Year 9 Mathematics | Victorian Curriculum 2.0
Measurement errors
Topic 11 | Measurement | Answer key

Tier 1

    1. Absolute error =∣3.35−3.2∣=0.15= |3.35 - 3.2| = 0.15=∣3.35−3.2∣=0.15 kg.
    2. Relative error =0.153.2=0.046875= \dfrac{0.15}{3.2} = 0.046875=3.20.15​=0.046875.
    3. Percentage error =0.046875×100%≈4.7%= 0.046875 \times 100\% \approx 4.7\%=0.046875×100%≈4.7%.
    4. Lower bound =15.75= 15.75=15.75 cm, upper bound =15.85= 15.85=15.85 cm.
    5. Lower bound =23.5= 23.5=23.5 s, upper bound =24.5= 24.5=24.5 s.
    6. Maximum possible error =102=5= \dfrac{10}{2} = 5=210​=5 g.
    7. Combined stated length =2.0= 2.0=2.0 m. Max error =0.05+0.05=0.10= 0.05 + 0.05 = 0.10=0.05+0.05=0.10 m. Lower bound =1.90= 1.90=1.90 m, upper bound =2.10= 2.10=2.10 m.
    8. Absolute error =∣3.14−3.14159∣=0.00159= |3.14 - 3.14159| = 0.00159=∣3.14−3.14159∣=0.00159. Percentage error =0.001593.14159×100%≈0.051%= \dfrac{0.00159}{3.14159} \times 100\% \approx 0.051\%=3.141590.00159​×100%≈0.051%.
    9. Lower bound =21.5°= 21.5°=21.5°C, upper bound =22.5°= 22.5°=22.5°C.
    10. Lower bound =45 229.5= 45\,229.5=45229.5 km, upper bound =45 230.5= 45\,230.5=45230.5 km.

Tier 2

    1. Student A: 0.412.0×100%=3.33%\dfrac{0.4}{12.0} \times 100\% = 3.33\%12.00.4​×100%=3.33%. Student B: 1.246.0×100%=2.61%\dfrac{1.2}{46.0} \times 100\% = 2.61\%46.01.2​×100%=2.61%. Student B has the smaller percentage error.
    2. Length bounds: 19.519.519.5 to 20.520.520.5 cm. Width bounds: 14.514.514.5 to 15.515.515.5 cm. Lower area =19.5×14.5=282.75= 19.5 \times 14.5 = 282.75=19.5×14.5=282.75 cm2^22. Upper area =20.5×15.5=317.75= 20.5 \times 15.5 = 317.75=20.5×15.5=317.75 cm2^22.
    3. Speed =40052.3≈7.65= \dfrac{400}{52.3} \approx 7.65=52.3400​≈7.65 m/s. Distance % error =0.5400×100%=0.125%= \dfrac{0.5}{400} \times 100\% = 0.125\%=4000.5​×100%=0.125%. Time % error =0.0552.3×100%≈0.096%= \dfrac{0.05}{52.3} \times 100\% \approx 0.096\%=52.30.05​×100%≈0.096%. Max % error in speed ≈0.125%+0.096%=0.22%\approx 0.125\% + 0.096\% = 0.22\%≈0.125%+0.096%=0.22%.
    4. Side bounds: 7.957.957.95 cm to 8.058.058.05 cm. Lower area =7.952=63.2025= 7.95^2 = 63.2025=7.952=63.2025 cm2^22. Upper area =8.052=64.8025= 8.05^2 = 64.8025=8.052=64.8025 cm2^22.
    5. Percentage error scales the error to the size of the quantity, enabling fair comparison. A 1 cm error on a 10 cm measurement (10%) is far more significant than a 1 cm error on a 1000 cm measurement (0.1%), but the absolute errors are identical.
    6. Volume =πr2h=π×1.22×3.0≈13.57= \pi r^2 h = \pi \times 1.2^2 \times 3.0 \approx 13.57=πr2h=π×1.22×3.0≈13.57 m3^33. Lower: π×1.152×2.95≈12.26\pi \times 1.15^2 \times 2.95 \approx 12.26π×1.152×2.95≈12.26 m3^33. Upper: π×1.252×3.05≈14.97\pi \times 1.25^2 \times 3.05 \approx 14.97π×1.252×3.05≈14.97 m3^33.
    7. Bounds: 36.75°36.75°36.75°C to 37.25°37.25°37.25°C. Max error =0.25°= 0.25°=0.25°C. Percentage error =0.2537.0×100%≈0.68%= \dfrac{0.25}{37.0} \times 100\% \approx 0.68\%=37.00.25​×100%≈0.68%.

