Date | November 2021 | Marks available | 2 | Reference code | 21N.2.SL.TZ0.2 |
Level | Standard Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Write down | Question number | 2 | Adapted from | N/A |
Question
The admissions team at a new university are trying to predict the number of student applications they will receive each year.
Let n be the number of years that the university has been open. The admissions team collect the following data for the first two years.
It is assumed that the number of students that apply to the university each year will follow a geometric sequence, un.
In the first year there were 10 380 places at the university available for applicants. The admissions team announce that the number of places available will increase by 600 every year.
Let vn represent the number of places available at the university in year n.
For the first 10 years that the university is open, all places are filled. Students who receive a place each pay an $80 acceptance fee.
When n=k, the number of places available will, for the first time, exceed the number of students applying.
Calculate the percentage increase in applications from the first year to the second year.
Write down the common ratio of the sequence.
Find an expression for un.
Find the number of student applications the university expects to receive when n=11. Express your answer to the nearest integer.
Write down an expression for vn .
Calculate the total amount of acceptance fees paid to the university in the first 10 years.
Find k.
State whether, for all n>k, the university will have places available for all applicants. Justify your answer.
Markscheme
12 669-12 30012 300×100 (M1)
3% A1
[2 marks]
1.03 A1
Note: Follow through from part (a).
[1 mark]
(un=) 12 300×1.03n-1 A1
[1 mark]
(u11=) 12 300×1.0310 (M1)
16530 A1
Note: Answer must be to the nearest integer. Do not accept 16500.
[2 marks]
(vn=) 10380+600(n-1) OR 600n+9780 M1A1
Note: Award M1 for substituting into arithmetic sequence formula, A1 for correct substitution.
[2 marks]
80×102(2(10380)+9(600)) (M1)(M1)
Note: Award (M1) for multiplying by 80 and (M1) for substitution into sum of arithmetic sequence formula.
$10 500 000 ($10 464 000) A1
[3 marks]
12 300×1.03n-1<10 380+600(n-1) or equivalent (M1)
Note: Award (M1) for equating their expressions from parts (b) and (c).
EITHER
graph showing y=12 300×1.03n-1 and y=10 380+600(n-1) (M1)
OR
graph showing y=12 300×1.03n-1-(10 380+600(n-1)) (M1)
OR
list of values including, (un=) 17537 and (vn=) 17580 (M1)
OR
12.4953… from graphical method or solving numerical equality (M1)
Note: Award (M1) for a valid attempt to solve.
THEN
(k=)13 A1
[3 marks]
this will not guarantee enough places. A1
EITHER
A written statement that un>vn, with range of n. R1
Example: “when n=24 (or greater), the number of applications will exceed the number of places again” (“un>vn, n≥24”).
OR
exponential growth will always exceed linear growth R1
Note: Accept an equivalent sketch. Do not award A1R0.
[2 marks]
Examiners report
The percentage increase proved more difficult than anticipated, with many using an incorrect denominator. Part (b) was accessible with many candidates earning at least three marks. The expression for the general term of the arithmetic sequence was found in part (c). A surprising number of candidates found the acceptance fees paid in the tenth year, rather than the required total acceptance fees found in the first ten years. Most candidates were able to earn one mark for multiplying their value by 80. In part (e), candidates were usually able to find the first point of intersection of un and vn, but did not always realize the answer must be an integer. Part (f) required candidates to support their answer through a comparison of growth rates, or by finding a range of values/single data point where un>vn for n>k. Though the answer "un is geometric, vn is arithmetic", inferred some understanding, this was insufficient justification and required a description that went a little bit further. It is recommended that teachers provide opportunities for candidates to explore the more advanced features of the GDC. An inappropriate choice of calculator window made it difficult for candidates to assess and appreciate the behaviour of un and vn.
The percentage increase proved more difficult than anticipated, with many using an incorrect denominator. Part (b) was accessible with many candidates earning at least three marks. The expression for the general term of the arithmetic sequence was found in part (c). A surprising number of candidates found the acceptance fees paid in the tenth year, rather than the required total acceptance fees found in the first ten years. Most candidates were able to earn one mark for multiplying their value by 80. In part (e), candidates were usually able to find the first point of intersection of un and vn, but did not always realize the answer must be an integer. Part (f) required candidates to support their answer through a comparison of growth rates, or by finding a range of values/single data point where un>vn for n>k. Though the answer "un is geometric, vn is arithmetic", inferred some understanding, this was insufficient justification and required a description that went a little bit further. It is recommended that teachers provide opportunities for candidates to explore the more advanced features of the GDC. An inappropriate choice of calculator window made it difficult for candidates to assess and appreciate the behaviour of un and vn.
