Mathematics · May / June 2023 · Paper 21 · Question 6
YMA01
As printed in the paper
The question as text
6. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. A software developer released an app to download. The numbers of downloads of the app each month, in thousands, for the first three months after the app was released were 2k – 15 k k + 4 where k is a constant. Given that the numbers of downloads each month are modelled as a geometric series,
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
6(a)[2]
k
k
=
k
2
=
2
k
2
+
8
k
−
15
k
−
60
k
2
−
7
k
−
60
=
0
*
2
k
−
15
k
k =
12
6(b)[4]
"12"
4
=
2 "12" 15
×
−
3
3
4
4
e.g. (
)
or (
)
"12" 4
+
×
=
...
2 "12" 15
×
−
×
" "
=
...
3
3
=
21300
6(c)[3]
4
N
9 1
−
3
4
N
1009
>
3000
e.g.
>
4
3
9
1
−
3
1009
log
4
N
1009
9
log
>
log
N
>
4
3
9
log 3
N
>
16.405...
N
=
17
(a)
M1
Forms a correct equation linking the three terms and attempts to rearrange to the given quadratic
(it cannot be awarded for just proceeding in one step from the starting equation to the given
answer and usually will involve attempting to multiply out brackets or dealing with any fractions).
Must proceed as far as a quadratic with terms collected.
Condone slips in their rearrangement and invisible brackets.
Possible starting equations will involve using the nth term formula or the sum of a geometric
series. The ones below are not exhaustive.
+
4
(
)
+
4
k
k
15) k
2
=
(
+
4)(2
−
15)
(
)
2
=
or
k
k
k
u
=
u
×
u
or
(2
k
−
=
2
1
3
2
k
−
15
k
k
3
k
(2
−
15) 1
−
k
2
−
15
=
k
k
4
or
2
15
4
(sum of 3 terms)
k
k
+
k
−
+
k
+
k
+
=
2
k
−
15
k
1
−
2
15
k
−
A1*
Rearranges to achieve
k
2
−
7
k
−
60
=
0
with no errors seen including brackets. There must
be at least some intermediate working (e.g. brackets multiplied out or multiplying to remove
any fractions) between their starting equation and the given answer.
(b) Note a candidate who uses k as r can only score maximum B1 in (b) and no marks in (c)
B1
12 (ignore 5
−) seen or used in (b) (may be implied by the value of the second term)
M1
Substitutes their 12 (must be positive) into the expressions for two of the terms and attempts
to find a value for r. Condone arithmetical slips but they the division the correct way round.
M1
"12" 4
+
(may be implied by awrt 21.3). Alternatively, they may find the first term a and use this with their
r to find the 4th term. Condone slips in finding a
Condone r to be rounded from their earlier work and condone missing brackets but not
incorrectly placed brackets.
A1
21300 or 21330 or 21333 only or their equivalences e.g. 21.3 thousand
(c) Typically, candidates will work in base 10, but other bases such as their r is acceptable
4
M1
Uses their
a =
"9"
or
a =
"9000"
and their positive
"
"
in the sum of a geometric formula
r =
3
4
N
"9" 1
" "
−
3
....3...
(may be = or any inequality in between) and proceeds to
A
" "
4
1 " "
−
3
(condone A B
×
to be negative), where A could be 1 and B could be unsimplified. Do not be
concerned with the number of 0s after the 3.
Do not be concerned by the mechanics of their rearrangement and may be implied by further work
if logarithms are used correctly. Condone r to be rounded from their earlier work and condone
missing brackets but not incorrectly placed brackets.
dM1
Attempts to find N using logarithms correctly. Do not be concerned with the use of equals or the
direction of any inequality used for this mark. May be implied by a correct expression (not a
value) for N. It is dependent on the previous method mark.
4
N
0
This mark can only be scored if
A
×
B
>
for their
A
" "
...
B
3
A1
17 cso (if an incorrect inequality sign is used at some point in their working then withhold this
final mark) This mark can only be scored provided there are no arithmetical errors and correct
log work is seen as part of their solution. Do not withhold the mark for invisible brackets around
their 4
N
. Condone if r is rounded from 4
3 . If a value for N is found before deducing
3
then it must be 16.4 or better. 17 with no working seen is 0 marks.
991
1000
Note e.g.
>
log
=
17
is A0 (arithmetical slip combining
1
)
N
N
+
4
9
9
3
N
4
9 1
−
3
4
N
1009
>
3000
>
=
17
scores M1dM0A0 (insufficient working)
N
4
3
9
1
−
3
N
4
1009
9 1
−
log
3
9
3000
17
scores M1dM1A0 (inequality sign errors)
<
N
<
N
=
4
4
1
−
log
3
3
4
N
9000 1
−
3
>
1009
3000000
>
log
=
17
scores M1dM1A1 (implied M marks)
N
N
4
4
9
1
−
3
3
Note: Methods relying on trial and improvement, or the equation solver score maximum M1dM0A0
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