Date | May 2015 | Marks available | 3 | Reference code | 15M.1.sl.TZ1.5 |
Level | SL only | Paper | 1 | Time zone | TZ1 |
Command term | Calculate and Give your answer | Question number | 5 | Adapted from | N/A |
Question
Assume the Earth is a perfect sphere with radius 6371 km.
Calculate the volume of the Earth in \({\text{k}}{{\text{m}}^3}\). Give your answer in the form \(a \times {10^k}\), where \(1 \leqslant a < 10\) and \(k \in \mathbb{Z}\).
The volume of the Moon is \(2.1958 \times {10^{10}}\;{\text{k}}{{\text{m}}^3}\).
Calculate how many times greater in volume the Earth is compared to the Moon.
Give your answer correct to the nearest integer.
Markscheme
\(\frac{4}{3}\pi {(6371)^3}\) (M1)
Note: Award (M1) for correct substitution into volume formula.
\( = 1.08 \times {10^{12}}\;\;\;(1.08320 \ldots \times {10^{12}})\) (A2) (C3)
Notes: Award (A1)(A0) for correct mantissa between 1 and 10, with incorrect index.
Award (A1)(A0) for \(1.08\rm{E}12\)
Award (A0)(A0) for answers of the type: \(108 \times {10^{10}}\).
\(\frac{{1.08320 \ldots \times {{10}^{12}}}}{{2.1958 \times {{10}^{10}}}}\) (M1)
Note: Award (M1) for dividing their answer to part (a) by \(2.1958 \times {10^{10}}\).
\( = 49.3308 \ldots \) (A1)(ft)
Note: Accept \(49.1848...\) from use of 3 sf answer to part (a).
\( = 49\) (A1) (C3)
Notes: Follow through from part (a).
The final (A1) is awarded for their unrounded non-integer answer seen and given correct to the nearest integer.
Do not award the final (A1) for a rounded answer of 0 or if it is incorrect by a large order of magnitude.
Examiners report
In part (a) many candidates correctly substituted the volume formula and wrote correctly their answer using scientific notation. The calculator notation E12 was very rarely used. A minority converted to metres, resulting in an incorrect exponent. Some candidates used an incorrect equation or used their calculator incorrectly.
In part (b) many candidates subtracted the values, where they should be divided, resulting in an answer of an unrealistic magnitude. Some reversed the numerator and denominator, leading to an answer of 0.02, which would have rounded to the unrealistic answer of 0. When a reasonable answer was found, the final mark for rounding was lost by some candidates when there was no rounding or when rounding was incorrect. There seemed to be little understanding of whether or not an answer was reasonable.