Mathematics · Oct / Nov 2024 · Paper 31 · Question 4
YMA01
As printed in the paper
The question as text
4. The number of bacteria on a surface is being monitored. The number of bacteria, N, on the surface, t hours after monitoring began is modelled by the equation log10 N = 0.35t + 2 Use the equation of the model to answer parts (a) to (c).
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Show the answer and mark scheme
4(a)[1]
log
=
2
=
100
N
N
10
4(b)[3]
0.35
t
+
2
log
N
=
0.35
t
+
2
N
=
10
10
0.35
t
2
=
10
×
10
N
t
N
=
100 2.24
×
4(c)[2]
dN
Rate of growth at 5 hours is
5
dt =
100 ln 2.24 2.24
×
×
Allow an answer in the range 4530 to 4550 (bacteria per hour)
(a)
B1: For 100 Do not accept just
10 and cannot be awarded from 100 in part (b)
2
(b) The demand of the question is ‘show that’, not just find values of a and b
Method 1: starting with
log
0.35
2
N
=
t
+
10
0.35
t
+
2
M1: Undoes the log. E.g.
log
=
0.35
+
2
=
10
N
t
N
10
0.35
t
2
Give BOD for those who go from
log
N
=
0.35
t
+
2
straight to
N =
10
×
10
10
dM1: And follows with a correct application of the index law.
t ×
0.35
+
2
0.35
2
0.35
Look for
log
=
0.35
+
2
=
10
t
=
10
t
×
10
o.e such as
N
t
N
10
100
10
t ×
0.35
t
2
0.35
But BOD for
log
=
0.35
+
2
=
10
×
10
o.e. such as
10
100
N
t
N
10
0.35
t
+
2
t
And BOD for
log
=
0.35
+
2
=
10
=
100 2.24
×
N
t
N
N
10
t
A1: CSO
N =
100 2.24
×
with no errors following M1, dM1. Allow awrt 2.24
Condone
a
=
100,
b
=
2.24
following the award of M1 dM1
You should expect to see at least one correct line of working and no errors
t
Method 2: starting with
=
N
ab
M1: Takes logs of both sides and applies one correct rule. E.g
=
t
log
=
log
+
log
N
ab
N
a
t
Give BOD for those who write
=
followed by log
=
log
+
log
N
ab
N
a
t
b
dM1: And follows with log
log
log
before setting
N
=
a
+
t
b
log
a
=
2 and log
b
=
0.35
and proceeding to values for a and b
A1: Fully shows and produces correct values following M1, dM1
CSO
100 2.24
t
with no errors. Allow awrt 2.24
N =
×
Condone
=
100,
=
2.24
following the award of M1 dM1
a
b
You should expect to see at least one correct line of working and no errors
Method 3: A simultaneous approach that finds values but does not show the result
M1: Sets up two simultaneous equations in a and b
E.g. Uses
=
0,
=
100
=
100
and
=
1,
=
10
2.35
1
=
223.9
t
N
a
t
N
ab
dM1: Solves to find values for a and b.
A0: Has not shown the result so cannot be awarded
Common error: Special Case
0.35
t
2
log
N
=
0.35
t
+
2
N
=
10
+
10
so
a =
100
and
b =
2.24
10
Or even
0.35
t
2
t
log
N
=
0.35
t
+
2
N
=
10
+
10
N
=
100 2.24
×
so
a =
100
and
b =
2.24
10
Such work resulting in the correct values for a and b scores SC 100
(c) Full marks in part (c) can be scored following the SC
dN
5
M1: Attempts
dt =
×
''2.24''
following through on their 2.24.
K
t
This cannot be scored from substituting t = 5 into their
ab
dN
5
A1: Calculates
dt =
100 ln 2.24 2.24
×
×
and achieves an answer in the range 4530 to 4550 (bacteria per
hour)
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