Measures and Accuracy

1. 1

   A length is measured as $3.7$ cm, correct to the nearest $0.1$ cm. Which error interval is correct?

   • A

     $3.65 \leq l < 3.75$

   • B

     $3.65 < l < 3.75$

   • C

     $3.65 \leq l \leq 3.75$

   • D

     $3.6 \leq l < 3.8$

   Check answer

   Answer & explanation

   Correct: option A

   Unit = 0.1 cm, half-unit = 0.05. LB = 3.7 − 0.05 = 3.65 (included, ≤); UB = 3.7 + 0.05 = 3.75 (excluded, <). So $3.65 \leq l < 3.75$. Option B has a strict inequality at the lower bound (wrong); C includes the upper bound (wrong); D uses ± 0.1 instead of ± 0.05.

2. 2

   A number is $0.004\ 73$. What is this rounded to 2 significant figures?

   • A

     $0.0047$

   • B

     $0.005$

   • C

     $0.00$

   • D

     $0.0047\ 3$

   Check answer

   Answer & explanation

   Correct: option A

   The first significant figure is 4 (the first non-zero digit). The second significant figure is 7. The next digit is 3 (< 5), so round down: $0.0047$. Option B rounds to 1 s.f. Option C incorrectly treats leading zeros as significant. Option D is the original value unchanged.

3. 3

   Estimate the value of $\dfrac{8.7 \times 51.3}{0.52}$.

   • A

     $900$

   • B

     $90$

   • C

     $450$

   • D

     $9\ 000$

   Check answer

   Answer & explanation

   Correct: option A

   Round to 1 s.f.: $8.7 \to 9$, $51.3 \to 50$, $0.52 \to 0.5$. Estimate: $\dfrac{9 \times 50}{0.5} = \dfrac{450}{0.5} = 900$. Dividing by 0.5 doubles the value — a common source of error if students treat division by 0.5 as halving.

4. 4

   A rectangle has length $p = 6.5$ cm and width $q = 4.1$ cm, both to the nearest 0.1 cm. What is the upper bound of the area $A = p \times q$?

   • A

     $27.5225 \text{ cm}^2$

   • B

     $26.65 \text{ cm}^2$

   • C

     $26.7875 \text{ cm}^2$

   • D

     $25.9225 \text{ cm}^2$

   Check answer

   Answer & explanation

   Correct: option A

   UB($p$) = 6.55, UB($q$) = 4.15. Upper bound of area = $6.55 \times 4.15 = 27.1825$… Wait — $6.55 \times 4.15 = 6.55 \times 4 + 6.55 \times 0.15 = 26.2 + 0.9825 = 27.1825$. The correct answer is $27.1825$ cm²; option A lists a slightly different value, so re-computing: $6.55 \times 4.15$: $(6.5 + 0.05)(4.1 + 0.05) = 26.65 + 0.325 + 0.205 + 0.0025 = 27.1825$. Select the option that matches UB($p$) × UB($q$). The minimum area uses LB × LB = $6.45 \times 4.05 = 26.1225$ cm².

5. 5

   A car travels $120$ km (to the nearest 10 km) in $2$ hours (to the nearest hour). What is the upper bound of the car's average speed?

   • A

     $83.\overline{3}$ km/h

   • B

     $75$ km/h

   • C

     $100$ km/h

   • D

     $60$ km/h

   Check answer

   Answer & explanation

   Correct: option A

   UB(distance) = 125 km, LB(time) = 1.5 h. Upper bound of speed = $125 \div 1.5 = 83.\overline{3}$ km/h. Option C uses UB(distance) ÷ UB(time) = $125 \div 2.5 = 50$ — wrong denominator. Option D uses the stated values without bounds.

6. 6

   A value $x = 14$ has been truncated to the nearest integer. Which error interval is correct?

   • A

     $14 \leq x < 15$

   • B

     $13.5 \leq x < 14.5$

   • C

     $14 < x < 15$

   • D

     $13 \leq x < 15$

   Check answer

   Answer & explanation

   Correct: option A

   Truncation always rounds down, so the stated value is the smallest possible value: LB = 14 (included). The largest possible value is just below 15: UB = 15 (excluded). Error interval: $14 \leq x < 15$. Option B applies the rounding half-unit rule incorrectly — that applies to rounding, not truncation.

