IB Mathematics SL 1.1 Operations with numbers AA SL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Part (a) [1 mark]
The diameter of the planet is \(6 \times 10^4\) km. The radius is half the diameter:
\[ \text{Radius} = \frac{6 \times 10^4}{2} = 3 \times 10^4 \, \text{km} \]
Part (b) [3 marks]
The volume of a sphere is given by \( V = \frac{4}{3} \pi r^3 \). Using \( r = 3 \times 10^4 \, \text{km} \):
\[ V = \frac{4}{3} \pi (3 \times 10^4)^3 \]
\[ = \frac{4}{3} \pi \times 27 \times 10^{12} \]
\[ = \frac{4 \times 27}{3} \pi \times 10^{12} = 36 \pi \times 10^{12} \]
\[ = \pi (36 \times 10^{12}) \]
Rewrite in the form \(\pi (a \times 10^k)\):
\[ 36 \times 10^{12} = 3.6 \times 10^{13} \]
Thus, \( a = 3.6 \), \( k = 13 \).
(a)
Correct radius: \(3 \times 10^4\) km [1]
(b)
Uses volume formula \( V = \frac{4}{3} \pi r^3 \) [1]
Computes \( r^3 = (3 \times 10^4)^3 = 27 \times 10^{12} \) [1]
Expresses volume as \(\pi (3.6 \times 10^{13})\), so \( a = 3.6 \), \( k = 13 \) [1]
▶️ Answer/Explanation
Part (a) [2 marks]
The value of the nickel is 8973 quadrillion euros, where 1 quadrillion = \(10^{15}\).
\[ 8973 \times 10^{15} = 8.973 \times 10^{3} \times 10^{15} = 8.973 \times 10^{18} \]
Thus, \( a = 8.973 \), \( k = 18 \).
Part (b) [2 marks]
The volume of a sphere is given by \( V = \frac{4}{3} \pi r^3 \). Using \( r = 113 \, \text{km} \):
\[ V = \frac{4}{3} \pi (113)^3 \]
\[ 113^3 = 113 \times 113 \times 113 = 1,442,897 \]
\[ V = \frac{4}{3} \pi \times 1,442,897 \approx 6,043,992.82 \, \text{km}^3 \]
Express in scientific notation: \( 6,043,992.82 \approx 6.04 \times 10^6 \, \text{km}^3 \).
Part (c) [2 marks]
Actual volume is \( 6.074 \times 10^6 \, \text{km}^3 \). Charlie’s estimate is \( 6.04399282 \times 10^6 \, \text{km}^3 \).
Percentage error = \(\left| \frac{\text{estimated value} – \text{actual value}}{\text{actual value}} \right| \times 100\):
\[ \left| \frac{6.04399282 \times 10^6 – 6.074 \times 10^6}{6.074 \times 10^6} \right| \times 100 \]
\[ = \left| \frac{6.04399282 – 6.074}{6.074} \right| \times 100 \approx \frac{0.03000718}{6.074} \times 100 \approx 0.494 \, \% \]
(a)
Correctly expresses 8973 quadrillion as \( 8.973 \times 10^{18} \) (EUR) [2]
(b)
Uses volume formula \( V = \frac{4}{3} \pi r^3 \) with \( r = 113 \, \text{km} \) [1]
Computes volume as \( 6.04 \times 10^6 \, \text{km}^3 \) (or equivalent, e.g., 6043992.82) [1]
(c)
Correctly sets up percentage error formula [1]
Calculates percentage error as 0.494% [1]
▶️ Answer/Explanation
Part (a) [3 marks]
Convert all numbers to a comparable form (scientific notation or decimal):
\(1.6 \times 10^{-19} = 0.00000000000000000016\)
\(9.8 \times 10^{-18} = 0.0000000000000000098\)
\(\pi \approx 3.14159\)
\(3.5 = 3.5\)
\(0.006073 \times 10^6 = 6073\)
\(60730 = 60730\)
\(6.073 \times 10^5 = 607300\)
Order from smallest to largest:
\[1.6 \times 10^{-19}, 9.8 \times 10^{-18}, \pi, 3.5, 0.006073 \times 10^6, 60730, 6.073 \times 10^5\]
Part (b) [1 mark]
There are 7 numbers. The median is the 4th number in the ordered list from part (a):
\[3.5\]
Part (c) [1 mark]
Among the numbers, \(\pi\) is irrational because it cannot be expressed as a fraction of two integers.
(a)
Correct order: \(1.6 \times 10^{-19}, 9.8 \times 10^{-18}, \pi, 3.5, 0.006073 \times 10^6, 60730, 6.073 \times 10^5\) [3]
Award (A1) for \(\pi\) before 3.5
Award (A1) for \(1.6 \times 10^{-19}\) before \(9.8 \times 10^{-18}\)
Award (A1) for the three numbers containing 6073 in the correct order
Award (A1) for the pair with negative indices placed before 3.5 and \(\pi\) and the remaining three numbers placed after
Award (A3) for numbers given in correct decreasing order
Award (A2) for decreasing order with at most 1 error
(b)
Correct median: 3.5 (follow through from candidate’s list) [1]
(c)
Correctly identifies \(\pi\) as irrational [1]