to balance – the very word ‘equation’ comes from ‘equal’ – so you have to make sure that anything you do to one side you also have to do to the other.

There are three main types of operation you need to do in the following order:

  1. Multiplying out any brackets
  2. Adding or subtracting
  3. Multiplying or dividing

Once you’ve multiplied out any brackets (see above), what you want to do is to simplify the equation by removing one expression at a time until you end up with something that says x = The Answer. It’s easier to start with adding and subtracting and then multiply or divide afterwards (followed by any square roots). To take the same example as before:

2(x + 3) = 8

Multiplying out the brackets gives us:

2x + 6 = 8

Subtracting 6 from both sides gives us:

2x = 2

Dividing both sides by 2 gives us the final answer:

x = 1

Simple!

Here are a few practice questions:

  1. b + 5 = 9
  2. 3y = 9
  3. 6(4 + c) = 36
  4. 3(a – 2) = 24
  5. 4(3 – p) = -8

Multiplying two expressions in brackets (‘FOIL’ method)

When you have to multiply something in brackets by something else in brackets, you should use what’s called the ‘FOIL’ method. FOIL is an acronym that stands for:

First
Outside
Inside
Last

This is simply a good way to remember the order in which to multiply the terms, so we start with the first terms in each bracket, then move on to the outside terms in the whole expression, then the terms in the middle and finally the last terms in each bracket. Just make sure that you use the same trick we saw earlier, combining the operators with the numbers and letters before multiplying them together. For example:

(a + 1)(a + 2)

First we multiply the first terms in each bracket:

a x a

…then the outside terms:

a x +2

…then the inside terms:

+1 x a

…and finally the last terms in each bracket:

+1 x +2

Put it all together and simplify:

(a + 1)(a + 2)

= a² + 2a + a + 2

=a² + 3a + 2

Here are a few practice questions:

  1. (a + 1)(b + 2)
  2. (a – 1)(a + 2)
  3. (b + 1)(a – 2)
  4. (p – 1)(q + 2)
  5. (y + 1)(y – 3)

Factorising quadratics (‘product and sum’ method)

This is just the opposite of multiplying two expressions in brackets. Normally, factorisation involves finding the Highest Common Factor (or HCF) and putting that outside a set of brackets containing the rest of the terms, but some expressions can’t be solved that way, eg a² + 3a + 2 (from the previous example). There is no combination of numbers and/or letters that goes evenly into a², 3a and 2, so we have to factorise using two sets of brackets. To do this, we use the ‘product and sum’ method. This simply means that we need to find a pair of numbers whose product equals the last number and whose sum equals the multiple of a. In this case, it’s 1 and 2 as +1 x +2 = +2 and +1 + +2 = +3. The first term in each bracket is just going to be a as a x a = a². Hence, factorising a² + 3a + 2 gives (a + 1)(a + 2). You can check it by using the FOIL method (see above) to multiply out the brackets:

(a + 1)(a + 2)

= a² + 2a + a + 2

=a² + 3a + 2

Subtracting one expression from another*

Here’s the reason why we don’t just write down operators as we come across them. Here’s a simple expression we need to simplify:

20 – 4(x – 3) = 16

If we use the ‘wrong’ method, then we get the following answer:

20 – 4(x – 3) = 16

20 – 4x – 12 = 16

8 – 4x = 16

4x = -8

x = -2

Now, if we plug our answer for x back into the original equation, it doesn’t balance:

20 – 4(-2 – 3) = 16

20 – 4 x -5 = 16

20 – -20 = 16

40 = 16!!

That’s why we have to use the other method, treating the operator as a negative or positive sign to be added to the number before we multiply it by whatever’s outside the brackets:

20 – 4(x – 3) = 16

20 – 4x + 12 = 16

32 – 4x = 16

4x = 16

x = 4

That makes much more sense, as we can see:

20 – 4(4 – 3) = 16

20 – 4 x 1 = 16

20 – 4 = 16

16 = 16

Thank Goodness for that!

Here are a few practice questions:

  1. 30 – 3(p – 1) = 0
  2. 20 – 3(a – 3) = 5
  3. 12 – 4(x – 2) = 4
  4. 24 – 6(x – 3) = 6
  5. 0 – 6(x – 2) = -12

Other tips to remember

  • If you have just one variable, leave out the number 1, eg 1x is just written as x.
  • When you have to multiply a number by a letter, leave out the times sign, eg 2 x p is written as 2p.
  • The squared symbol only relates to the number or letter immediately before it, eg 3m² means 3 x m x m, NOT (3 x m) x (3 x m).

Divisibility rules OK!

Divisibility rules OK

And, no, this is nothing to do with Harry Potter’s invisibility cloak…

Times tables can be tricky, and there’s no substitute for learning them by heart. However, the divisibility rules can at least tell you whether an answer is definitely wrong. I’m a great believer in ‘sanity checking’ your work. Just ask yourself, “Is this crazy?” If it is, you’ll have to do the question again!

The divisibility rules are quite simple (except for the ones for 7 and 8). They tell you whether a number can be divided by any number from 1 to 10. They’re most useful when simplifying fractions…or when you’re struggling to remember your times tables!

  1. Must be a whole number, eg 2, but not 2.5.
  2. Must be an even number ending in 0, 2, 4, 6 or 8, eg 22, but not 23.
  3. The sum of the digits must be divisible by 3, eg 66, but not 67.
  4. The number formed by the last two digits must be ’00’ or divisible by 4, eg 500 or 504, but not 503.
  5. Must end in 5 or 0, eg 60, but not 61.
  6. Must be divisible by both 2 and 3, eg 18, but not 23.
  7. If you double the last digit and take this away from the number formed by the rest of the digits, the result must be 0 or divisible by 7, eg 672 (2 x 2 = 4, and 67 – 4 = 63, which is divisible by 7), but not 674.
  8. The number formed by the last three digits must be ’00’ or divisible by 8, eg 5,000 or 5,008, but not 5,003.
  9. The sum of the digits must be divisible by 9, eg 666, but not 667.
  10. Must end in 0, eg 110, but not 111.

Tips for the QTS numeracy test

“If I’d known I’d have to go back to school, I’d never have become a teacher!” Back-to-school-blackboard-chalk

The QTS numeracy and literacy tests are not very popular, but trainee teachers still have to pass them before they can start teaching in the state sector, so I thought I’d try and help out. There is always more than one way of doing a Maths question, but I hope I’ll demonstrate a few useful short cuts and describe when and how they should be used. The point of short cuts is that, even though you may have to do more sums, they’ll be easier sums that can be done faster and more accurately. The numeracy test consists of two sections – mental Maths and interpreting charts – and I’m going to focus on the first of these.

