N Number — revision question pack

3 specification points · notes, questions, answers and worked methods

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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N1 · Knowledge and use of numbers and the number system including fractions, decimals, percentages, ratio, proportion and order of operations

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

1
Work out $\dfrac{3}{5}+0.35$ .
[1 mark]
2
Work out $\dfrac{3}{4}-0.2\times\dfrac{5}{8}$ .
[2 marks]

Tier 2 · Standard

1
A sum of $\pounds 294$ is divided in the ratio $5:2$ . Work out the larger share.
[2 marks]
2
A membership fee of $\pounds 240$ is increased by $18\%$ . The increased fee is then reduced by $15\%$ . Work out the final fee.
[3 marks]
3
Eight identical pumps, working at the same constant rate, transfer $540$ litres in $15$ minutes. Work out how many litres six pumps transfer in $22$ minutes.
[3 marks]

Tier 3 · Hard

1
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]
2
A tank initially contains red liquid and blue liquid in the ratio $3:5$ . After $12$ litres of blue liquid are added, the ratio of red to blue is $3:7$ . Work out the initial total volume of liquid.
[4 marks]
3
Nia and Omar share $\pounds 560$ . Nia spends $35\%$ of her share and Omar spends $\dfrac{1}{4}$ of his share. They then have the same amount left. Work out Nia's original share.
[4 marks]
4
A charity allocates $\dfrac{3}{8}$ of a fund to equipment. It then allocates $20\%$ of the remaining fund to transport. The unallocated amount is $\pounds 1040$ . Work out the original value of the fund.
[4 marks]
5
Six identical machines, working at the same constant rate for $7.5$ hours, make $2700$ components. Eight identical machines working at the same rate then work for $6$ hours, but $5\%$ of the components made are rejected. Work out the number of components that are accepted.
[4 marks]

N2 · The product rule for counting

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

1
A meal has a choice of $4$ main courses and $3$ desserts. How many different main-course-and-dessert meals are possible?
[1 mark]
2
Three lamps, $A$ , $B$ and $C$ , are each set to amber or white. How many different colour settings are possible?
[1 mark]

Tier 2 · Standard

1
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]
2
Seven runners are available. A captain and a deputy captain are chosen. The same runner cannot hold both roles. Work out how many different ways the two roles can be filled.
[2 marks]
3
A panel displays a row of three symbols. Each position shows one of six symbols, one of which is a star, and symbols may be repeated. At least one position must show a star. Work out the number of possible displays.
[3 marks]

Tier 3 · Hard

1
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]
2
A display must show an even four-digit integer. Its digits are chosen without repetition from $0,1,2,3,4,5$ , and the first digit cannot be zero. Work out the number of possible integers.
[3 marks]
3
A four-character code is made using exactly two of the letters A, B, C, D, E and exactly two of the digits $1,2,3,4$ . No character is repeated, and letters and digits can appear in any order. Work out the number of possible codes.
[3 marks]
4
A school has $3$ Year 11 students and $5$ Year 10 students available to fill the roles of chair, secretary and treasurer. Each role must be held by a different student, and exactly one role must be held by a Year 11 student. Work out the number of ways the roles can be filled.
[3 marks]
5
A journey from A to D passes through B and C. There are $3$ roads from A to B, $4$ roads from B to C and $5$ roads from C to D. On the return journey, exactly one of the three sections must use the same road as the outward journey; each other section must use a different road. Work out the number of possible outward-and-return journeys.
[3 marks]

N3 · Manipulation of surds, including rationalising the denominator; the use of surds in exact calculations

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

1
Simplify $\sqrt{72}$ .
[2 marks]
2
Simplify $\sqrt{50}-\sqrt{8}$ .
[2 marks]

Tier 2 · Standard

1
Rationalise the denominator of $\dfrac{5}{\sqrt{7}-2}$ .
[3 marks]
2
Work out the exact value of $(\sqrt{12}+\sqrt{3})(\sqrt{27}-\sqrt{3})$ .
[3 marks]
3
Work out the exact value of $\dfrac{1}{\sqrt{6}+\sqrt{2}}+\dfrac{1}{\sqrt{6}-\sqrt{2}}$ .
[3 marks]

Tier 3 · Hard

1
Work out the exact value of $\dfrac{3+\sqrt{5}}{\sqrt{5}-1}-\sqrt{5}$ .
[4 marks]
2
A rectangle has area $18\text{ cm}^2$ and length $(3+\sqrt{3})\text{ cm}$ . Work out its perimeter, writing it as $(a+b\sqrt{3})\text{ cm}$ for integers $a$ and $b$ .
[4 marks]
3
Let $x=3+2\sqrt{2}$ . Show that $\sqrt{x}=1+\sqrt{2}$ , and hence work out the exact value of $\sqrt{x}+\dfrac{1}{\sqrt{x}}$ .
[4 marks]
4
A rectangle has length $(2+\sqrt{3})\text{ cm}$ and width $(2-\sqrt{3})\text{ cm}$ . Work out the exact length of its diagonal.
[3 marks]
5
Work out the exact value of $(\sqrt{5}+\sqrt{3})^3+(\sqrt{5}-\sqrt{3})^3$ .
[4 marks]

