N Number — printable coverage pack

N1 · Order positive and negative integers, decimals and fractions; use the symbols =, ≠, <, >, ≤, ≥

Numbers can be compared reliably after converting them to a common form, such as decimals or fractions with a common denominator.
On a number line, values increase from left to right; among negative numbers, the value farther from zero is the smaller one.
For example, $-1.7<-\dfrac{3}{2}$ because $-1.7<-1.5$ , while $0.4=\dfrac{2}{5}$ .
A common error is to compare only the digits and claim that $-8>-3$ ; the inequality is reversed because $-8$ lies farther left.

Tier 1 · Easy

1. Insert either $<$ , $>$ or $=$ between $-0.62$ and $-\dfrac{3}{5}$ .[1 mark]
Answer
- $-0.62<-\dfrac{3}{5}$
Method: Convert $-\dfrac{3}{5}$ to $-0.6$ . Since $-0.62$ is farther left on the number line than $-0.6$ , $-0.62<-\dfrac{3}{5}$ .

Tier 2 · Standard

1. Write $-\dfrac{7}{4}$ , $-1.68$ , $\dfrac{5}{3}$ and $1.7$ in ascending order.[2 marks]
Answer
- $-\dfrac{7}{4},\ -1.68,\ \dfrac{5}{3},\ 1.7$
Method: Use decimal comparisons: $-\dfrac{7}{4}=-1.75$ and $\dfrac{5}{3}=1.666\ldots$ . Therefore $-1.75<-1.68<1.666\ldots<1.7$ .

Tier 3 · Hard

1. An integer $k$ satisfies $-2.4<\dfrac{k}{3}\leq1.7$ . Write down every possible value of $k$ .[3 marks]
Answer
- $k=-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5$
Method: Multiply every part by positive $3$ , so the inequality signs stay unchanged: $-7.2<k\leq5.1$ . The integers in this interval are $-7$ through $5$ inclusive.

N2 · Apply the four operations, including formal written methods, to integers, decimals and simple fractions (proper and improper), and mixed numbers, positive and negative; understand and use place value

The four operations apply to integers, decimals, fractions and mixed numbers; convert mixed numbers to improper fractions before multiplying or dividing.
For addition or subtraction of fractions, use a common denominator; for division by a fraction, multiply by its reciprocal.
For example, $-2\dfrac{1}{4}\div\dfrac{3}{5}=-\dfrac{9}{4}\times\dfrac{5}{3}=-\dfrac{15}{4}$ .
A common error is to ignore place value in decimal work, so align decimal points in written addition and subtraction rather than aligning final digits.

Tier 1 · Easy

1. Work out $-17+29$ .[1 mark]
Answer
- $12$
Method: The signs are different, so find $29-17=12$ and use the sign of the number with the larger magnitude. The result is $12$ .

Tier 2 · Standard

1. Work out $-2\dfrac{1}{4}\div\dfrac{3}{5}$ .[2 marks]
Answer
- $-\dfrac{15}{4}$
- $-3\dfrac{3}{4}$
Method: Write $-2\dfrac{1}{4}=-\dfrac{9}{4}$ and multiply by the reciprocal of $\dfrac{3}{5}$ . This gives $-\dfrac{9}{4}\times\dfrac{5}{3}=-\dfrac{45}{12}=-\dfrac{15}{4}=-3\dfrac{3}{4}$ .

Tier 3 · Hard

1. Work out $3.6-(-1.75)+\dfrac{5}{6}\div\left(-\dfrac{5}{9}\right)$ . Give your answer as a decimal.[3 marks]
Answer
- $3.85$
Method: First $\dfrac{5}{6}\div\left(-\dfrac{5}{9}\right)=\dfrac{5}{6}\times\left(-\dfrac{9}{5}\right)=-\dfrac{3}{2}=-1.5$ . Then $3.6+1.75-1.5=5.35-1.5=3.85$ .