Fractions to percentages – type 1

There are a number of typical types of questions in the numeracy test, and a lot of them involve multiplication – so knowing your times tables is an absolute must! One of the most common kinds of question involves converting fractions to percentages. These are just two ways of showing the same thing, but to answer these questions you’ll need to try different approaches. First of all, have a look to see if the denominator (or the number on the bottom of the fraction) is a factor or a multiple of 100. If it is, you can simply multiply or divide the numerator (the number on the top) and the denominator by whatever it takes to leave 100 on the bottom. Any fraction over 100 is just a percentage in disguise, so you just need to put the percentage sign after the numerator, eg what is the percentage mark if:

  1. a pupil scores 7 out of a possible 20?
    Answer: 20 x 5 = 100, so 7 x 5 = 35%.
  2. a pupil scores 18 out of a possible 25?
  3. a pupil scores 7 out of a possible 10?
  4. a pupil scores 9 out of a possible 20?
  5. a pupil scores 130 out of a possible 200?

Fractions to percentages – type 2

If the denominator is not a factor of 100, check if it’s a multiple of 10. If it is, you can convert the fraction into tenths and then multiply the top and bottom by 10 to get a fraction over 100, which, again, is just a percentage in disguise, eg what is the percentage mark if:

  1. A pupil scores 24 marks out of a possible 40?
    Answer: 40 ÷ 4 = 10, so 24 ÷ 4 = 6 and 6 x 10 = 60%.
  2. A pupil scores 12 marks out of a possible 30?
  3. A pupil scores 32 marks out of a possible 80?
  4. A pupil scores 49 marks out of a possible 70?
  5. A pupil scores 24 marks out of a possible 60?

Fractions to percentages – type 3

If neither of the first two methods works, that means you have to simplify the fraction. Once you’ve done that, you should be able to convert any common fraction into a percentage in your head. The most commonly used fractions are halves, quarters, fifths and eighths, so it’s worth learning the decimal and percentage equivalents off-by-heart, ie

  • ½ = 0.5 = 50%
  • ¼ = 0.25 = 25%
  • ¾ = 0.75 = 75%
  • ⅕ = 0.2 = 20%
  • ⅖ = 0.4 = 40%
  • ⅗ = 0.6 = 60%
  • ⅘ = 0.8 = 80%
  • ⅛ = 0.125 = 12.5%
  • ⅜ = 0.375 = 37.5%
  • ⅝ = 0.625 = 62.5%
  • ⅞ = 0.875 = 87.5%

To simplify the fractions, check first to see if the numerator goes into the denominator. If it does, you can simply divide both numbers by the numerator to get what’s called a ‘unit fraction’, in other words a fraction with a one on top. By definition, a unit fraction can’t be simplified, so then you just have to convert it into a percentage. If the numerator doesn’t go exactly, try the first few prime numbers, ie 2, 3, 5, 7 and perhaps 11. Keep dividing both numbers in the fraction by the lowest possible prime number, and you’ll eventually show the fraction in its lowest terms. (If you happen to find a bigger number you can use, that’s great, as it means you won’t need to do as many sums.) When you’re left with one of the common fractions in the list above, you just have to convert it into the correct percentage, eg what is the percentage mark if:

  1. a pupil scores 7 out of a possible 28?
    Answer: 7 goes into 28 four times, so the fraction is 1/4, which is 25%.
  2. a pupil scores 27 out of a possible 36?
    Answer: 27 doesn’t go into 36, but 3 does, so 27/36 = 9/12, but 9 and 12 are also divisible by 3, so that makes 3/4, which is 75%.
  3. a pupil scores 24 out of a possible 48?
  4. a pupil scores 8 out of possible 32?
  5. a pupil scores 9 out of a possible 24?

Multiplying three numbers involving money

There is often a ‘real world’ money problem in the QTS numeracy test. That usually means multiplying three numbers together. The first thing to say is that it doesn’t matter in which order you do it, eg 1 x 2 x 3 is the same as 3 x 2 x 1. The next thing to bear in mind is that you will usually have to convert from pence to pounds. You could do this at the end by simply dividing the answer by 100, but a better way is to divide one of the numbers by 100 (or two of the numbers by 10) at the beginning and then multiply the remaining three numbers together, eg a number of pupils in a class took part in a sponsored spell to raise money for charity. The pupils were expected to get a certain number of correct spellings, and the average amount of sponsorship is shown for each. How many pounds would the class expect to raise for charity if the basic sum is:

  1. 20 x 30 x 5p?
    Answer: 2 x 3 x 5 = 6 x 5 = £30.
  2. 40 x 500 x 7p?
  3. 30 x 400 x 6p?
  4. 50 x 40 x 8p?
  5. 60 x 20 x 9p?

Division by single-digit numbers

This is what I call the ‘wedding planner problem’. There are three ways of doing this type of question:

  • Method A: Use the ‘bus stop’ method to divide the total number of guests by the number of seats per table – remembering to add one if there is a remainder.
  • Method B: Go straight to the end of your times tables by multiplying the number of seats by 12, then calculating the remainder and dividing by the number of seats per table, again remembering to add one if there is another remainder.
  • Method C: Use trial and error by estimating the number of tables needed using a nice, round number such as 5, 10 or 20 and working out the remainder as before.
  1. Dining tables seat 7 children. How many tables are needed to seat 100 children?
    Answer:
    Method A) 100 ÷ 7 = 14 r 2, so 14 + 1 = 15 tables are needed.

    Method B) 7 x 12 = 84, 100 – 84 = 16, 16 ÷ 7 = 2 remainder 2, 12 + 2 + 1 = 15 tables.
    Method C) 10 x 7 = 70, which is too small, 20 x 7 = 140, which is too big, 15 x 7 = 70 + 35 = 105, which is just right as there are only 5 seats to spare.
  2. Dining tables seat 6 children. How many tables are needed to seat 92 children?
  3. Dining tables seat 5 children. How many tables are need to seat 78 children?
  4. Dining tables seat 9 children. How many tables are needed to seat 120 children?
  5. Dining tables seat 6 children. How many tables are needed to seat 75 children?

Division by two-digit numbers

If the number of seats is outside your times tables, the best option is just to use trial and error, starting with 5, 10 or 20, eg

  1. It is possible to seat 40 people in a row across the hall. How many rows are needed to seat 432 people?
    Answer: 40 x 10 = 400, 432 – 400 = 32, so one more row is needed, making a total of 10 + 1 = 11 rows.
  2. It is possible to seat 32 people in a row across the hall. How many rows are needed to seat 340 people?
  3. It is possible to seat 64 people in a row across the hall. How many rows are needed to 663 people?
  4. It is possible to seat 28 people in a row across the hall. How many rows are needed to seat 438 people?
  5. It is possible to seat 42 people in a row across the hall. How many rows are needed to seat 379 people?

Percentages to fractions

This is a type of question that looks hard at first but becomes dead easy with the right short cut. All you need to do is to work out 10% first and then multiply by the number of tens in the percentage. Another way of saying that is just to knock one zero off each number and multiply them together, eg a test has a certain number of questions, each worth one mark. For the stated pass mark, how many questions had to be answered correctly to pass the test?

