3.3

Estimation

We often have to estimate when we do not know an actual value. Estimation is the process of finding an approximate value of a number or a measurement. In this section, we look at different ways to make use of estimation in our daily lives.

AEstimation in Numbers

There are many instances in our daily lives where we need to estimate a sum when a calculator is not readily available. Let us explore some estimation strategies which can help us make estimates close to the actual values.

Activity 2

Objective: To explore estimation strategies for numbers.

Case 1. Mrs Lim has $30. She wants to buy: milk powder $22.50, cereals $3.45, and 4 buns $2.80. Without using a calculator, show a method to estimate if she has enough money.

自评：

Case 2. 1180 students go on an excursion by bus. Each bus carries 40 students. Without a calculator, estimate the number of buses needed.

Buses ≈

Case 3. Monthly salaries of 5 employees: $3150, $2980, $3040, $2890, $2950. Estimate the total monthly salary.

所有数字都接近——可以用集群值（cluster value）：

$3000

Total ≈ 5 × $3000 = $15 000

One method of estimating the value of a number is to round off the number by either rounding up or rounding down.

When the numbers involved in an estimation of their sum are close in value to each other, we can select a cluster value to represent each number. By averaging the given numbers and using an estimate, we can select the cluster value.

📖 WORKED EXAMPLE 10

An employer invites 194 staff to watch a movie. The price of a movie ticket is $9.80. Is $2000 enough to pay for all the movie tickets? Explain your answer.

Solution

We can round up the numbers as follows: $9.80 ≈ $10, 194 ≈ 200.

Estimated total cost = 200 × $10 = $2000.

The number of adults AND the ticket price are both rounded up, so the actual total cost < $2000. Hence $2000 is enough to pay for all the movie tickets.

✏️ TRY IT YOURSELF 10

There were 56 people who went for a company buffet lunch that cost $39.90 per person. Without using a calculator, determine whether the total bill exceeded $2500.

Total exceeded $2500?

Explain:

自评：

📖 WORKED EXAMPLE 11

Joyce buys items from an online supermarket with free delivery if she spends at least $20:

$3.95 mixed nuts $1.20 juice $13.90 chocolate box $4.50 grapes

Will she enjoy free delivery? Explain.

Solution

Round DOWN each price to the nearest dollar:

$3.95 ≈ $3, $1.20 ≈ $1, $13.90 ≈ $13, $4.50 ≈ $4

Estimated total = $3 + $1 + $13 + $4 = $21 > $20.

Since each was rounded DOWN, the actual cost > $21 > $20 → Joyce qualifies for free delivery.

💬 DISCUSS

Danny wants two items costing $10.95 and $9.95. He rounds DOWN: $10 + $9 = $19 < $20. Does this mean he can't enjoy free delivery? Explain.

自评：

✏️ TRY IT YOURSELF 11

Joanna bought 110 handphone casings at $3.60 each to sell online. She sold all of them at $8.95 each. Without a calculator, determine whether she made a profit of more than $500.

Profit > $500?

Explain:

自评：

📖 WORKED EXAMPLE 12

The heights of five basketball players are 179, 183, 186, 175, 182 cm. Using a cluster value, estimate:

(a) total height of the five players, (b) average height.

Solution

Cluster value = 180 cm (all numbers cluster around 180).

(a) Total ≈ 5 × 180 = 900 cm.

(b) Average ≈ 180 cm (which is the cluster value itself).

NOTE: The cluster value is the estimate of the average of the given numbers.

THINK²

In which of these cases is it suitable to use a cluster value to estimate the total sum? Explain.

(a) 31, 32, 28, 33, 27 (b) 4, 15, 3, 59, 30

(a) suitable? (b) suitable?

自评：

✏️ TRY IT YOURSELF 12

The masses of six boys are 53, 56, 61, 52, 59, 54 kg. Using a cluster value, estimate:

(a) total mass ≈ kg (b) average mass ≈ kg

BEstimation in Measurements

We often use measuring tools to measure lengths and distances. If measuring tools are not available, how can we get a reasonable estimate? Let us explore some estimation strategies for measurements.

We often make use of benchmarks as reference measures in our estimations. Examples:

Diameter of a $1 coin ≈ 2.5 cm
Arm span / footstep (varies per person — calibrate against ruler once, then reuse)
Typical storey height (residential building) ≈ 3 m

Without an appropriate benchmark, an estimation can be made by dividing the object to be measured into several parts. Estimate each part, then add them up. This is the decomposition-recomposition strategy.

