1A_Ch5(1). 5.2Estimation Methods and Strategies Relating to Numbers A Estimation Methods B...

1A_Ch5(1)

5.2 Estimation Methods and Strategies Relating to Numbers

A Estimation Methods

B Estimation Strategies

Index

1A_Ch5(2)

Index

A Benchmark Strategies

Decomposition-recomposition Strategies

B

Measuring Groups of ObjectsC

Using FormulasD

5.3 Estimation Strategies in Measurement

1A_Ch5(3)

Using Graph PapersE

The Significance of Estimation1. Meaning of Estimation

Index

5.1 The Significance of Estimation 1A_Ch5(4)

i. Estimation is the way of finding an approximate

value of a number. In the case of a numerical

expression, the value of the entire expression is

estimated by taking an approximate value of each

number in the expression for computation.

ii. The result of estimating a numerical expression is

called an estimated value or an estimate.

The Significance of Estimation

2. Reasons for Estimation

Index


There are many reasons for making estimation. Listed

below are some of the main ones.

i. To simplify calculations.

ii. Limitations in measurement.

iii. Value unknown.

iv. Value varies.

v. Estimation helps to understand the numbers better.

vi. Checking reasonableness of results.

Example

In each of the following, which option A, B, C or D do you think

will give the best estimated value for the given exact value

(underlined)?

Index Key Concept 5.1.1


1. The length of the bridge is 17.8999 km.

A. 17.8 B. 18 C. 17 D. 16

2. The population of a city is 5 988 974.

A. 5 988 000 B. 5 900 000 C. 6 000 000 D. 5 500 000

3. The travelling time from Central to North Point is 11.024 min. A. 12 B. 11.1 C. 11.5 D. 11

Estimation Methods

Index

5.2 Estimation Methods and Strategies Relating to Numbers 1A_Ch5(7)

E.g. i. 573 570 (correct to the nearest ten)

ii. 573 600 (correct to the nearest

hundred)

E.g. i. 34.5 30 ii. 185 100

iii. 1.24 1 iv. 0.89 0

1. Rounding off Example

2. Front-end Example

A)

Estimation Methods

Index


3. Special Cases

When we need to apply estimation to real-life

problems, we have to adjust the degree of accuracy

of our estimation according to the situation. There

are two ways we often use to make the adjustment.

A)

Estimation Methods

Index


i. Rounding Up

Sometimes we have to make an estimated value

that is slightly larger than the exact value. We

always advance 1 to the place which is to be

corrected to, regardless of the digit that follows.

ii. Rounding Down

Sometimes we have to make an estimated value

that is slightly smaller than the exact value. All the

digits after the place which is to be corrected to are

replaced by 0.

3. Special Cases Example

A)

Index 5.2

Index


Estimate the following numbers by rounding off.

(a) 4 620 (correct to the nearest hundred)

(b) 599 (correct to the nearest ten)

(c) 0.237 (correct to the nearest 0.01)

(d) 0.0064 (correct to the nearest 0.01)

(e) 0.712 (correct to the nearest 0.1)

4600

600

0.24

0.01

0.7

Key Concept 5.2.1

Index Key Concept 5.2.1


Estimate the following numbers by the Front-end method.

(a) 1 108

(b) 4 253

(c) 321

(d) 12.8

(e) 0.334

1000

4 000

300

10

0

Index


In each of the following,

i. determine whether the underlined value in the expression

should be estimated by ‘rounding up’ or ‘rounding down’;

ii. write down the estimated value of the expression.

(a) The volume of a box of apple juice is 254 mL. How many boxes

of apple juice are needed to fill up a 1000 mL plastic box?

Expression : 1000 ÷ 254

(b) Ann baked 164 cakes. She put 5 cakes in a box, how many

boxes she needed to pack up all the cakes?

Expression : 164 ÷ 5

Index


(a) i. Rounding down

ii. Estimated value : 1000 ÷ 250 = 4

∴ 4 boxes of apple juice are needed t

o fill up a 1000 mL plastic box.

(b) i. Rounding up

ii. Estimated value : 170 ÷ 5 = 34

∴ She needed 34 boxes to pack up al

l the cakes.

Back to Question

Key Concept 5.2.2

Estimation Strategies

Index


‧ Estimation strategies help us to decide which estimation

method we can use to make a better estimate.

B)

E.g. Estimate 11 – 2.3 – 2.1 – 2.05 – 1.91 – 2.51.

