Date | May 2015 | Marks available | 1 | Reference code | 15M.1.sl.TZ1.12 |
Level | SL only | Paper | 1 | Time zone | TZ1 |
Command term | Write down | Question number | 12 | Adapted from | N/A |
Question
An iron bar is heated. Its length, \(L\), in millimetres can be modelled by a linear function, \(L = mT + c\), where \(T\) is the temperature measured in degrees Celsius (°C).
At 150°C the length of the iron bar is 180 mm.
Write down an equation that shows this information.
At 210°C the length of the iron bar is 181.5 mm.
Write down an equation that shows this second piece of information.
At 210°C the length of the iron bar is 181.5 mm.
Hence, find the length of the iron bar at 40°C.
Markscheme
\(180 = 150m + c\;\;\;\)(or equivalent) (A1) (C1)
\(181.5 = 210m + c\;\;\;\)(or equivalent) (A1) (C1)
\(m = 0.25,{\text{ }}c = 176.25\) (A1)(A1)(ft)
Note: Follow through from part (a) and part (b), irrespective of working shown.
\(L = 0.025(4) + 176.25\) (M1)
Note: Award (M1) for substitution of their \(m\), their \(c\) and 40 into equation.
\(L = 177\;\;\;(177.25){\text{ (mm)}}\) (A1)(ft) (C4)
Note: Follow through, within part (c), from their \(m\) and \(c\) only if working shown.
Examiners report
The equations in part (a) and (b) were given correctly by the vast majority of the candidates.
The equations in part (a) and (b) were given correctly by the vast majority of the candidates.
Part (c) was in most cases either completely correct or awarded no marks at all. Only few were able to find the values of m and c, and therefore the length at 40°C. Part (c) was often left open or answered incorrectly. A common answer was L = 40m + c . Very few partial correct responses were given. Some candidates managed a correct 3 sf answer by intelligent guessing. As the question was not structured asking for the m and c values explicitly, not many candidates made an attempt to find those values. Very few seemed to realize they could find those values using their GDC. An attempt to use simultaneous equations was the most common approach.