Approximation and accuracy - Year 10 Mathematics

What you will learn

distinguish between exact and approximate representations of numbers,
understand how rounding and truncation introduce error,
calculate absolute and relative (percentage) error,
analyse how errors accumulate when approximate values are used in repeated calculations,
decide when an exact form (surd, fraction) is preferable to a decimal approximation.

Worked example 0 Real-world example: accumulated rounding in finance

A bank account earns 3.7% interest per year, compounded annually. The principal is $10 000. Compare the balance after 5 years using (a) the exact multiplier and (b) the multiplier rounded to 2 decimal places.

Exact multiplier: $1.037$ . After 5 years: $10\,000 \times 1.037^5 = 10\,000 \times 1.198\,691\ldots = 11\,986.91$ (to nearest cent).
Rounded multiplier: $1.04$ . After 5 years: $10\,000 \times 1.04^5 = 10\,000 \times 1.216\,653 = 12\,166.53$ .
The difference is $12\,166.53 - 11\,986.91 = 179.62$ — a significant error from rounding just one decimal place.

Key idea: rounding before repeated multiplication compounds the error at every step.

1. Exact vs approximate representations

An exact representation leaves no ambiguity about the true value:

Fractions: $\dfrac{1}{3}$ is exact; $0.333$ is an approximation.
Surds: $\sqrt{2}$ is exact; $1.414$ is an approximation.
Multiples of $\pi$ : $2\pi$ is exact; $6.283$ is an approximation.

A decimal approximation is often more practical for measurement and communication, but it always introduces some error.

Worked example 1 Exact vs decimal form

Express $\dfrac{5}{7}$ as a decimal and state the error when rounded to 3 decimal places.

$\dfrac{5}{7} = 0.714\,285\,714\ldots$ (recurring).
Rounded to 3 d.p.: $0.714$ .
Error $= 0.714\,285\ldots - 0.714 = 0.000\,285\ldots$

The exact form $\dfrac{5}{7}$ carries no error at all.

2. Rounding and truncation

Rounding replaces a number with the nearest value at a given precision (e.g. to 2 decimal places, 3 significant figures).

Truncation simply cuts off digits beyond a given place — it always rounds toward zero.

Formula reference

\text{Absolute error} = |\text{approximate value} - \text{exact value}|

\text{Relative error} = \frac{\text{absolute error}}{|\text{exact value}|} \times 100\%

Worked example 2 Comparing rounding and truncation

The exact value is $3.6478$ . Find the absolute and relative error when this is (a) rounded to 2 d.p. and (b) truncated to 2 d.p.

Rounded to 2 d.p.: $3.65$ . Absolute error $= |3.65 - 3.6478| = 0.0022$ . Relative error $= \dfrac{0.0022}{3.6478} \times 100\% \approx 0.060\%$ .
Truncated to 2 d.p.: $3.64$ . Absolute error $= |3.64 - 3.6478| = 0.0078$ . Relative error $= \dfrac{0.0078}{3.6478} \times 100\% \approx 0.214\%$ .

Truncation produced a larger error in this case.

Worked example 3 Significant figures

Round $0.004\,562\,3$ to 3 significant figures.

The first three significant figures are $4$ , $5$ , and $6$ (leading zeros are not significant).
The next digit is $2 < 5$ , so round down.
Result: $0.004\,56$ .

3. Error accumulation in repeated calculations

When an approximate value is used repeatedly — for example, multiplied by itself — the error grows with each step.

Worked example 4 Error growth through repeated multiplication

A length is measured as $x = 4.3$ cm (rounded to 1 d.p., so the true value is between $4.25$ and $4.35$ ). Find the range of possible values for $x^3$ .

Minimum: $4.25^3 = 76.765\,625$ .
Maximum: $4.35^3 = 82.312\,875$ .
Using the rounded value: $4.3^3 = 79.507$ .
The result could be anywhere from about $76.8$ to $82.3$ — a range of $5.5$ cm $^3$ , even though the original measurement error was only $\pm 0.05$ cm.