  1. ?/30 = 40%
    Answer: 3 x 4 = 12 questions (ie 10% of 30 is 3 questions, but we need 40%, which is 4 x 10%, so we need four lots of three, which is the same as 3 x 4).
  2. ?/40 = 70%
  3. ?/50 = 90%
  4. ?/80 = 70%
  5. ?/300 = 60%

Ratio – distance

There are two ways of converting between different units of distance from the metric and imperial systems:

  • Method A: Make the ratio into a fraction and multiply the distance you need to find out by that same fraction, ie multiply it by the numerator and divide it by the denominator. (Start with multiplication if doing the division first wouldn’t give you a whole number.)
  • Method B: Draw the numbers in a little 2 x 2 table, with the figures in the ratio in the top row and the distance you need to find out in the column with the appropriate units, then find out what you need to multiply by to get from the top row to the bottom row and multiply the distance you have to find out by that number to fill in the final box.
  1. 8km is about 5 miles. How many kilometres is 40 miles?
    Answer:
    Method A) 8:5 becomes 8/5, and 40 x 8/5 = 40 ÷ 5 x 8 = 8 x 8 = 64km.
    Method B)
    Miles                 km
    5                          8
    x 8
    40               8 x 8 = 64km
  2. 6km is about 4 miles. How many kilometres is 36 miles?
  3. 4km is about 3 miles. How many kilometres is 27 miles?
  4. 9km is about 7 miles. How many miles is 63 kilometres?
  5. 7km is about 4 miles. How many kilometres is 32 miles?

Ratio – money

You can use the same methods when converting money, except that the exchange rate is now a decimal rather than a fraction. Just remember that the pound is stronger than any other major currency, so there will always be fewer of them. It’s easy to get things the wrong way round, so it’s worth spending a couple of seconds checking, eg

  1. £1 = €1.70. How much is £100 in euros?
    Method A) 100 x 1.70 = €170.
    Method B)
    £                                      €
    1.00                              1.70
    x 100
    100                    1.70 x 100 = €170
  2. £1 = €1.60. How much is £200 in euros?
  3. £1 = €1.50. How much is €150 in pounds?
  4. £1 = €1.80. How much is €90 in pounds?
  5. £2 = €3.20. How much is £400 in euros?

Time – find the end time

The most useful trick to use here is rounding. If the length of a lesson is 45 minutes or more, then just round up to the full hour and take the extra minutes off at the end. This avoids having to add or subtract ‘through the hour’, which is more difficult. If the lessons are less than 45 minutes long, just work out the total number of minutes, then convert into hours and minutes and add to the start time, eg

  1. A class starts at 9:35. The class lasts 45 minutes. What time does the class finish?
    Answer: 9:35 + 1 hour – 15 minutes = 10:35 – 15 minutes = 10:20.
  2. A class starts at 11:45. There are three consecutive classes each lasting 25 minutes and then half an hour for lunch. What time does lunch finish?
    Answer: 11:45 + 3 x 25 + 30 = 11:45 + 75 + 30 = 11:45 + 1 hour and 15 minutes + 30 minutes = 13:30.
  3. Lessons start at 11:15. There are two classes each lasting 40 minutes and then lunch. What time does lunch start?
  4. Lessons start at 2:00 in the afternoon. There are four 50-minute classes with a 15-minute break in the middle. What time does the day finish?
  5. Lessons start at 9:40. There are two classes of 50 minutes each with a break of 15 minutes in between. What time do the classes finish?

Time – find the start time

It’s even more important to use rounding when working backwards from the end of an event, as subtraction is that bit more difficult, eg

  1. A school day finishes at 3:15. There are two classes of 50 minutes each after lunch with a break of 15 minutes in the middle. What time does lunch end?
    Answer: 3:15 – 2 hours + 2 x 10 minutes – 15 minutes = 1:15 + 20 minutes -15 minutes = 1:20.
  2. A school day finishes at 4:30. There are two classes of 40 minutes each after lunch. What time does lunch finish?
    Answer: 4:30 – 2 x 40 = 4:30 – 80 minutes = 4:30 – 1 hour and 20 minutes = 3:10.
  3. Lunch starts at 1:05. There are two classes before lunch of 55 minutes each. What time do the classes start?
  4. Lunch starts at 1:15. There are three classes before lunch of 45 minutes each. What time do the classes start?
  5. A school bus arrives at school at 8:45. It picks up 20 children, and it takes an average of four minutes to pick up each child. What time is the first child picked up?

Percentage to decimal

A decimal is a fraction of one unit, but a percentage is a fraction of 100 units, so, to convert from a percentage to a decimal, you just need to divide by 100, eg

  1. What is 20% as a decimal?
    Answer: 20 ÷ 100 = 0.2.
  2. What is 30% as a decimal?
  3. What is 17% as a decimal?
  4. What is 6% as a decimal?
  5. What is 48% as a decimal?

Multiplying decimals

Decimal points can be confusing, so the best way to do these sums is to take out the decimal point and put it back at the end. You just need to remember to make sure there are the same number of decimal places in the answer as in both numbers in the question, eg

  1. 1.5 x 1.5
    Answer: 15 x 15 = 10 x 15 + 5 x 15 = 150 + 75 = 225, but there are two decimal places in the numbers you’re multiplying together, so the answer must be 2.25.
  2. 3 x 4.5
  3. 4.7 x 8
  4. 7.5 x 7.5
  5. 2.5 x 6.5

Multiplying decimals by a power of 10

Because we have 10 fingers, we’ve ended up with a ‘decimal’ number system based on the number 10. That makes it really easy to multiply by powers of 10, because all you have to do is to move the decimal point to the right by a suitable number of places, eg one place when multiplying by 10, two when multiplying by 100 etc. (You can also think of it as moving the digits in the opposite direction.) This type of question is therefore one of the easiest, eg

  1. 4.5 x 10
    Answer: 45.
  2. 3.8 x 100
  3. 7.6 x 1000
  4. 4.6 x 100
  5. 3.5 x 10

Percentage of quantity

Finding a percentage is easy if it ends with a zero, as you can start by finding 10% (Method A). If you happen to know what the fraction is, you can also divide by the numerator of that fraction (Method B), so 20% is 1/5, so you just need to divide by five, eg

  1. Find 20% of 360
    Answer:
    Method A) 360/10 x 2 = 36 x 2 = 72.
    Method B) 360 ÷ 5 = 72 (or 360 x 2 ÷ 10 = 720 ÷ 10 = 72).
  2. Find 20% of 45
  3. Find 30% of 320
  4. Find 60% of 60
  5. Find 80% of 120

Multiplication

Just because this is the ‘mental Maths’ section of the test doesn’t mean that you can’t work things out on paper, and these simple multiplication sums can be done like that. Alternatively, you can use ‘chunking’, which means multiplying the tens and units separately and adding the results together, and the short cut for multiplying by five is to multiply by 10 and then divide by two, eg

  1. 23 x 7
    Answer: 20 x 7 + 3 x 7 = 140 + 21 = 161.
  2. 42 x 5
    Answer: 42 x 10 ÷ 2 = 420 ÷ 2 = 210
  3. 34 x 6
  4. 56 x 8
  5. 34 x 8

Short division

Again, working these sums out on paper is probably quicker (and more reliable), although the easiest way to divide by four is probably to halve the number twice, and the short cut for dividing by five is to multiply by two and then divide by 10.