Take 2 as the cluster and the expression can be

rewritten as 11 – 2 – 2 – 2 – 2 – 2 or 11 – (2 × 5).

1. Clustering Example


Index


B)

E.g. Estimate 963 ÷ 2.89 + 7.【 Take 3 as an approximate value of 2.89, then 963 and 3 form a

pair of compatible numbers under division. 】

2. Compatible Numbers Example

963 ÷ 2.89 + 7 963 ÷ 3 + 7

= 321 + 7

= 328


Index


B)

3. Translation Example

E.g. Estimate .43

16251.0

【 Change the order of the numbers or the operations in a given

expression so that the entire expression can be estimated more

easily. 】 )425.0(3

164

3

16251.0

13

16

3

16


Index


B)

4. Compensation Example

E.g. Estimate 28.4 + 31.3 + 21.6 + 67.9.【 Make adjustment to initial estimates so as to raise the degree of

accuracy of the estimation. 】

28.4 + 31.3 + 21.6 + 67.9 20 + 30 + 20 +

60

= 130

∴

8.4 + 1.3 10

1.6 + 7.9 10

∴ Estimated value after compensation = 130 + 10 + 10

= 150

Index 5.2

Estimate (204.1 + 200.9 + 192.7) ÷ 5.

Index


【 The approximate value of all the 3 numbers inside the

brackets is 200, so 200 can be regarded as the cluster. 】

(204.1 + 200.9 + 192.7) ÷ 5 200 × 3 ÷ 5

= 120

Index


Locate a cluster in each of the following numerical expressions

and find an estimated value of the expression.

(a) 83 + 79 + 82 + 80 + 76

(b) (18 + 21 + 17 + 17) × 3

(a) 83 + 79 + 82 + 80 + 76 80 × 5

= 400

(b) (18 + 21 + 17 + 17) × 3 20 × 4 x 3

= 240

Key Concept 5.2.3

Estimate 0.325 × 933 – 11.

Index


0.325 × 933 – 11

= 311 – 11

= 300

【 If we take as an approximate value of 0.325, then and 933

will form a pair of compatible numbers under

multiplication. 】

31

31

1193331

Index


Find an estimated value of each of the following numerical

expressions using compatible numbers.

(a) 92 × 0.188 × 2 (b) 179 ÷ 0.113 ÷ 2

(a) 92 × 0.188 × 2

= 18 × 2

= 36

251

90

(b) 179 ÷ 0.113 ÷ 2

= 1620 ÷ 2

= 810

291

180

Key Concept 5.2.4

Estimate (878 × 7.4) ÷ 63.

Index


【 (878 × 7.4) ÷ 63 can be written as 878 × (7.4 ÷ 63), then use

the approximate values of 878 and 7.4 to do the

estimation. 】(878 × 7.4) ÷ 63 = 878 × (7.4 ÷ 63)

878 × (7 ÷ 63)

=91

878

91

900

= 100

Index


Estimate (37 × 4.6) ÷ 18.

(37 × 4.6) ÷ 18 = 37 × (4.6 ÷ 18)

37 × (5 ÷ 20)

=41

37

41

40

= 10

Key Concept 5.2.5

Estimate 1.16 + 4.79 + 0.99 + 1.27 + 2.82.

Index


【 At first, the value of the expression is estimated to be 8 using

the method of Front-end. Then, we adjust this initial estimate

by compensating for the sum of the digits after the decimal

place so that the degree of accuracy of the estimate is

increased. 】

Front-end : 1.16 + 4.79 + 0.99 + 1.27 + 2.82

1 + 4 + 0 + 1 +

2

= 8

Index


Adjust the digits after the decimal place:

0.16 + 0.79

1

0.99

1

0.27 + 0.82

1

Estimated value after compensation

= 8 + 1 + 1 + 1

= 11

Back to Question

Index


Estimate 5.82 + 3.43 + 4.26 +

7.67.

Key Concept 5.2.6

Front-end : 5.82 + 3.43 + 4.26 + 7.67 5 + 3 + 4 + 7

= 19

Adjust the digits after the decimal place:

0.82 + 0.26 1

0.43 + 0.67 1

Estimated value after compensation

= 19 + 1 + 1

= 21

Benchmark Strategies

Index

5.3 Estimation Strategies in Measurement 1A_Ch5(27)

A)

‧ In benchmark strategies, the quantity to be

measured is estimated by comparing it with a known

benchmark.