Key idea: cubing amplified the original $\pm 0.05$ error into a range of $5.5$ — over 100 times larger.

Worked example 5 Rounding at each step vs rounding at the end

Calculate $1.7 \times 2.3 \times 4.1$ two ways: (a) round each intermediate product to 1 d.p., (b) keep full precision and round only the final answer.

Method (a): $1.7 \times 2.3 = 3.91 \approx 3.9$ . Then $3.9 \times 4.1 = 15.99 \approx 16.0$ .
Method (b): $1.7 \times 2.3 = 3.91$ . Then $3.91 \times 4.1 = 16.031$ . Rounded to 1 d.p.: $16.0$ .
In this case the answers agree, but with longer calculation chains or less friendly numbers the early-rounding method can diverge significantly.

Key idea: always keep full precision through intermediate steps and round only the final answer.

Practice

Fluency

Tier 1: representations and basic error

State whether each is exact or approximate: (a) $\sqrt{5}$ (b) $2.236$ (c) $\dfrac{22}{7}$ (d) $3.14$ .
Round $7.3496$ to (a) 1 d.p., (b) 2 d.p., (c) 3 significant figures.
Truncate $7.3496$ to (a) 1 d.p., (b) 2 d.p.
Find the absolute error when $\pi$ is approximated by $3.14$ .
Find the relative (percentage) error when $\sqrt{3}$ is approximated by $1.73$ .
A calculator shows $0.666\,666\,7$ for $\dfrac{2}{3}$ . What is the absolute error?
Round $0.005\,049$ to 2 significant figures.
Express $\dfrac{5}{6}$ as a decimal rounded to 4 d.p. and state the absolute error.
A length of $\dfrac{7}{3}$ m is recorded as $2.33$ m. Find the absolute and relative error.
True or false: truncation always gives a smaller error than rounding. Justify your answer.

Reasoning

Tier 2: accumulated error

A square has side length $5.2$ cm (rounded to 1 d.p.). Find the range of possible values for the area.
An investment of $1000 grows by a factor of $1.065$ each year. Compare the amount after 10 years using the exact multiplier vs the multiplier rounded to $1.07$ .
A recipe calls for $\dfrac{2}{3}$ cup of sugar. A cook measures $0.67$ cups. If the recipe is tripled, what is the total error?
The radius of a circle is $3.4$ cm (rounded to 1 d.p.). Find the range of possible values for the circumference. Use $C = 2\pi r$ .
A student calculates $2.45 \times 3.15 \times 1.85$ by rounding each factor to 1 d.p. first. Find the error compared to the exact product.
Explain why keeping values in surd form during intermediate steps gives a more accurate final answer.
The value $x = 1.4$ is used to compute $x^5$ . If the true value is $1.41$ , find the absolute error in $x^5$ .
A measurement of $12.0$ cm has a relative error of $0.5\%$ . Find the range of possible true values.

Reasoning

Tier 3: analysis and explanation

Prove that when a value rounded to $\pm\epsilon$ is squared, the maximum absolute error in the square is approximately $2x\epsilon$ (for small $\epsilon$ ). Hint: compare $(x + \epsilon)^2$ with $x^2$ .
A GPS unit reports a distance of $142$ km, rounded to the nearest kilometre. This distance is used to calculate fuel needed at $8.3$ L per 100 km. Find the maximum error in the fuel estimate.
A scientist measures the sides of a cuboid as $3.2$ cm, $4.5$ cm and $6.1$ cm (each to 1 d.p.). Calculate the maximum and minimum possible volumes and the percentage range.
Explain, with an example, why subtraction of nearly equal approximate numbers is particularly dangerous for accuracy.

