Foundation June 2023 Paper 1 Q10
10
(a) Finley is asked to solve the equation \(5x + 4 = 19\).
Finley’s working is shown below.
\[\begin{aligned} 5x + 4 &= 19 \\ 5x &= 19 + 4 \\ 5x &= 23 \\ x &= 4.6 \end{aligned}\]Write down the error that Finley has made. [1]
(b) Charlie is asked to use the formula\[v = u + at\]
to find the initial velocity, when
- the acceleration is 5 m/s2
- the final velocity is 29 m/s
- the time is 3 seconds.
Charlie’s working is shown below.
\[\begin{aligned} v &= 29 + (5 \times 3) \\ v &= 29 + 15 \\ v &= 44 \end{aligned}\]Write down the error that Charlie has made. [1]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| He has added 4 | 1 | Do not allow contradictions See appendices | |
Appendix
Exemplar responses for 10a
| Response | Mark | |
|---|---|---|
| Added 4 | 1 | |
| The error Finley has made is he has added 4 whereas he needed to do the opposite | 1 | |
| He moved the equation to the other side \(5x = 19 + 4\) he should have taken 4 off 19 | 1 | |
| \(5x = 19 + 4\) | 1 | |
| He has added 19+4 instead of taking 4 off 19 | 1 | |
| \(5x\) isn’t equal to 19+4, \(5x\) plus 4 is equal to 19 | first part is correct, second part is irrelevant not a contradiction | 1 |
| Finley hasn’t worked out the right way as he has added 4 onto the 19 | 1 | |
| It should be \(5x = 19 - 4\) | 1 | |
| On the 2nd part where it says \(5x=19+4\) is wrong he shouldn’t have added 4 to 19 | 1 | |
| He should have subtracted 4 from 19 | 1 bod | |
| Finley did not subtract 4 | 0 | |
| \(5x = 19 - 4 \quad 5x = 15 \quad x = 3\) with no further explanation | 0 | |
| He should have subtracted 4 | 0 | |
| You need to do the same to both sides | 0 | |
| Finley got the answer wrong because it is \(x = 3\) | need to see a reason | 0 |
| He’s added 4 on to 19 when it’s supposed to equal 19 | 0 | |
| He wasn’t supposed to add them | do they mean 5x+4 or 19+4 | 0 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| He has used 29 as the initial velocity | 1 | Accept \(u\) and \(v\) if clear \(v = 29\) not \(u = 29\) | See appendices |
Appendix
Exemplar responses for 10b
| Response | Mark |
|---|---|
| \(v = 29\) not \(u\) | 1 |
| He put 29 in the place of the u and not the v. Its meant to be 29 = u + (5x3) not v=29+(5x3) | 1 |
| Initial velocity not 29 | 1 |
| v not initial velocity | 1 |
| He didn’t substitute properly he put 29m/s for initial velocity | 1 |
| He has written the final velocity down in the equation in the wrong place which therefore makes his calculation wrong | 1 bod |
| He has used the 29m/s when that is the velocity, he was using the wrong numbers | 0 |
| He substituted the formula wrong not enough | 0 |
| It should equal 29 not v ‘it’ isn’t enough | 0 |