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IB Mathematics AI SL Approximation decimal places, significant figures Study Notes - New Syllabus

IB Mathematics AI SL Approximation decimal places, significant figures Study Notes

LEARNING OBJECTIVE

  • Approximation: decimal places, significant figures.
  • Upper and lower bounds of rounded numbers.
  • Percentage errors.
  • Estimation.

Key Concepts: 

  • Approximation & Estimation

MAI HL and SL Notes – All topics

 Rounding Numbers

♦ Rounding to the Nearest 10, 100, 1000

To round to a specific place value:
1. Identify the digit to round to.
2. Look at the digit to its right.
3. If the digit is 5 or more, round up. Otherwise, round down.
4. Replace digits to the right with zeros.

Example

Round \( 4853 \) to:
Nearest 10: ?
Nearest 100: ?
Nearest 1000: ?

▶️Answer/Explanation

Solution:

 Nearest 10: \( \rm{4850} \)
Nearest 100: \( \rm{4900} \)
 Nearest 1000: \( \rm{5000} \)

♦  Rounding to Decimal Places (dp)

To round a number to \( n \) decimal places:
 Identify the \( n \)-th digit after the decimal point.
Check the digit immediately after it to round accordingly.

Example

Round \( 78.45831 \) to:
 1 dp: ?
 2 dp: ?
 3 dp: ?

▶️Answer/Explanation

Solution:

 1 dp: \( \rm{78.5} \)
 2 dp: \( \rm{78.46} \)
 3 dp: \( \rm{78.458} \)

 

♦  Rounding to Significant Figures (sf)

1. Start counting from the first non-zero digit.
2. Keep the number of significant figures required.
3. Use rounding rules to decide the final digit.

Example

Round \( 53{,}879 \) to:  1 sf: ,  2sf:

Round \( 0.0004996 \) to:  1 sf: ,  2sf:

▶️Answer/Explanation

Solution:

Round \( 53{,}879 \) to:
 1 sf: \( \rm{50{,}000} \)
2 sf: \( \rm{54{,}000} \)

Round \( 0.0004996 \) to:
1 sf: \( \rm{0.0005} \)
 2 sf: \( \rm{0.00050} \)

 

Upper and Lower Bounds

When a number is rounded, its true value lies within a range (or bound).

$
\text{Lower Bound} = \text{Rounded Value} – \frac{\text{Smallest Unit}}{2}
$
$
\text{Upper Bound} = \text{Rounded Value} + \frac{\text{Smallest Unit}}{2}
$

Example

A value rounded to the nearest 10 is \( 250 \):

▶️Answer/Explanation

Solution:

$
\text{Lower Bound} = 250 – 5 = \rm{245}
$
$
\text{Upper Bound} = 250 + 5 = \rm{255}
$
Hence, actual value \( \in [245, 255) \)

 

 Errors in Measurement

♦  Absolute Error
$
\text{Absolute Error} = |\text{Measured Value} – \text{True Value}|
$

♦  Relative Error
$
\text{Relative Error} = \frac{\text{Absolute Error}}{\text{True Value}}
$

♦ Percentage Error
$
\text{Percentage Error} = \left( \frac{\text{Error}}{\text{True Value}} \right) \times 100\%
$

Example

Measured: \( 1.15 \), True: \( 1.25 \)

▶️Answer/Explanation

Solution:

$
\text{Error} = 0.10 \quad \Rightarrow \quad \text{Percentage Error} = \frac{0.10}{1.25} \times 100\% = \rm{8\%}
$

Example

Correct 2 sf value: \( 0.025 \), Student gave \( 0.03 \)

▶️Answer/Explanation

Solution:

$
\text{Percentage Error} = \frac{0.03 – 0.025}{0.025} \times 100\% = \rm{20\%}
$

 Estimation

♦ Estimation involves using rounded values to perform approximate calculations.

Example

A car travels \( 100 \text{ km} \) in \( 1 \text{ hr } 39 \text{ min} = \frac{99}{60} = 1.65 \text{ hrs} \)

▶️Answer/Explanation

Solution:

$
\text{Exact Speed} = \frac{100}{1.65} \approx \rm{60.6 \text{ km/h}}
$

Rounded time: \( 2 \text{ hours} \Rightarrow \text{Estimated Speed} = \frac{100}{2} = \rm{50 \text{ km/h}} \)

 

♦ Estimating Sums

Values:
$
\begin{align}
34.25 &\pm 0.005 \\
26.0 &\pm 0.05 \\
10.0 &\pm 0.5 \\
\end{align}
$

Estimated total:
$
70.25 \pm (0.005 + 0.05 + 0.5) = 70.25 \pm 0.555
$

Answer to reasonable accuracy: \( \rm{70} \) (2 sf)

Truncation vs Rounding

♦ Truncation:
Simply remove digits beyond a certain point (no rounding).

Example

\( 7.98 \) truncated to 1 dp: 
\( 7.98 \) rounded to 1 dp: 

▶️Answer/Explanation

Solution:

\( 7.98 \) truncated to 1 dp: \( \rm{7.9} \)
\( 7.98 \) rounded to 1 dp: \( \rm{8.0} \)

 

♦ Useful Approximations in IB Problems

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