Approximation AI HL

IB Mathematics AI HL · Unit 1

Number and Algebra · Lesson 1.6

Approximation

Round with purpose, describe possible values, and judge whether an answer is reasonable.

Unit 1 theme · God’s love

✦ Wisdom

Romans 5:8 says, “But God commendeth his love toward us, in that, while we were yet sinners, Christ died for us.” Our measurements and estimates have limits, but this verse gives a clear account of God’s love shown through Jesus.

You are invited to come to Jesus as your Savior and Lord. You can bring him your doubts, ask for forgiveness, and trust in his love for you.

🎯 Learning objectives

By the end of this lesson, you should be able to:

round numbers to a specified number of decimal places or significant figures;
choose a sensible degree of accuracy for an answer;
find lower and upper bounds for rounded measurements;
use bounds to find possible values of calculated quantities;
calculate and interpret percentage error; and
estimate to check whether a result is reasonable.

01 · Decimal places

Rounding to a stated number of decimal places tells us how many digits to keep after the decimal point. Look at the next digit to decide whether to round up.

Worked example

Round 18.7463 to two decimal places.

$18.7463\approx\boxed{18.75}$

We keep the 7 and 4 in the first two decimal places. The next digit is 6, so the 4 rounds up to 5.

In context: Money is commonly reported to the nearest cent, while a count of people is usually reported as a whole number.

02 · Significant figures

Significant figures count meaningful digits beginning with the first nonzero digit. Zeros between nonzero digits are significant; zeros at the beginning are not.

Worked example

Round 0.006785 to three significant figures.

$0.006785\approx\boxed{0.00679}$

The first three significant digits are 6, 7, and 8. The next digit is 5, so the 8 rounds up to 9.

Worked example

Write 45 980 to three significant figures without making the number of significant figures ambiguous.

$45\,980\approx\boxed{4.60\times10^4}$

The zero in 4.60 clearly shows that the answer has three significant figures.

03 · Choosing appropriate accuracy

A calculator may display many digits, but the situation and the precision of the data determine how many digits are useful. Keep extra digits during calculations and round your final answer appropriately.

Worked example

A calculator gives a ticket price of €17.8462. If customers pay to the nearest cent, state the price.

$€17.8462\approx\boxed{€17.85}$

Reporting €17.8462 would suggest a level of precision that cannot be used in an ordinary cash payment.

Important: “Round to three significant figures” and “round to three decimal places” are different instructions.

04 · Bounds of a rounded number

If a value is rounded to the nearest unit of size h, its actual value is within half a unit of the reported value. The lower bound is included; the upper bound is excluded when the usual rule rounds halfway values up.

Worked example

A length x is reported as 4.1 cm to one decimal place. What values could x have?

$\boxed{4.05\leq x<4.15\text{ cm}}$

The rounding unit is 0.1 cm, so half of that unit is 0.05 cm.

Worked example

A distance is reported as 380 m to the nearest 10 m. State its bounds.

$\boxed{375\leq d<385\text{ m}}$

Notation: The upper bound is often used as the greatest possible value in a calculation. Strictly, an actual measurement below an excluded upper bound can get arbitrarily close to it but cannot equal it.

05 · Bounds in a calculation

For positive lengths, the smallest rectangular area comes from multiplying the lower bounds. The upper area bound comes from multiplying the upper bounds.

Worked example

A rectangle is measured as 12.4 cm by 7.8 cm, with both lengths rounded to one decimal place. Find the bounds for its area A.

$12.35\leq L<12.45$

$7.75\leq W<7.85$

$12.35(7.75)\leq A<12.45(7.85)$

$\boxed{95.7125\leq A<97.7325\text{ cm}^2}$

Multiplying the reported measurements gives 96.72 cm², which lies within the possible area interval.

06 · Percentage error

When an accepted or actual value is known, percentage error is usually calculated relative to that value:

$\boxed{\text{Percentage error} =\frac{|\text{approximate value}-\text{accepted value}|} {|\text{accepted value}|}\times100\%}$

Worked example

A mass is estimated as 18.4 g. A more accurate measurement gives 17.9 g. Find the percentage error of the estimate relative to the more accurate measurement.

