Date | May 2015 | Marks available | 7 | Reference code | 15M.2.sl.TZ2.6 |
Level | SL only | Paper | 2 | Time zone | TZ2 |
Command term | Explain | Question number | 6 | Adapted from | N/A |
Question
Ramiro walks to work each morning. During the first minute he walks 80 metres. In each subsequent minute he walks 90% of the distance walked during the previous minute.
The distance between his house and work is 660 metres. Ramiro leaves his house at 08:00 and has to be at work by 08:15.
Explain why he will not be at work on time.
Markscheme
METHOD 1
recognize that the distance walked each minute is a geometric sequence (M1)
egr=0.9, valid use of 0.9
recognize that total distance walked is the sum of a geometric sequence (M1)
egSn, a(1−rn1−r)
correct substitution into the sum of a geometric sequence (A1)
eg80(1−0.9n1−0.9)
any correct equation with sum of a geometric sequence (A1)
eg80(0.9n−10.9−1)=660, 1−0.9n=6680
attempt to solve their equation involving the sum of a GP (M1)
eggraph, algebraic approach
n=16.54290788 A1
since n>15 R1
he will be late AG N0
Note: Do not award the R mark without the preceding A mark.
METHOD 2
recognize that the distance walked each minute is a geometric sequence (M1)
egr=0.9, valid use of 0.9
recognize that total distance walked is the sum of a geometric sequence (M1)
egSn, a(1−rn1−r)
correct substitution into the sum of a geometric sequence (A1)
eg80(1−0.9n1−0.9)
attempt to substitute n=15 into sum of a geometric sequence (M1)
egS15
correct substitution (A1)
eg80(0.915−10.9−1)
S15=635.287 A1
since S<660 R1
he will not be there on time AG N0
Note: Do not award the R mark without the preceding A mark.
METHOD 3
recognize that the distance walked each minute is a geometric sequence (M1)
egr=0.9, valid use of 0.9
recognize that total distance walked is the sum of a geometric sequence (M1)
egSn, a(1−rn1−r)
listing at least 5 correct terms of the GP (M1)
15 correct terms A1
80,72,64.8,58.32,52.488,47.2392,42.5152,38.2637,34.4373,30.9936,27.8942,25.1048,22.59436,20.3349,18.3014
attempt to find the sum of the terms (M1)
egS15, 80+72+64.8+58.32+52.488+…+18.301433
S15=635.287 A1
since S<660 R1
he will not be there on time AG N0
Note: Do not award the R mark without the preceding A mark.
[7 marks]
Examiners report
Many found this question accessible, although the most common approach was to calculate each term by brute force, which at times contained small errors or inaccuracies that affected the overall sum. Although this was a valid method, it meant an inefficient use of time that could have affected the performance on other questions.
Those who applied the formula for geometric series were typically more successful and far more efficient in answering the question.