IB Mathematics SL 1.1 Operations with numbers AI SL Paper 1 - Exam Style Questions - New Syllabus
Question
Mira is designing a bridge to cross a river. She estimates that the weight of the steel needed for this bridge is approximately \(53\,632\,000\ \text{kg}\).
The precise weight of the steel needed for the bridge is \(55\,625\,000\ \text{kg}\).
(a) Find the percentage error in Mira’s approximation. [2]
Mira’s design is used to construct five the same bridges.
(b) (i) Find the weight of the steel needed for these five bridges, to three significant figures. [1]
(ii) Write down your answer to part (b)(i) in the form \(a \times 10^{k}\), where \(1 \le a < 10\), \(k \in \mathbb{Z}\). [2]
▶️ Answer/Explanation
Markscheme
(a)
Percentage error formula (using exact as reference):
\[ \text{Percentage error} \;=\; \frac{\lvert \text{approx} – \text{exact} \rvert}{\text{exact}} \times 100 \;=\; \frac{\lvert 53\,632\,000 – 55\,625\,000 \rvert}{55\,625\,000} \times 100. \] Compute the difference: \(55\,625\,000 – 53\,632\,000 = 1\,993\,000\).
Divide by exact value: \(\dfrac{1\,993\,000}{55\,625\,000} \approx 0.0358292\ldots\).
Convert to percent: \(0.0358292\ldots \times 100 \approx 3.58\%\). M1 A1
\[ \text{Percentage error} \;=\; \frac{\lvert \text{approx} – \text{exact} \rvert}{\text{exact}} \times 100 \;=\; \frac{\lvert 53\,632\,000 – 55\,625\,000 \rvert}{55\,625\,000} \times 100. \] Compute the difference: \(55\,625\,000 – 53\,632\,000 = 1\,993\,000\).
Divide by exact value: \(\dfrac{1\,993\,000}{55\,625\,000} \approx 0.0358292\ldots\).
Convert to percent: \(0.0358292\ldots \times 100 \approx 3.58\%\). M1 A1
[2 marks]
(b)
(i) Total steel for five bridges (use multiplication with a cross):
\[ 5 \times 55\,625\,000\ \text{kg} \;=\; 278\,125\,000\ \text{kg}. \] To three significant figures: \(\boxed{278\,000\,000\ \text{kg}}\). A1
\[ 5 \times 55\,625\,000\ \text{kg} \;=\; 278\,125\,000\ \text{kg}. \] To three significant figures: \(\boxed{278\,000\,000\ \text{kg}}\). A1
[1 mark]
(ii) Scientific notation of the rounded value:
\[ 278\,000\,000\ \text{kg} \;=\; 2.78 \times 10^{8}\ \text{kg}. \] Mantissa \(a=2.78\) (with \(1 \le a < 10\)) and exponent \(k=8\). A1 A1
\[ 278\,000\,000\ \text{kg} \;=\; 2.78 \times 10^{8}\ \text{kg}. \] Mantissa \(a=2.78\) (with \(1 \le a < 10\)) and exponent \(k=8\). A1 A1
[2 marks]
Total Marks: 5
Question
Amara is positioned at point \( K \), the top of a \( 218 \, \text{m} \) vertical cliff. The base of the cliff is located at point \( B \). A ship is located at point \( S \), \( 1200 \, \text{m} \) from Amara.
This information is shown in the following diagram (not to scale).
(a) Obtain the angle of elevation from the ship to Amara. [2]
(b) Obtain the horizontal distance from the base of the cliff to the ship. [2]
(c) Write down your answer to part (b) in the form \( a \times 10^k \), where \( 1 \leq a < 10 \) and \( k \in \mathbb{Z} \). [2]
▶️ Answer/Explanation
Markscheme
(a)
In right triangle \( \triangle BSK \), \( KB = 218 \, \text{m} \) (opposite), \( KS = 1200 \, \text{m} \) (hypotenuse). \[ \begin{aligned} \sin(\angle B\hat{S}K) &= \frac{\text{opposite}}{\text{hypotenuse}} = \frac{218}{1200} = \frac{109}{600} \approx 0.1816666667 \quad \text{(M1)} \\ \angle B\hat{S}K &= \sin^{-1}\left( \frac{109}{600} \right) \approx 10.4668\ldots^\circ \end{aligned} \] Answer: \( 10.5^\circ \) A1
[2 marks]
(b)
Using the Pythagorean theorem in \( \triangle BSK \)
\(SB^2 + KB^2 = KS^2 \)
\(SB^2 + 218^2 = 1200^2 \quad \text{(M1)} \)
\(SB^2 + 47524 = 14400010\)
\(SB^2 = 1440000 – 47524 = 1392476 \)
\(SB = \sqrt{1392476} \approx 1180.032203\)
\(\text{Answer: } 1180\,\mathrm{m}\)
A1
[2 marks]
(c)
Convert \( 1180 \) to scientific notation (\( a \times 10^k \), \( 1 \leq a < 10 \), \( k \in \mathbb{Z} \)): \[ \begin{aligned} 1180 &= 1.18 \times 1000 \quad \text{(A1)} \\ &= 1.18 \times 10^3 \quad \text{(A1)} \end{aligned} \] Answer: \( 1.18 \times 10^3 \)
[2 marks]
Total Marks: 6