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Secondary 3 · Mathematics
Rational numbers · Numbers

Approximation and Errors

Before you start, make sure you know: Approximation and rounding (Secondary Year 2 (first cycle))
Before you start, make sure you know: Percents (Secondary Year 2 (first cycle))

A video with 3,894,215 views shows up in the app as "3.9 M". The scale at the colmado reads 58.4 kg even though the real weight has more decimal places. And nobody pays 2,345.6789 pesos for a purchase: the bill is always rounded to cents. We use approximate numbers all the time. The important question is: how much are we off by?

Truncating and Rounding

There are two ways to keep fewer digits:

  • Truncating: just cut off the digits. $\pi = 3.14159\ldots$ truncated to the hundredths is $3.14$.

  • Rounding: keep the closest value. Look at the first digit you drop: if it is 5 or more, add 1 to the last digit you keep; if it is less than 5, leave it as it is.

Number line between 2.7 and 2.8 split at 2.75, showing that 2.7846 rounds to 2.8 because it is closer to that endpoint. 2.72.82.75 rounds to 2.7rounds to 2.8 2.7846 → 2.8

Move the slider and compare. Notice that the rounding error is at most half a unit of the place you round to.

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The irrational number $\sqrt{10} = 3.16227766...$ appears on your calculator screen. Copy the following table into your exercise book and complete it by carrying out the…

The irrational number $\sqrt{10} = 3.16227766...$ appears on your calculator screen.

Copy the following table into your exercise book and complete it by carrying out the indicated approximations for that number:

Type of approximation To the nearest unit To the nearest tenth To the nearest hundredth To the nearest thousandth
By truncation
By rounding
Solve
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You are helping to organize a charity run at your school. When calculating the exact distance of a section of the route with the GPS, you get 2.3458 km. To make the…

You are helping to organize a charity run at your school. When calculating the exact distance of a section of the route with the GPS, you get 2.3458 km.

To make the information posters, you need to simplify this number. Write what this distance would be if we approximate by rounding:

  1. a) To the units (km).
  1. b) To the tenths.
  1. c) To the hundredths.

For part c), use the following number line, which shows the section between 2.34 and 2.35 and where the number falls:

Number line between 2,34 and 2,35 with marks every thousandth, the midpoint 2,345 marked and the point 2,34582.342.352.345Midpoint2.3458Each small mark: 0.001 km
Solve
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You have gone to the supermarket to buy ingredients for dinner. The actual prices of the products are: Pasta: 111 pesos Tomato sauce: 129 pesos Grated cheese: 208.80…

You have gone to the supermarket to buy ingredients for dinner. The actual prices of the products are:

  • Pasta: 111 pesos

  • Tomato sauce: 129 pesos

  • Grated cheese: 208.80 pesos

  • Soft drink: 75 pesos

  1. a) Make a quick estimate of the total cost by rounding each price to the nearest unit before adding.
  2. b) Calculate the exact sum of the prices.
  3. c) Calculate the difference (absolute error) between your estimate and the actual price. Do you think it was a good estimate to know if you had 600 pesos in your pocket?
Solve
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At many gas stations, the price of fuel is displayed with three decimal places (thousandths of a pesos), although we pay with coins that only go down to cents…

At many gas stations, the price of fuel is displayed with three decimal places (thousandths of a pesos), although we pay with coins that only go down to cents (hundredths).

The price of diesel today is $1.489$ pesos/liter.

You are going to refuel with exactly $35$ liters.

  1. a) Calculate the exact actual price by multiplying the $35$ liters by the price with three decimal places.
  2. b) Calculate the price if we first round the price per liter to two decimal places ($1.49$ pesos/L).
  3. c) What is the difference in pesos between both methods? Who benefits from rounding first, the customer or the gas station?
Solve

Significant Figures

Significant figures are the digits that give information about precision. When we write 3,894,215 as "3.9 million," we are using 2 significant figures.

