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 | Justify | 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.
Find the value of the bicycle during the 5th year. Give your answer to two decimal places.
Calculate, in years, when the bicycle value will be less than 50 USD.
Find the total amount John has paid to insure his bicycle for the first 5 years.
John purchased the bicycle in 2008.
Justify why John should not insure his bicycle in 2019.
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]
OR (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 .
[3 marks]
(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 as a valid method.
OR
AND (M1)
Note: Award (M1) for their and both seen. If the student states , without 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]
(M1)(A1)
Note: Award (M1) for substitution into arithmetic series formula, (A1) for correct substitution.
565 (USD) (A1)(G2)
[3 marks]
2019 is the 12th year/term (M1)
Note: Award (M1) for 12 seen.
75.59 (value of bicycle) AND 81.5 (cost of insurance policy) (A1)(ft)
Note: Award (A1) for both sequences’ 12th term seen. The value of the bicycle will follow through from their common ratio in part (b). Do not award (M0)(A1).
the cost of the insurance policy is greater than the value of the bicycle (R1)(ft)
Note: Award (R1)(ft) for a reason consistent with their cost of insurance policy and their value of the bicycle. Follow through within this part. Award (R0) if the correct values are not explicitly seen. Accept the following contextualized reasons: “the insurance is not worth it", "the values are too close", "insurance is as much as the value of the bike", but only if their cost of insurance is greater than the value of the bicycle.
OR
75.59 < 81.5 (R1)(ft)
Note: Award (R1)(ft) for a correct numerical comparison showing their cost of insurance policy is greater than their value of the bicycle. Follow through within this part.
[3 marks]