4MA0 Higher June 2014 Paper 4R Q14
14 Rob is making a scale model of the Solar System on the school field.
He wants the distance from the Sun to Jupiter to be 8 metres on his scale model.
The real distance from the Sun to Jupiter is \(7.8 \times 10^8\) kilometres.
Give your answer in the form \(1 : n\), where \(n\) is written in standard form. (3)
Rob wants to put the position of a space probe on the scale model.
The real distance of the space probe from the Sun is \(1.9 \times 10^{10}\) kilometres, correct to 2 significant figures.
Give your answer in metres. (3)
| Scheme | Marks |
|---|---|
| \(7.8 \times 10^8 \times 1000\) or \(7.8 \times 10^{11}\) oe or 8 ÷ 1000 or 0.008 | M1 |
| \(7.8 \times 10^8 \times 1000 \div 8\) or \(7.8 \times 10^8 \div 0.008\) | M1 |
| \(9.75 \times 10^{10}\) | A1 |
| (3) |
Notes
M1: for correct conversion from m to km or from km to m
M1: (indep) award for digits 975
(eg. an answer of \(9.75 \times 10^7\) gets M0 M1 A0)
A1: cao
| Scheme | Marks |
|---|---|
| \(1.95 \times 10^{10}\) km | B1 |
\(\dfrac{1.95 \times 10^{10}}{\text{“}9.75 \times 10^{10}\text{”}}\) (=0.2(km)) or \(\dfrac{1.95 \times 10^{13}}{\text{“}9.75 \times 10^{10}\text{”}}\) (=200(m)) or \(\dfrac{1.95 \times 10^{10}}{7.8 \times 10^8} \times 8\) (=200(m)) or \(\dfrac{1.95 \times 10^{13}}{7.8 \times 10^{11}} \times 8\) (=200(m)) NB: 1.95 may be the candidate’s upper bound | M1 |
| 200 | A1 |
| (3) | |
| (6 marks) |
Notes
B1: cao
A1: cao must be from correct figures used in a correct calculation
The mark scheme states that for this question the correct answer alone does not imply a correct method.