  1. 292 ÷ 4
    Answer: 292 ÷ 2 ÷ 2 = 146 ÷ 2 = 73.
  2. 345 ÷ 5
    Answer: 345 x 2 ÷ 10 = 690 ÷ 10 = 69.
  3. 282 ÷ 3
  4. 565 ÷ 5
  5. 432 ÷ 4

Ratios

Hundreds of years ago, it was traditional to put dragons on maps in places that were unknown, dangerous or poorly mapped. Ratios are one of those places…

Here be ratios...!

Here be ratios…!

A ratio is just a model of the real world, shown in the lowest terms, but answering ratio questions can be just as scary as meeting dragons if you don’t know what you’re doing. The key to understanding ratios is to work out the scale factor. This is just like the scale on a map. If a map is drawn to a scale of 1:100,000, for instance, you know that 1cm on the map is the same as 100,000cm (or 1km) in the real world. To convert distances on the map into distances in the real world, you just need to multiply by the scale factor, which is 100,000 in this case. (You can also go the other way – from the real world to the map – by dividing by the scale factor instead.)

To work out the scale factor in a Maths question, you need to know the matching quantities in the real world and in the model (or ratio). Once you know those two numbers, you can simply divide the one in the real world by the one in the ratio to get the scale factor. For example:

If Tom and Katie have 32 marbles between them in the ratio 3:1, how many marbles does Tom have?

To answer this question, here are the steps to take:

  1. Work out the scale factor. The total number of marbles in the real world is 32, and the total in the ratio can be found by adding the amounts for both Tom and Katie, which means 3 + 1 = 4. Dividing the real world total by the ratio total gives 32 ÷ 4 = 8, so the scale factor is 8.
  2. Multiply the number you want in the ratio by the scale factor. If Tom’s share of the marbles in the ratio is 3, then he has 3 x 8 = 24 marbles.

The matching numbers in the real world and the ratio are sometimes the totals and sometimes the individual shares, but it doesn’t matter what they are. All you need to do is find the same quantity in both places and divide the real world version by the ratio version to get the scale factor. Once you’ve done that, you can multiply any of the ratio numbers to get to the real world number (or divide any real world number to get to the ratio number). Different questions might put the problem in different ways, but the principle is the same.

One complication might be having two ratios that overlap. In that case you need to turn them into just one ratio that includes all three quantities and answer the question as you normally would. For example:

If there are 30 black sheep, and the ratio of black to brown sheep is 3:2, and the ratio of brown to white sheep is 5:4, how many white sheep are there?

This is a bit more complicated, but the basic steps are the same once you’ve found out the ratio for all three kinds of sheep. To do this, we need to link the two ratios together somehow, but all the numbers are different, so how do we do it? The answer is the same as for adding fractions with different denominators (or for solving the harder types of simultaneous equations, for that matter): we just need to multiply them together. If we were adding fifths and halves, we would multiply the denominators together to convert them both into tenths. Here, the type of sheep that is in both ratios is the brown one, so we simply have to make sure the numbers of brown sheep in each ratio (2 and 5) are the same by multiplying them together (to give 10). Once we’ve done that, we can combine the two ratios into one and answer the question. Here goes:

  • Ratio of black sheep to brown sheep = 3:2

Multiply by 5

  • Ratio of black sheep to brown sheep = 15:10
  • Ratio of brown to white sheep = 5:4

Multiply by 2

  • Ratio of brown to white sheep = 10:8
  • Therefore, ratio of black sheep to brown sheep to white sheep = 15:10:8

Now that we have just one ratio, we can answer the question by following exactly the same steps as before:

  1. Work out the scale factor. The total number of black sheep in the real world is 30, and the total in the ratio is 15. Dividing the real world total by the ratio total gives 30 ÷ 15 = 2, so the scale factor is 2.
  2. Multiply the number you want in the ratio by the scale factor. If the number of white sheep in the ratio is 8, then there are 8 x 2 = 16 white sheep.

Simple!

Here are a few practice questions:

  1. One hundred paintings have to be selected for an art exhibition. If the ratio of amateur paintings to professional paintings has to be 2:3, how many amateur paintings and professional paintings have to be selected?
  2. The ratio of brown rats to black rats is 3:2. If there are 16 black rats, how many brown rats are there?
  3. Peter has 20 blue pens. How many red pens must he buy if the ratio of blue to red pens has to be 2:3?
  4. There are 35 children in a class and 15 are boys. What is the ratio of boys to girls?
  5. There are 15 girls and 12 boys in a class. What is the ratio of girls to boys? Give your answer in its simplest form.
  6. A newspaper includes 12 pages of sport and 8 pages of TV. What is the ratio of sport to TV? Give your answer in its simplest form.
  7. Anna has 75p, and Fiona has £1.20. What is the ratio of Anna’s money to Fiona’s money in its simplest form?
  8. Sam does a scale drawing of his kitchen. He uses a scale of 1:100. He measures the length of the kitchen as 5.9m. How long is the kitchen on the scale drawing? Give your answer in mm.
  9. A recipe to make lasagne for 6 people uses 300 grams of minced beef. How much minced beef would be needed to serve 8 people?
  10. A recipe for flapjacks requires 240g of oats. This makes 18 flapjacks. What quantity of oats is needed to make 24 flapjacks?
  11. Amit is 12 years old. His brother, Arun, is 9. Their grandfather gives them £140, which is to be divided between them in the ratio of their ages. How much does each of them get?
  12. The angles in a triangle are in the ratio 1:2:9. Find the size of the largest angle.
  13. In a certain town, the ratio of left-handed people to right-handed people is 2:9. How many right-handed people would you expect to find in a group of 132 people?
  14. Twelve pencils cost 72p. Find the cost of 30 pencils.
  15. Jenny buys 15 felt-tip pens. It costs her £2.85. How much would 20 pens have cost?
  16. If three apples cost 45p, how much would five apples cost?
  17. Sam is 16 years old. His sister is 24 years old. What’s the ratio of Sam’s age to his sister’s age? Give your answer in its simplest form.
  18. A map scale is 1:20000. A distance on the map is measured to be 5.6cm. What’s the actual distance in real life? Give your answer in metres.
  19. A recipe for vegetable curry needs 300 grams of rice, and it feeds 4 people. How much rice would be needed for 7 people?
  20. £60 is to be divided between Brian and Kate in the ratio 2:3. How much does Kate get?

Working with fractions

Me again...

Me again…

People don’t like fractions. I don’t know why. They’re difficult to begin with, I know, but a few simple rules will help you add, subtract, multiply and divide.