Example. Estimate the height of a 25-storey block of flats. Assume each storey ≈ 3 m. Estimated height = 25 × 3 = 75 m.

storey 1 ≈ 3 m

storey 2 ≈ 3 m

...

storey 25 ≈ 3 m

75 m

📖 WORKED EXAMPLE 13

In the diagram, the door knob is 1 m above the ground. Estimate:

(a) the height of the door, (b) the height of the ceiling.

Solution

We take the height of the door knob from the ground as a benchmark for measurement.

(a) Height of the door ≈ 2 × (height of door knob) = 2 × 1 = 2 m.

(b) Height of the ceiling ≈ 3 × 1 = 3 m.

✏️ TRY IT YOURSELF 13

In a diagram, the boy is 1.2 m tall. Estimate:

(a) the height of the tree ≈ m (b) the height of the lamp post ≈ m

💡 思考后展开提示

依靠视觉比例——树是男孩的几倍？灯柱呢？常见答案 (a) 2.4 m 或 3.6 m, (b) 3.6 m 或 4.8 m。

📖 WORKED EXAMPLE 14

There are 3 adjacent stalls along one side of a school canteen. If each stall is about 4 m wide, estimate the length of the side of the canteen.

Solution

Length of the side = total width of the stalls ≈ 3 × 4 = 12 m.

✏️ TRY IT YOURSELF 14

A stack of 10 books on a table; each book is about 2 cm thick. Estimate the height of the stack.

Stack height ≈ cm

CEstimating Results of Computation

We may get a wrong answer when using a calculator (pressing wrong keys, etc.). We can use estimation to check if our answers are reasonable.

📖 WORKED EXAMPLE 15

Miriam calculates the value of each expression and obtains the answers shown. Determine if each answer is reasonable.

(a) 62.9 − 4.67 × 18.62 = 542.0446 (Miriam's)

(b) √83 ÷ ∛130 = 1.798 420 573 (Miriam's)

Solution

(a) Round each to 1 s.f.: 62.9 ≈ 60; 4.67 ≈ 5; 18.62 ≈ 20.
Estimate = 60 − 5 × 20 = 60 − 100 = −40. Miriam's 542 ≠ −40. NOT reasonable.
(Actual ≈ −24.06 — close to −40 in sign and magnitude.)

(b) Use approximate values whose square root and cube root are easy: 83 ≈ 81 (= 9²); 130 ≈ 125 (= 5³).
Estimate = √81 ÷ ∛125 = 9 ÷ 5 = 1.8. Miriam's 1.798… ≈ 1.8 → Reasonable.

✏️ TRY IT YOURSELF 15

Jason calculates the value of each expression and obtains the answers shown. Determine whether each answer is reasonable.

(a) 79.5 − 33.21 × 29.52 = −185.3592 Reasonable?

(b) ∛7900 ÷ √50 = 12.569 805 09 Reasonable?

💡 思考后展开估算过程

(a) 1 s.f. 估算：80 − 30 × 30 = −820。Jason 的 −185 ≠ −820 → 不合理。实际 ≈ −900.86。
(b) 估算：∛8000 ÷ √49 = 20/7 ≈ 2.86。Jason 的 12.57 ≠ 2.86 → 不合理。实际 ≈ 2.82。

DFollow-through Errors

In the intermediate steps, we may encounter non-exact values like √2 = 1.414 213 562 37…. Suppose we need to express our final answer correct to 3 s.f., how many s.f. should we round 1.414 213 562 37 to?

Activity 4

Objective: To compare follow-through errors.

A teacher goes through a proof to show 0.9̇ = 1:

Since 0.3̇ + 0.3̇ + 0.3̇ = 0.9̇, and 1/3 = 0.3̇,
then 1/3 + 1/3 + 1/3 = 0.9̇.
Also 1/3 + 1/3 + 1/3 = 1, so we have 0.9̇ = 1.

Amy observes she gets a different answer when she calculates:

1/3 ≈ 0.3. Hence 1/3 + 1/3 + 1/3 ≈ 0.3 + 0.3 + 0.3 = 0.9 (correct to 1 s.f.).

1. How many significant figures does Amy round 1/3 to? s.f.

2. If Amy rounds off 1/3 to 2 s.f., what will her final answer be?

1/3 ≈ (correct to 2 s.f.)

Hence 1/3 + 1/3 + 1/3 ≈ + + =

= (correct to 1 s.f.)