N3 · Recognise and use relationships between operations, including inverse operations; use conventional notation for priority of operations, including brackets, powers, roots and reciprocals

Inverse operations undo each other: addition and subtraction, multiplication and division, squaring and taking a square root are inverse pairs in suitable domains.
Use brackets first, then powers and roots, then multiplication and division, then addition and subtraction; work left to right within the same level.
For example, $18-3^2\times2=18-9\times2=0$ , not $162$ or $18$ .
A common error is to treat a reciprocal as a negative: the reciprocal of $\dfrac{a}{b}$ is $\dfrac{b}{a}$ , whereas its additive inverse is $-\dfrac{a}{b}$ .

Tier 1 · Easy

1. Work out $18-3^2\times2$ .[1 mark]
Answer
- $0$
Method: Evaluate the power first: $3^2=9$ . Then multiply, $9\times2=18$ , and subtract: $18-18=0$ .

Tier 2 · Standard

1. Work out $\left(\sqrt{81}-5\right)^2\div2$ .[2 marks]
Answer
- $8$
Method: Inside the brackets, $\sqrt{81}-5=9-5=4$ . Then $4^2=16$ and $16\div2=8$ .

Tier 3 · Hard

1. A positive number is squared, $11$ is subtracted, and the reciprocal of the result is $\dfrac{1}{14}$ . Find the original number.[3 marks]
Answer
- $5$
Method: Undo the reciprocal first: the result before taking the reciprocal was $14$ . Add $11$ to undo the subtraction, giving $25$ . The positive square root of $25$ is $5$ .

N4 · Prime numbers, factors (divisors), multiples, common factors and multiples, highest common factor, lowest common multiple, prime factorisation with product notation and unique factorisation theorem

A prime number has exactly two positive factors, while every integer greater than $1$ has a unique prime factorisation apart from the order of its factors.
Write each number as a product of prime powers; the HCF uses the smaller shared powers and the LCM uses the largest powers present.
For example, $84=2^2\times3\times7$ and $126=2\times3^2\times7$ , so their HCF is $2\times3\times7=42$ .
A common error is to count $1$ as prime or to find the LCM by multiplying the numbers without removing repeated prime factors.

Tier 1 · Easy

1. Write $756$ as a product of its prime factors.[2 marks]
Answer
- $756=2^2\times3^3\times7$
Method: Divide successively by primes: $756=2\times378=2^2\times189=2^2\times3^3\times7$ .

Tier 2 · Standard

1. Find both the HCF and the LCM of $84$ and $126$ .[3 marks]
Answer
- HCF $=42$
- LCM $=252$
Method: Use $84=2^2\times3\times7$ and $126=2\times3^2\times7$ . The smaller common powers give $2\times3\times7=42$ . The largest powers give $2^2\times3^2\times7=252$ .

Tier 3 · Hard

1. $180n$ is a cube number, where $n$ is a positive integer. Find the smallest possible value of $n$ .[4 marks]
Answer
- $n=150$
Method: Prime factorise $180=2^2\times3^2\times5$ . A cube needs every exponent to be a multiple of $3$ , so multiply by $2\times3\times5^2$ . Therefore $n=2\times3\times25=150$ , and $180n=27000=30^3$ .

N5 · Apply systematic listing strategies, including use of the product rule for counting (m ways of doing one task and n ways of doing another gives m × n ways in total)

A systematic list uses a fixed order so that every possible outcome appears once and no outcome is repeated.
Fix one choice while cycling through every permitted value of the next choice, then move to the next case.
For example, pairing A and B with $1$ , $2$ and $3$ gives A₁, A₂, A₃, B₁, B₂, B₃.
A common error is to change two choices at once, which can omit an outcome or list the same outcome twice.

Tier 1 · Easy

1. A badge uses one of the letters A, B or C and one of the numbers $1$ , $2$ or $3$ . List every possible badge code in a systematic order.[2 marks]
Answer
- A₁, A₂, A₃, B₁, B₂, B₃, C₁, C₂, C₃
Method: Hold the letter fixed while cycling through the numbers: A₁, A₂, A₃; then B₁, B₂, B₃; then C₁, C₂, C₃. This gives all $9$ codes once each.

