Mathematics · Oct / Nov 2020 · Paper 31 · Question 2
YMA01
As printed in the paper
The question as text
2. A scientist monitored the growth of bacteria on a dish over a 30‑day period. The area, N mm2, of the dish covered by bacteria, t days after monitoring began, is modelled by the equation log10 N = 0.0646 t + 1.478 0 t 30
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
2(a)[4]
10
1.478
10
0.0646
10
0.0646
t+
1.478
Sight of
or
or
(
)awrt 30
or (
)awrt 1.16
a =
b =
0.0646
t
+
1.478
0.0646
t
1.478
log
N
=
0.0646
t
+
1.478
→
N
=
10
=
10
10
10
t
=
"30
"
×
"1.16
"
t
N =
30 1.16
×
30
2(b)[2]
Attempts
N =
30 1.16
×
=
awrt 2600
(a)
NB This shows as MMAA on ePEN but is being marked as MAMA.
M1
Sight of
10
1.478
or
10
0.0646
or
10
0.0646
t+
1.478
(allowing slips copying the values) anywhere in their solution.
This mark is implied by seeing awrt 30 or awrt 1.16
Sight of either awrt 30 or awrt 1.16
A1
(
)
t
0.0646
1.478
dM1
Applies correct index laws and proceeds to find values for a and b.
N
=
10
×
10
=
10
0.0464
t+
For this mark there must be evidence of correct index work so expect to see at least
answer and no incorrect index work.
A1
a
=
awrt
30
and
b
=
awrt
1.16
as long as there is no contrary work, or states that
N =
30 1.16
×
(b)
30
N =
30 1.16
×
M1
Attempts
with their values of a and b.
3.416
log
N
=
0.0646 30 1.478
×
+
⇒
N
=
10
Alternatively
10
A1
awrt 2600, isw after a correct answer.
...................................................................................................................................................................................
t
N
=
ab
Alt (a) using
as a starting point.
M1
As main scheme.
a =
awrt 30
b =
awrt 1.16
A1
or
(must be correctly assigned)
dM1
Takes
log
of both sides and proceeds to at least
log
N
=
log
a
+
t
log
b
and attempts to find a value for
10
10
10
10
both of the constants with no incorrect log work.
A1
a
=
awrt
30
and
b
=
awrt
1.16
as long as there is no contrary work, or states that
N =
30 1.16
×
1
1
(
)
(
)
(
)
4
x
−
1
× −
2
2
x
+
3
×
2 4
x
−
1
−
d
y
2
2
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