$\text{Percentage error} =\frac{|18.4-17.9|}{17.9}\times100\%$

$\boxed{\text{Percentage error}\approx2.79\%}$

Check the denominator: The question should say which value is the reference value. Use that value in the denominator.

07 · Maximum percentage error from a rounded measurement

When the actual value is unknown, bounds tell us how far a calculated result could be from the result obtained using the reported measurement. State clearly what you use as the reference value.

Worked example

A circle’s radius is measured as 2.5 cm to one decimal place. Find the greatest possible percentage difference between its actual area and the area calculated using the reported radius, relative to the reported area.

$2.45\leq r<2.55$

$A_{\text{reported}}=\pi(2.5)^2=6.25\pi$

$A_{\text{lower}}=\pi(2.45)^2=6.0025\pi$

$A_{\text{upper}}=\pi(2.55)^2=6.5025\pi$

The percentage difference at the lower end is:

$\frac{6.25\pi-6.0025\pi}{6.25\pi}\times100\% =3.96\%$

The upper-end percentage difference approaches:

$\frac{6.5025\pi-6.25\pi}{6.25\pi}\times100\% =\boxed{4.04\%}$

Therefore, the percentage difference relative to the reported area is less than 4.04%, but can be arbitrarily close to it.

Interpret carefully: If a question instead defines percentage error relative to the unknown actual area, the denominator changes. At the lower radius bound, that error is approximately 4.12%.

08 · Estimation and reasonableness

Round numbers to convenient values before calculating. An estimate helps detect calculator-entry errors and impossible answers.

Worked example

Estimate the value of the following calculation without a calculator:

$\frac{49.8(19.6)}{5.1}$

Use 50, 20, and 5 as convenient nearby numbers:

$\frac{49.8(19.6)}{5.1} \approx\frac{50(20)}{5} =\boxed{200}.$

A calculator answer of 2000 would therefore be suspicious.

Context check: A calculated length, area, or time cannot be negative. A population estimate may need to be rounded to a whole person.

✏️ IB-style Paper 1 practice

A graphic display calculator may be used. Show appropriate working and include units when required.

Round 0.00084756 to three significant figures.
[2 marks]
Round 73.9864 to:

(a) two decimal places; [1 mark]

(b) three significant figures. [1 mark]
A mass is recorded as 6.3 kg to one decimal place. State the lower
and upper bounds for the actual mass m.
[2 marks]
A distance is recorded as 460 m to the nearest 10 m. Write an
inequality for the actual distance d.
[2 marks]
A rectangle has measured sides of 12.4 cm and 7.8 cm, each
correct to one decimal place. Find the lower and upper bounds
for its area. [4 marks]
An estimate is 18.4 g and the accepted value is 17.9 g.
Find the percentage error of the estimate, relative to the
accepted value. [3 marks]
Estimate

$\frac{49.8(19.6)}{5.1}$

using convenient rounded
numbers. Explain how your estimate would help you detect an
incorrect calculator result of 2000. [3 marks]

Reveal Paper 1 solutions

$0.00084756\approx\boxed{0.000848}.$
(a)

$73.9864\approx\boxed{73.99}$

to two decimal places.

(b)

$73.9864\approx\boxed{74.0}$

to three significant figures.
$\boxed{6.25\leq m<6.35\text{ kg}}.$
$\boxed{455\leq d<465\text{ m}}.$
$12.35(7.75)\leq A<12.45(7.85)$

$\boxed{95.7125\leq A<97.7325\text{ cm}^2}.$
$\frac{|18.4-17.9|}{17.9}\times100\% \approx\boxed{2.79\%}.$
$\frac{49.8(19.6)}{5.1} \approx\frac{50(20)}5=\boxed{200}.$

A result of 2000 is roughly ten times the estimate, suggesting
an entry or place-value error.

🌍 Applications and advanced challenges

Maximum area error. A circular garden has a reported radius of 2.5 m, measured to one decimal place.

(a) State the bounds for the actual radius.
[2 marks]

(b) Find bounds for the actual area.
[3 marks]

(c) Find the greatest possible percentage difference relative to the area calculated using the reported radius.
[4 marks]
Average speed. A journey’s distance is reported as 145 km to the nearest kilometre. Its duration is reported
as 2.3 hours to one decimal place.