Example

Index 5.3

In the figure, the height of the door is

known to be around 2 m. Estimate the

height of the ceiling above the ground.

Index


【 By observation, we see that AB is higher than the door by about

half the door’s height. 】

The height of the ceiling m 21

22

= (2 + 1) m

= 3 m

The figure shows a measuring cup with a 600 mL mark.

Index


(a) Estimate the volume of the liquid inside the

measuring cup.

(b) The measuring cup is now completely filled with

liquid and not overflowing. Estimate the volume of

the liquid inside the measuring cup.

Soln

Soln

Index

Back to Question


(a) We see that the liquid inside the measuring cup occupies

about of 600 mL.31

∴ The volume of the liquid inside the measuring cup

mL 31

600

= 200 mL

Index

Back to Question


(b) When the measuring cup is completely

filled with liquid, the volume of the

liquid is about 4 times what we

presently have.

∴ The volume of the liquid inside the measuring cup

4 × 200 mL

= 800 mL

Fulfill Exercise Objective

Use benchmark strategies to estimate. Key Concept 5.3.1

Decomposition-recomposition Strategies

Index


B)

‧ In decomposition-recomposition strategies, the

quantity to be measured is broken down into smaller

parts which are estimated first. Then all these smaller

parts are recomposed and the original quantity can be

estimated.

Index 5.3

Example

In the figure, the length of the corridor can be decomposed into

the widths of 6 classrooms. It is estimated that the width of each

classroom is about 7.5 m. Estimate the length of the entire

corridor.

Index


The length of the entire corridor 6 × 7.5 m

= 45 m

Index


In the figure, a bookcase is decomposed into 8

compartments. Books each about 6 cm thick are now

put into the bookcase. Estimate

(a) the width of each compartment,

(b) the width of the entire bookcase,

(c) the number of books when the bookcase is full.

Width = ?

Index

Back to Question


【 By observation, about 5 books of the same thickness can be

put into each compartment. 】

(a) The width of each compartment 5 × 6 cm

= 30 cm

(b) The width of the entire bookcase 4 × 30 cm

= 120 cm

(c) The number of books 8 × 5

= 40Fulfill Exercise Objective

Use decomposition-recomposition strategies to estimate.

Key Concept 5.3.2

Measuring Groups of Objects

Index


C)

‧ In measuring objects of very small sizes, instead of

making the measurement directly, we often measure

a large number of such objects. The measurement we

need for the small object is then obtained by division.

This is called the measuring groups of objects

strategy in estimation.

Index 5.3

Example

Suppose the total volume of 1 000 water droplets is measured to

be 270 mL,

Index


then the volume of a water droplet mL 000 1270

= 0.27 mL

Index


Bobby measures that the total weight

of 200 rubber bands is 52 g. Estimate

the weight of a rubber band.

The weight of a rubber band g 20052

= 0.26 gFulfill Exercise Objective

Measure groups of objects to estimate. Key Concept 5.3.3

Using Formulas

Index


D)

‧ Using formulas, the required quantity is obtained

indirectly when related quantities are measured and

the values are substituted into the formula for

calculation.

Index 5.3

Example

If the length of each side of the cube in the

figure is measured to be 3.5 cm, estimate

the volume of the cube.

Index


By formula, volume of cube = length × length × length.

∴ The volume 3.5 × 3.5 × 3.5 cm3

= 42.875 cm3

43 cm3

Index


The figure shows a cardboard box

for storing canned food. It is known

that two layers of cans can be

stacked up in the box and each layer

contains 12 cans.

The height and the diameter of each can have been

measured to be 11 cm and 8 cm respectively. Estimate

the volume of the cardboard box.

Index

Back to Question


Fulfill Exercise Objective

Use formulas to estimate.

Height 2 × 11 cm

= 22 cm

Width 3 × 8 cm

= 24 cm

Length 4 × 8 cm

= 32 cm

∴ Volume of the cardboard box 32 × 24 × 22 cm3

= 16 896 cm3

17 000 cm3

Key Concept 5.3.4

Using Graph Papers

Index


E)

‧ For figures with irregular shapes, we use graph

papers to estimate their areas.

Index 5.3

Example

Index

Estimate the area of the following figure.

Key Concept 5.3.5


Full square: counted as 1

More than half: counted as 1

Less than half: counted as 0

Area = 42 cm2

1 cm

1 cm

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