3. If she rounds off 1/3 to 3 s.f. instead, will her final answer be the same as the teacher's proof? Explain.

自评：

Follow-through errors occur when intermediate values are rounded off to different degrees of accuracy. When the final answer is required to be correct to n significant figures, we should round off the values in the intermediate steps to at least (n + 1) significant figures.

📖 WORKED EXAMPLE 16

The area of a square tile is 2 m². Find the perimeter of the square tile, correct to 3 s.f.

Solution

Area of square tile = 2 m²
Length of square tile = √2 = 1.414 mcorrect to 4 s.f. (one more than final 3 s.f.)
Perimeter = 1.414 × 4 = 5.656
= 5.66 mcorrect to 3 s.f.

SPOTLIGHT: If we round √2 to 3 s.f. (1.41) instead, perimeter = 1.41 × 4 = 5.64 m — which is wrong to 3 s.f. By calculating with √2 × 4 directly, the answer is 5.66 m. Always carry one extra s.f. through intermediate steps.

✏️ TRY IT YOURSELF 16

The volume of a cube is 10 cm³. Find the perimeter of one face of the cube, correct to 4 s.f.

Perimeter ≈ cm

📝 PRACTICE EXERCISE 3.3

BASIC MASTERY

Without a calculator, estimate each expression by rounding off each number to 1 s.f.

(a) 125 + 3.91 × 27.48 ≈ (b) 13.24 × 4.83 × 6.09 ≈

(c) 38 467 × 0.002 96 ≈ (d) 1878 ÷ 19.2 ÷ 4.14 ≈

(e) (16.9 − 5.47) × 7.09 ≈ (f) 6.25² ÷ 4.38 ≈
Using a calculator, work out the value of each expression. Give your answers correct to 1 s.f.

(a) (159 − 60.3) × 7.58 → (b) 6.03² ÷ 1.48 →

(c) √(14.76 × 80.2 / 2.99) → (d) 23.33 × 1.991 / 0.202³ →
Use estimation to check if the given answer is reasonable.

(a) 347 − 482 + 659 = −500 →

(b) 23.92 × 4.801 = 1200 →

(c) √45 ÷ ∛9 × √99 = 87.442 359 1 →

(d) ∛64 ÷ 0.418 = 10 →

💡 思考后展开各题估算

(a) 估算 350 − 500 + 700 = 550，与 −500 差很远 → 不合理（实际 = 524）。
(b) 24 × 5 = 120，与 1200 差 10 倍 → 不合理。
(c) √49 / ∛8 × √100 = 7/2 × 10 = 35，与 87 差很远 → 不合理。
(d) 4 / 0.4 = 10 → 合理（实际 ≈ 9.57）。

INTERMEDIATE

Joyce had $3458 in her wallet. She spent $361, $86, $405, $299 on four items. Estimate the amount she had left by rounding each amount to the nearest $100.

Estimated amount left = $
In a long jump event, an athlete made four attempts with distances 7.58 m, 7.62 m, 7.54 m, 7.63 m. Without a calculator, estimate using a cluster value:

(a) Total distance ≈ m (b) Average distance ≈ m
Consider the expression 4.5012 − 3.2167 − 6.7183.

(a) Evaluate the expression correct to 3 s.f. →

(b) Evaluate after rounding each number to 3 s.f. first; give answer to 3 s.f. →

(c) Are (a) and (b) the same? If not, which is more accurate?

ENDifferent. (a) = −5.43; (b) = −5.44. (a) is more accurate — rounding at the END preserves precision through the computation. Rounding each term FIRST introduces follow-through error.

⏱ 15 秒后查看中文版

中文不同。(a) = −5.43；(b) = −5.44。(a) 更精确——最后一步才舍入保留计算过程中的精度；中间先舍入会引入"累积误差"。

自评：
The diagram shows a floor plan of a living room drawn to scale. Given the sofa is 2 m, estimate:

(a) cabinet length ≈ m (b) table width ≈ m

💡 思考后展开提示

用 sofa 长度 2 m 做基准比较——视觉估算即可。常见答案 (a) 3 m, (b) 1 m。
A rectangular field is 64.19 m long and 25.68 m wide.

(a) Find its area correct to 2 s.f.: m²

(b)(i) 64.19 to 2 s.f. = m (ii) 25.68 to 2 s.f. = m

(b)(iii) Using the rounded values, estimate the area correct to 2 s.f. = m²

(c) Are (a) and (b)(iii) the same?

