IB Mathematics SL 1.6 Approximation decimal places, significant figures AI SL Paper 1- Exam Style Questions- New Syllabus
Question
The Great Pyramid in Egypt is the oldest of the Seven Wonders of the Ancient World. When it was built, 4500 years ago, the measurements of the pyramid were in Royal Egyptian Cubits ( REC ).
Oliver checks online that 1 REC is equal to 0.52 metres, rounded to two decimal places.
(a) Write down the upper and lower bounds of 1 REC in metres. [2]
The Great Pyramid in Egypt has a square footprint with side lengths of 440 REC and a height of 280 REC . Oliver assumes that these two measurements are exact and that the Great Pyramid can be modelled as a square-based pyramid with flat faces.
(b) Find the minimum possible volume of the pyramid in cubic metres. [4]
▶️Answer/Explanation
Markscheme (with detailed working)
(a)
Given 1 REC is rounded to two decimal places as 0.52 m, so the half-step is \(0.005\) m.
Lower bound \(= 0.52 – 0.005 = \boxed{0.515\ \text{m}}\). A1
Upper bound \(= 0.52 + 0.005 = \boxed{0.525\ \text{m}}\). A1
Lower bound \(= 0.52 – 0.005 = \boxed{0.515\ \text{m}}\). A1
Upper bound \(= 0.52 + 0.005 = \boxed{0.525\ \text{m}}\). A1
(b) Minimum possible volume
To minimise the volume in metres, use the lower bound conversion \(1\ \text{REC}=0.515\ \text{m}\). M1
Convert side and height:
\(\text{base side}=440\times 0.515 = 226.6\ \text{m}\), \(\quad \text{height}=280\times 0.515 = 144.2\ \text{m}\). M1
Square-based pyramid volume: \[ V=\frac{1}{3}\times(\text{side})^2\times(\text{height}) =\frac{1}{3}\,(226.6)^2\,(144.2)\ \text{m}^3. \] Compute: \((226.6)^2=51338.56\), so \[ V=\frac{1}{3}\times 51338.56 \times 144.2 = \boxed{2\,468\,106.051\ldots\ \text{m}^3} \approx \boxed{2.47\times 10^6\ \text{m}^3}\ (\text{to 3 s.f.}). \] Award A1 for the substituted expression, A1 for a correct numerical result (exact/3 s.f.). A1 A1
Convert side and height:
\(\text{base side}=440\times 0.515 = 226.6\ \text{m}\), \(\quad \text{height}=280\times 0.515 = 144.2\ \text{m}\). M1
Square-based pyramid volume: \[ V=\frac{1}{3}\times(\text{side})^2\times(\text{height}) =\frac{1}{3}\,(226.6)^2\,(144.2)\ \text{m}^3. \] Compute: \((226.6)^2=51338.56\), so \[ V=\frac{1}{3}\times 51338.56 \times 144.2 = \boxed{2\,468\,106.051\ldots\ \text{m}^3} \approx \boxed{2.47\times 10^6\ \text{m}^3}\ (\text{to 3 s.f.}). \] Award A1 for the substituted expression, A1 for a correct numerical result (exact/3 s.f.). A1 A1
Total Marks: 6
Question
Katya approximates \( \pi \), correct to four decimal places, using the expression \( 3 + \dfrac{1}{6 + \dfrac{13}{16}} \).
(a) Find Katya’s approximation of \( \pi \), correct to four decimal places. [2]
(b) Find the percentage error of Katya’s approximation, correct to four decimal places, compared to the exact value of \( \pi \). [2]
▶️ Answer/Explanation
Markscheme
(a)
Evaluate: \( 6 + \dfrac{13}{16} = \dfrac{96}{16} + \dfrac{13}{16} = \dfrac{109}{16} \). M1
Then: \( \dfrac{1}{\dfrac{109}{16}} = \dfrac{16}{109} \), so \( 3 + \dfrac{16}{109} = \dfrac{327 + 16}{109} = \dfrac{343}{109} \approx 3.14678899 \).
Rounded to four decimal places: 3.1468. A1
[2 marks]
Evaluate: \( 6 + \dfrac{13}{16} = \dfrac{96}{16} + \dfrac{13}{16} = \dfrac{109}{16} \). M1
Then: \( \dfrac{1}{\dfrac{109}{16}} = \dfrac{16}{109} \), so \( 3 + \dfrac{16}{109} = \dfrac{327 + 16}{109} = \dfrac{343}{109} \approx 3.14678899 \).
Rounded to four decimal places: 3.1468. A1
[2 marks]
(b)
Approximation: 3.1468. Exact \( \pi \approx 3.1415926535 \).
Absolute error: \( |3.1468 – 3.1415926535| \approx 0.0052073465 \). M1
Percentage error: \( \dfrac{0.0052073465}{3.1415926535} \times 100 \approx 0.165754\% \approx 0.166\% \). A1
[2 marks]
Approximation: 3.1468. Exact \( \pi \approx 3.1415926535 \).
Absolute error: \( |3.1468 – 3.1415926535| \approx 0.0052073465 \). M1
Percentage error: \( \dfrac{0.0052073465}{3.1415926535} \times 100 \approx 0.165754\% \approx 0.166\% \). A1
[2 marks]
Total Marks: 4