Diagram showing how to count significant figures: in 0.0045 the zeros on the left do not count, and in 3.050 four figures count. 0.0045 0.00: do not count 45: 2 significant figures 3.050 4 significant figures (the last 0 shows precision)

When rounding to significant figures, the result is often written in scientific notation: $14{,}592{,}301$ to 3 significant figures is $14{,}600{,}000 = 1.46 \times 10^7$.

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A famous streamer has uploaded a video that has exactly 3 894 215 views. a) Write down this quantity rounded to the nearest million. b) Write down the quantity rounded…

A famous streamer has uploaded a video that has exactly 3 894 215 views.

  1. a) Write down this quantity rounded to the nearest million.
  2. b) Write down the quantity rounded to the nearest hundred thousand.
  3. c) How many significant figures have you used in part b?
Solve
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In science and technology, it is crucial to know how precisely we are working. Round the following numbers to three significant figures: a) The number of views of a…

In science and technology, it is crucial to know how precisely we are working. Round the following numbers to three significant figures:

  1. a) The number of views of a viral video: $14\,592\,301$.
  2. b) The mass of a dust particle: $0.00040961$ grams.
  3. c) The number $\pi$: $3.141592...$
  4. d) The capacity of a hard drive in bytes: $1\,024\,576\,000$.

Afterwards, express each result using standard form.

Solve

Absolute Error and Relative Error

You are off by 1 cm when measuring your table and 1 cm when measuring the thickness of a coin. The error is the same... or is it? On the table it is trivial; on the coin it is a disaster. That is why there are two kinds of error.

Comparison of a 1 cm error on a 120 cm table with a 1 mm error on a 2 mm coin, to illustrate relative error. Table: 120 cm, you measure 121 cm Eₐ = 1 cm · Eᵣ = 1/120 ≈ 0.008 = 0.8% → very good Coin: 2 mm, you measure 3 mm Eₐ = 1 mm · Eᵣ = 1/2 = 0.5 = 50% → terrible
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When measuring the length of your study desk, the actual factory value is 120 cm. You use a tape measure and get a measurement of 119 cm. a) Calculate the absolute error…

When measuring the length of your study desk, the actual factory value is 120 cm. You use a tape measure and get a measurement of 119 cm.

  1. a) Calculate the absolute error ($E_a$) made in your measurement. Remember that the absolute error is the positive difference between the actual value and the approximate value: $E_a = |V_{real} - V_{aprox}|$.
  2. b) If your classmate measures the school corridor, which actually measures 5 000 cm, and gets 4 999 cm, their absolute error is 1 cm, the same as yours. Does this mean both measurements have the same precision? Reason your answer (you can use the concept of relative error to support your answer).
Solve
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We are going to calculate the relative error ($E_r$) to compare two different weighings. Remember that the formula is $E_r = \frac{E_a}{|V_{real}|}$. 1. Case A: You…

We are going to calculate the relative error ($E_r$) to compare two different weighings. Remember that the formula is $E_r = \frac{E_a}{|V_{real}|}$.

  1. Case A: You weigh an apple whose actual weight is 200 g and your scales show 205 g.

  2. Case B: You weigh your dog whose actual weight is 20 000 g (20 kg) and the scales show 20 005 g.

Calculate the relative error in both cases (you can express it as a decimal or percentage) and state which measurement was more precise (higher quality).

Solve
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Two students carry out different measurements for a science project. Lucía measures the length of a football pitch and gets $105$ m, with an absolute error of $0.5$ m.…

Two students carry out different measurements for a science project.

  • Lucía measures the length of a football pitch and gets $105$ m, with an absolute error of $0.5$ m.

  • Marcos measures the length of a mobile phone screen and gets $15$ cm, with an absolute error of $0.1$ cm.

Which of the two measurements is more precise (has higher quality)? Justify your answer by calculating the relative errors of both measurements.

