Higher November 2019 Paper 5 Q3
3 Martina has answered some questions on algebra.
In each question, she has made an error.
Describe her error and give the correct answer to each problem.
(a) Question 1
Simplify. \(2a \times a \times a\)
Martina’s answer \(4a\)
[2]Martina’s answer \(4a\)
(b) Question 2
Simplify. \(\dfrac{x^{10}}{x^2}\)
Martina’s answer \(x^5\)
[2]Martina’s answer \(x^5\)
(c) Question 3
\(s = ut + \frac{1}{2}at^2\)
Find \(s\) when \(u = 0\), \(t = 5\) and \(a = 6\).
Martina’s solution \(s = 0 \times 5 + \frac{1}{2} \times 6 \times 5^2\)
\(s = 0 + 15^2\)
\(s = 225\)
[2]Find \(s\) when \(u = 0\), \(t = 5\) and \(a = 6\).
Martina’s solution \(s = 0 \times 5 + \frac{1}{2} \times 6 \times 5^2\)
\(s = 0 + 15^2\)
\(s = 225\)
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| She added the terms oe | 1 | In all 3 parts any incorrect statement treat as choice Allow correct descriptions of what Martina should have done in each part See AG | |
| \(2a^3\) | 1 | ||
Appendix: Exemplar responses for Q3a
| Response | Mark | |
|---|---|---|
| 1 | She did not multiply [the terms] | 1 |
| 2 | She added up all the a | 1 |
| 3 | She added [to] the 2a’s | 1BOD |
| 4 | She hasn’t multiplied the a’s | 1 |
| 5 | it is × not + | 1 |
| 6 | she would be right if they were plus signs and not times | 1 |
| 7 | Not 2a + a + a | 1 |
| 8 | a would equal 2a + a + a. When you times you add them 2a x a x a = 2a3 | 1 |
| 9 | She added a to the multiplication instead of using index laws | 0 |
| 10 | She has added the 2a’s then times by the 2a | 0 |
| 11 | She added the a’s to the 2 instead of multiplying them | 0 |
| 12 | She added the 2a to the a x a | 0 |
| 13 | 2a is different to a times a = a2 | 0 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| She divided the powers oe | 1 | See AG | |
| \(x^8\) | 1 | ||
Appendix: Exemplar responses for Q3b
| Response | Mark | |
|---|---|---|
| 1 | She divided the 10 and the 2/she divided the powers (must refer to ‘numbers’ or ‘indices’) | 1 |
| 2 | she done 10 ÷ 2 = 5 | 1 |
| 3 | She should have done 10 – 2 | 1 |
| 4 | She should take away the indices | 1 |
| 5 | She divided 10 by 2 instead of subtracting 2 | 1 |
| 6 | Laws of indices it should be taken away | 1 |
| 7 | Divided the numbers | 1 |
| 8 | She divided instead of taking away (‘indices’ implied by referring to division and subtraction) | 1 |
| 9 | She has used division | 0 |
| 10 | She divided it when it should be timesd | 0 |
| 11 | She didn’t use the laws of indices | 0 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| She squared (½ × 6 × 5) oe | 1 | See AG | |
| 75 | 1 | ||
Appendix: Exemplar responses for Q3c
| Response | Mark | |
|---|---|---|
| 1 | didn’t do the power first | 1 |
| 2 | She did 5 x 6 and then x ½ but the ² is near the 5² not all of it | 1 |
| 3 | She should have squared 5 first and then divide 6 by 2 and times them together | 1 |
| 4 | She didn’t do the 5² bit first | 1 |
| 5 | She didn’t square 5 and times 6 by 25 | 1 |
| 6 | She hasn’t squared the 5 | 1 |
| 7 | Doing 15² she needs to do 5² | 1 |
| 8 | She did 15² | 0 |
| 9 | She halved 30 when she meant to halve 6 earlier in the equation | 0 |
| 10 | She didn’t square the number 5 correctly nor did she times by 6 or halve it by half | 0 |
| 11 | She square after, when you square before | 0 |
| 12 | She halved 6 before multiplying by 5² | 0 |
| 13 | Didn’t use BODMAS | 0 |
| 14 | 6 x 25 = 150 | 0 |