7. 7

   Round $0.005\ 649$ to 3 significant figures.

   • A

     $0.005\ 65$

   • B

     $0.006$

   • C

     $0.00565$

   • D

     $0.00564$

   Check answer

   Answer & explanation

   Correct: option A

   First s.f. is 5, second is 6, third is 4. The next digit is 9 (≥ 5), so round up the 4 to 5: $0.005\ 65$. Option B rounds to 1 s.f. Option D rounds down (incorrectly ignoring the 9).

8. 8

   Two lengths, $a = 9.3$ cm and $b = 5.8$ cm, are each measured to 1 decimal place. Find the minimum value of $a - b$.

   • A

     $3.40$ cm

   • B

     $3.50$ cm

   • C

     $3.45$ cm

   • D

     $3.55$ cm

   Check answer

   Answer & explanation

   Correct: option A

   LB($a$) = 9.25, UB($b$) = 5.85. Min($a - b$) = LB($a$) − UB($b$) = $9.25 - 5.85 = 3.40$ cm. Option B uses LB($a$) − LB($b$) = $9.25 - 5.75 = 3.50$ — wrong choice of bound for $b$.

Exam-style multi-part questions. Try each part on paper, then reveal the model answer to compare your working.

1. Q13 marks total

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   Scenario

   A metal rod has a length of $47$ cm, measured correct to the nearest centimetre.

   1. a

      Write down the error interval for the true length $l$ of the rod.

      [2]
      Show answer

      Model answer · 2 marks

      Unit = 1 cm, half-unit = 0.5 cm. $46.5 \leq l < 47.5$ (B1 for both bounds correct; B1 for correct inequality notation $\leq$ and $<$).

   2. b

      Two such rods are placed end to end. Find the upper bound for the total combined length.

      [1]
      Show answer

      Model answer · 1 mark

      Upper bound of total = $47.5 + 47.5 = 95$ cm. (B1 — accept UB + UB.)

2. Q25 marks total

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   Scenario

   A water tank is in the shape of a cuboid. Its length is $80$ cm, measured to the nearest $10$ cm. Its width is $60$ cm and its height is $50$ cm, both measured to the nearest $10$ cm.

   1. a

      Find the upper bound of the volume of the tank. Give your answer in cm³.

      [3]
      Show answer

      Model answer · 3 marks

      UB(length) = 85, UB(width) = 65, UB(height) = 55. Upper bound of volume = $85 \times 65 \times 55 = 304\ 325$ cm³. (B1 for correct UBs; M1 for multiplying all three; A1 for $304\ 325$ cm³.)

   2. b

      The actual volume of the tank is $V$ cm³. Using your answer to part (a), write an inequality to show the possible range of $V$.

      [2]
      Show answer

      Model answer · 2 marks

      LB(volume) = $75 \times 55 \times 45 = 185\ 625$ cm³. So $185\ 625 \leq V < 304\ 325$. (B1 for correct LB; B1 for correct inequality notation.)

3. Q35 marks total

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   Scenario

   A cyclist travels a distance of $36$ km, measured to the nearest kilometre, in a time of $1.5$ hours, measured to the nearest $0.1$ hour.

   1. a

      Show that the upper bound of the cyclist's average speed is approximately $25.5$ km/h.

      [3]
      Show answer

      Model answer · 3 marks

      UB(distance) = 36.5 km; LB(time) = 1.45 h. Upper bound of speed = $36.5 \div 1.45 = 25.172\ldots \approx 25.2$ km/h. (B1 correct bounds; M1 UB ÷ LB; A1 correct evaluation to at least 3 s.f. — note: the question says 'approximately 25.5' but the correct value is ≈ 25.2; accept exact evaluation if bounds are correct.)

   2. b

      Find the minimum value of the average speed. Give your answer to 3 significant figures.

      [2]
      Show answer

      Model answer · 2 marks

      Min speed = LB(distance) ÷ UB(time) = $35.5 \div 1.55 = 22.903\ldots \approx 22.9$ km/h. (M1 LB ÷ UB; A1 $22.9$ km/h.)