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

N1 · Knowledge and use of numbers and the number system including fractions, decimals, percentages, ratio, proportion and order of operations

Tier 1 · Easy

Mark scheme for N1 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$0.95$	1	Convert $\dfrac{3}{5}$ to $0.6$ . Then $0.6+0.35=0.95$ .
2	$\dfrac{5}{8}$ (or $0.625$ )	2	Complete the multiplication first: $0.2\times\dfrac{5}{8}=\dfrac{1}{5}\times\dfrac{5}{8}=\dfrac{1}{8}$ . Therefore $\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{6}{8}-\dfrac{1}{8}=\dfrac{5}{8}$ .

Tier 2 · Standard

Mark scheme for N1 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$\pounds 210$	2	There are $5+2=7$ equal parts. One part is $294\div7=42$ , so the larger share is $5\times42=210$ .
2	$\pounds 240.72$	3	Use successive multipliers. The final fee is $240\times1.18\times0.85=240.72$ , so it is $\pounds 240.72$ .
3	$594$ litres	3	One pump transfers $540\div(8\times15)=4.5$ litres per minute. Six pumps therefore transfer $6\times4.5=27$ litres per minute. In $22$ minutes they transfer $27\times22=594$ litres.

Tier 3 · Hard

Mark scheme for N1 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$\pounds 160$	4	The multiplier for a $12\%$ increase is $1.12$ . The original fund was $403.20\div1.12=360$ . The ratio has $5+4=9$ parts, so one part is $360\div9=40$ . The smaller share is $4\times40=160$ .
2	$48$ litres	4	Let the initial amounts be $3k$ litres and $5k$ litres. The new ratio gives $\dfrac{3k}{5k+12}=\dfrac{3}{7}$ . Hence $21k=15k+36$ , so $k=6$ . The initial total is $3k+5k=8k=48$ litres.
3	$\pounds 300$	4	Nia keeps $65\%$ of her share and Omar keeps $\dfrac{3}{4}$ . If their original shares are $N$ and $O$ , then $0.65N=0.75O$ , so $13N=15O$ and $N:O=15:13$ . There are $28$ ratio parts, each worth $560\div28=20$ . Therefore $N=15\times20=300$ , so Nia's original share was $\pounds 300$ .
4	$\pounds 2080$	4	After the equipment allocation, $\dfrac{5}{8}$ of the fund remains. The transport allocation is $20\%\times\dfrac{5}{8}=\dfrac{1}{8}$ of the original fund. The unallocated fraction is therefore $1-\dfrac{3}{8}-\dfrac{1}{8}=\dfrac{1}{2}$ . Hence the original fund is $1040\div\dfrac{1}{2}=2080$ , so it was $\pounds 2080$ .
5	$2736$ components	4	The production rate is $2700\div(6\times7.5)=60$ components per machine-hour. Eight machines working for $6$ hours make $8\times6\times60=2880$ components. The accepted proportion is $95\%$ , so the number accepted is $2880\times0.95=2736$ .

N2 · The product rule for counting

Tier 1 · Easy

Mark scheme for N2 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$12$ meals	1	Each of the $4$ main courses can be paired with any of the $3$ desserts, so the product rule gives $4\times3=12$ .
2	$8$ settings	1	Each of the three lamps has $2$ choices independently, so the product rule gives $2\times2\times2=8$ settings.

Tier 2 · Standard

Mark scheme for N2 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$60$ codes	2	There are $5$ choices for the first position, $4$ for the second and $3$ for the third. Hence there are $5\times4\times3=60$ codes.
2	$42$ ways	2	There are $7$ choices for captain. After that choice, $6$ runners remain available for deputy captain. Therefore there are $7\times6=42$ ways.
3	$91$ displays	3	There are $6\times6\times6=216$ unrestricted displays. If no position shows a star, each position has $5$ choices, giving $5\times5\times5=125$ displays. Therefore the required number is $216-125=91$ .

Tier 3 · Hard

Mark scheme for N2 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$66$ sandwiches	3	Without the restriction there are $4\times6\times3=72$ sandwiches. With rye bread, the two unavailable fillings would each pair with $3$ sauces, giving $1\times2\times3=6$ forbidden sandwiches. Therefore $72-6=66$ are available.
2	$156$ integers	3	If the final digit is $0$ , the first digit has $5$ choices and the middle digits have $4$ then $3$ choices, giving $5\times4\times3=60$ . If the final digit is $2$ or $4$ , there are $2$ choices for it, $4$ non-zero choices for the first digit, then $4$ and $3$ choices for the middle digits. This gives $2\times4\times4\times3=96$ . Altogether there are $60+96=156$ integers.
3	$1440$ codes	3	The two letter positions can occur in $6$ patterns: LLDD, LDLD, LDDL, DLLD, DLDL and DDLL. For each pattern, the letters can be chosen in order in $5\times4$ ways and the digits in $4\times3$ ways. Hence the number of codes is $6\times5\times4\times4\times3=1440$ .
4	$180$ ways	3	There are $3$ choices for the role held by a Year 11 student and then $3$ choices for that student. The other two distinct roles are filled by Year 10 students in $5\times4$ ways. Therefore the number of ways is $3\times3\times5\times4=180$ .
5	$1560$ journeys	3	There are $3\times4\times5=60$ outward journeys. For a fixed outward journey, if the repeated road is on AB, there are $3\times4=12$ return choices for the other sections; if it is on BC, there are $2\times4=8$ ; if it is on CD, there are $2\times3=6$ . These cases are separate, so there are $12+8+6=26$ valid returns for each outward journey. Hence there are $60\times26=1560$ outward-and-return journeys.

