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Date	May 2019	Marks available	3	Reference code	19M.2.SL.TZ2.T_6
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 2
Command term	Determine	Question number	T_6	Adapted from	N/A

Question

Tommaso plans to compete in a regional bicycle race after he graduates, however he needs to buy a racing bicycle. He finds a bicycle that costs 1100 euro (EUR). Tommaso has 950 EUR and invests this money in an account that pays 5 % interest per year, compounded monthly.

The cost of the bicycle, $C$ , can be modelled by $C = 20x + 1100$ , where $x$ is the number of years since Tommaso invested his money.

Determine the amount that he will have in his account after 3 years. Give your answer correct to two decimal places.

[3]

Find the difference between the cost of the bicycle and the amount of money in Tommaso’s account after 3 years. Give your answer correct to two decimal places.

[3]

After $m$ complete months Tommaso will, for the first time, have enough money in his account to buy the bicycle.

Find the value of $m$ .

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$950 \times {\left( {1 + \frac{5}{{12 \times 100}}} \right)^{12 \times 3}}$ (M1)(A1)

Note: Award (M1) for substitution in the compound interest formula: (A1) for correct substitution.

N = 3
I% = 5
PV = 950
P/Y = 1
C/Y = 12 (A1)(M1)

Note: Award (A1) for C/Y = 12 seen, (M1) for other correct entries.

N = 36
I% = 5
PV = 950
P/Y = 12
C/Y = 12 (A1)(M1)

Note: Award (A1) for C/Y = 12 seen, (M1) for other correct entries.

1103.40 (EUR) (A1)(G3)

Note: Answer must be given to 2 decimal places.

[3 marks]

(20 × 3 + 1100) − 1103.40 (M1)(M1)

Note: Award (M1) for correct substitution into cost of bike function, (M1) for subtracting their answer to part (a). This subtraction may be implied by their final answer (follow through from their part (a) for this implied subtraction).

55.60 (EUR) (A1)(ft)(G3)

Note: Follow through from part (a). The answer must be two decimal places.

[3 marks]

METHOD 1

$950 \times {\left( {1 + \frac{5}{{12 \times 100}}} \right)^{12x}} = 20x + 1100$ (M1)(M1)

Note: Award (M1) for their correct substitution in the compound interest formula with a variable in the exponent; (M1) for comparing their expressions provided variables are the same (not an expression with $x$ for years and another with $x$ representing months). Award at most (M0)(M1)(A0)(M1)(A0) for substitution of an integer in both expressions and comparison of the results. Accept inequality.

( $x$ =) 4.52157… (years) (A1)(ft)

4.52157… × 12 (= 54.2588…) (M1)

Note: Award (M1) for multiplying their value for $x$ by 12. This may be implied.

$m$ = 55 (months) (A1)(ft)(G4)

METHOD 2

$950 \times {\left( {1 + \frac{5}{{12 \times 100}}} \right)^m} = 20 \times \frac{m}{{12}} + 1100$ (M1)(M1)(M1)

Note: Award (M1) for their correct substitution in the compound interest formula with a variable in the exponent to solve; (M1) for comparing their expressions provided variables are the same; (M1) for converting years to months in these expressions. Award at most (M0)(M1)(A0)(M1)(A0) for substitution of an integer in both expressions and comparison of the results. Accept inequality.

$m$ = 54.2588… (months) (A1)(ft)

$m$ = 55 (months) (A1)(ft)(G4)

METHOD 3

(M1)(M1)

Note: Award (M1) for each graph drawn.

( $x$ =) 4.52157… (years) (A1)(ft)

4.52157… × 12 (= 54.2588…) (M1)

Note: Award (M1) for multiplying their value for $x$ by 12. This may be implied.

If the graphs drawn are in terms of months, leading to a value of 54.2588…, award (M1)(M1)(M1)(A1), consistent with METHOD 2.

$m$ = 55 (months) (A1)(ft)(G4)

Note: Follow through for a compound interest formula consistent with their part (a). The final (A1)(ft) can only be awarded for correct answer, or their correct answer following through from previous parts and only if value is rounded up. For example, do not award (M0)(M0)(A0)(M1)(A1)(ft) for an unsupported “5 years × 12 = 60” or similar.

[5 marks]

Examiners report

[N/A]

Syllabus sections

Topic 2—Functions » SL 2.4—Key features of graphs, intersections using technology

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