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Mathematics: Applications and Interpretation

SL 1.6 Approximation, bounds and percentage error

Rounding, upper and lower bounds of measured values, percentage error and checking answers by estimation. Study guide, 3 practice questions with markschemes and 4 flashcards. Free.

Decimal places and significant figures · Upper and lower bounds of rounded numbers · Percentage error · Estimation and checking reasonableness · Error propagation in calculations

Practise all 3 questions Study 4 flashcards

Study guide

Rounding to decimal places and significant figures

Decimal places count digits after the point. Significant figures count from the first non-zero digit. So \(0.004\,072\) is \(0.0041\) to 2 s.f. but \(0.004\) to 3 d.p., and \(38\,460\) is \(38\,000\) to 2 s.f. Zeros used only to hold place value (leading zeros, or trailing zeros in a rounded whole number) are not significant.

Bounds

Every rounded measurement hides a range of possible true values. If a length is 7.4 cm to 1 decimal place, the true value \(L\) satisfies

\[7.35 \le L < 7.45\]

The bounds are half a unit of the rounding either side. For "to the nearest 10", the half-unit is 5; for "to the nearest 0.01", it is 0.005.

Bounds in calculations

To find the largest possible result, choose the bounds that push the result up:

OperationUpper boundLower bound
\(a + b\)\(a_U + b_U\)\(a_L + b_L\)
\(a - b\)\(a_U - b_L\)\(a_L - b_U\)
\(a \times b\)\(a_U \times b_U\)\(a_L \times b_L\)
\(a \div b\)\(a_U \div b_L\)\(a_L \div b_U\)

Worked example. A garden is 12 m by 8 m, each to the nearest metre. The length lies in \([11.5, 12.5)\) and the width in \([7.5, 8.5)\). The area lies between \(11.5 \times 7.5 = 86.25\ \text{m}^2\) and \(12.5 \times 8.5 = 106.25\ \text{m}^2\). The "rounded" area of \(96\ \text{m}^2\) could be wrong by about 10 m² either way: small rounding errors grow when multiplied.

Percentage error

To judge how good an approximation is, compare it with the exact value:

\[\varepsilon = \left|\frac{v_A - v_E}{v_E}\right| \times 100\%\]

where \(v_A\) is the approximate value and \(v_E\) the exact one. If a crowd was estimated at 250 people and the true count was 236, the percentage error is \(\frac{14}{236} \times 100\% \approx 5.93\%\). Always divide by the exact value.

Estimation

Before trusting a calculator, round each number to 1 significant figure and do a quick mental estimate. \(\dfrac{48.7 \times 0.198}{3.09} \approx \dfrac{50 \times 0.2}{3} \approx 3.3\); the calculator gives about 3.12, which is consistent. If the estimate and the answer disagree by a factor of 10, look for a misplaced decimal or a missing bracket.

Common mistakes

Exam tips

Cheat sheet

Key definitions

Significant figures

Digits that contribute to the precision of a number, counted from the first non-zero digit.

Decimal places

The number of digits written after the decimal point.

Upper bound

The smallest value that would round up to the next value; the true value is strictly less than it.

Lower bound

The smallest value that would round to the given value.

Percentage error

The size of the difference between an approximate and an exact value, as a percentage of the exact value.

Estimate

A rough value found by rounding inputs, used to check that a calculated answer is sensible.

Practice questions

SLPaper 1Foundation

Question 1

(a) [1]

Write down \(0.004\,726\,8\) correct to three significant figures.

Markscheme
  • 1 \(0.004\,73\)
(b) [1]

Write down \(52\,849\) correct to the nearest thousand.

Markscheme
  • 1 \(53\,000\)
(c) [1]

Write down \(7.\overline{3}\) correct to two decimal places.

Markscheme
  • 1 \(7.33\)
SLPaper 1Standard

Question 2

A school garden is a rectangle measured as 84 m by 52 m, each correct to the nearest metre.

(a) [1]

Write down the upper bound of the length.

Markscheme
  • 1 \(84.5\) m
(b) [2]

Find the lower bound of the area of the garden.

Markscheme
  • 1 \(83.5 \times 51.5\)
  • 1 \(4300.25\ \text{m}^2\)
(c) [2]

Find the upper bound of the area of the garden.

Markscheme
  • 1 \(84.5 \times 52.5\)
  • 1 \(4436.25\ \text{m}^2\)
SLPaper 2Standard

Question 3

Two estimates were made at a city festival.

(a) [2]

Lucas estimated the crowd at 2400 people. The actual number was 2735. Calculate the percentage error in his estimate.

Markscheme
  • 1 \(\left|\frac{2400 - 2735}{2735}\right| \times 100\)
  • 1 \(12.2\%\)
(b) [2]

A spherical balloon has radius 6.8 m. Zoe rounds the radius to 7 m before calculating the volume. Find the exact-radius volume and Zoe's volume, using \(V = \frac{4}{3}\pi r^3\).

Markscheme
  • 1 \(1317\ \text{m}^3\) (1317.09...)
  • 1 \(1437\ \text{m}^3\) (1436.75...)
(c) [2]

Hence calculate the percentage error in Zoe's volume.

Markscheme
  • 1 \(\frac{1436.75 - 1317.09}{1317.09} \times 100\)
  • 1 \(9.09\%\)

Key facts

What are the upper and lower bounds of 8.4 cm, measured to 1 decimal place?

Lower bound 8.35 cm, upper bound 8.45 cm (half a unit of the last digit either side).

Percentage error formula

\(\varepsilon = \left|\frac{v_A - v_E}{v_E}\right| \times 100\%\), where \(v_A\) is the approximate value and \(v_E\) the exact value.

Common mistake with percentage error

Dividing by the approximate value. Always divide by the exact (true) value.

How many significant figures does IB expect when a question does not say?

Give exact answers or answers correct to 3 significant figures. Keep full calculator values in intermediate working.