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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(1rn1r)

correct substitution into the sum of a geometric sequence     (A1)

eg80(10.9n10.9)

any correct equation with sum of a geometric sequence     (A1)

eg80(0.9n10.91)=660, 10.9n=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(1rn1r)

correct substitution into the sum of a geometric sequence     (A1)

eg80(10.9n10.9)

attempt to substitute n=15 into sum of a geometric sequence     (M1)

egS15

correct substitution     (A1)

eg80(0.91510.91)

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(1rn1r)

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.

Syllabus sections

Topic 1 - Algebra » 1.1 » Arithmetic sequences and series; sum of finite arithmetic series; geometric sequences and series; sum of finite and infinite geometric series.
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