Tier 2 · Standard

1. A three-digit number is made using three different digits from $2$ , $4$ , $5$ and $7$ . List all the possible even numbers.[3 marks]
Answer
- $452$ , $472$ , $542$ , $572$ , $742$ , $752$ , $254$ , $274$ , $524$ , $574$ , $724$ , $754$
Method: List by the final digit. Ending in $2$ gives $452$ , $472$ , $542$ , $572$ , $742$ , $752$ . Ending in $4$ gives $254$ , $274$ , $524$ , $574$ , $724$ , $754$ . The fixed final digit makes the list systematic and complete.

Tier 3 · Hard

1. Two different numbers are selected from $1$ , $2$ , $3$ , $4$ and $5$ . The order of selection does not matter. List every pair whose product is even and whose sum is greater than $5$ .[3 marks]
Answer
- $(2,4)$ , $(2,5)$ , $(3,4)$ , $(4,5)$
Method: List unordered pairs in rows beginning with $1$ , then $2$ , then $3$ , then $4$ . Reject pairs with an odd product or sum at most $5$ . The pairs left are $(2,4)$ , $(2,5)$ , $(3,4)$ and $(4,5)$ .

N6 · Use positive integer powers and associated real roots (square, cube and higher), recognise powers of 2, 3, 4, 5; estimate powers and roots of any given positive number

A positive integer power is repeated multiplication, and an associated root reverses that power, such as $\sqrt[3]{125}=5$ because $5^3=125$ .
Learn common powers of $2$ , $3$ , $4$ and $5$ ; match a root to the power that it reverses.
For example, $5^3=125$ , so $\sqrt[3]{125}=5$ .
A common error is to multiply the base by the exponent, so remember that $3^4=3\times3\times3\times3$ , not $3\times4$ .

Tier 1 · Easy

1. Work out $2^5$ .[1 mark]
Answer
- $32$
Method: $2^5=2\times2\times2\times2\times2=32$ .

Tier 2 · Standard

1. Work out $\sqrt[3]{343}$ .[1 mark]
Answer
- $7$
Method: Since $7^3=7\times7\times7=343$ , the associated cube root is $\sqrt[3]{343}=7$ .

Tier 3 · Hard

1. Work out $\sqrt[4]{1296}+\sqrt[3]{512}$ .[3 marks]
Answer
- $14$
Method: Because $6^4=1296$ , $\sqrt[4]{1296}=6$ . Because $8^3=512$ , $\sqrt[3]{512}=8$ . Their sum is $6+8=14$ .

N7 · Calculate with roots, and with integer and fractional indices

Integer indices include positive, zero and negative values: $a^0=1$ and $a^{-n}=1/a^n$ for non-zero $a$ .
Evaluate a root by identifying the number whose corresponding power gives the radicand.
For example, $\sqrt[3]{-216}=-6$ because $(-6)^3=-216$ .
A common error is to make a negative index produce a negative value; it produces a reciprocal, so $4^{-2}=1/16$ , not $-16$ .

Tier 1 · Easy

1. Work out $4^{-2}$ .[1 mark]
Answer
- $\dfrac{1}{16}$
Method: A negative index means take the reciprocal: $4^{-2}=\dfrac{1}{4^2}=\dfrac{1}{16}$ .

Tier 2 · Standard

1. Work out $\sqrt[3]{-216}$ .[1 mark]
Answer
- $-6$
Method: The cube root is the number whose cube is $-216$ . Since $(-6)^3=-216$ , $\sqrt[3]{-216}=-6$ .

Tier 3 · Hard

1. Work out $\sqrt{144}+5^{-2}$ . Give an exact answer.[3 marks]
Answer
- $\dfrac{301}{25}$
Method: $\sqrt{144}=12$ and $5^{-2}=1/5^2=1/25$ . Therefore $12+\dfrac{1}{25}=\dfrac{300}{25}+\dfrac{1}{25}=\dfrac{301}{25}$ .

