| Date | May 2007 | Marks available | 3 | Reference code | 07M.1.sl.TZ0.1 |
| Level | SL only | Paper | 1 | Time zone | TZ0 |
| Command term | Calculate and Find | Question number | 1 | Adapted from | N/A |
Question
Five pipes labelled, “6 metres in length”, were delivered to a building site. The contractor measured each pipe to check its length (in metres) and recorded the following;
5.96, 5.95, 6.02, 5.95, 5.99.
(i) Find the mean of the contractor’s measurements.
(ii) Calculate the percentage error between the mean and the stated, approximate length of 6 metres.
Calculate \(\sqrt {{{3.87}^5} - {{8.73}^{ - 0.5}}} \), giving your answer
(i) correct to the nearest integer,
(ii) in the form \(a \times 10^k\), where 1 ≤ a < 10, \(k \in {\mathbb{Z}}\) .
Markscheme
(i) Mean = (5.96 + 5.95 + 6.02 + 5.95 + 5.99) / 5 = 5.974 (5.97) (A1)
(ii) \({\text{% error}} = \frac{{error}}{{actualvalue}} \times 100\% \)
\( = \frac{{6 - 5.974}}{{5.974}} \times 100\% = 0.435\% \) (M1)(A1)(ft)
(M1) for correctly substituted formula.
Allow 0.503% as follow through from 5.97
Note: An answer of 0.433% is incorrect. (C3)
[3 marks]
number is 29.45728613
(i) Nearest integer = 29 (A1)
(ii) Standard form = 2.95 × 101 (accept 2.9 × 101) (A1)(ft)(A1)
Award (A1) for each correct term
Award (A1)(A0) for 2.95 × 10 (C3)
[3 marks]
Examiners report
a) Almost all candidates calculated the mean correctly but less than half were able to find the % error, many dividing by 6. This was despite the boldening of 'approximate' in the question.
b) Main errors were giving the answer correct to 1 significant figure (30) or 1 decimal place. Some candidates just counted the number of figures on the calculator to determine the index for the standard form, giving 109 instead of 101.
