Park School started in January 2000 with $100$ students. Every full year, there is an increase of $6\%$ in the number of students.

Find the number of students attending Park School in

(i)     January 2001;

(ii)    January 2003.

Park School started in January 2000 with $100$ students. Every full year, there is an increase of $6\%$ in the number of students.

Show that the number of students attending Park School in January 2007 is $150$.

Grove School had $110$ students in January 2000. Every full year, the number of students is $10$ more than in the previous year.

Find the number of students attending Grove School in January 2003.

Grove School had $110$ students in January 2000. Every full year, the number of students is $10$ more than in the previous year.

Find the year in which the number of students attending Grove School will be first $60\%$ more than in January 2000.

Each January, one of these two schools, the one that has more students, is given extra money to spend on sports equipment.

(i)     Decide which school gets the money in 2007. Justify your answer.

(ii)    Find the first year in which Park School will be given this extra money.

(i)     $100 \times 1.06 = 106$     (M1)(A1)(G2)

Note: (M1) for multiplying by $1.06$ or equivalent. (A1) for correct answer.

(ii)    $100 \times {1.06^3} = 119$     (M1)(A1)(G2)

Note: (M1) for multiplying by ${1.06^3}$ or equivalent or for list of values. (A1) for correct answer.

[4 marks]

$100 \times {1.06^7} = 150.36 \ldots  = 150$ correct to the nearest whole     (M1)(A1)(AG)

Note: (M1) for correct formula or for list of values. (A1) for correct substitution or for $150$ in the correct position in the list. Unrounded answer must be seen for the (A1).

[2 marks]

$110 + 3 \times 10 = 140$     (M1)(A1)(G2)

Note: (M1) for adding $30$ or for list of values. (A1) for correct answer.

[2 marks]

In (d) and (e) follow through from (c) if consistent wrong use of correct AP formula.

$110 + (n - 1) \times 10 > 176$     (A1)(M1)

$n = 8\therefore {\text{year 2007}}$     (A1)(A1)(ft)(G2)

Note: (A1) for $176$ or $66$ seen. (M1) for showing list of values and comparing them to $176$ or for equating formula to $176$ or for writing the inequality. If $n = 8$ not seen can still get (A2) for 2007. Answer $n = 8$ with no working gets (G1).

OR

$110 + n \times 10 > 176$     (A1)(M1)

$n = 7\therefore {\text{year 2007}}$     (A1)(A1)(ft)(G2)

[4 marks]

In (d) and (e) follow through from (c) if consistent wrong use of correct AP formula.

(i)     $180$     (A1)(ft)

Grove School gets the money.     (A1)(ft)

Note: (A1) for $180$ seen. (A1) for correct answer.

(ii)    ${\text{100}} \times {\text{1}}{\text{.0}}{{\text{6}}^{n - 1}} > 110 + (n - 1) \times 10$     (M1)   

$n = 20\therefore {\text{year 2019}}$     (A1)(A1)(ft)(G2)

Note: (M1) for showing lists of values for each school and comparing them or for equating both formulae or writing the correct inequality. If $n = 20$ not seen can still get (A2) for 2019. Follow through with ratio used in (b) and/or formula used in (d).

OR

$100 \times {1.06^n} > 110 + n \times 10$     (M1)

$n = 19\therefore {\text{year 2019}}$     (A1)(A1)(ft)(G2)

OR

graphically

Note: (M1) for sketch of both functions on the same graph, (A1) for the intersection point, (A1) for correct answer.

[5 marks]

This question was well answered by the majority of the candidates. Most of the candidates were able to distinguish between the arithmetic and the geometric progression. A number of candidates worked out term by term by hand for which they needed more time than those that used the formulae to find the requested terms. Some of the students that found the terms the long way also lost a mark for premature rounding. It was pleasing to see how the last part of the question was answered using different methods. Those candidates that worked throughout the question using AP and GP formulae used either the solver or a graph to find the solution of the inequality. Those candidates that worked throughout the question in the long way also managed to compare the terms and find the correct year.

This question was well answered by the majority of the candidates. Most of the candidates were able to distinguish between the arithmetic and the geometric progression. A number of candidates worked out term by term by hand for which they needed more time than those that used the formulae to find the requested terms. Some of the students that found the terms the long way also lost a mark for premature rounding. It was pleasing to see how the last part of the question was answered using different methods. Those candidates that worked throughout the question using AP and GP formulae used either the solver or a graph to find the solution of the inequality. Those candidates that worked throughout the question in the long way also managed to compare the terms and find the correct year.

This question was well answered by the majority of the candidates. Most of the candidates were able to distinguish between the arithmetic and the geometric progression. A number of candidates worked out term by term by hand for which they needed more time than those that used the formulae to find the requested terms. Some of the students that found the terms the long way also lost a mark for premature rounding. It was pleasing to see how the last part of the question was answered using different methods. Those candidates that worked throughout the question using AP and GP formulae used either the solver or a graph to find the solution of the inequality. Those candidates that worked throughout the question in the long way also managed to compare the terms and find the correct year.

This question was well answered by the majority of the candidates. Most of the candidates were able to distinguish between the arithmetic and the geometric progression. A number of candidates worked out term by term by hand for which they needed more time than those that used the formulae to find the requested terms. Some of the students that found the terms the long way also lost a mark for premature rounding. It was pleasing to see how the last part of the question was answered using different methods. Those candidates that worked throughout the question using AP and GP formulae used either the solver or a graph to find the solution of the inequality. Those candidates that worked throughout the question in the long way also managed to compare the terms and find the correct year.

This question was well answered by the majority of the candidates. Most of the candidates were able to distinguish between the arithmetic and the geometric progression. A number of candidates worked out term by term by hand for which they needed more time than those that used the formulae to find the requested terms. Some of the students that found the terms the long way also lost a mark for premature rounding. It was pleasing to see how the last part of the question was answered using different methods. Those candidates that worked throughout the question using AP and GP formulae used either the solver or a graph to find the solution of the inequality. Those candidates that worked throughout the question in the long way also managed to compare the terms and find the correct year.