Foundation June 2019 Paper 2R Q11
11 Karl has 5700 bricks.
He wants to put all the bricks into crates.
Diagram NOT accurately drawn
Each brick is a cuboid measuring 9 cm by 3 cm by 5 cm.
Each crate is a cuboid measuring 72 cm by 36 cm by 75 cm.
Karl has 4 crates.
Is there enough room in the 4 crates for 5700 bricks?
Show your working clearly.
(4)
| Scheme | Marks |
|---|---|
| Capacity of 1 brick = 9 × 3 × 5 (= 135) Capacity of 5700 bricks = 5700 × “135” (= 769500) Capacity of 1 crate = 72 × 36 × 75 (= 194400) Capacity of 4 crates = 4 × “194400” (= 777600) Bricks needed in 1 crate = 5700 ÷ 4 (= 1425) Max no: of bricks in 1 crate = 8 × 12 × 15 (= 1440) or 194400 ÷ 135 (= 1440) | M3 |
| Yes as 777600 > 769500 or Yes as 1440 > 1425 | A1 |
| (4) | |
| (4 marks) |
Notes
M3: for calculations leading to any 3 of
135, 769500, 194400, 777600, 1425 or 1440
M2 for any 2 of the above
M1 for any 1 of the above
NB: sight of 769500 implies 135 and sight of 777600 implies 194400
A1: Comparing 777600 with 769500 or
Comparing 1440 with 1425
NB. To get A1 they have to state that there is enough room for the bricks or “Yes”) and justify this by referring explicitly to 2 values e.g 777600 – 769500 (= 8100)
Alternative scheme: max number of bricks in 4 crates v 5700
| Scheme | Marks |
|---|---|
| 72 ÷ 9 (= 8) and 36 ÷ 3 (= 12) and 75 ÷ 5 (= 15) | M1 |
| “8” × “12” × “15” (= 1440) | M1 |
| “1440” × 4 (= 5760) or 5700 ÷ 4 (= 1425) | M1 |
| Yes as 5760 > 5700 or Yes as 1440 > 1425 | A1 |
Notes
M1: Dividing lengths, widths & heights
M1: Max number of bricks in 1 crate
M1: Max number of bricks in 4 crates
A1: Yes + comparison of 2 numbers
NB. Ditto comments above