N8 · Calculate exactly with fractions, surds and multiples of π; simplify surd expressions involving squares (e.g. √12 = √(4 × 3) = √4 × √3 = 2√3) and rationalise denominators

An exact answer keeps fractions and multiples of $\pi$ rather than replacing them with rounded decimals.
Use a common denominator for exact fraction calculations and leave a circle result as a coefficient multiplied by $\pi$ .
For example, a circle of radius $7$ has exact area $\pi\times7^2=49\pi$ .
A common error is to replace $\pi$ with $3.14$ when an exact answer is requested.

Tier 1 · Easy

1. Work out $\dfrac{3}{4}+\dfrac{5}{6}$ . Give an exact answer.[2 marks]
Answer
- $\dfrac{19}{12}$
- $1\dfrac{7}{12}$
Method: Use denominator $12$ : $\dfrac{3}{4}=\dfrac{9}{12}$ and $\dfrac{5}{6}=\dfrac{10}{12}$ . Their sum is $\dfrac{19}{12}=1\dfrac{7}{12}$ .

Tier 2 · Standard

1. A circle has radius $7$ cm. Work out its exact area.[2 marks]
Answer
- $49\pi\text{ cm}^2$
Method: Use $A=\pi r^2$ . Then $A=\pi\times7^2=49\pi\text{ cm}^2$ .

Tier 3 · Hard

1. Work out $\dfrac{3\pi}{5}+\dfrac{7\pi}{8}-\dfrac{\pi}{4}$ . Give an exact answer.[3 marks]
Answer
- $\dfrac{49\pi}{40}$
Method: Use denominator $40$ : $\dfrac{3\pi}{5}=\dfrac{24\pi}{40}$ , $\dfrac{7\pi}{8}=\dfrac{35\pi}{40}$ and $\dfrac{\pi}{4}=\dfrac{10\pi}{40}$ . Therefore the result is $\dfrac{(24+35-10)\pi}{40}=\dfrac{49\pi}{40}$ .

N9 · Calculate with and interpret standard form A × 10^n, where 1 ≤ A < 10 and n is an integer

Standard form is $A\times10^n$ with $1\leq A<10$ and integer $n$ ; positive powers represent large values and negative powers represent small values.
When multiplying or dividing, operate on the decimal factors and the powers of $10$ separately, then adjust the result so its first factor is in the required range.
For example, $(6\times10^7)(4\times10^{-3})=24\times10^4=2.4\times10^5$ .
A common error is to leave a result such as $24\times10^4$ uncorrected; it has the right value but is not in standard form because $24\geq10$ .

Tier 1 · Easy

1. Write $0.000072$ in standard form.[1 mark]
Answer
- $7.2\times10^{-5}$
Method: Move the decimal point $5$ places right to make $7.2$ . Moving right gives a negative power, so $0.000072=7.2\times10^{-5}$ .

Tier 2 · Standard

1. Work out $(6\times10^7)(4\times10^{-3})$ . Give your answer in standard form.[2 marks]
Answer
- $2.4\times10^5$
Method: Multiply the factors and add the indices: $6\times4=24$ and $10^7\times10^{-3}=10^4$ . Thus the product is $24\times10^4=2.4\times10^5$ .

Tier 3 · Hard

1. Work out $\dfrac{3.6\times10^{-4}+7.5\times10^{-5}}{1.5\times10^3}$ . Give your answer in standard form.[4 marks]
Answer
- $2.9\times10^{-7}$
Method: Express the numerator with a common power: $3.6\times10^{-4}+0.75\times10^{-4}=4.35\times10^{-4}$ . Divide the factors and subtract the indices: $(4.35/1.5)\times10^{-4-3}=2.9\times10^{-7}$ .

