3.X
Chapter 3 — Let's Sum Up · Review Exercise · Maths Journal
🎯 LET'S SUM UP!
APPROXIMATION
To approximate is to find a suitable value for a quantity within a specified degree of accuracy.
Rules for rounding off numbers to specified place value / decimal places
- Look for the digit in the specified place value / decimal place.
- Consider the next digit to its right. If it is 5 or more, round up; if less than 5, round down.
34 267 = 34 000 (correct to the nearest 1000)
6.0784 = 6.08 (correct to 2 d.p.)
Rules to identify significant digits
- All non-zero digits are significant.
- All zeros between non-zero digits are significant.
- For any integer, the zeros at the end may or may not be significant depending on how the number is approximated.
- For any decimal, all zeros before the first non-zero digit are NOT significant.
- For any decimal, zeros after the first non-zero digit ARE significant.
Rules for rounding off to specified s.f.
- Look for the specified significant figure starting from the left.
- Consider the next digit to its right. If 5 or more, round up; if less than 5, round down.
ESTIMATION
Estimation is the process of finding an approximate value of a number or a measurement.
Estimation strategies
(a) In numbers
- Round off numbers
- Use cluster values
(b) In measurements
- Use benchmarks
- Use decomposition-recomposition
Follow-through errors
Follow-through errors in numerical calculations might occur when intermediate values are rounded off to different degrees of accuracy.
If the final answer is required to be correct to 3 s.f., then all intermediate values should be given to at least 4 s.f.
Reasons for using approximation and estimation
- Difficult to obtain exact values due to limitations of measuring instruments.
- Unnecessary to use the exact value.
- Too difficult to obtain the exact value.
- Useful to check the reasonableness of the answers in computations.
本章 8 道综合复习题——结合 3.1(d.p.)、3.2(s.f.)、3.3(estimation)的所有规则。
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Evaluate 3 1/2 × 7 1/3 + (−1 1/8) ÷ 2 1/4, giving your answer:
(a) in the exact value:
(b) correct to 2 decimal places:
(c) correct to 3 significant figures:
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Three items cost $16.95, $23.40, $5.15.
(a) Estimate the total by rounding each to 1 s.f.: $
(b) Mr Cai has only $50 in his wallet. Does he have enough?
Explain:
自评: -
A car park has 6 identical parking lots arranged in 3 columns × 2 rows. A car of 4.5 m × 1.7 m fits in one lot.
(a) Each lot is approximately m by m
(b) The area of the car park = m²
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Edmund uses two different weighing scales. Scale A displays his mass as 53.0 kg; Scale B displays his mass as 53.00 kg. Do the two scales measure to the same degree of accuracy? If not, which is more accurate?
自评: -
The radius of a circle is 18 cm, correct to the nearest cm. The actual radius is a value with one decimal place. Find the largest possible error when we calculate the area using the rounded value of 18 cm. Give your answer correct to 3 s.f.
Largest possible error ≈ cm²
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Haoquan and his parents plan to visit Singapore for 4 days, 3 nights from Beijing.
- 3 air tickets, round trip BEIJING ↔ SINGAPORE: SGD $562 each.
- Accommodation (1 room for 3 nights): 4 Apr $198, 5 Apr $189, 6 Apr $209.
(a) Estimate the total cost in SGD (rounded to the nearest $100): $
(b) Convert to Renminbi (RMB). Given 1 SGD = 4.89 RMB, estimate total RMB to the nearest 1000 RMB: RMB
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A high-sensitivity balance measures masses to 0.0001 g.
(a) Mass of object A is 40.0005 g. Round to 3 s.f.: g
(b) Mass of object B is 2.01 g, correct to 3 s.f. Find the smallest and greatest possible actual mass.
Smallest = g Greatest = g
💡 思考后展开解释
3 s.f. 的 2.01 g 表示真值在 [2.005, 2.015) g。最小 = 2.005;最大就看测量精度——若用 4 s.f. 表示是 2.0149(因为 2.0150 会进位到 2.015 → 3 s.f. = 2.02,不再是 2.01)。所以 greatest < 2.015 → 实际取 2.0149 g(4 dp)或近似 2.014 g(3 dp)。
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The sum of 3 consecutive even numbers is estimated to be 200. If 200 is rounded off to 1 s.f., what are:
(a) the smallest possible values of the 3 numbers:
(b) the largest possible values of the 3 numbers:
💡 思考后展开推导
1 s.f. 的 200 表示真值范围 [150, 250)。设 3 个连续偶数为 n, n+2, n+4,和为 3n+6。
3n + 6 ∈ [150, 250) → n ∈ [48, 81.33),且 n 是偶数。
最小:n = 48 → 48, 50, 52(和 150)
最大:n = 80 → 80, 82, 84(和 246)
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