NavGanitMateby NavGreen Compute, AustraliaBased on Australian Curriculum: Mathematics v9.0
← Back to practice

Accuracy and measurement errors

Wide tables scroll horizontally. For detailed working, you may prefer a larger screen.

Required skills (click to reveal)

A community garden in Canberra is planning a new rectangular vegetable plot. The design specifies that the length must be exactly 92 metres and the width exactly 72 metres.

However, the measuring tape used has a known maximum error of ±0.15 metres. The gardeners calculate the area using the measured values, but due to measurement error, the actual area could vary.

What is the exact maximum possible area of the plot, in square metres, when both length and width are at their maximum measured values? Express your answer as a simplified fraction.

Concept illustrations

These are separate worked examples of related concepts, not a diagram of this question. Their values and dimensions do not describe your question.

Measurement compared with a reference. Measured length 49 cm differs from the reference 50 cm by 1 cm. Absolute error is 1 cm and percentage error is 2%. The axis is a close-up from 48 to 51 cm.48 cm49 cm50 cm51 cmMeasured: 49Reference: 50|49 − 50| = 1 cm
Measured length 49 cm differs from the reference 50 cm by 1 cm. Absolute error is 1 cm and percentage error is 2%. The axis is a close-up from 48 to 51 cm.
Nearest hundredth on a number line. 6.347 lies between 6.34 and 6.35. It is beyond the midpoint 6.345, so rounding to the nearest hundredth gives 6.35.6.346.3456.356.3476.347 ≈ 6.35
6.347 lies between 6.34 and 6.35. It is beyond the midpoint 6.345, so rounding to the nearest hundredth gives 6.35.

Press Start when ready. Time includes time spent in another tab. Submit your answer before opening a help panel for a timed rank.

00:00
0 wordsOnly your numerical final answer or selected MCQ option is checked. Scratch working is not graded.

Report a problem or suggest an improvement

My notes — optional, not graded 0 words

Notes and scratch working stay in this browser and are linked to this question version. They are not graded or included in feedback. On shared devices, other people using this browser may see them.

For this question only. Your full notebook is on the dashboard.

Notes save in this browser. Download a copy to keep them elsewhere.

Report a problem or suggest an improvement

Your observations help improve this question. You can report an issue or recommend a better explanation, visual or experience.

0 words

The report includes your message and question/version details. Saved profile details, answers, notes and attempt history are not automatically attached. Leave out names and contact details, and check screenshots for personal information and embedded metadata.

Attachments may contain personal information. Review them before sharing. Logs collect only event codes and timestamps after you enable them. Download the ZIP to keep a copy, then attach it to an email if you choose.

Formula and explanation (click to reveal)

Opening this before your first submission makes the attempt practice only.

Formula

Explanation

Worked solution and answer (click to reveal)

Opening this before your first submission makes the attempt practice only.

Worked solution

  1. The design specifies a length of 92 metres and a width of 72 metres, but the measuring tape has a maximum error of ±0.15 metres. To find the maximum possible area, both length and width must be at their maximum measured values.
  2. The maximum measured length is 92 + 0.15 = 92.15 metres.
  3. The maximum measured width is 72 + 0.15 = 72.15 metres.
  4. The area of a rectangle is given by length × width. Substitute the maximum values: 92.15 × 72.15.
  5. Convert the decimal values to fractions to maintain exactness. 92.15 = 9215/100 = 1843/20. 72.15 = 7215/100 = 1443/20.
  6. Multiply the two fractions: (1843/20) × (1443/20) = (1843 × 1443) / (20 × 20).
  7. Calculate 1843 × 1443 using vertical and crosswise multiplication (Urdhva Tiryagbhyam). First, compute 1843 × 1400 = 1843 × 14 × 100. 1843 × 14 = 1843 × (10 + 4) = 18430 + 7372 = 25802. Then 25802 × 100 = 2,580,200.
  8. Now compute 1843 × 43. 1843 × 40 = 73,720; 1843 × 3 = 5,529. Sum: 73,720 + 5,529 = 79,249.
  9. Add the two parts: 2,580,200 + 79,249 = 2,659,449.
  10. The denominator is 20 × 20 = 400. So the area is 2,659,449 / 400 square metres.
  11. Simplify the fraction. Check if 2,659,449 and 400 share any common factors. 400 = 2^4 × 5^2. 2,659,449 is odd, so not divisible by 2. It does not end in 0 or 5, so not divisible by 5. Thus, the fraction is already in simplest form.
  12. The maximum possible area is 2,659,449 / 400 square metres. This is the exact value when both length and width are at their maximum measured values.

Reference answer: 2659449/400

Alternative methods (click to reveal)

Shortcuts are shown only when applicable. Check the conditions before using one.

Earlier attempts for this question

#Date / timeYour answerResultSecondsHelp revealedRank
History stays in this browser when storage is available. Export it to your shared Drive folder and import it on another computer.
Worksheet details

Naming pattern: Topic_Grade_Complexity_YYYYMMDD_HH24MISS.html

Validation compares your answer with the generated reference. The model's reference can be wrong. Time ranks use an estimated student solving time, not a measured benchmark or a comparison with other students. An incorrect answer receives no speed rank.

NavGanitMate privacy & saved work

Saved in this browser. Your profile preferences, answers, practice history and notes are stored here when browser storage is available. They are not automatically uploaded to our servers. People sharing this browser may see saved work; changing the learner name does not create a separate private account. A nickname is enough.

Keep copies of important work. Clearing browser data, private browsing or changing browser or device can mean losing saved work. Download notes and export history separately. Downloads are copies of the selected content, not a complete account backup; not every format can be imported to restore your work. Keep these files somewhere private.

Visiting the website. Cloudflare helps deliver and protect this site and processes connection information such as IP addresses and requested pages. Server error logs may also contain request details. Security features may use cookies. This is separate from your browser-saved learning records. Cloudflare privacy information.

A parent-connection reminder uses a local list of up to five completed question IDs for this browser profile and a tab-session dismissal marker. Reminder activity is not uploaded and practice remains available.

Sharing feedback is your choice. A report includes your message and question/version details. Your saved profile, answers and notes are not automatically attached. Files you select can reveal personal details or contain embedded metadata: check and redact them before sharing. Diagnostic event codes and timestamps are included only after you enable logging. Sending email shares your message and attachments with the recipient and email providers. A download alone sends nothing to us. Website feedback delivery may be unavailable; only a confirmed receipt indicates it was received.