Adding and subtracting

Adding and subtracting are usually the easiest sums, but not when it comes to fractions. If fractions have the same denominator (the number on the bottom), then you can simply add or subtract the second numerator from the first, eg 4/5 – 3/5 = 1/5. If not, it would be like adding apples and oranges. They’re just not the same, so you first have to convert them into ‘pieces of fruit’ – or a common unit. The easiest way of doing that is by multiplying the denominators together. That guarantees that the new denominator is a multiple of both the others. Once you’ve found the right denominator, you can multiply each numerator by the denominator from the other fraction (because whatever you do to the bottom of the fraction you have to do to the top), add or subtract them and then simplify and/or convert into a mixed number if necessary, eg 2/3 + 4/5 = (2 x 5 + 4 x 3) / (3 x 5) = (10 + 12) / 15 = 22/15 = 1 7/15.

  1. Multiply the denominators together and write the answer down as the new denominator
  2. Multiply the numerator of the first fraction by the denominator of the second and write the answer above the new denominator
  3. Multiply the numerator of the second fraction by the denominator of the first and write the answer above the new denominator (after a plus or minus sign)
  4. Add or subtract the numerators and write the answer over the new denominator
  5. Simplify and/or turn into a mixed number if necessary

Note that you can often use a simpler method. If one of the denominators is a factor of the other, you can simply multiply the numerator and denominator of that fraction by 2, say, so that you get matching denominators, eg 1/5 + 7/10 = 2/10 + 7/10 = 9/10. This means fewer steps in the calculation and lower numbers, and that probably means less chance of getting it wrong.

Sample questions

  1. 1/5 + 2/3
  2. 2/7 + 3/5
  3. 4/5 – 2/3
  4. 7/8 – 3/4
  5. 5/8 – 2/3

Multiplication

This is the easiest thing to do with fractions. You simply have to multiply the numerators together, multiply the denominators together and then put one over the other, simplifying and/or converting into a mixed number if necessary, eg 2/3 x 4/5 = (2 x 4) / (3 x 5) = 8/15.

  1. Multiply the numerators together
  2. Multiply the denominators together
  3. Put the result of Step 1 over the result of Step 2 in a fraction
  4. Simplify and/or turn into a mixed number if necessary

Sample questions

  1. 1/5 x 2/3
  2. 2/7 x 3/5
  3. 4/5 x 2/3
  4. 7/8 x 3/4
  5. 5/8 x 2/3

Division

Dividing by a fraction must have seemed like a nightmare to early mathematicians, because nobody ever does it! That’s right. Nobody divides by a fraction, because it’s so much easier to multiply. That’s because dividing by a fraction is the same as multiplying by the same fraction once it’s turned upside down, eg 2/3 ÷ 4/5 = 2/3 x 5/4 = (2 x 5) / (3 x 4) = 10/12 = 5/6. You can even cut out the middle step and simply multiply each numerator by the denominator from the other fraction, eg 2/3 ÷ 4/5 = (2 x 5) / (3 x 4) = 10/12 = 5/6.

  1. Multiply the numerator of the first fraction by the denominator of the second
  2. Multiply the numerator of the second fraction by the denominator of the first
  3. Put the result of Step 1 over the result of Step 2 in a fraction
  4. Simplify and/or turn into a mixed number if necessary

Sample questions

  1. 1/5 ÷ 2/3
  2. 2/7 ÷ 3/5
  3. 4/5 ÷ 2/3
  4. 7/8 ÷ 3/4
  5. 5/8 ÷ 2/3

Simplifying fractions

By the way, to simplify a fraction, try dividing the denominator by the numerator first, eg 9/18 = 1/2. If that works, you don’t have to do anything else. If not, try dividing by the first few prime numbers, ie 2, 3, 5, 7 and 11. You don’t need to try the other numbers, because they’re all multiples of the primes, so they won’t work if the others don’t, eg 4 won’t work if 2 doesn’t work. Ideally, the quickest way would be to divide the numerator and denominator by the highest common factor (or HCF), but you don’t know what that is at the beginning, so it would take time to work it out. This way is a good compromise.

  1. If possible, divide the numerator and denominator by the numerator
  2. If the numerator doesn’t go exactly, start dividing by the smallest prime number that will go into both numbers, starting with 2, 3, 5, 7 and 11. (If you happen to see a larger number that will work, then use that to save time.)
  3. Repeat Step 2 until the only number that goes into the numerator and denominator is 1

Sample questions

  1. Simplify 14/28
  2. Simplify 8/24
  3. Simplify 4/12
  4. Simplify 27/36
  5. Simplify 30/50

Turning improper fractions into mixed numbers

To turn an improper fraction into a mixed number, simply divide the numerator by the denominator to find the whole number and then put the remainder over the original denominator and simplify if necessary, eg 9/6 = 1 3/6 = 1 1/2.

  1. Divide the numerator by the denominator
  2. Write down the answer to Step 1 as a whole number
  3. Put any remainder into a new fraction as the numerator, using the original denominator
  4. Simplify the fraction if necessary

Sample questions

  1. What is 22/7 as a mixed number?
  2. What is 16/5 as a mixed number?
  3. What is 8/3 as a mixed number?
  4. What is 18/8 as a mixed number?
  5. What is 13/6 as a mixed number?

Turning mixed numbers into improper fractions

To turn a mixed number into an improper fraction, multiply the whole number by the denominator of the fraction and add the existing numerator to get the new numerator while keeping the same denominator, eg 2 2/5 = (10 + 2)/5 = 12/5.

  1. Multiply the whole number by the denominator of the fraction
  2. Add the answer to the existing numerator to get the new numerator
  3. Write the answer over the original numerator
  4. Simplify if necessary

Sample questions

  1. What is 2 2/7 as an improper fraction?
  2. What is 3 2/3 as an improper fraction?
  3. What is 4 1/4 as an improper fraction?
  4. What is 5 1/5 as an improper fraction?
  5. What is 3 2/9 as an improper fraction?

There you go. Easy peasy lemon squeezy!

Number sequences

2-4-6-8 ain't never too late...

2-4-6-8 ain’t never too late…

Number sequences appear in Nature all over the place, from sunflowers to conch shells. They can also crop up either in Maths or Verbal Reasoning, and both are essential parts of 11+ and other school examinations. The trick is to be able to recognise the most common sequences and, if you find a different one, to work out the pattern so that you can find the missing values (or ‘terms’).

Common sequences

Here are a few of the commonest number sequences. For each one, I’ve given the rule for working out the nth term, where n stands for its position in the sequence.

Even numbers: 2, 4, 6, 8 etc… Rule: 2n
Odd numbers: 1, 3, 5, 7 etc… Rule: 2n – 1
Powers of 2: 2, 4, 8, 16 etc… Rule: 2ⁿ
Prime numbers: 2, 3, 5, 7 etc… Rule: n/a (each number is only divisible by itself and one)
Square numbers: 1, 4, 9, 16 etc… Rule: n²
Triangular numbers: 1, 3, 6, 10 etc… Rule: sum of the numbers from 1 to n
Fibonacci sequence
: 1, 1, 2, 3 etc… Rule: n₋₂ + n₋₁ (ie each successive number is produced by adding the previous two numbers together, eg 1 + 1 = 2, 1 + 2 = 3)

Here are a few questions for you to try. What are the next two numbers in each of the following sequences?