N10 · Work interchangeably with terminating decimals and their corresponding fractions (such as 3.5 and 7/2 or 0.375 or 3/8); change recurring decimals into their corresponding fractions and vice versa

A terminating decimal has a finite number of decimal places and can be written as a fraction over a power of $10$ .
To convert a fraction to a decimal, divide the numerator by the denominator; to convert back, use place value and simplify.
For example, $0.375=375/1000=3/8$ , while $37/80=0.4625$ .
A common error is to stop simplifying a fraction before the numerator and denominator have no common factor greater than $1$ .

Tier 1 · Easy

1. Write $0.375$ as a fraction in its simplest form.[1 mark]
Answer
- $\dfrac{3}{8}$
Method: $0.375=\dfrac{375}{1000}$ . Divide numerator and denominator by $125$ to obtain $\dfrac{3}{8}$ .

Tier 2 · Standard

1. Write $\dfrac{37}{80}$ as a decimal.[1 mark]
Answer
- $0.4625$
Method: Make the denominator a power of $10$ : $\dfrac{37}{80}=\dfrac{4625}{10000}=0.4625$ .

Tier 3 · Hard

1. Which is greater, $0.56$ or $\dfrac{9}{16}$ ? Work out the difference as a fraction.[3 marks]
Answer
- $\dfrac{9}{16}$ is greater.
- The difference is $\dfrac{1}{400}$ .
Method: $\dfrac{9}{16}=0.5625$ , so it is greater than $0.56$ . The difference is $0.5625-0.56=0.0025=\dfrac{25}{10000}=\dfrac{1}{400}$ .

N11 · Identify and work with fractions in ratio problems

A ratio describes relative parts, so a fraction of the whole can be converted into a part-to-part ratio by using the remaining fraction.
Find the total number of ratio parts, calculate one part, and then apply any stated fraction to the relevant share.
For example, if red counters are $3/5$ of the total, blue counters are $2/5$ , so red : blue is $3:2$ .
A common error is to use the denominator as the other ratio part; if one group is $3/5$ , the remainder is $2/5$ , not $5/5$ .

Tier 1 · Easy

1. $\dfrac{3}{5}$ of the beads in a bag are red and the rest are blue. Write the ratio of red beads to blue beads.[1 mark]
Answer
- $3:2$
Method: The blue fraction is $1-\dfrac{3}{5}=\dfrac{2}{5}$ . Therefore red : blue is $\dfrac{3}{5}:\dfrac{2}{5}=3:2$ .

Tier 2 · Standard

1. The ratio of Ava's tokens to Ben's tokens is $5:7$ . Ben has $84$ tokens. Work out $\dfrac{3}{4}$ of Ava's number of tokens.[3 marks]
Answer
- $45$ tokens
Method: Seven ratio parts represent $84$ , so one part is $84\div7=12$ . Ava has $5\times12=60$ tokens. Then $\dfrac{3}{4}\times60=45$ .

Tier 3 · Hard

1. $\pounds330$ is shared between Imran and Jo in the ratio $4:7$ . Imran spends $\dfrac{3}{8}$ of his share and Jo spends $\dfrac{2}{7}$ of her share. Work out the total amount they have left.[4 marks]
Answer
- $\pounds225$
Method: There are $11$ parts, so one part is $330\div11=30$ . Imran receives $120$ and keeps $\dfrac{5}{8}\times120=75$ . Jo receives $210$ and keeps $\dfrac{5}{7}\times210=150$ . Together they keep $75+150=225$ .

N12 · Interpret fractions and percentages as operators

A fraction or percentage acts as an operator meaning multiplication, so $3/5$ of a quantity is found by multiplying it by $3/5$ .
Convert a percentage to a fraction over $100$ or a decimal multiplier; for repeated operations, apply the multipliers in the stated order.
For example, $17.5\%$ of $240$ is $0.175\times240=42$ .
A common error is to divide by the numerator and multiply by the denominator; for $3/5$ of an amount, divide by $5$ and then multiply by $3$ .