Challenge

Reasoning

Harder reasoning

A bank compounds interest monthly at a nominal rate of $4.8\%$ per annum. Compare the balance after 20 years on a $50 000 deposit using (a) the exact monthly multiplier $1 + \dfrac{0.048}{12}$ and (b) the multiplier rounded to 4 d.p. How large is the discrepancy?
The golden ratio is $\phi = \dfrac{1 + \sqrt{5}}{2}$ . A student approximates $\phi \approx 1.618$ and calculates $\phi^{10}$ . Find the percentage error compared to the exact value of $\phi^{10}$ .
Two measurements are $a = 10.0 \pm 0.05$ and $b = 9.9 \pm 0.05$ . Show that the relative error of $a - b$ can exceed $100\%$ while the relative errors of $a$ and $b$ individually are each under $1\%$ .
A computer stores numbers in floating point with 7 significant digits. Explain how computing $\sqrt{10\,000\,001} - \sqrt{10\,000\,000}$ could lose almost all significant figures, and describe a rearrangement that avoids this problem.

Answers

Answer key

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Year 10 core - answers

Fluency

Tier 1: representations and basic error

(a) Exact (b) Approximate (c) Exact (d) Approximate.
(a) $7.3$ (b) $7.35$ (c) $7.35$ (3 s.f.).
(a) $7.3$ (b) $7.34$ .
$|\pi - 3.14| = 0.001\,592\ldots \approx 0.0016$ .
$\sqrt{3} = 1.732\,050\ldots$ Absolute error $= |1.73 - 1.732\,050\ldots| = 0.002\,05\ldots$ Relative error $= \dfrac{0.002\,05}{1.732\,05} \times 100\% \approx 0.118\%$ .
$\dfrac{2}{3} = 0.666\,666\ldots$ repeating. Absolute error $= |0.666\,666\,7 - 0.666\,666\ldots| \approx 3.3 \times 10^{-8}$ .
$0.0050$ .
$\dfrac{5}{6} = 0.8333\ldots$ Rounded to 4 d.p.: $0.8333$ . Absolute error $= 0.0000\overline{3} \approx 3.3 \times 10^{-5}$ .
$\dfrac{7}{3} = 2.333\ldots$ Absolute error $= |2.33 - 2.333\ldots| = 0.003\ldots$ Relative error $= \dfrac{0.003\overline{3}}{2.333\ldots} \times 100\% \approx 0.143\%$ .
False. For example, truncating $3.6478$ to 2 d.p. gives $3.64$ (error $0.0078$ ), while rounding gives $3.65$ (error $0.0022$ ). Truncation gave the larger error.

Reasoning

Tier 2: accumulated error

Side is between $5.15$ and $5.25$ . Area range: $5.15^2 = 26.5225$ to $5.25^2 = 27.5625$ . Range is approximately $26.52$ to $27.56$ cm $^2$ .
Exact: $1000 \times 1.065^{10} = 1000 \times 1.877\,14\ldots = 1877.14$ . Rounded: $1000 \times 1.07^{10} = 1000 \times 1.967\,15\ldots = 1967.15$ . Difference $\approx 90.01$ .
Exact amount per batch: $\dfrac{2}{3} = 0.666\ldots$ cup. Measured: $0.67$ . Error per batch: $0.67 - 0.666\ldots = 0.003\overline{3}$ . Tripled: total error $= 3 \times 0.003\overline{3} = 0.01$ cup.
Radius between $3.35$ and $3.45$ cm. Circumference range: $2\pi(3.35) = 6.7\pi \approx 21.05$ to $2\pi(3.45) = 6.9\pi \approx 21.68$ cm.
Exact: $2.45 \times 3.15 \times 1.85 = 14.276\,625$ . Rounded factors: $2.5 \times 3.2 \times 1.9 = 15.2$ . Error $= |15.2 - 14.276\,625| = 0.923\,375$ .
Surds are exact representations. Rounding introduces error that compounds through further operations. For example, $\sqrt{2} \times \sqrt{2} = 2$ exactly, but $1.414 \times 1.414 = 1.999\,396$ .
$1.4^5 = 5.378\,24$ . $1.41^5 = 5.584\,059\ldots$ Absolute error $= |5.584\,06 - 5.378\,24| \approx 0.206$ .
$0.5\%$ of $12.0 = 0.06$ . True value is between $12.0 - 0.06 = 11.94$ and $12.0 + 0.06 = 12.06$ cm.

