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Date May 2021 Marks available 2 Reference code 21M.2.SL.TZ2.3
Level Standard Level Paper Paper 2 Time zone Time zone 2
Command term Find Question number 3 Adapted from N/A

Question

A new concert hall was built with 14 seats in the first row. Each subsequent row of the hall has two more seats than the previous row. The hall has a total of 20 rows.

Find:

The concert hall opened in 2019. The average number of visitors per concert during that year was 584. In 2020, the average number of visitors per concert increased by 1.2%.

The concert organizers use this data to model future numbers of visitors. It is assumed that the average number of visitors per concert will continue to increase each year by 1.2%.

the number of seats in the last row.

[3]
a.i.

the total number of seats in the concert hall.

[2]
a.ii.

Find the average number of visitors per concert in 2020.

[2]
b.

Determine the first year in which this model predicts the average number of visitors per concert will exceed the total seating capacity of the concert hall.

[5]
c.

It is assumed that the concert hall will host 50 concerts each year.

Use the average number of visitors per concert per year to predict the total number of people expected to attend the concert hall from when it opens until the end of 2025.

[4]
d.

Markscheme

recognition of arithmetic sequence with common difference 2        (M1) 

use of arithmetic sequence formula        (M1) 

14+220-1

52                  A1

 

[3 marks]

a.i.

use of arithmetic series formula      (M1) 

14+522×20

660                  A1

 

[2 marks]

a.ii.

584+584×0.012  OR  584×1.0121      (M1) 

591  591.008                  A1


Note: Award M0A0 if incorrect r used in part (b), and FT with their r in parts (c) and (d).

[2 marks]

b.

recognition of geometric sequence         (M1) 

equating their nth geometric sequence term to their 660         (M1) 


Note: Accept inequality.


METHOD 1

EITHER

600=584×1.012x-1                  A1

x-1= 10.3  10.2559

x=11.3  11.2559                  A1

2030                  A1


OR

600=584×1.012x                  A1

x=10.3  10.2559                  A1

2030                  A1

 

METHOD 2

11th term 658  657.987         (M1)A1 

12th term 666  666.883         (M1)A1 

2030                  A1


Note: The last mark can be awarded if both their 11th and 12th correct terms are seen.

[5 marks]

c.

7 seen        (A1) 


EITHER

5841.0127-11.012-1         (M1) 

multiplying their sum by 50         (M1) 


OR

sum of the number of visitors for their r and their seven years         (M1) 

multiplying their sum by 50         (M1) 


OR

292001.0127-11.012-1        (M1)(M1) 


THEN

212000  211907.3                  A1


Note: Follow though from their r from part (b).

[4 marks]

d.

Examiners report

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

Syllabus sections

Topic 1—Number and algebra » SL 1.2—Arithmetic sequences and series
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Topic 1—Number and algebra

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