AQA Level 2 Further Maths revision notes

Number

Section N

3 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against AQA 8365 section N

Checked against AQA 8365 section N. Review basis: the qualification registry sourced from the AQA Level 2 Certificate in Further Mathematics (8365) specification; registry verification recorded 11 July 2026.

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Knowledge and use of numbers and the number system including fractions, decimals, percentages, ratio, proportion and order of operations

Notes

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Explanation

Fractions, decimals and percentages are equivalent ways to express a proportion, so conversion should put quantities into a common form before comparison or calculation. A ratio $a:b$ contains $a+b$ equal parts; dividing by the total number of parts finds one part before either share is scaled.
Percentage change uses a multiplier: $1+\frac{p}{100}$ for an increase and $1-\frac{p}{100}$ for a decrease.
Reverse percentage divides the final amount by the multiplier.
Order of operations applies brackets, powers and roots, multiplication and division, then addition and subtraction, with equal-priority operations completed from left to right.
Examiners expect exact arithmetic, a clear base quantity and an ordered chain of operations.

Worked example

After a $15\%$ increase, a fund is worth $\pounds 621$ . The original fund is shared in the ratio $7:5$ . Work out the smaller share.

1.Use the multiplier $1.15$ : original fund $=621\div1.15=540$ .
2.The ratio contains $7+5=12$ equal parts, so one part is $540\div12=45$ .
3.The smaller share is $5\times45=225$ .

Answer: The smaller share is $\pounds 225$ .

Common mistakes

Don't multiply $621$ by $0.85$ instead of dividing by $1.15$ in the reverse-percentage step.
Don't divide the fund by $7$ or $5$ instead of by the total $7+5=12$ ratio parts.
Don't perform addition before a multiplication that is not enclosed in brackets.

Exam tip

For a multi-step percentage-and-ratio question, state the multiplier and the value of one ratio part so each method mark is visible.

Tier 1 · Easy

ORIGINAL

Work out

\dfrac{3}{5}+0.35

[1 mark]

Tier 2 · Standard

ORIGINAL

A sum of

\pounds 294

is divided in the ratio

5:2

. Work out the larger share.

[2 marks]

Tier 3 · Hard

ORIGINAL

After an increase of

12\%

, a fund contains

\pounds 403.20

. Its original value is divided in the ratio

5:4

. Work out the smaller share.

[4 marks]

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The product rule for counting

Notes

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Your confidence:

Explanation

The product rule counts outcomes made by successive choices. If a first stage has $a$ possible choices and each can be followed by $b$ choices, there are $ab$ outcomes; further stages add further factors.
The number of choices must be reconsidered at each stage when repetition is forbidden or a restriction applies. A tree diagram, labelled slots or a systematic list can expose these stage counts.
Sometimes it is shorter to count every unrestricted outcome and subtract forbidden cases, provided those cases do not overlap.
Examiners expect a product whose factors correspond to the available choices, not merely a final number.
Addition applies to separate, mutually exclusive routes, whereas multiplication applies to choices made together.

Worked example

A code contains two different letters chosen from A, B, C, D and E, followed by three different digits chosen from $1,2,3,4$ . How many codes are possible?

1.The two letter positions have $5$ choices followed by $4$ choices.
2.The three digit positions have $4$ , then $3$ , then $2$ choices because repetition is forbidden.
3.Apply the product rule: $5\times4\times4\times3\times2=480$ .

Answer: $480$ codes are possible.

Common mistakes

Don't use $5^2\times4^3$ even though a letter or digit cannot be repeated.
Don't reduce the number of digit choices after choosing letters, even though the two sets are separate.
Don't add the stage counts $5+4+4+3+2$ instead of multiplying them.

Exam tip

Place the most restricted position first, then show one factor for the number of choices remaining at every stage.

Tier 1 · Easy

ORIGINAL

A meal has a choice of

4

main courses and

3

desserts. How many different main-course-and-dessert meals are possible?

[1 mark]

Tier 2 · Standard

ORIGINAL

A code consists of one of

5

letters, followed by one of

4

digits, followed by one of

3

symbols. Repetition is allowed. Work out the number of possible codes.

[2 marks]

Tier 3 · Hard

ORIGINAL

A sandwich uses one of

4

breads, one of

6

fillings and one of

3

sauces. For rye bread, two of the fillings are unavailable. Work out the number of available sandwiches.

[3 marks]

Manipulation of surds, including rationalising the denominator; the use of surds in exact calculations

Notes

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Your confidence:

Explanation

A surd is an irrational root kept in exact form. Simplification uses $\sqrt{ab}=\sqrt a\sqrt b$ for non-negative $a$ and $b$ , extracting the largest square factor where possible.
Only like surds can be collected, just as only like algebraic terms can be collected. Exact calculations should retain surds rather than rounded decimals.
To rationalise a denominator containing one surd, multiply numerator and denominator by that surd.
For a two-term denominator $a+b\sqrt c$ , multiply by the conjugate $a-b\sqrt c$ ; the denominator becomes rational through the difference of two squares.
Examiners expect the final answer simplified, with no surd left in the denominator and no unnecessary common factor.

Worked example

Simplify $\dfrac{3\sqrt{8}+\sqrt{18}}{\sqrt{3}}$ and give the exact value.

1.Simplify the numerator: $3\sqrt8+\sqrt{18}=6\sqrt2+3\sqrt2=9\sqrt2$ .
2.Rationalise: $\dfrac{9\sqrt2}{\sqrt3}\times\dfrac{\sqrt3}{\sqrt3}=\dfrac{9\sqrt6}{3}$ .
3.Cancel the common factor to obtain $3\sqrt6$ .

Answer: $3\sqrt6$ .

Common mistakes

Don't write $\sqrt8=4\sqrt2$ instead of extracting the square factor to get $2\sqrt2$ .
Don't collect unlike surds, for example treating $\sqrt2+\sqrt3$ as $\sqrt5$ .
Don't multiply only the denominator by the rationalising surd and change the value of the expression.

Exam tip

In an exact-value question, simplify every root and rationalise the denominator before presenting the final line.

Tier 1 · Easy

ORIGINAL

Simplify

\sqrt{72}

[2 marks]

Tier 2 · Standard

ORIGINAL

Rationalise the denominator of

\dfrac{5}{\sqrt{7}-2}

[3 marks]

Tier 3 · Hard

ORIGINAL

Work out the exact value of

\dfrac{3+\sqrt{5}}{\sqrt{5}-1}-\sqrt{5}

[4 marks]

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