IB Applications and Interpretation SL · Basic algebra and arithmetic · Section 2 of 5

Approximation, scientific notation and percentage error

3 worked examples · By Álvaro López

Key idea

In IB exams almost all numerical answers must be given to a specific precision, and many questions ask you to compare an approximate value with the exact one. In this section you learn to round (to decimal places and to significant figures), to write very large or very small numbers in scientific notation and to measure how far an estimate is off using percentage error.

Rounding: decimal places and significant figures

Rounding means replacing a number with a shorter one that is close to it, with a precision fixed in advance. There are two ways of fixing that precision.

Decimal places (dp): how many digits are kept to the right of the decimal point. Rounding to 2 dp means keeping the hundredths.
Significant figures (sf): how many useful digits we want, counting from the first digit that is not zero, regardless of where the decimal point is.

Rounding to the nearest ten, hundred or thousand follows the same method, counting to the left of the decimal point: $4736$ to the nearest hundred is $4700$ , and $4750$ to the nearest hundred is $4800$ .

For example, $6205$ has 4 sf; $0.0068$ has 2 sf (the first three zeros do not count); $0.5400$ has 4 sf (the two final zeros show that this precision was measured) and $12.0$ has 3 sf.

For rounding to $n$ significant figures, count $n$ digits from the first non-zero digit and apply the previous method using the $(n + 1)$ -th digit. The zeros that appear when rounding the integer part are placeholders, not significant figures: $48317$ to 2 sf is $48000$ .

Limits of a rounded value

A rounded measurement is not an exact number: it represents a whole interval of values. If a measurement has been rounded to a certain place, the true value may be up to half a unit of that place below or above. The lower end of the interval is the lower bound and the upper end is the upper bound.

For example, a length given as $7.4$ m (rounded to tenths, so half a unit is $0.05$ m) satisfies $7.35 \leq L < 7.45$ . A mass of $250$ g rounded to the nearest unit satisfies $249.5 \leq m < 250.5$ .

Estimating

Estimating means working out, by approximation, a value close enough to the true one, rounding the data first so that the calculations are simple. In the IB, when an estimate is asked for, each piece of data is rounded to 1 significant figure and the calculation is carried out with those values.

Scientific notation

Scientific notation (or standard form) lets us write very large or very small quantities without a long row of zeros.

To convert to scientific notation, move the decimal point until just one non-zero digit is to its left; the exponent counts how many places it has moved: to the left, positive exponent; to the right, negative. Thus, $93000000 = 9.3 \times 1 0^{7}$ and $0.00072 = 7.2 \times 1 0^{- 4}$ .

Be careful: $25 \times 1 0^{3}$ and $0.4 \times 1 0^{6}$ are equal to the numbers they represent, but they are not in scientific notation, because $a$ is not between $1$ and $10$ .

Error and percentage error

When a quantity is approximated, it matters to know how far we have strayed.

The absolute value makes $ε$ always positive: it tells us how far off we are, not in which direction. The denominator is always the exact value, never the approximation, because the error is measured relative to the real quantity.

Intuition: an error of 2 units is huge if the exact value is 10 (a $20 %$ ) and almost invisible if the exact value is 5,000 (a $0.04 %$ ). The percentage error captures that difference.

Worked examples

Example 1 Basic

Rounding to different precisions

Round the number $5 287.046$ in three ways:

a) to 3 significant figures;

b) to 2 decimal places;

c) to the nearest hundred.

d) Round $0.004052$ to 2 significant figures.

Solution

a) The first three digits are $5$ , $2$ and $8$ , and the next is $7 \geq 5$ , so $528$ becomes $529$ . The digits missing from the integer part are replaced by zeros: $5290$ .

b) The first two decimal digits are $0$ and $4$ , and the next digit is $6 \geq 5$ , so $0.04$ becomes $0.05$ : $5 287.05$ .

c) The hundreds digit is $2$ (from $52 \dots$ ) and the next digit is $8 \geq 5$ , so $52$ hundreds become $53$ : $5300$ .

d) The leading zeros are not significant. The first significant digit is $4$ , the second is $0$ and the next is $5 \geq 5$ , so $0.0040$ becomes $0.0041$ .