Tier 1 · Easy

1. Work out $\dfrac{3}{5}$ of $70$ .[1 mark]
Answer
- $42$
Method: Divide by $5$ and multiply by $3$ : $70\div5\times3=14\times3=42$ .

Tier 2 · Standard

1. Work out $17.5\%$ of $240$ .[2 marks]
Answer
- $42$
Method: Use the decimal operator $17.5\%=0.175$ . Then $0.175\times240=42$ .

Tier 3 · Hard

1. A machine costs $\pounds640$ . Its price is reduced by $15\%$ , then a customer pays $\dfrac{3}{8}$ of the reduced price as a deposit. Work out the balance still to pay.[4 marks]
Answer
- $\pounds340$
Method: After the reduction, the price is $0.85\times640=544$ . The deposit is $\dfrac{3}{8}\times544=204$ . Therefore the remaining balance is $544-204=340$ .

N13 · Use standard units of mass, length, time, money and other measures (including standard compound measures) using decimal quantities where appropriate

Standard measures include length, area, volume, mass, time and money, while compound measures combine units, such as km/h, g/cm $^3$ or litres per $100$ km.
Convert every quantity to compatible units before calculating, using place value carefully for squared or cubed conversion factors.
For example, $2.75\text{ kg}=2750\text{ g}$ because $1\text{ kg}=1000\text{ g}$ .
A common error is to treat decimal time as minutes: $1.5$ hours is $1$ hour $30$ minutes, not $1$ hour $50$ minutes.

Tier 1 · Easy

1. Change $2.75$ kg to grams.[1 mark]
Answer
- $2750$ g
Method: There are $1000$ grams in a kilogram, so $2.75\times1000=2750$ g.

Tier 2 · Standard

1. A workshop starts at $09{:}38$ and lasts for $1$ hour $47$ minutes. Work out the finishing time.[2 marks]
Answer
- $11{:}25$
Method: Adding $1$ hour gives $10{:}38$ . Adding $47$ minutes gives $11{:}25$ because $22$ minutes reach $11{:}00$ and $25$ minutes remain.

Tier 3 · Hard

1. A car travels $54$ km and uses fuel at a rate of $7.5$ litres per $100$ km. Fuel costs $\pounds1.68$ per litre. Work out the fuel cost for the journey, giving your answer to the nearest penny.[4 marks]
Answer
- $\pounds6.80$
Method: The fuel used is $54\times\dfrac{7.5}{100}=4.05$ litres. The cost is $4.05\times1.68=6.804$ , which rounds to $\pounds6.80$ to the nearest penny.

N14 · Estimate answers; check calculations using approximation and estimation, including answers obtained using technology

An estimate replaces awkward values with nearby numbers that are easy to calculate, often using one significant figure.
Round each input before calculating and use the estimate to check the order of magnitude and position of the decimal point in an exact or calculator answer.
For example, $19.8\times0.49\approx20\times0.5=10$ .
A common error is to round only the final calculator display; estimation must simplify the input values and provide an independent reasonableness check.

Tier 1 · Easy

1. Estimate the value of $19.8\times0.49$ .[1 mark]
Answer
- $10$
Method: Use convenient one-significant-figure values: $19.8\approx20$ and $0.49\approx0.5$ . Then $20\times0.5=10$ .

Tier 2 · Standard

1. Estimate $\dfrac{48.7\times0.203}{0.098}$ .[2 marks]
Answer
- $100$
Method: Round to convenient values: $48.7\approx50$ , $0.203\approx0.2$ and $0.098\approx0.1$ . Then $\dfrac{50\times0.2}{0.1}=\dfrac{10}{0.1}=100$ .