  1. 14, 16, 18, 20…
  2. 9, 16, 25, 36…
  3. 3, 6, 12, 24…
  4. 7, 11, 13, 17…
  5. 5, 8, 13, 21…

Working out the pattern

The best way to approach an unfamiliar sequence is to calculate the gaps between the terms. Most sequences involve adding or subtracting a specific number, eg 4 in the case of 5, 9, 13, 17 etc. Sometimes, the difference will rise or fall, as in 1, 2, 4, 7 etc. If you draw a loop between each pair of numbers and write down the gaps (eg +1 or -2), the pattern should become obvious, enabling you to work out the missing terms.

  • If the missing terms are in the middle of the sequence, you can still work out the pattern by using whatever terms lie next to each other, eg 1, …, 5, 7, …, 11 etc. You can confirm it by checking that the gap between every other term is double that between the ones next to each other.
  • If the gaps between terms are not the same and don’t go up (or down) by one each time, it may be that you have to multiply or divide each term by a certain number to find the next one, eg 16, 8, 4, 2 etc.
  • If the gaps go up and down, there may be two sequences mixed together, which means you’ll have to look at every other term to spot the pattern, eg 1, 10, 2, 8 etc. Here, every odd term goes up by 1 and every even term falls by two.

Generating a formula

At more advanced levels, you may be asked to provide the formula for a number sequence.

Arithmetic sequences

If the gap between the terms is the same, the sequence is ‘arithmetic’. The formula for the nth term of an arithmetic sequence is xn ± k, where x is the gap, n is the position of the term in the sequence and k is a constant that is added or subtracted to make sure the sequence starts with the right number, eg the formula for 5, 8, 11, 14 etc is 3n + 2. The gap between each term is 3, which means you have to multiply n by 3 each time and add 2 to get the right term, eg for the first term, n = 1, so 3n would be 3, but it should be 5, so you have to add 2 to it. Working out the formula for a sequence is particularly useful at 13+ or GCSE level, when you might be given a drawing of the first few patterns in a sequence and asked to predict, say, the number of squares in the 50th pattern. You can also work out the position of the pattern in the sequence if you are given the number of elements. You do this by rearranging the formula, ie by adding or subtracting k to the number of elements and dividing by 𝒳. For example, if 3n +2 is the formula for the number of squares in a tiling pattern, and you have 50 squares in a particular pattern, the number of that pattern in the sequence = (50-2) ÷ 3 = 48 ÷ 3 = 16.

Quadratic sequences

If the gap between the terms changes by the same amount each time, the sequence is ‘quadratic’, which just means there is a square number involved. The formula for a quadratic sequence is 𝒳n² ± k, where 𝒳 is half the difference between the gaps (or ‘second difference’), n is the position of the term in the sequence and k is a constant that is added or subtracted to make sure the sequence starts with the right number, eg the formula for 3, 9, 19, 33 etc is 2n² + 1. The differences between the terms are 6, 10, 14, so the second difference is 4, which means you need to multiply the square of n by 4 ÷ 2 = 2 and add 1, eg for the first term, n = 1, so 2n² would be 2, but it should be 3, so you have to add 1 to it.

Geometric sequences

If each term is calculated by multiplying the previous term by the same number each time, the sequence is ‘geometric’. The formula for the nth term of a geometric sequence (or progression) is ar(n-1), where a is the first term, r is the multiplier (or ‘common ratio’) and n is the position of the term in the sequence, eg the formula for 2, 8, 32, 128 etc is 2 x 4(n-1). The first term is 2, and each term is a power of 4 multiplied by 2, eg the fourth term = 2 x 4(4-1) = 2 x 43 = 2 x 64 = 128.

Here are a few questions for you to try. What is the formula for the nth term in each of the following sequences?

  1. 14, 16, 18, 20…
  2. -1, 3, 7, 11…
  3. 4, 6, 10, 16…
  4. 9, 7, 5, 3…
  5. 2, 6, 18, 54…

Fractions, decimals and percentages

Fractions, decimals and percentages

“Now, who wants 5/6 of this…?”

Pizzas are very useful, mathematically speaking. However much we hate fractions, we all know what half a pizza looks like, and that’s the point. Numbers don’t have any intrinsic meaning, and we can’t picture them unless they relate to something in the real world, so pizzas are just a useful way of illustrating fractions, decimals and percentages. They all do the same job of showing what share of something you have, and a common question involves converting from one to another, so here are a few tips…

Fractions to decimals

Calculator

  • Simply divide the numerator by the denominator, eg 3/4 = 3 ÷ 4 = 0.75.

Non-calculator

You can always use the standard ‘bus stop’ method to divide the numerator by the denominator on paper (or in your head), but the numbers may be easy enough for you to use a short cut.

  • If the denominator is a power of 10 (eg 10 or 100), write the numerator down straight away as a decimal. You just have to make sure you end up with the digits in the right columns, eg a fraction involving hundredths needs to end in the second column after the decimal point, so 29/100 = 0.29.
  • If the denominator ends in zero, you may be able to simplify the fraction into tenths first and then convert it into a decimal, eg 16/20 = 8/10 = 0.8.
  • If you express the fraction in its lowest terms by simplifying it (ie dividing the numerator and denominator by the same numbers until you can’t go any further), you may  recognise a common fraction that you can easily convert, eg 36/45 = 4/5 = 0.8.

Fractions to percentages

Calculator

  • Simply divide the numerator by the denominator, multiply by 100 and add the ‘%’ sign, eg 3/4 = 3 ÷ 4 x 100 = 0.75 x 100 = 75%.

Non-calculator

You can always convert the fraction into a decimal (see above) and then multiply by 100 and add the ‘%’ sign. Otherwise, try these short cuts in order.

  • If the denominator is a factor of 100 (eg 10, 20, 25 or 50), multiply the numerator by whatever number will turn the denominator into 100 and add the ‘%’ sign, eg 18/25 = 18 x 4 = 72%.
  • If the denominator is a multiple of 10 (eg 30, 40 or 70), divide the numerator by the first digit(s) of the denominator to turn the fraction into tenths, multiply the numerator by 10 and add the ‘%’ sign, eg 32/80 = 32 ÷ 8 x 10 = 4 x 10 = 40%.
  • If you express the fraction in its lowest terms by simplifying it (ie dividing the numerator and denominator by the same numbers until you can’t go any further), you may  recognise a common fraction that you can easily convert from memory, eg 8/64 = 1/8 = 12.5%.

Decimals to fractions

Every decimal is really a fraction in disguise, so the method is the same whether you’re allowed a calculator or not.

Calculator/non-calculator

  • Check the final column of the decimal (eg tenths or hundredths) and place all the digits over the relevant power of 10 (eg 100 or 1000) before simplifying if necessary, eg 0.625 = 625/1000 = 5/8.

Decimals to percentages

Again, this is an easy one, so the method is the same whether you’re allowed a calculator or not.

Calculator/non-calculator

  • Multiply by 100 and add the ‘%’ sign, eg 0.375 x 100 = 37.5%.

