Mathematics · Oct / Nov 2025 · Paper 31 · Question 3
YMA01
As printed in the paper
The question as text
3. The share price, £V, of a company is being monitored. A graph is drawn of log10 V against t, where t is the number of years after monitoring began. The graph is a straight line passing through the points (0, 2) and (5, 2.25) Using this information,
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
3(a)[2]
2.25
2
m
=
5
log
V
=
0.05
t
+
2
10
3(b)[3]
=
+
=
0.05
t
+
2
log
V
0.05
t
2
V
10
10
a =
100
or
b =
1.12
(
)
t
V =
100
×
1.12
3(c)[4]
50
100 ln1.12 (1.12)
×
×
T
=
50
(1.12)
T
=
100ln1.12
50
(
)
=
=
T
log
...
1.12
100ln1.12
(
)13
T =
(a)
M1:
Attempts to find the gradient of the line, must be change in “y”/change in
“x”. May be found via simultaneous equations – look for an attempt to
solve the correct equations.
=
+
A1:
log
V
0.05
t
2
Accept log in place of log10 here and throughout but
10
must be correct variables.
(b)
M1:
Correct attempt to make V the subject, or alternatively takes log of
1
and finds an equation in just a or b ie either
=
or
=
2
log a
log
10
20
is no incorrect working allow for obtaining an expression for a or b, e.g.
a =
2
or
b =
0.05
.
10
10
A1:
a =
100
or
b =
(may be stated or embedded in an equation of the
awrt1.12
correct form).
A1:
. The equation must be given not just values.
SC Award M1A1A0 for candidates who start with an incorrect step of
but recover to
in subsequent
work (and similar for variants).
(c)
×
B1ft: Differentiates to the form "100"ln"1.12" ("1.12")t
follow through their
values for a and b from part (b). Accept any unsimplified equivalents and
values correct to 3sf.
Sets their d
V
M1:
t which must be a changed function of the form
d
T =
(K may be 1) equal to 50 and rearranges to the form ("1.12")
...
T =
dM1: Correctly proceeds from and equation of form ("1.12")
...
to
valid means, you may need to check their value if no method is shown).
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