Higher June 2018 Paper 6 Q14
14
(a) Standard bricks have dimensions 21.5 cm by 10.3 cm by 6.5 cm, correct to 1 decimal place.
A house is built using 4663 standard bricks.
Joslin says
Placed end to end, the bricks from the house would definitely reach over 1 km.
Show that Joslin’s statement is correct. [4]
(b) A standard brick should weigh 2.8 kg, correct to 1 decimal place.
A truck can carry a maximum load of 20 tonnes.
A truck can carry a maximum load of 20 tonnes.
(i) Calculate the maximum number of standard bricks that the truck should be able to carry. [3]
(ii) Explain why your answer to (b)(i) may not be possible to achieve. [1]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 21.45 × 4663 ÷ 100 000 = 1.000 2[1..] (km) or 21.45 × 4663 = 100 020 to 100 021.4 > 100 000 (cm) or. 100 000 ÷ 21.45 = 4662[.0..] < 4663 or 100 000 ÷ 4663 = 21.44[5..] < 21.45 Note the first method does not require a comparison against 1 (km) | 4 | B1 for (minimum length =) 21.45 seen | Allow access to all marks if brick and 1 km are in consistent units. |
| B1 for 1 km = 100 000 cm soi oe such as ÷ 100 then ÷ 1000 or use of 1m = 100cm and 1km = 1000m if working in metres. | Allow these conversions even with their volume or surface area. eg 21.5 × 10.3 × 6.5 = 1439.425 cm/cm\(^2\)/cm\(^3\) = 0.014 394 25 km | ||
| M1 for their 21.45 × 4663 (÷ 100 000) or 100 000 ÷ their 21.45 or 100 000 ÷ 4663 | their 21.45 must be in the range 21.45 to 21.55 but accept equivalent if attempting the unit conversion first eg B0B0M1 for 21.5 cm = 0.0215 km followed by 0.0215 × 4663 | ||
| If M0 scored, allow SC1 for \(k\) × 4663 (÷ 100 000) or 100 000 ÷ \(k\) with \(k\) in the range 10.25 to 10.35 or 6.45 to 6.55 | Thus, use of width or height of the brick may score B0,B1,SC1 whereas use of volume may score B0/1,B1,SC0 Accept equivalent if working in m or km | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) 7017 to 7020 | 3 | B1 for 20 000 or 2.849[…] or 2.85 or 0.0028[…] seen M1 for their 20 000 ÷ their 2.85 or 20 ÷ their 0.00285 | Ignore other bound ie a division after an attempt to reach consistent units their 2.85 must be in the range 2.75 to 2.85 inc.; their 0.00285 must be in the range 0.00275 to 0.00285. B0M0 for 20 ÷ 2.8 as no attempt to reach consistent units |
| (ii) The truck may not have enough room oe Safety regulations may not allow it | 1 | Mark their best reason. 0 for we do not know the exact weight of the bricks oe 0 for because the truck may need to carry other loads 0 there may not be enough bricks available | |