Percentages to fractions

You can think of a percentage as simply a fraction over 100, so the method is easy enough whether you’re allowed a calculator or not.

Calculator/non-calculator

  • If the percentage is a whole number, remove the ‘%’ sign, place the percentage over 100 and simplify if necessary, eg 75% = 75/100 = 3/4. If not, turn the fraction into a whole number as you go by multiplying the numerator and denominator by whatever number you need to (usually 2, 3 or 4), eg 37.5% = (37.5 x 2) / (100 x 2) = 75/200 = 3/8.

Percentages to decimals

This is easy enough, so the method is the same whether you’re allowed a calculator or not.

Calculator/non-calculator

  • Remove the ‘%’ sign and divide by 100, eg 70% ÷ 100 = 0.7.

Useful formulas

Useful formulas

Which one of these is it again…?

What is a problem? A problem = a fact + a judgment. That is a simple formula that tells us something about the way the world works. Maths is full of formulas, and that can intimidate some people if they don’t understand them or can’t remember the right one to use. However, formulas should be our friends, as they help us to do sometimes complex calculations accurately and repeatably in a consistent and straightforward way. The following is a list of the most useful ones I’ve come across while teaching Maths to a variety of students at a variety of ages and at a variety of stages in their education.

Averages

  • The mean is found by adding up all the values and dividing the total by how many there are, eg the mean of the numbers 1-10 is 5.5, as 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55, and 55 ÷ 10 = 5.5.
  • The mode is the most common value (or values), eg the mode of 1, 2, 2, 3, 4, 5 is 2.
  • The median of an odd number of values sorted by size is the one in the middle, eg the median of the numbers 1-5 is 3. The median of an even number of values is the mean of the two numbers in the middle, eg the median of the numbers 1-10 is 5.5, as 5 and 6 are the numbers in the middle, and 11 ÷ 2 = 5.5.
  • The range is the highest value minus the lowest, eg the range of the numbers 1-10 = 10 – 1 = 9.

Geometry

  • Angles around a point add up to 360º
  • Angles on a straight line add up to 180º
  • Opposite angles are equal, ie the two pairs of angles opposite each other when two straight lines bisect (or cross) each other
  • Alternate angles are equal, ie the angles under the arms of a ‘Z’ formed by a line (or ‘transversal’) bisecting two parallel lines
  • Corresponding angles are equal, ie the angles under the arms of an ‘F’ formed by a line (or ‘transversal’) bisecting two parallel lines
  • Complementary angles add up to 90º
  • Any straight line can be drawn using y = mx + c, where m is the gradient and c is the point where the line crosses the y-axis (the ‘y-intercept’)
  • The gradient of a straight line is shown by δy/δx (ie the difference in the y-values divided by the difference in the x-values of any two points on the line, usually found by drawing a triangle underneath it)

Polygons

  • Number of diagonals in a polygon = (n-3)(n÷2) where n is the number of sides
  • The sum of the internal angles of a polygon = (n-2)180º, where n is the number of sides
  • Any internal angle of a regular polygon = (n-2)180º ÷ n, where n is the number of sides

Rectangles

  • Perimeter of a rectangle = 2(l + w), where l = length and w = width
    Note that this is the same formula for the perimeter of an L-shape, too.
  • Area of a rectangle = lw, where l = length and w = width

Trapeziums

  • Area of a trapezium = lw, where l = average length and w = width (in other words, you have to add both lengths together and divide by two in order to find the average length)

Triangles (Trigonometry)

  • Area of a triangle = ½bh, where b = base and h = height
  • Angles in a triangle add up to 180º
  • Pythagoras’s Theorem: in a right-angled triangle, a² + b² = c², ie the area of a square on the hypotenuse (or longest side) is equal to the sum of the areas of squares on the other two sides

 

 

 

 

 

 

Circles

  • Circumference of a circle = 2πr, where r = radius
  • Area of a circle = πr², where r = radius
  • π = 3.14 to two decimal places and is sometimes given as 22/7

Spheres

  • Volume of a sphere = 4/3πr³, where r = radius
  • Surface area of a sphere = 4πr², where r = radius

Cuboids

  • Volume of a cuboid = lwh, where l is length, w is width and h is height
  • Surface area of a cuboid = 2(lw + lh + wh), where l is length, w is width and h is height

Number sequences

  • An arithmetic sequence (with regular intervals) = xn ± k, where x is the interval (or difference) between the values, n is the value’s place in the sequence and k is a constant that is added or subtracted to make sure the sequence starts at the right number, eg the formula for 5, 8, 11, 14…etc is 3n + 2
  • The sum of n consecutive numbers is n(n + 1)/2, eg the sum of the numbers 1-10 is 10(10 + 1)/2 = 110/2 = 55

Other

  • Speed = distance ÷ time
  • Profit = sales – cost of goods sold
  • Profit margin = profit ÷ sales
  • Mark-up = profit ÷ cost of goods sold

Short cuts

Short cuts

It’s just through here…

There is always more than one way of solving a Maths problem. That can be confusing, but it can also be an opportunity – if only you can find the right trade-off between speed and accuracy. I’ve taught a lot of QTS numeracy candidates recently, and the Maths itself isn’t particularly difficult, particularly in the mental arithmetic section. The trick is to be familiar with all the possible short cuts and capable of using the right one at the right time. It may mean having to do more sums, but it will be much simpler and quicker in the long run. You don’t have to use all of these all the time, but it is useful to know what they are just in case you need them.