The percentage increase proved more difficult than anticipated, with many using an incorrect denominator. Part (b) was accessible with many candidates earning at least three marks. The expression for the general term of the arithmetic sequence was found in part (c). A surprising number of candidates found the acceptance fees paid in the tenth year, rather than the required total acceptance fees found in the first ten years. Most candidates were able to earn one mark for multiplying their value by 80. In part (e), candidates were usually able to find the first point of intersection of un and vn, but did not always realize the answer must be an integer. Part (f) required candidates to support their answer through a comparison of growth rates, or by finding a range of values/single data point where un>vn for n>k. Though the answer "un is geometric, vn is arithmetic", inferred some understanding, this was insufficient justification and required a description that went a little bit further. It is recommended that teachers provide opportunities for candidates to explore the more advanced features of the GDC. An inappropriate choice of calculator window made it difficult for candidates to assess and appreciate the behaviour of un and vn.
The percentage increase proved more difficult than anticipated, with many using an incorrect denominator. Part (b) was accessible with many candidates earning at least three marks. The expression for the general term of the arithmetic sequence was found in part (c). A surprising number of candidates found the acceptance fees paid in the tenth year, rather than the required total acceptance fees found in the first ten years. Most candidates were able to earn one mark for multiplying their value by 80. In part (e), candidates were usually able to find the first point of intersection of un and vn, but did not always realize the answer must be an integer. Part (f) required candidates to support their answer through a comparison of growth rates, or by finding a range of values/single data point where un>vn for n>k. Though the answer "un is geometric, vn is arithmetic", inferred some understanding, this was insufficient justification and required a description that went a little bit further. It is recommended that teachers provide opportunities for candidates to explore the more advanced features of the GDC. An inappropriate choice of calculator window made it difficult for candidates to assess and appreciate the behaviour of un and vn.
The percentage increase proved more difficult than anticipated, with many using an incorrect denominator. Part (b) was accessible with many candidates earning at least three marks. The expression for the general term of the arithmetic sequence was found in part (c). A surprising number of candidates found the acceptance fees paid in the tenth year, rather than the required total acceptance fees found in the first ten years. Most candidates were able to earn one mark for multiplying their value by 80. In part (e), candidates were usually able to find the first point of intersection of un and vn, but did not always realize the answer must be an integer. Part (f) required candidates to support their answer through a comparison of growth rates, or by finding a range of values/single data point where un>vn for n>k. Though the answer "un is geometric, vn is arithmetic", inferred some understanding, this was insufficient justification and required a description that went a little bit further. It is recommended that teachers provide opportunities for candidates to explore the more advanced features of the GDC. An inappropriate choice of calculator window made it difficult for candidates to assess and appreciate the behaviour of un and vn.
The percentage increase proved more difficult than anticipated, with many using an incorrect denominator. Part (b) was accessible with many candidates earning at least three marks. The expression for the general term of the arithmetic sequence was found in part (c). A surprising number of candidates found the acceptance fees paid in the tenth year, rather than the required total acceptance fees found in the first ten years. Most candidates were able to earn one mark for multiplying their value by 80. In part (e), candidates were usually able to find the first point of intersection of un and vn, but did not always realize the answer must be an integer. Part (f) required candidates to support their answer through a comparison of growth rates, or by finding a range of values/single data point where un>vn for n>k. Though the answer "un is geometric, vn is arithmetic", inferred some understanding, this was insufficient justification and required a description that went a little bit further. It is recommended that teachers provide opportunities for candidates to explore the more advanced features of the GDC. An inappropriate choice of calculator window made it difficult for candidates to assess and appreciate the behaviour of un and vn.
The percentage increase proved more difficult than anticipated, with many using an incorrect denominator. Part (b) was accessible with many candidates earning at least three marks. The expression for the general term of the arithmetic sequence was found in part (c). A surprising number of candidates found the acceptance fees paid in the tenth year, rather than the required total acceptance fees found in the first ten years. Most candidates were able to earn one mark for multiplying their value by 80. In part (e), candidates were usually able to find the first point of intersection of un and vn, but did not always realize the answer must be an integer. Part (f) required candidates to support their answer through a comparison of growth rates, or by finding a range of values/single data point where un>vn for n>k. Though the answer "un is geometric, vn is arithmetic", inferred some understanding, this was insufficient justification and required a description that went a little bit further. It is recommended that teachers provide opportunities for candidates to explore the more advanced features of the GDC. An inappropriate choice of calculator window made it difficult for candidates to assess and appreciate the behaviour of un and vn.
The percentage increase proved more difficult than anticipated, with many using an incorrect denominator. Part (b) was accessible with many candidates earning at least three marks. The expression for the general term of the arithmetic sequence was found in part (c). A surprising number of candidates found the acceptance fees paid in the tenth year, rather than the required total acceptance fees found in the first ten years. Most candidates were able to earn one mark for multiplying their value by 80. In part (e), candidates were usually able to find the first point of intersection of un and vn, but did not always realize the answer must be an integer. Part (f) required candidates to support their answer through a comparison of growth rates, or by finding a range of values/single data point where un>vn for n>k. Though the answer "un is geometric, vn is arithmetic", inferred some understanding, this was insufficient justification and required a description that went a little bit further. It is recommended that teachers provide opportunities for candidates to explore the more advanced features of the GDC. An inappropriate choice of calculator window made it difficult for candidates to assess and appreciate the behaviour of un and vn.