| Date | May 2008 | Marks available | 1 | Reference code | 08M.2.sl.TZ1.4 |
| Level | SL only | Paper | 2 | Time zone | TZ1 |
| Command term | Draw | Question number | 4 | Adapted from | N/A |
Question
Mal is shopping for a school trip. He buys \(50\) tins of beans and \(20\) packets of cereal. The total cost is \(260\) Australian dollars (\({\text{AUD}}\)).
The triangular faces of a square based pyramid, \({\text{ABCDE}}\), are all inclined at \({70^ \circ }\) to the base. The edges of the base \({\text{ABCD}}\) are all \(10{\text{ cm}}\) and \({\text{M}}\) is the centre. \({\text{G}}\) is the mid-point of \({\text{CD}}\).
Write down an equation showing this information, taking \(b\) to be the cost of one tin of beans and \(c\) to be the cost of one packet of cereal in \({\text{AUD}}\).
Stephen thinks that Mal has not bought enough so he buys \(12\) more tins of beans and \(6\) more packets of cereal. He pays \(66{\text{ AUD}}\).
Write down another equation to represent this information.
Stephen thinks that Mal has not bought enough so he buys \(12\) more tins of beans and \(6\) more packets of cereal. He pays \(66{\text{ AUD}}\).
Find the cost of one tin of beans.
(i) Sketch the graphs of the two equations from parts (a) and (b).
(ii) Write down the coordinates of the point of intersection of the two graphs.
Using the letters on the diagram draw a triangle showing the position of a \({70^ \circ }\) angle.
Show that the height of the pyramid is \(13.7{\text{ cm}}\), to 3 significant figures.
Calculate
(i) the length of \({\text{EG}}\);
(ii) the size of angle \({\text{DEC}}\).
Find the total surface area of the pyramid.
Find the volume of the pyramid.
Markscheme
\(50b + 20c = 260\) (A1)
[1 mark]
\(12b + 6c = 66\) (A1)
[1 mark]
Solve to get \(b = 4\) (M1)(A1)(ft)(G2)
Note: (M1) for attempting to solve the equations simultaneously.
[2 marks]
(i)
(A1)(A1)(A1)
Notes: Award (A1) for labels and some idea of scale, (A1)(ft)(A1)(ft) for each line.
The axis can be reversed.
(ii) \((4,3)\) or \((3,4)\) (A1)(ft)
Note: Accept \(b = 4\), \(c = 3\)
[4 marks]
(A1)
[1 mark]
\(\tan 70 = \frac{h}{5}\) (M1)
\(h = 5\tan 70 = 13.74\) (A1)
\(h = 13.7{\text{ cm}}\) (AG)
[2 marks]
Unit penalty (UP) is applicable in this part of the question where indicated in the left hand column.
(i) \({\text{E}}{{\text{G}}^2} = {5^2} + {13.7^2}\) OR \({5^2} + {(5\tan 70)^2}\) (M1)
(UP) \({\text{EG}} = 14.6{\text{ cm}}\) (A1)(G2)
(ii) \({\text{DEC}} = 2 \times {\tan ^{ - 1}}\left( {\frac{5}{{14.6}}} \right)\) (M1)
\( = {37.8^ \circ }\) (A1)(ft)(G2)
[4 marks]
Unit penalty (UP) is applicable in this part of the question where indicated in the left hand column.
\({\text{Area}} = 10 \times 10 + 4 \times 0.5 \times 10 \times 14.619\) (M1)
(UP) \( = 392{\text{ c}}{{\text{m}}^2}\) (A1)(ft)(G2)
[2 marks]
Unit penalty (UP) is applicable in this part of the question where indicated in the left hand column.
\({\text{Volume}} = \frac{1}{3} \times 10 \times 10 \times 13.7\) (M1)
(UP) \( = 457{\text{ c}}{{\text{m}}^3}\) (\(458{\text{ c}}{{\text{m}}^3}\)) (A1)(G2)
[2 marks]
Examiners report
Most candidates managed to write down the equation.
Most candidates managed to write down the equation.
Many managed to find the correct answer and the others tried their best but made some mistake in the process.
(i) Few candidates sketched the graphs well. Few used a ruler.
(ii) Many candidates could not be awarded ft from their graph because the answer they gave was not possible.
Very few correct drawings.
Some managed to show this more by good fortune and ignoring their original triangle than by good reasoning.
(i) Many found this as ft from the previous part. Some lost a UP here.
(ii) This was not well done. The most common answer was \({40^ \circ }\).
Many managed this or were awarded ft points.
This was well done and most candidates also remembered their units on this part.
