IB Mathematics SL 1.6 Approximation decimal places, significant figures AI HL Paper 1- Exam Style Questions- New Syllabus
Five pipes labelled “6 metres in length” were delivered to a building site. The contractor measured each pipe to check its length (in metres) and recorded the following: 5.96, 5.95, 6.02, 5.95, 5.99.
a) (i) Find the mean of the contractor’s measurements.
(ii) Calculate the percentage error between the mean and the stated approximate length of 6 metres.
b) Calculate \( \sqrt{3.87^5 – 8.73^{-0.5}} \), giving your answer
(i) correct to the nearest integer,
(ii) in the form \( a \times 10^k \), where \( 1 \leq a < 10 \), \( k \in \mathbb{Z} \).
▶️ Answer/Explanation
a) (i) To find the mean of the measurements:
Compute:
\( \text{Mean} = \frac{5.96 + 5.95 + 6.02 + 5.95 + 5.99}{5} = 5.974 \)
Thus:
The mean is \( 5.974 \).
a) (ii) To calculate the percentage error:
Use the formula:
\( \text{% error} = \frac{|6 – 5.974|}{5.974} \times 100\% \)
Compute:
\( \text{% error} = \frac{0.026}{5.974} \times 100\% \approx 0.435\% \)
Thus:
The percentage error is \( 0.435\% \).
b) To calculate \( \sqrt{3.87^5 – 8.73^{-0.5}} \):
Compute the expression:
\( 3.87^5 \approx 3452.59208024 \)
\( 8.73^{-0.5} = \frac{1}{\sqrt{8.73}} \approx 0.338505098 \)
\( 3.87^5 – 8.73^{-0.5} \approx 3452.59208024 – 0.338505098 = 3452.25357514 \)
\( \sqrt{3452.25357514} \approx 29.45728613 \)
(i) Round to the nearest integer:
\( 29.45728613 \approx 29 \)
Thus:
The answer is 29.
(ii) Express in the form \( a \times 10^k \):
\( 29.45728613 = 2.945728613 \times 10^1 \approx 2.95 \times 10^1 \)
Thus:
The answer is \( 2.95 \times 10^1 \).