# Maths GCSE Top Topics tutored

At MyTutor, we’ve got lots of dedicated Maths tutors across the UK who love helping teens achieve their best when exams come around. Since we started in 2013, we’ve given more than 250,000 one-to-one lessons, and over 1 million school pupils have used our online resource centre. Over time we’ve been able to get a strong understanding of exactly the topics in each subject that kids tend to need that extra help with.

Here are the top five Maths GCSE topics our students struggle most with, and some example answers written by our tutors. These can double as handy study notes for your child to get them revision-ready.

## 1. Lowest common multiples and highest common factors.

Top tip: As there are different methods to approach these questions, students need to fully understand the purpose of each step before they begin to tackle the question.

Example question: What is the difference between LCM and HCF?

LCM stands for Lowest Common Multiple, and HCF stands for Highest Common Factor.

The key to telling the difference between these two things is knowing the difference between a multiple and a factor.

A multiple of an integer (whole number) is any integer that appears in its times table. For example, the multiples of 3 are 3, 6, 9, 12, and so on.

A factor of an integer is any integer that divides the integer with no remainder. For example, the factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36.

We use LCM and HCF to compare two (or more) integers.

The LCM of two integers is the smallest whole number that appears in both of their times tables, that is, the smallest integer that is a multiple of both numbers.

For example, the LCM of 4 and 5 is 20. To see this, look at the multiples (times table) of 4:

4, 8, 12, 16, 20, 24, 28, …

and of 5:

5, 10, 15, 20, 25, …

The LCM is 20 because this is the first number that appears in both lists.

The HCF of two integers is the largest whole number that divides both numbers without leaving a remainder.

For example, the HCF of 16 and 24 is 8. Again, we can look at both sets of factors and compare. The factors of 16 are:

1, 2, 4, 8, 16

and the factors of 24 are:

1, 2, 3, 4, 6, 8, 12, 24.

Although 2 and 4 are common factors (that is, they appear in both lists), we are looking for the highest common factor. The answer is 8.

## 2. Expand and simplify equations in brackets

Top tip: Don’t forget the rules of ‘signs’ when multiplying the number outside the bracket with each term inside the bracket. E.g – 2X (4-3x)= (-2x*4) + (-2x*-3x).

Example question: Expand and simplify 3(2x + 5) – 2(x – 4)

Firstly, to expand an equation like this, you must multiply the brackets by the number outside of the brackets. Make sure that you multiply every number inside the bracket by the number directly outside, and remember the signs:

3(2x + 5) – 2(x-4) becomes

(3 * 2x) + (3 * 5) + (-2 * x) + (-2 * -4) =

6x + 15 – 2x + 8 (remember that ‘-‘ * ‘-‘ = ‘+’)

Then you need to do something called ‘collecting the like terms’. This means collecting together all the ‘x’ terms and all of the ‘number’ terms, like this:

6x + 15 – 2x + 8 becomes

(6x – 2x) + (15 + 8), working this out means: The answer is: 4x + 23.

## 3. Converting decimals to fractions

Example question: How would you convert the decimal 0.125 into a fraction in its simplest form?

0.125 is the same as 125/1000. We have to remove the decimal point before creating a fraction.

0.125 multiplied by 1000 is 125 and at this point the decimal point has been removed, so 0.125 is equal to 125/1000. But 125/1000 is not the simplest form that the fraction could be in.

If we divide both the top and the bottom by 5 since we know 125 is a multiple of 5, we get 25/200.

Divide both by 5 again to get 5/40.

Divide by 5 one more time to get 1/8, and since the numerator is 1, the fraction cannot be simplified any further.

Therefore, the answer in its simplest form is 1/8.

## 4. Finding “nth terms”

Top tip: It’s easy to forget the different methods for linear sequences and quadratic sequences. As it’s a quadratic sequence it needs extra steps!

Example question: How do you find the nth term formula for a sequence with non-constant difference?

Take the sequence;
9, 12, 19, 30, …
(1) The first step is always to look at the difference between the terms;

9, 12, 19, 30, …
+3, +7, +11, …
+4, +4, …

We can see the difference is not constant, (2) so we looked at the change in the difference each term.

This gives a constant change in the difference of an extra +4 each term. The fact that we needed to take 2 turns to find the constant difference means we are dealing with a quadratic sequence.

(3) Furthermore, because the difference is +4, we are dealing with a 2n2 sequence.
If the change in the difference is (a) then the nth term follows a (1/2a)n2 pattern.

(4) Now we can rewrite the sequence as follows;
n n2 2n2
9 1 1 2
12 2 4 8
19 3 9 18
30 4 16 32

(5) We need to find the difference between the sequence and 2n2.
2n2 d
9 2 -7
12 8 -4
19 18 -1
30 32 +2

(6) The difference here will either be a constant number, in which case the nth term is (1/2a)n2 +d. Or like in this case, will itself follow a linear sequence with constant difference, which we should know how to solve.
1 2 3 4
-7, -4, -1, +2
+3 +3 +3
This gives 3n – 10.

Therefore, the whole formula for the nth term is;

(7) 2n2 + 3n – 10

Example question: Solve the quadratic equation x^2 + 4x +1 = 0 by completing the square.

Completing the square means to put our equation into a slightly different form which looks like this, where a and b are real numbers:

(x+a)2 + b = 0

From here, we can rearrange the equation and directly solve for x. Let’s have a look at our specific example:

x2 +4x +1 = 0

The first step is to divide the coefficient of x by 2, and add this to x (this is our value of ‘a’ to go inside our bracket). We then square this value of a and subtract it outside the bracket.

In our example it will look like this:

(x+2)2 – 4 + 1 = 0
(x+2)2 – 3 = 0

We have our equation in completed square form.

[There is a quick way to check we’ve got this right by expanding out this equation quickly:

(x+2)(x+2) – 3 = 0
x2 + 4x + 4 – 3 = 0
x2 + 4x +1 = 0
We’re back to our original equation, so we know we’ve got it right. Let’s go and solve our equation in completed square form.]

We simply rearrange for x:

(x+2)2 – 3 = 0

(x+2)2 = 3

Take the square root of both sides. This splits into two possible cases:

Case 1: Positive square root of 3
x+2 = + sqrt(3)
x = – 2 + sqrt(3)

Case 2: Negative square root of 3
x+2 = – sqrt(3)
x = – 2 – sqrt(3)

x = – 2 + sqrt(3)
x = – 2 – sqrt(3)

Good luck! You can find your own Maths tutor for extra one-to-one support today.

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