(a) State bounds for the actual distance and duration.
[2 marks]

(b) Find bounds for the average speed.
[4 marks]

(c) Explain why the smallest possible speed uses the smallest distance and greatest duration.
[2 marks]
Designing a container. A spherical container has a measured radius of 3.6 cm, correct to one decimal place.
Its volume is modelled by

$V=\frac43\pi r^3.$

(a) Find the volume calculated using the reported radius.
[2 marks]

(b) Find lower and upper bounds for the actual volume.
[4 marks]

(c) Determine the greatest possible percentage difference relative to the volume calculated in part (a).
[3 marks]
Communicating a result. A calculation produces 27.4381 students per classroom. Explain why this should not be reported as 27.4381 students. Give a sensible way to communicate the result and describe one question you would need to answer before deciding how to round it.
[3 marks]

Reveal application solutions

(a)

$\boxed{2.45\leq r<2.55\text{ m}}.$

(b)

$\boxed{\pi(2.45)^2\leq A<\pi(2.55)^2},$

or approximately

$\boxed{18.8574\leq A<20.4282\text{ m}^2}.$

(c) Using the reported area

$\pi(2.5)^2$

as the reference, the lower-end difference is 3.96% and the upper-end difference approaches 4.04%. The greatest possible percentage difference is therefore less than, but arbitrarily close to,
4.04%.
(a)

$144.5\leq d<145.5\text{ km}$

and

$2.25\leq t<2.35\text{ h}.$

(b) Since speed is distance divided by time,

$\frac{144.5}{2.35}<v<\frac{145.5}{2.25}.$

Therefore,

$\boxed{61.489\ldots<v<64.666\ldots \text{ km h}^{-1}}.$

(c) A smaller distance reduces the quotient, while a larger time also reduces it.
(a)

$V_{\text{reported}} =\frac43\pi(3.6)^3 \approx\boxed{195.432\text{ cm}^3}.$

(b) The radius bounds are

$3.55\leq r<3.65.$

Hence

$\frac43\pi(3.55)^3\leq V<\frac43\pi(3.65)^3,$

giving approximately

$\boxed{187.402\leq V<203.689\text{ cm}^3}.$

(c) Relative to the reported volume, the upper-end difference approaches

$\left[\left(\frac{3.65}{3.6}\right)^3-1\right]100\% \approx\boxed{4.22\%}.$

This exceeds the percentage difference at the lower bound.
A number of students must be a whole number, so 27.4381 suggests inappropriate precision. One might report
about 27 students per classroom as an average.
If planning classroom capacity, one might instead need 28 places per classroom. We would need to
know whether the number represents an average or the minimum capacity required.

✓ Lesson recap

Decimal places count digits after the decimal point; Significant figures count meaningful digits from the first
nonzero digit.
A value rounded to the nearest unit h has bounds half a unit h on either side.
Use the correct upper and lower bounds when calculating possible areas, volumes, or rates.
For a percentage error, identify the value used as the denominator.
Estimate before or after calculating to check whether an answer makes sense.

Approximation

✦ Wisdom

🎯 Learning objectives

01 · Decimal places

02 · Significant figures

03 · Choosing appropriate accuracy

04 · Bounds of a rounded number

05 · Bounds in a calculation

06 · Percentage error

07 · Maximum percentage error from a rounded measurement

08 · Estimation and reasonableness

✏️ IB-style Paper 1 practice

🌍 Applications and advanced challenges

✓ Lesson recap

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IB Mathematics: Applications and Interpretation HL 6 Sections80 LessonsLifetime

Curriculum

Approximation AI HL

✦ Wisdom

🎯 Learning objectives

01 · Decimal places

02 · Significant figures

03 · Choosing appropriate accuracy

04 · Bounds of a rounded number

05 · Bounds in a calculation

06 · Percentage error

07 · Maximum percentage error from a rounded measurement

08 · Estimation and reasonableness

✏️ IB-style Paper 1 practice

🌍 Applications and advanced challenges

✓ Lesson recap

Related

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Company

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Recommend

alphy school publishers

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