Higher June 2019 Unit 2 Q1
1.
(a) Write down the \(n\)th term of the following sequence. [2]
8, 11, 14, 17, ……
(b) Make \(t\) the subject of the formula \(r = 3t - 8\). [2]
(c) A rectangle has a length of \((x + 5)\) cm and a width of \((2x - 3)\) cm.
Its perimeter is 46 cm.
Calculate the value of \(x\). [4]
Its perimeter is 46 cm.
Calculate the value of \(x\). [4]
| Scheme | Marks | Comments |
|---|---|---|
| 3n + 5 or equivalent | B2 | B1 for sight of 3n. B0 for −3n Mark final answer. |
| Scheme | Marks | Comments |
|---|---|---|
| \(3t = r + 8\) or \(r + 8 = 3t\) or \(-3t = -r - 8\) \(t = \dfrac{r + 8}{3}\) or \(\dfrac{r + 8}{3} = t\) or \(t = \dfrac{-r - 8}{-3}\) | B1 B1 | F.T. only from \(3t = \pm r \pm 8\), stated or implied. (\(3t = r + 8\) will have already gained the previous B1.) B1B0 for \(-\,t = \dfrac{-r - 8}{3}\) or equivalent. Mark final answer. Note Allow B1B0 for \(t = (r + 8) \div 3\) with or without brackets. Allow B1B0 for \(\dfrac{r + 8}{3}\) (‘\(t\)’ missing) |
| Scheme | Marks | Comments |
|---|---|---|
| 6x + 4 = 46 OR 3x + 2 = 23 | B2 | B1 for (x + 5) + (2x – 3) + (x + 5) + (2x – 3) = 46 or equivalent e.g. (x + 5) + (2x – 3) = 23 |
| 6x = 42 OR 3x = 21 | B1 | FT collection of ‘their terms’ if of equivalent difficulty. (linear equation only.) |
| ( x =) 7 | B1 | FT only from ax = b. Allow a fraction from a FT value unless x is a whole number. (x =) 7 gains all four marks. Each B mark implies all previous B marks. Mark final answer. |
1.(c) Alternative method
| Scheme | Marks | Comments |
|---|---|---|
| A trial showing correct values and understanding of perimeter. (e.g. 2(4 + 5) + 2(2×4 – 3) = 28 ) | B1 | Consistent use of x AND correct evaluation. |
| An improved trial. | B1 | Dependent on first B1. If 1st trial is using ‘7’ award B1B1 followed by B1 if left embedded but B2 if shown as x = 7. |
| (x = )7 | B2 | B1 for an implied / embedded ‘x = 7’ but not shown as x = 7. (x =) 7 gains all four marks. Mark final answer. |