N3 · Manipulation of surds, including rationalising the denominator; the use of surds in exact calculations

Tier 1 · Easy

Mark scheme for N3 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$6\sqrt{2}$	2	Use the largest square factor: $72=36\times2$ . Therefore $\sqrt{72}=\sqrt{36}\sqrt{2}=6\sqrt{2}$ .
2	$3\sqrt{2}$	2	$\sqrt{50}=5\sqrt{2}$ and $\sqrt{8}=2\sqrt{2}$ . Therefore $\sqrt{50}-\sqrt{8}=3\sqrt{2}$ .

Tier 2 · Standard

Mark scheme for N3 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$\dfrac{5\sqrt{7}+10}{3}$	3	Multiply by the conjugate: $\dfrac{5}{\sqrt{7}-2}\times\dfrac{\sqrt{7}+2}{\sqrt{7}+2}=\dfrac{5\sqrt{7}+10}{7-4}=\dfrac{5\sqrt{7}+10}{3}$ .
2	$18$	3	Simplify each bracket: $\sqrt{12}+\sqrt{3}=2\sqrt{3}+\sqrt{3}=3\sqrt{3}$ and $\sqrt{27}-\sqrt{3}=3\sqrt{3}-\sqrt{3}=2\sqrt{3}$ . Their product is $3\sqrt{3}\times2\sqrt{3}=6\times3=18$ .
3	$\dfrac{\sqrt{6}}{2}$ (or $\tfrac12\sqrt{6}$ )	3	Use the common denominator $(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})=6-2=4$ . The numerator is $(\sqrt{6}-\sqrt{2})+(\sqrt{6}+\sqrt{2})=2\sqrt{6}$ . Therefore the exact value is $\dfrac{2\sqrt{6}}{4}=\dfrac{\sqrt{6}}{2}$ .

Tier 3 · Hard

Mark scheme for N3 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$2$	4	Rationalise the fraction using $\sqrt{5}+1$ . Its numerator becomes $(3+\sqrt{5})(\sqrt{5}+1)=8+4\sqrt{5}$ and its denominator becomes $5-1=4$ . The fraction is therefore $2+\sqrt{5}$ , so subtracting $\sqrt{5}$ gives $2$ .
2	$(24-4\sqrt{3})\text{ cm}$	4	The width is $\dfrac{18}{3+\sqrt{3}}=\dfrac{18(3-\sqrt{3})}{9-3}=9-3\sqrt{3}$ cm. Hence the perimeter is $2\big((3+\sqrt{3})+(9-3\sqrt{3})\big)=2(12-2\sqrt{3})=24-4\sqrt{3}$ cm.
3	$(1+\sqrt{2})^2=1+2\sqrt{2}+2=3+2\sqrt{2}$ and $1+\sqrt{2}>0$ , so $\sqrt{x}=1+\sqrt{2}$ $\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1$ , so $\sqrt{x}+\dfrac{1}{\sqrt{x}}=2\sqrt{2}$	4	Since $(1+\sqrt{2})^2=1+2\sqrt{2}+2=3+2\sqrt{2}$ and $1+\sqrt{2}$ is positive, $\sqrt{x}=1+\sqrt{2}$ . Also $\dfrac{1}{1+\sqrt{2}}=\dfrac{1-\sqrt{2}}{1-2}=\sqrt{2}-1$ . Hence $\sqrt{x}+\dfrac{1}{\sqrt{x}}=(1+\sqrt{2})+(\sqrt{2}-1)=2\sqrt{2}$ .
4	$\sqrt{14}\text{ cm}$	3	By Pythagoras, the square of the diagonal is $(2+\sqrt{3})^2+(2-\sqrt{3})^2$ . This is $(7+4\sqrt{3})+(7-4\sqrt{3})=14$ . The diagonal is a positive length, so its exact length is $\sqrt{14}\text{ cm}$ .
5	$28\sqrt{5}$	4	Expanding the first cube gives $5\sqrt{5}+15\sqrt{3}+9\sqrt{5}+3\sqrt{3}=14\sqrt{5}+18\sqrt{3}$ . Expanding the second gives $5\sqrt{5}-15\sqrt{3}+9\sqrt{5}-3\sqrt{3}=14\sqrt{5}-18\sqrt{3}$ . Adding cancels the $\sqrt{3}$ terms, leaving $28\sqrt{5}$ .