| Date | May 2009 | Marks available | 2 | Reference code | 09M.1.sl.TZ1.13 |
| Level | SL only | Paper | 1 | Time zone | TZ1 |
| Command term | Find | Question number | 13 | Adapted from | N/A |
Question
Tennis balls are sold in cylindrical tubes that contain four balls. The radius of each tennis ball is 3.15 cm and the radius of the tube is 3.2 cm. The length of the tube is 26 cm.
Find the volume of one tennis ball.
Calculate the volume of the empty space in the tube when four tennis balls have been placed in it.
Markscheme
Unit penalty (UP) applies
\({\text{Volume of tennis ball}} = \frac{{4}}{{3}} \pi 3.15^3\) (M1)
Note: Award (M1) for correct substitution into correct formula.
(UP) Volume of tennis ball = 131 cm3 (A1) (C2)
[2 marks]
Unit penalty (UP) applies
\({\text{Volume of empty space}} = \pi 3.2^2 \times 26 - 4 \times 130.9\) (M1)(M1)(M1)
Note: Award (M1) for correct substitution into cylinder formula, (M1) 4 × their (a), (M1) for subtracting appropriate volumes.
(UP) Volume of empty space = 313 cm3 (A1)(ft) (C4)
Note: Accept 312 cm3 with use of 131.
[4 marks]
Examiners report
This question was poorly answered by many but perfectly well by many others, there being little in between. Volume seemed to be little understood and this part of the course is perhaps overlooked. A (candidate drawn) diagram helped visualise the situation and this, in general, is to be encouraged.
This question was poorly answered by many but perfectly well by many others, there being little in between. Volume seemed to be little understood and this part of the course is perhaps overlooked. A (candidate drawn) diagram helped visualise the situation and this, in general, is to be encouraged. Many found (b) difficult due to it not being broken up into “one stage” parts in the question. Practice in multi-stage questions is recommended.