Tier 3 · Hard

1. A calculator display gives $931.24$ for $\dfrac{598.4\times0.0317}{0.204}$ . Use an estimate to decide whether this display is reasonable. Give a reason.[3 marks]
Answer
- The display is not reasonable.
- An estimate is $90$ , so the displayed answer is about ten times too large.
Method: Use $598.4\approx600$ , $0.0317\approx0.03$ and $0.204\approx0.2$ . This gives $\dfrac{600\times0.03}{0.2}=\dfrac{18}{0.2}=90$ . Since $931.24$ is near $900$ rather than $90$ , it is not reasonable and likely has a decimal-place error.

N15 · Round numbers and measures to an appropriate degree of accuracy (decimal places or significant figures); use inequality notation to specify simple error intervals due to truncation or rounding

Decimal places count digits after the decimal point, while significant figures begin at the first non-zero digit.
For rounding to a unit $u$ , the error interval extends half a unit below and above; include the lower bound but exclude the upper bound.
For example, $12.6$ correct to $1$ decimal place means $12.55\leq x<12.65$ .
A common error is to include the upper endpoint; that value would round to the next stated number, so the upper inequality is strict.

Tier 1 · Easy

1. Write $0.007846$ correct to $2$ significant figures.[1 mark]
Answer
- $0.0078$
Method: The first two significant digits are $7$ and $8$ . The next digit is $4$ , so the $8$ stays unchanged and the rounded value is $0.0078$ .

Tier 2 · Standard

1. A number $x$ is $12.6$ correct to $1$ decimal place. Write the error interval for $x$ .[2 marks]
Answer
- $12.55\leq x<12.65$
Method: The rounding unit is $0.1$ , so half a unit is $0.05$ . Subtract and add $0.05$ to get the boundaries $12.55$ and $12.65$ ; include the lower boundary only.

Tier 3 · Hard

1. A positive number $y$ is truncated to $4.37$ at $2$ decimal places. Write its error interval and find the greatest possible integer value of $100y$ .[3 marks]
Answer
- $4.37\leq y<4.38$
- Greatest possible integer value of $100y$ is $437$ .
Method: Truncation to $2$ decimal places keeps every value from $4.37$ up to but not including $4.38$ , so $4.37\leq y<4.38$ . Multiplying by $100$ gives $437\leq100y<438$ , whose greatest possible integer value is $437$ .

N16 · Apply and interpret limits of accuracy, including upper and lower bounds

A rounded measurement represents a range of possible true values, determined by half of the rounding unit.
Use the smallest and largest possible input values to decide whether a claimed result is possible, without treating a rounded value as exact.
For example, a length shown as $12$ cm to the nearest centimetre can differ from $12$ cm by at most $0.5$ cm.
A common error is to allow the top endpoint of a rounding interval even though it would round to the next displayed value.

Tier 1 · Easy

1. A length is recorded as $12$ cm to the nearest centimetre. Write down the maximum possible rounding error.[1 mark]
Answer
- $0.5$ cm
Method: The rounding unit is $1$ cm, so the true length can differ from the recorded value by half of this: $1\div2=0.5$ cm.

Tier 2 · Standard

1. Higher only: Two lengths are recorded as $4.2$ cm and $3.7$ cm, each to the nearest $0.1$ cm. Could their exact total be less than $7.8$ cm? Give a reason.[3 marks]
Answer
- No.
- The smallest possible total is $4.15+3.65=7.80$ cm.
Method: The first length is at least $4.15$ cm and the second is at least $3.65$ cm. Their smallest possible total is $4.15+3.65=7.80$ cm, so the exact total cannot be less than $7.8$ cm.

Tier 3 · Hard

1. Higher only: A scale records the mass of each of $8$ identical boxes as $2.4$ kg to the nearest $0.1$ kg. Could the exact total mass of the boxes be $20$ kg? Justify your answer.[3 marks]
Answer
- No.
- The exact total must be less than $19.6$ kg.
Method: A displayed mass of $2.4$ kg means one box has mass less than $2.45$ kg. Therefore $8$ boxes have total mass less than $8\times2.45=19.6$ kg. Since $20>19.6$ , a total of $20$ kg is impossible.