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Date	November Example questions	Marks available	1	Reference code	EXN.2.SL.TZ0.7
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 0
Command term	State	Question number	7	Adapted from	N/A

Question

Helen and Jane both commence new jobs each starting on an annual salary of $$ 70, 000$ $$ 70, 000$ . At the start of each new year, Helen receives an annual salary increase of $$ 2400$ $$ 2400$ .

Let $$ H_{n}$ $$ H_{n}$ represent Helen’s annual salary at the start of her $n$ $n$ th year of employment.

At the start of each new year, Jane receives an annual salary increase of $3 %$ $3 %$ of her previous year’s annual salary.

Jane’s annual salary, $$ J_{n}$ $$ J_{n}$ , at the start of her $n$ $n$ th year of employment is given by $J_{n} = 70000 {(1.03)}^{n - 1}$ $J_{n} = 70 000 {(1.03)}^{n - 1}$ .

At the start of year $N$ $N$ , Jane’s annual salary exceeds Helen’s annual salary for the first time.

Show that $H_{n} = 2400 n + 67600$ $H_{n} = 2400 n + 67 600$ .

[2]

Given that $J_{n}$ $J_{n}$ follows a geometric sequence, state the value of the common ratio, $r$ $r$ .

[1]

Find the value of $N$ $N$ .

[3]

c.i.

For the value of $N$ $N$ found in part (c) (i), state Helen’s annual salary and Jane’s annual salary, correct to the nearest dollar.

[2]

c.ii.

Find Jane’s total earnings at the start of her $10$ $10$ th year of employment. Give your answer correct to the nearest dollar.

[4]

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

uses $H_{n} = H_{1} + (n - 1) d$ $H_{n} = H_{1} + (n - 1) d$ with $H_{1} = 70000$ $H_{1} = 70 000$ and $d = 2400$ $d = 2400$ (M1)

$H_{n} = 70000 + 2400 (n - 1)$ $H_{n} = 70 000 + 2400 (n - 1)$ A1

so $H_{n} = 2400 n + 67600$ $H_{n} = 2400 n + 67 600$ AG

[2 marks]

$r = 1.03$ $r = 1.03$ A1

[1 mark]

evidence of use of an appropriate table or graph or GDC numerical solve feature to find the value of $N$ $N$ such that $J_{n} > H_{n}$ $J_{n} > H_{n}$ (M1)

EITHER

for example, an excerpt from an appropriate table

(A1)

for example, use of a GDC numerical solve feature to obtain $N = 10.800 \dots$ $N = 10.800 \dots$ (A1)

Note: Award A1 for an appropriate graph. Condone use of a continuous graph.

THEN

$N = 11$ $N = 11$ A1

[3 marks]

c.i.

$H_{11} = 94000 ($)$ $H_{11} = 94 000 ($)$ A1

$J_{11} = 94074 ($)$ $J_{11} = 94 074 ($)$ A1

Helen’s annual salary is $$ 94000$ $$ 94 000$ and Jane’s annual salary is $$ 94074$ $$ 94 074$

Note: Award A1 for a correct $H_{11}$ $H_{11}$ value and A1 for a correct $J_{11}$ $J_{11}$ value seen in part (c) (i).

[2 marks]

c.ii.

at the start of the $10$ $10$ th year, Jane will have worked for $9$ $9$ years so the value of $S_{9}$ $S_{9}$ is required R1

Note: Award R1 if $S_{9}$ $S_{9}$ is seen anywhere.

uses $S_{n} = \frac{J_{1} (r^{n} - 1)}{r - 1}$ $S_{n} = \frac{J_{1} (r^{n} - 1)}{r - 1}$ with $J_{1} = 70000$ $J_{1} = 70 000$ , $r = 1.03$ $r = 1.03$ and $n = 9$ $n = 9$ (M1)

Note: Award M1 if $n = 10$ $n = 10$ is used.

$S_{9} = \frac{70000 ({(1.03)}^{9} - 1)}{1.03 - 1} = 711137.42 \dots$ $S_{9} = \frac{70 000 ({(1.03)}^{9} - 1)}{1.03 - 1} = 711 137.42 \dots$ (A1)