IB Mathematics SL 1.1 Operations with numbers AI HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)
Diameter: \(6 \times 10^4 \, \text{km}\).
Radius: \(\frac{6 \times 10^4}{2} = 3 \times 10^4 \, \text{km}\).
Result: \(3 \times 10^4 \, \text{km}\) [1].
(b)
Volume of a sphere: \(V = \frac{4}{3} \pi r^3\).
Radius: \(r = 3 \times 10^4 \, \text{km}\).
Compute: \(r^3 = (3 \times 10^4)^3 = 27 \times 10^{12}\).
Volume: \(V = \frac{4}{3} \pi \times 27 \times 10^{12} = \pi \times 36 \times 10^{12} = \pi \times 3.6 \times 10^{13}\).
Form: \(\pi (a \times 10^k)\), so \(a = 3.6\), \(k = 13\).
Result: \(a = 3.6\), \(k = 13\) [3].
▶️ Answer/Explanation
(a)
Value: 8973 quadrillion euros = \(8973 \times 10^{15}\).
Convert: \(8973 = 8.973 \times 10^3\), so \(8973 \times 10^{15} = 8.973 \times 10^{18}\).
Result: \(8.973 \times 10^{18} \, \text{EUR}\) [2].
(b)
Volume of a sphere: \(V = \frac{4}{3} \pi r^3\).
Radius: \(r = 113 \, \text{km}\).
Compute: \(r^3 = 113^3 = 1442897\), so \(V = \frac{4}{3} \pi \times 1442897 \approx 6.04 \times 10^6 \, \text{km}^3\).
Result: \(6.04 \times 10^6 \, \text{km}^3\) [2].
(c)
Estimated volume: \(6.04 \times 10^6 \, \text{km}^3 \approx 6043992.82 \, \text{km}^3\).
Actual volume: \(6.074 \times 10^6 \, \text{km}^3\).
Percentage error: \(\left| \frac{6043992.82 – 6074000}{6074000} \right| \times 100 \approx 0.494\%\).
Result: \(0.494\%\) [2].