Reasoning

Tier 3: analysis and explanation

$(x + \epsilon)^2 = x^2 + 2x\epsilon + \epsilon^2$ . For small $\epsilon$ , $\epsilon^2$ is negligible, so the error $\approx 2x\epsilon$ . Similarly $(x - \epsilon)^2 \approx x^2 - 2x\epsilon$ , confirming maximum absolute error $\approx 2x\epsilon$ .
Distance between $141.5$ and $142.5$ km. Fuel: $\dfrac{141.5 \times 8.3}{100} = 11.7445$ to $\dfrac{142.5 \times 8.3}{100} = 11.8275$ L. Using $142$ : $11.786$ L. Maximum error $\approx 0.083$ L.
Min volume: $3.15 \times 4.45 \times 6.05 = 84.835\ldots$ Max volume: $3.25 \times 4.55 \times 6.15 = 90.966\ldots$ Using rounded values: $3.2 \times 4.5 \times 6.1 = 87.84$ . Percentage range $= \dfrac{90.97 - 84.84}{87.84} \times 100\% \approx 6.98\%$ .
Catastrophic cancellation: if $a \approx 1000.3$ and $b \approx 1000.1$ (each accurate to 4 s.f.), then $a - b \approx 0.2$ , which has only 1 significant figure. The relative error jumps from $\sim 0.01\%$ in each value to potentially $50\%$ in their difference.

Reasoning

Challenge

Exact monthly rate: $r = 1 + \dfrac{0.048}{12} = 1.004$ . After 240 months: $50\,000 \times 1.004^{240} = 50\,000 \times 2.6051\ldots = 130\,255.20$ . Rounded to 4 d.p. the multiplier is already $1.0040$ , so there is no rounding error at 4 d.p. in this case. With a cruder rounding (e.g. 3 d.p. giving $1.004$ ) the result is the same. If rounded to 2 d.p. as $1.00$ , the balance would be $50 000 — a discrepancy of roughly $80 255.
$\phi^{10} = \left(\dfrac{1+\sqrt{5}}{2}\right)^{10} = \dfrac{123 + 55\sqrt{5}}{2^{10}} \cdot 2^{10}$ . Exact value $= 122.991\,869\ldots$ Using $1.618^{10} = 122.966\ldots$ Percentage error $= \dfrac{|122.992 - 122.966|}{122.992} \times 100\% \approx 0.021\%$ .
$a - b$ ranges from $(10.0 - 0.05) - (9.9 + 0.05) = 0.0$ to $(10.0 + 0.05) - (9.9 - 0.05) = 0.2$ . Best estimate of $a - b = 0.1$ . Error up to $\pm 0.1$ , relative error $= \dfrac{0.1}{0.1} = 100\%$ , while individual relative errors are $\dfrac{0.05}{10.0} = 0.5\%$ and $\dfrac{0.05}{9.9} \approx 0.505\%$ .
With 7 significant digits: $\sqrt{10\,000\,001} = 3162.277\,8\ldots$ and $\sqrt{10\,000\,000} = 3162.277\,7\ldots$ The difference $\approx 0.000\,016$ , but both stored values agree in their first 7 digits, so the subtraction leaves at most 1-2 correct digits. Rearrangement: multiply by the conjugate: $\sqrt{10\,000\,001} - \sqrt{10\,000\,000} = \dfrac{1}{\sqrt{10\,000\,001} + \sqrt{10\,000\,000}} \approx \dfrac{1}{6324.555} \approx 1.581 \times 10^{-4}$ , which retains full precision.

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