  • Multiplying and dividing by 5
    The most useful short cut I’ve come across is very simple. To multiply by 5, try multiplying by 10 and then dividing by 2 (or vice versa), eg
    13 x 5
    = 13 x 10 ÷ 2
    = 130 ÷ 2
    = 65
    You have to do two sums rather than one, but the point is that you should be able to save time and improve the chances of getting the answer right by doing both in your head rather than having to work out a more difficult sum on paper.
    You can do divide by 5 in a similar way by multiplying by 2 and dividing by 10 (or vice versa), eg
    65 ÷ 5
    = 65 x 2 ÷ 10
    = 130 ÷ 10
    = 13
    You can do a similar trick with 50, 500 etc simply by multiplying or dividing by a higher power of ten.
  • Chunking
    If you have to multiply by a two-digit number outside your times tables, chunking is an easy way to do the sum in your head. Instead of writing it down on paper and using long multiplication (which would take a long time and is easy to get wrong!), try multiplying by the tens and the units separately and adding up the results, eg 16 x 15 = 10 x 15 + 6 x 15 = 150 + 90 = 240. The numbers might still be too tricky to do it comfortably, but it’s often worth a try.
  • Rounding
    To avoid sums with ‘tricky’ numbers, try rounding them up to the nearest ‘easy’ figure and adjusting at the end. This is particularly useful when working out start and end times, eg
    ‘The morning session in a school began at 09:25. There were three lessons of 50
    minutes each and one break of 20 minutes. At what time did the morning session end? Give your answer using the 24-hour clock.’
    If you assume the lessons last an hour, you can add three hours to 09:25 to get 12:25. You would normally then knock off 3 x 10 = 30 minutes, but the 20-minute break means you only need to subtract 10 minutes, which means the session ended at 12:15.
  • Money problems
    There is often a ‘real world’ money problem in the QTS numeracy test. That usually means multiplying three numbers together. The first thing to say is that it doesn’t matter in which order you do it – 1 x 2 x 3 is just the same as 3 x 2 x 1. The next thing to bear in mind is that you will usually have to convert from pence to pounds. You could do this at the end by simply dividing the answer by 100, but a better way is to divide one of the numbers by 100 (or two of the numbers by 10) at the beginning or turn multiplication by a fraction of a pound into a division sum, eg
    ‘All 30 pupils in a class took part in a sponsored spell to raise money for charity. The pupils were expected to get an average of 18 spellings correct each. The average amount of sponsorship was 20 pence for each correct spelling. How many pounds would the class expect to raise for charity?’
    The basic sum is 30 x 18 x 20p, and there are a couple of ways you could do this:
    1) Knock off the zeroes in two of the numbers, change the order of the numbers to make it easier and double and halve the last pair to give yourself a sum in your times tables, ie
    30 x 18 x 20p
    = 3 x 18 x 2
    = 3 x 2 x 18
    = 6 x 18
    = 12 x 9
    = £108
    2) Convert pence into pounds, turn it into a fraction, change the order of the numbers, divide by the denominator and, again, double and halve the last pair to give yourself a sum in your times tables, ie
    30 x 18 x 20p
    = 30 x 18 x £0.20
    = 30 x 18 x ⅕
    = 30 x 18 ÷ 5
    = 30 ÷ 5 x 18
    = 6 x 18
    = 12 x 9
    = £108
  • Percentages
    Many students get intimidated by percentages, fractions and decimals, but they are all just different ways of showing what share you have of something. You will often by asked to add or subtract a certain percentage. The percentage will usually end in zero (eg 20%, 30% or 40%), so the easiest way is probably to find 10% first. That just means dividing by 10, which means moving the decimal point one place to the left or, if you can, knocking off a zero. Once you know what 10% is, you can simply multiply by 2, 3 or 4 etc and add or subtract that number to find the answer, eg
    ‘As part of the numeracy work in a lesson, pupils were asked to stretch a spring to extend its length by 40 per cent. The original length of the spring was 45 centimetres. What should be the length of the extended spring? Give your answer in centimetres.’
    You need to find 40% of 45cm, so you can start by finding 10%, which is 45 ÷ 10 or 4.5cm. You can then multiply it by 4 to find 40%, which is best done by doubling twice, ie 4.5 x 2 x 2 = 9 x 2 = 18. Finally, you just add 18cm to the original length of the spring to find the answer, which is 45 + 18 = 63cm.
  • Common fractions
    An awful lot of questions involve converting between fractions, percentages and decimals. There is a proper technique for doing any of those, but it’s very useful if you learn the most common fractions and their decimal and percentage equivalents by heart, eg
    ½ = 0.5 = 50%
    ¼ = 0.25 = 25%
    ¾ = 0.75 = 75%
    ⅕ = 0.2 = 20%
    ⅖ = 0.4 = 40%
    ⅗ = 0.6 = 60%
    ⅘ = 0.8 = 80%
    ⅛ = 0.125 = 12.5%
    ⅜ = 0.375 = 37.5%
    ⅝ = 0.625 = 62.5%
    ⅞ = 0.875 = 87.5%
  • Times tables
    There are far more multiplication questions in the QTS numeracy test than any other kind, so it’s very important to know your times tables inside out. Some pupils are taught to memorise only the results, eg 4, 8, 12… etc. This is catastrophic! If you have to go through the whole table to find the answer, counting off the number of fours on your fingers, you can’t save yourself any time at all. The proper way is to learn the whole sum, eg 1 x 4 is 4, 2 x 4 is 8, etc (or 1 4 is 4, 2 4s are 8, etc). That way, the answer to any question in your times tables will pop into your head as soon as you’ve heard it. One good way of learning your tables is to time yourself using the stopwatch function on your iPhone. If you press ‘Lap’ after you’ve recited each table, you can write down your times and work out which tables you need to practise. Once you’re confident, you can make certain sums fit into your times tables by doubling one number and halving the other, eg
    3 x 24
    = (3 x 2) x (24 ÷ 2)
    = 6 x 12
    = 72
    Alternatively, you can halve just one of the numbers and double the result, eg
    24 x 9
    = 12 x 9 x 2
    = 108 x 2
    = 216
  • Multiplying by 4
    If you have to multiply by 4 and the number is not in your times tables, a simple way to do it is to double it twice, eg
    26 x 4
    = 26 x 2 x 2
    = 52 x 2
    = 104
  • Multiplying by a multiple of 10
    If you have to multiply by a multiple of 10 such as 20 or 30, try knocking the zero off and adding it in again afterwards. That way, you don’t have to do any long multiplication and, with any luck, the sum will be in your times tables, eg
    12 x 30
    = 12 x 3 x 10
    = 36 x 10
    = 360
  • Multiplying decimals
    This can be a bit confusing, so the best way of doing it is probably to ignore any decimal points, multiply the numbers together and then add back the decimal point to the answer so that you end up with the same number of decimal places as you had in the beginning, eg
    0.5 x 0.5
    = 5 x 5 ÷ 100
    = 25 ÷ 100
    = 0.25
  • Using the online calculator
    The second section of the QTS numeracy test consists of on-screen questions that can be answered using an online calculator. This obviously makes working out the answer a lot easier, and short cuts are therefore less useful. However, just because the calculator’s there doesn’t mean you have to use it, particularly for multiple-choice questions. If you have to add up a column of cash values, for example, and compare it with a number of options, you could simply tot up the number of pence and pick the option with the right amount. Alternatively, the level of accuracy needed in the answer may give you a helping hand if it rules out all but one of the possible answers, eg 6 ÷ 21 to one decimal place is always going to be 0.3. Why? Well, it’s a bit less than 7 ÷ 21, which would be a third or 0.3 recurring. An answer of 0.4 would be more than that, and 0.2 would be a fifth, which is far too small, so it must be 0.3.
  • Don’t do more than you have to!
    There are several types of question that could tempt you into doing more work than you need to do. If you’re trying to work out how many tables you need at a wedding reception for a given number of guests, the answer is always going to need rounding up to the next whole number, so you don’t need to spend any time working out the exact answer to one or two decimal places. Equally, some numbers are so close to being an ‘easy’ number that you don’t need to add or subtract anything after rounding up or down to make the basic sum easier, eg
    ‘For a science experiment a teacher needed 95 cubic centimetres of vinegar for each pupil. There were 20 pupils in the class. Vinegar comes in 1000 cubic centimetre
    bottles. How many bottles of vinegar were needed?
    If you round 95cc to 100cc, the answer is 20 x 100 ÷ 1000 or 2 bottles, and the remainder consisting of 20 lots of 5cc of vinegar can safely be ignored.