(i)     Find the number of taxis in the city at the end of 2005.

(ii)    Find the year in which the number of taxis is double the number of taxis there were at the end of 2000.

At the end of 2000 there were \(25600\) people in the city who used taxis.

After n years the number of people, P, in the city who used taxis is given by\[P = \frac{{2560000}}{{10 + 90{{\rm{e}}^{ - 0.1n}}}} .\](i)     Find the value of P at the end of 2005, giving your answer to the nearest whole number.

(ii)    After seven complete years, will the value of P be double its value at the end of 2000? Justify your answer.

Let R be the ratio of the number of people using taxis in the city to the number of taxis. The city will reduce the number of taxis if \(R < 70\) .

(i)     Find the value of R at the end of 2000.

(ii)    After how many complete years will the city first reduce the number of taxis?

(i) \(n = 5\)     (A1)

\(T = 280 \times {1.12^5}\)

\(T = 493\)     A1     N2

(ii) evidence of doubling     (A1)

e.g. 560

setting up equation     A1

e.g. \(280 \times {1.12^n} = 560\), \({1.12^n} = 2\)

\(n = 6.116 \ldots \)     (A1)

in the year 2007     A1     N3

[6 marks]

(i) \(P = \frac{{2560000}}{{10 + 90{{\rm{e}}^{ - 0.1(5)}}}}\)     (A1)

\(P = 39635.993 \ldots \)     (A1)

\(P = 39636\)     A1     N3

(ii) \(P = \frac{{2560000}}{{10 + 90{{\rm{e}}^{ - 0.1(7)}}}}\)

\(P = 46806.997 \ldots \)     A1

not doubled     A1     N0

valid reason for their answer     R1

e.g. \(P < 51200\)

[6 marks]

(i) correct value     A2     N2

e.g. \(\frac{{25600}}{{280}}\) , 91.4, \(640:7\)

(ii) setting up an inequality (accept an equation, or reversed inequality)     M1

e.g. \(\frac{P}{T} < 70\) , \(\frac{{2560000}}{{(10 + 90{{\rm{e}}^{ - 0.1n}})280 \times {{1.12}^n}}} < 70\)

finding the value \(9.31 \ldots \)     (A1)

after 10 years     A1     N2

[5 marks]

A number of candidates found this question very accessible. In part (a), many correctly solved for n, but often incorrectly answered with the year 2006, thus misinterpreting that 6.12 years after the end of 2000 is in the year 2007.

Many found correct values in part (b) and often justified their result by simply noting the value after seven years is less than 51200. A common alternative was to divide 46807 by 25600 and note that this ratio is less than two. There were still a good number of candidates who failed to provide any justification as instructed.

Part (c) proved more challenging to candidates. Many found the correct ratio for R, however few candidates then created a proper equation or inequality by dividing the function for P by the function for T and setting this equal (or less) than 70. Such a function, although unfamiliar, can be solved using the graphing or solving features of the GDC. Many candidates chose a tabular approach but often only wrote down one value of the table, such as \(n = 10\) , \(R = 68.3\) . What is essential is to include the two values between which the correct answer falls. Sufficient evidence would include \(n = 9\) , \(R = 70.8\) so that it is clear the value of \(R = 70\) has been surpassed.