ENDifferent. (a) = 1600; (b)(iii) = 1700. (a) is more accurate because the multiplication was done on full-precision values before final rounding. (b)(iii) rounds inputs first → follow-through error → final answer differs by ~6%.

⏱ 15 秒后查看中文版

中文不同。(a) = 1600；(b)(iii) = 1700。(a) 更精确——用完整数值相乘后再舍入。(b)(iii) 先把输入舍入再乘，累积误差导致最终差距约 6%。

自评：

ADVANCED

[OPEN] There were 96 cars in a queue at a traffic congestion along the AYE. Estimate the length of the queue, supposing each car was about 4.5 m long. What additional assumption would you make?

Estimated length ≈ m

Additional assumption:

ENRound: 100 cars × 5 m = 500 m. Assumption: cars are bumper-to-bumper (no gaps between cars). In reality, queued cars maintain 0.5–2 m gaps, so the true queue length is somewhat longer (perhaps 600 m).

⏱ 15 秒后查看中文版

中文取整：100 × 5 = 500 m。假设：汽车首尾相接，没有间隙。实际排队车之间有 0.5–2 m 间隔，所以真实长度可能更长（约 600 m）。

自评：
The diagram shows a $1 coin (diameter 24.65 mm) on an ez-link card. Nathan and Farah propose two methods to estimate the area of the card:

Nathan: Step 1: find area of one coin. Step 2: estimate the total number of coins needed to cover the card. Step 3: multiply.

Farah: Step 1: estimate length & width of the card using the coin. Step 2: multiply length × width.

Whose method is more accurate?

Explain:

ENFarah's method is more accurate. Circles do not tile (pack) a rectangle perfectly — there are always gaps between circles, even in the densest packing (only ~78% coverage). So Nathan's "count × area-per-coin" significantly over-counts the actual coverage. Farah measures the actual rectangular footprint of the card directly.

⏱ 20 秒后查看中文版

中文Farah 的方法更准。圆形铺不满矩形——再密的圆形排列也只有 ~78% 覆盖率，硬币之间一定有缝隙。Nathan "硬币数 × 单枚面积"会显著高估卡片面积。Farah 直接量长宽，避开了这个问题。

自评：
[OPEN] Maximum load of a lift is 600 kg. There are 8 people whose masses are 71.8, 73.4, 75.3, 74.8, 76.4, 73.9, 78.4, 77.4 kg.

(a) Estimate the total mass by rounding each to 2 s.f. (i.e., nearest integer kg):

Estimated total = kg

(b) Based on this estimation, is it safe to let all 8 people take the lift at the same time?

Safe?

(c) Verify by calculating the actual total mass:

Actual total = kg

(d) Devise an estimation strategy that ensures the lift is not overloaded when the estimated total mass is not greater than 600 kg.

ENRound each mass UP, not to nearest. If estimated total (with upper-bound masses) is ≤ 600 kg, then actual total ≤ estimate ≤ 600 — guaranteed safe. Example: 72 + 74 + 76 + 75 + 77 + 74 + 79 + 78 = 605 kg > 600 → warn "may be overloaded; verify or reduce people".

General rule: when checking a "≤ threshold" safety constraint, always round UP the inputs. Symmetrically, when checking "≥ threshold", always round DOWN.

⏱ 30 秒后查看中文版

中文每个体重都"向上取整"（不是四舍五入）。这样估算总额是真实总额的上界，估算 ≤ 600 就保证实际 ≤ 600。例：72 + 74 + 76 + 75 + 77 + 74 + 79 + 78 = 605 kg > 600 → 警告"可能超载"。

通用规则：检查"≤ 阈值"安全约束时，输入向上取整；检查"≥ 阈值"时，输入向下取整。

自评：

🎯 Chapter 3 SUM UP — Approximation & Estimation

本章三件套：

3.1 小数位（d.p.）：看下一位 5 进 4 舍。1 d.p. = 0.1 精度。
3.2 有效数字（s.f.）：5 条规则 → 从第一个非零数字数起。s.f. ≥ d.p. 更能反映精度。
3.3 估算：cluster value / benchmarks / decomposition；向上取证小于、向下取证大于；中间步骤多保留 1 s.f.。

核心心法：所有测量都是近似——选合适的精度，把累积误差控制好。

章末概念检查 · Concept Checkpoints

5 道封闭题。本章核心：估算 + 累积误差 + 选合适精度。