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Date May 2019 Marks available 3 Reference code 19M.2.SL.TZ1.T_5
Level Standard Level Paper Paper 2 Time zone Time zone 1
Command term Find Question number T_5 Adapted from N/A

Question

John purchases a new bicycle for 880 US dollars (USD) and pays for it with a Canadian credit card. There is a transaction fee of 4.2 % charged to John by the credit card company to convert this purchase into Canadian dollars (CAD).

The exchange rate is 1 USD = 1.25 CAD.

John insures his bicycle with a US company. The insurance company produces the following table for the bicycle’s value during each year.

The values of the bicycle form a geometric sequence.

During the 1st year John pays 120 USD to insure his bicycle. Each year the amount he pays to insure his bicycle is reduced by 3.50 USD.

Calculate, in CAD, the total amount John pays for the bicycle.

[3]
a.

Find the value of the bicycle during the 5th year. Give your answer to two decimal places.

[3]
b.

Calculate, in years, when the bicycle value will be less than 50 USD.

[2]
c.

Find the total amount John has paid to insure his bicycle for the first 5 years.

[3]
d.

Markscheme

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

1.042 × 880 × 1.25  OR  (880 + 0.042 × 880) × 1.25      (M1)(M1)

Note: Award (M1) for multiplying 880 by 1.042 and (M1) for multiplying 880 by 1.25.

1150 (CAD)  (1146.20 (CAD))      (A1)(G2)

Note: Accept 1146.2 (CAD)

[3 marks]

a.

704 880   OR  563.20 704       (M1)

Note: Award (M1) for correctly dividing sequential terms to find the common ratio, or 0.8 seen.

880(0.8)5−1      (M1)

Note: Award (M1) for correct substitution into geometric sequence formula.

360.45 (USD)      (A1)(G3)

Note: Do not award the final (A1) if the answer is not correct to 2 decimal places. Award at most (M0)(M1)(A0) if r = 1.25 .

[3 marks]

b.

880 ( 0.8 ) n 1 < 50      (M1)

Note: Award (M1) for correct substitution into geometric sequence formula and (in)equating to 50. Accept weak or strict inequalities. Accept an equation. Follow through from their common ratio in part (b). Accept a sketch of their GP with y = 50 as a valid method.

OR

u 13 = 60.473   AND   u 14 = 48.379       (M1)

Note: Award (M1) for their u 13 and u 14 both seen. If the student states u 14 = 48.379 < 50 , without u 13 = 60.473 seen, this is not sufficient to award (M1).

14 or “14th year” or “after the 13th year”     (A1)(ft)(G2)

Note: The context of the question requires the final answer to be an integer. Award at most (M1)(A0) for a final answer of 13.9 years. Follow through from their 0.8 in part (b).

[2 marks]

c.

5 2 ( ( 2 × 120 ) + ( 3.5 ( 5 1 ) ) )     (M1)(A1)

Note: Award (M1) for substitution into arithmetic series formula, (A1) for correct substitution.

565 (USD)    (A1)(G2)

[3 marks]

d.

Examiners report

[N/A]
a.
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b.
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c.
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d.

Syllabus sections

Topic 1—Number and algebra » SL 1.2—Arithmetic sequences and series
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Topic 1—Number and algebra » SL 1.3—Geometric sequences and series
Topic 1—Number and algebra
Prior learning

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