8. Heledd is the captain of a cargo ship. She is planning her next voyage.
(a) Heledd has been employed to deliver 3 × 105 tonnes of sand.
Heledd needs to know the volume of the sand before the sand can be loaded on to the ship. She has been given the following information about the sand:
Mass of a grain of sand
Volume of a grain of sand
1·2 × 10−3 grams
0·32 mm3
(i) Calculate the number of grains of sand in 3 × 105 tonnes of sand. Give your answer in standard form. [3]
(ii) Calculate the volume of the 3 × 105 tonnes of sand in m3. [3]
…………… m3
(b) Heledd has been given the following instructions for her voyage:
From port A, sail 200 km due south to port B.
From port B, sail due east to port C.
From port C, sail on a bearing of 318º back to port A.
Use the space below to draw a sketch of the ship’s voyage.
Calculate the distance from port C directly back to port A. [4]
(c) The last time the cargo ship was unloading sand in port B, it took 3 cranes 45 hours to unload 250 000 tonnes of sand.
Port B now has 5 cranes. Calculate the time it will take 5 cranes to unload 300 000 tonnes of sand.
You can assume that all cranes unload sand at the same rate, and that all other conditions remain the same.
Give your answer in hours and minutes. You must show all your working. [4]
Time it will take is …………… hours and …………… minutes
Mark scheme (a)
Scheme
Marks
Comments
(i) (Number of grains of sand =) \(\underset{\text{g}}{\dfrac{3 \times 10^{11}}{1{\cdot}2 \times 10^{-3}}} \;\text{ or }\; \underset{\text{kg}}{\dfrac{3 \times 10^{8}}{1{\cdot}2 \times 10^{-6}}} \;\text{ or }\; \underset{\text{t}}{\dfrac{3 \times 10^{5}}{1{\cdot}2 \times 10^{-9}}}\)
M2
Or equivalent M1 for 3/1∙2 with inconsistent place value
\(= 2{\cdot}5 \times 10^{14}\)
A1
CAO
(ii) (Vol of sand =) \(2{\cdot}5 \times 10^{14} \times 0{\cdot}32 \div 10^9\) or equivalent
M2
FT ‘their \(2{\cdot}5 \times 10^{14}\)’ from (a)(i) M1 for \(2{\cdot}5 \times 10^{14} \times 0{\cdot}32\) with place value errors M1 for \(0{\cdot}32 \div 10^9\)
= 80 000 or equivalent (m3)
A1
FT from M2 only Be aware of 80 000 coming from incorrect workings
Mark scheme (b)
Scheme
Marks
Comments
Sight or use of a correctly placed 48º or 42º
B1
Either in their diagram OR in their sin equation OR in their cos equation
(distance =) \(\dfrac{200}{\sin48(^\circ)}\) OR \(\dfrac{200}{\cos42(^\circ)}\)
M2
FT 'their 48' or 'their 42' provided <90 AND clearly indicated on their diagram or stated M1 for \(\sin48(^\circ) = \dfrac{200}{x}\) OR \(\cos42(^\circ) = \dfrac{200}{x}\) OR M2 for alternative full method (trigonometry followed by Pythagoras) M1 for trigonometry followed by Pythagoras with values substituted correctly
= 269(∙126… km)
A1
FT from M2 only Allow 270 (km) from correct workings
If no marks, and no diagram seen, or a correct diagram with no angles seen, SC1 for sight of \(\dfrac{200}{\sin42(^\circ)}\) OR \(\dfrac{200}{\cos48(^\circ)}\)
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