Example 2 Medium

Operations in scientific notation

Let $m = 3.6 \times 1 0^{6}$ and $n = 4.5 \times 1 0^{- 2}$ . Calculate, with the result in scientific notation:

a) $m \cdot n$

b) $\frac{m}{n}$

c) $m^{2}$

Then write each result as an integer.

Solution

a) Multiply the numbers and add the exponents:

$m \cdot n = (3.6 \cdot 4.5) \times 1 0^{6 + (- 2)} = 16.2 \times 1 0^{4}$

As $16.2$ is not between $1$ and $10$ , move the decimal point one place and add $1$ to the exponent: $m \cdot n = 1.62 \times 1 0^{5} = 162000$ .

b) Divide the numbers and subtract the exponents:

$\frac{m}{n} = \frac{3.6}{4.5} \times 1 0^{6 - (- 2)} = 0.8 \times 1 0^{8}$

Now the number $0.8$ is less than $1$ : move the decimal point one place to the right and subtract $1$ from the exponent. $\frac{m}{n} = 8 \times 1 0^{7} = 80000000$ .

c) Square the number and double the exponent:

$m^{2} = 3. 6^{2} \times 1 0^{12} = 12.96 \times 1 0^{12} = 1.296 \times 1 0^{13}$

That is, $12960000000000$ .

Example 3 Hard

Estimating and measuring error

A rectangular room measures $4.8$ m long, $6.3$ m wide and $2.6$ m high.

a) Calculate its exact volume and give it to 3 significant figures.

b) To estimate the volume, Marta rounds each dimension to 1 significant figure and multiplies. What volume does she estimate?

c) Calculate, to 3 significant figures, the percentage error Marta makes in her estimate.

d) Pablo calculates the volume with the exact measurements and rounds only the result to 1 significant figure. What is the percentage error of his value? Compare it with Marta's.

Solution

a) $V = 4.8 \cdot 6.3 \cdot 2.6 = 30.24 \cdot 2.6 = 78.624 m^{3}$ . To 3 sf: $78.6 m^{3}$ .

b) Round each dimension to 1 sf: $4.8 \to 5$ , $6.3 \to 6$ and $2.6 \to 3$ . The estimate is $5 \cdot 6 \cdot 3 = 90 m^{3}$ .

c) The exact value is $v_{E} = 78.624$ (unrounded) and the approximate value is $v_{A} = 90$ :

$ε = \frac{90 - 78.624}{78.624} \times 100 % = \frac{11.376}{78.624} \times 100 % \approx 14.47 %$

To 3 sf: $ε \approx 14.5 %$ .

d) Rounded to 1 sf, $78.624$ becomes $80$ . Then

$ε = \frac{80 - 78.624}{78.624} \times 100 % = \frac{1.376}{78.624} \times 100 % \approx 1.75 %$

Pablo's error is about eight times smaller. This is what happens when you round at the beginning rather than at the end: the rounding errors of the three measurements accumulate in the product.

In the exam: always use the exact, unrounded value in the denominator and in the subtraction. If the question says "estimate", round the data to 1 sf before operating; if not, round only at the end. Write the formula before substituting: it usually earns marks.

Common mistakes

Summary

Rounding: if the next digit is less than 5, the kept digit does not change; if it is 5 or more, it increases by 1. You can round to decimal places or to significant figures.
Significant figures: count the non-zero digits, the zeros between digits and the final zeros after the decimal point; leading zeros are not counted.
If the question does not state a precision, the IB expects the exact answer or 3 sf; round only the final result.
A rounded value allows half a unit of the last place above and below (lower and upper bounds). Estimating means operating with data rounded to 1 sf.
Scientific notation: $a \times 1 0^{k}$ with $1 \leq a < 10$ and $k \in Z$ . After operating, renormalise the result.
Percentage error: $ε = \frac{v _{A} - v _{E}}{v _{E}} \times 100 %$ , with the exact value in the denominator.
On the GDC, the E on the screen is not copied: write $\times 1 0^{k}$ .

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