Tier 3

    1. Distance percentage error =∣84.3−85.0∣85.0×100%=0.785.0×100%≈0.82%= \dfrac{|84.3 - 85.0|}{85.0} \times 100\% = \dfrac{0.7}{85.0} \times 100\% \approx 0.82\%=85.0∣84.3−85.0∣​×100%=85.00.7​×100%≈0.82%. Area using measured distance =84.32=7106.49= 84.3^2 = 7106.49=84.32=7106.49 m2^22. Actual area =85.02=7225= 85.0^2 = 7225=85.02=7225 m2^22. Area percentage error =∣7106.49−7225∣7225×100%≈1.64%= \dfrac{|7106.49 - 7225|}{7225} \times 100\% \approx 1.64\%=7225∣7106.49−7225∣​×100%≈1.64%, which is approximately double the distance error (since area ∝d2\propto d^2∝d2).
    2. Density =12050=2.4= \dfrac{120}{50} = 2.4=50120​=2.4 g/cm3^33. Mass % error =0.5120×100%≈0.42%= \dfrac{0.5}{120} \times 100\% \approx 0.42\%=1200.5​×100%≈0.42%. Volume % error =150×100%=2.0%= \dfrac{1}{50} \times 100\% = 2.0\%=501​×100%=2.0%. Max % error in density ≈0.42%+2.0%=2.42%\approx 0.42\% + 2.0\% = 2.42\%≈0.42%+2.0%=2.42%. Density =2.4±0.06= 2.4 \pm 0.06=2.4±0.06 g/cm3^33.
    3. The error in A−BA - BA−B is maximised when AAA is at its upper bound and BBB is at its lower bound (or vice versa). This gives the largest possible spread: (A+δA)−(B−δB)=(A−B)+(δA+δB)(A + \delta_A) - (B - \delta_B) = (A - B) + (\delta_A + \delta_B)(A+δA​)−(B−δB​)=(A−B)+(δA​+δB​). The individual errors add regardless of whether we add or subtract the measurements.
    4. Max error =3+3=6= 3 + 3 = 6=3+3=6 m. For 200 m: % error =6200×100%=3%= \dfrac{6}{200} \times 100\% = 3\%=2006​×100%=3%. For 10 m: % error =610×100%=60%= \dfrac{6}{10} \times 100\% = 60\%=106​×100%=60%. The GPS is unreliable for short distances.

Challenge

    1. Radius =7.0= 7.0=7.0 cm, so area =π×7.02=49π≈153.94= \pi \times 7.0^2 = 49\pi \approx 153.94=π×7.02=49π≈153.94 cm2^22. Diameter % error =0.114.0×100%≈0.714%= \dfrac{0.1}{14.0} \times 100\% \approx 0.714\%=14.00.1​×100%≈0.714%. Since area ∝d2\propto d^2∝d2, area % error ≈2×0.714%=1.43%\approx 2 \times 0.714\% = 1.43\%≈2×0.714%=1.43%.
    2. g=2×1.2000.4952=2.4000.245025≈9.795g = \dfrac{2 \times 1.200}{0.495^2} = \dfrac{2.400}{0.245025} \approx 9.795g=0.49522×1.200​=0.2450252.400​≈9.795 m/s2^22. Distance % error =0.0051.200×100%≈0.42%= \dfrac{0.005}{1.200} \times 100\% \approx 0.42\%=1.2000.005​×100%≈0.42%. Time % error =0.0050.495×100%≈1.01%= \dfrac{0.005}{0.495} \times 100\% \approx 1.01\%=0.4950.005​×100%≈1.01%. Since g∝t−2g \propto t^{-2}g∝t−2, time contributes 2×1.01%=2.02%2 \times 1.01\% = 2.02\%2×1.01%=2.02%. Total % error ≈0.42%+2.02%=2.44%\approx 0.42\% + 2.02\% = 2.44\%≈0.42%+2.02%=2.44%. This gives g=9.80±0.24g = 9.80 \pm 0.24g=9.80±0.24 m/s2^22. The accepted value 9.819.819.81 falls within this range, so the result is acceptable.
    3. Result =50.0−49.0=1.0= 50.0 - 49.0 = 1.0=50.0−49.0=1.0. Max error =0.5+0.5=1.0= 0.5 + 0.5 = 1.0=0.5+0.5=1.0. Percentage error =1.01.0×100%=100%= \dfrac{1.0}{1.0} \times 100\% = 100\%=1.01.0​×100%=100%. When two nearly equal quantities are subtracted, the result is small but the absolute error remains the sum of the original errors, leading to a very large relative error. This catastrophic cancellation should be avoided in practice by redesigning the measurement approach.
    4. Map distance =4.2= 4.2=4.2 cm, so actual distance =4.2×25 000=105 000= 4.2 \times 25\,000 = 105\,000=4.2×25000=105000 cm =1050= 1050=1050 m. Lower bound: (4.2−0.1)×25 000=4.1×25 000=102 500(4.2 - 0.1) \times 25\,000 = 4.1 \times 25\,000 = 102\,500(4.2−0.1)×25000=4.1×25000=102500 cm =1025= 1025=1025 m. Upper bound: (4.2+0.1)×25 000=4.3×25 000=107 500(4.2 + 0.1) \times 25\,000 = 4.3 \times 25\,000 = 107\,500(4.2+0.1)×25000=4.3×25000=107500 cm =1075= 1075=1075 m. Actual distance is between 1025 m and 1075 m.
Year 9 Mathematics study companion | Answer key