Lucía Football pitch L = 105 m Eₐ = 0.5 m Marcos Mobile phone screen L = 15 cm Eₐ = 0.1 cm
Solve
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Calculate the relative error made in the following approximations and state which of them is the worst approximation (the one with the largest relative error): a)…

Calculate the relative error made in the following approximations and state which of them is the worst approximation (the one with the largest relative error):

  1. a) Approximate the fraction $\frac{1}{3}$ as $0.33$.
  2. b) Approximate the number $\sqrt{2}$ as $1.41$.
  3. c) Approximate a budget of $9\,000\,000$ pesos as $8\,400\,000$ pesos.
Solve

Error Bounds

Often you don't know the exact value, but you do know the precision of the instrument. If a scale rounds to 100 grams and reads $54.2$ kg, the real value is between $54.15$ and $54.25$ kg. This bound only describes the rounding; a real instrument can have other errors.

This also works for large quantities: "25,000 people" rounded to the thousands means a value between $24{,}500$ and $25{,}500$, and the error bound is $500$.

When you do operations with approximate values, the errors accumulate. To find the bound of the result, calculate the minimum and maximum possible values using the bounds of each piece of data. For example, if the length is $20$ dm and the width $10$ dm with bound $0.5$ dm, the area is between $19.5 \times 9.5$ and $20.5 \times 10.5$.

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Look at the following image showing the digital speedometer of an e-bike: The speedometer only displays integers (no decimals) and reads 24 km/h rounded to the nearest…

Look at the following image showing the digital speedometer of an e-bike:

E-BIKE ±0.5 24 km/h

The speedometer only displays integers (no decimals) and reads 24 km/h rounded to the nearest unit.

This means that the actual speed is not exactly 24 km/h, but falls within an interval. Between which two values (lower bound and upper bound) does the actual speed of the bike lie?

Solve
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Digital bathroom scales usually have a precision of $100$ grams ($0.1$ kg). This means that if the scales read $54.2$ kg, the actual weight could be slightly above or…

Digital bathroom scales usually have a precision of $100$ grams ($0.1$ kg). This means that if the scales read $54.2$ kg, the actual weight could be slightly above or slightly below, but rounded to that decimal place.

  1. a) Write down the interval of possible actual values for a measured weight of $54.2$ kg.
  2. b) What is the absolute error bound for these scales?
Solve
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On the news, it was announced that "approximately 25 000 people"* attended a demonstration. a) If we know that this figure was obtained by rounding to the nearest…

On the news, it was announced that "approximately 25 000 people" attended a demonstration.

  1. a) If we know that this figure was obtained by rounding to the nearest thousand, what is the maximum and minimum number of people that could have actually attended?
  2. b) Represent this interval on the number line.
  3. c) What is the maximum absolute error that could be made with that headline?
24 000 ? 25 000 ? 26 000 Minimum Approximation Upper limit ? ? Maximum absolute error = ?
Solve
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A digital bicycle speedometer shows the speed with one decimal place. At a given moment, it displays $24.5 \text{ km/h}$. We know that this device approximates by…

A digital bicycle speedometer shows the speed with one decimal place. At a given moment, it displays $24.5 \text{ km/h}$.

We know that this device approximates by rounding to the nearest tenth.

  1. a) Write the interval of possible actual values in which the exact speed of the bicycle lies. That is, what is the minimum and maximum speed the cyclist could be traveling at for the device to display 24.5?
  2. b) If the cyclist maintains that speed for 2 exact hours, what is the maximum difference in kilometers (absolute error bound) between the distance calculated using the value 24.5 and the possible actual distance?
Solve

Back to the beginning: the video with 3,894,215 views is shown as 3.9 M (3,900,000). The absolute error is only 5,785 views and the relative error is $E_r = \frac{5{,}785}{3{,}894{,}215} \approx 0.0015 = 0.15\%$: a negligible error that makes the number much easier to read.

Summary

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