Foundation November 2022 Paper 1 Q13
13 Emma tries to simplify \(\quad cd \times 2\)
Here is her method.
| \(c \times 2 = 2c\) \(d \times 2 = 2d\) \(2c \times 2d = 4cd\) |
What is wrong with her method? [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| Valid explanation referencing the multiplication by 2 twice | B1 | eg she has multiplied by 2 twice |
Additional guidance
| She multiplied 2 by 2 but there was only one 2 to start with | B1 |
| \(2 \times 2\) should not be calculated | B1 |
| She doubled everything | B1 |
| There should only be one 2 | B1 |
| There should be a 2 | B0 |
| She’s adding up the 2s, whereas it should be \(cd \times 2 = 2cd\) | B0 |
| She multiplied by 4 (instead of 2) | B1 |
| She has 4 instead of 2 | B0 |
| The 4 is wrong | B0 |
| She should not have both \(2c\) and \(2d\) | B1 |
| She has multiplied (each of) \(c\) and \(d\) by 2 separately | B1 |
| She has multiplied (each of) \(c\) and \(d\) separately | B0 |
| She did \(2c \times 2d\) | B0 |
| Answer is \(2c + 2d\) | B0 |
| She shouldn’t separate the \(c\) and \(d\), it’s just \(2c\) | B0 |
| You don’t times each letter by 2 | B1 |
| She has multiplied each letter by 2 | B1 |
| She has multiplied each letter by 2, it should be \(2cd^2\) | B0 |
| It is \(c \times d \times 2\) not \(2c \times 2d\) | B1 |
| She shouldn’t do all that it is just \(cd \times 2 = 2cd\) | B0 |
| Answer is \(2cd\) | B0 |
| Her answer is wrong | B0 |