iGCSE Mathematics (0580) - C1.10 Limits of accuracy- Exam Style Questions Paper 1- New Syllabus
▶️ Answer/Explanation
The measurement is correct to the nearest \(1 \text{ cm}\).
To find the bounds, halve the degree of accuracy: \(1 \div 2 = 0.5 \text{ cm}\).
Lower bound: \(180 – 0.5 = 179.5\).
Upper bound: \(180 + 0.5 = 180.5\).
✅ Answer: \(179.5 \le h < 180.5\)
▶️ Answer/Explanation
Rounding to the nearest metre means the degree of accuracy is 1 m. To find the lower and upper limits, we subtract and add half of this unit ($0.5\text{ m}$) from our measurement: Lower bound $= 635 – 0.5 = 634.5\text{ m}$ and Upper bound $= 635 + 0.5 = 635.5\text{ m}$.
✅ Answer: $634.5 \le h < 635.5$
✅ Answer: $634.5 \le h < 635.5$
▶️ Answer/Explanation
Answer:
(a) 4900 could result from rounding to different place values (e.g., nearest 10 or 100). Since 4896 rounds to 4900 in both cases, Rahul can’t be certain which method Tanvi used.
(b) Solution steps:
- Calculate numerator: 6.4 × 4² = 6.4 × 16 = 102.4
- Calculate denominator: 17.9 – 6.1 = 11.8
- Divide: 102.4 ÷ 11.8 ≈ 8.678 (to 3 decimal places)
Final answer: 8.678