Mathematics · May / June 2023 · Paper 31 · Question 7
YMA01
As printed in the paper
The question as text
7. A scientist is studying two different populations of bacteria. The number of bacteria N in the first population is modelled by the equation kt t 0 N = Ae where A and k are positive constants and t is the time in hours from the start of the study. Given that • there were 2500 bacteria in this population at the start of the study • there were 10 000 bacteria 8 hours later
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Show the answer and mark scheme
7(a)[3]
States or implies that
k
×
8
10000
=
2500e
8
k
=
ln4
k
=
...
1 ln 4 or awrt 0.1733
k
=
8
d
N
−
0.6 5
×
60000
0.6e
1792
7(b)[2]
=
×−
= −
So decrease is 1790
d
t
−
0.6
t
0.1733
t
60000e
2500e
7(c)[3]
=
0.1733
t
+
0.6
t
2
4
=
e
0.1733
t
+
0.6
t
=
l
n
2
4
t
=
.
.
.
T =
4.11
Notes
(a)
kt
N =
2500e
B1
States or implies that
A =
2 500
. E.g award for
kt
M1
Attempts to use
N
=
A
e
with
t
=
8,
N
=
10 000
and their A to set up and solve an equation in k.
Correct ln work must be used to solve their equation.
Allow this mark for attempts to find k first by solving simultaneously if they use t = 1 for the start
k
8
k
7
k
of the study:
2
500
=
A
e
,10000
=
A
e
e
=
4
7
k
=
l 4
n
k
=
..
.
but the index and ln work
must be correct.
. Accept the exact value 1 ln 4
A1
k =
awrt 0.1733
and isw after seen.
8
(b)
d
N
−
0.6 5
×
−
0.6
t
−
0.6
t
M1
=
C
e
=
...
where C is a constant. Condone
60000e
→
60000e
as long as it is clear
d
t
they think they have found d
N
t . Must be correct index (not kt).
d
A1
Awrt 1790 from a correct derivative. Condone awrt 1790
−
(c)
−
0.6
t
"0.1733"
t
M1
Sets
60000e
=
their 2500e
May use T or another variable instead. Allow a slip on e.g. the
60000 as 6000. Allow with k in place of their “0.1733” as long as they have a value for k in (a).
dM1
Proceeds to rearrange to
e
mt
=
D D
(
>
0)
and applies ln to find t. The ln work must be correct,
though there may be slips in the coefficients or index work reaching
e
mt
=
D D
(
>
0)
implied by a correct answer for their
e
mt
=
D
Alternatively, takes ln of both sides first and applies correct ln laws to proceed to make t the
−
0.6
t
0.1733
t
l
n 60000e
=
ln 2
500e
ln 60000
−
0.6
t
=
l
n 250
0
+
0.17 3
3
t
subject:
A1
awrt 4.11 Must be a value, not an expression in ln terms for this mark.
Note Answer only scores no marks, method must be shown and the dM1 must be achieved in order to
access the A mark.
(
)
2
(
)
3
f ( )
′
x
=
6 2
x
+
1 e
−
4
x
−
4 2
x
+
1 e
−
4
x
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