Mathematics · May / June 2022 · Paper 21 · Question 9
YMA01
As printed in the paper
The question as text
9. A scientist is using carbon‑14 dating to determine the age of some wooden items. The equation for carbon‑14 dating an item is given by N = kλt where • N grams is the amount of carbon‑14 currently present in the item • k grams was the initial amount of carbon‑14 present in the item • t is the number of years since the item was made • λ is a constant, with 0 < λ < 1
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
9(a)[2]
N
1
t
Correct shape from or passing through a point on positive vertical axis. May extend
to the left of the vertical axis and allow to pass into quadrant 4. There must be no
(obvious) turning points. Labels not required on axes and ignore any that are given.
Shape and position correct, accept 1 or k as intercept on the positive vertical axis
and allow to extend to the left of the vertical axis as shown. Condone
(1, 0) or (k, 0) as long as it is in the correct position. The curve should approach a
horizontal asymptote that is halfway between the horizontal axis and the intercept
or below. Be tolerant with “wobbles” as it approaches the asymptote. May just
“touch” the horizontal axis but not go below it. Labels not required on axes and
ignore any that are given.
1
1
9(b)[2]
(see notes for method via substitution)
2 k
=
kλ
5700
λ
5700
=
2
1
1
5700
λ
=
=
0.999878
to 6 d.p.*
2
9(c)[2]
When t = 3250,
N
=
15
×
0.999
878
3250
=
..
.
(
)
=
awrt 10.1 grams
1
8
=
25 0 998 8t
×
.
9
7
9(d)[3]
18
log 25
18
0.999878
t
...
=
=
t
=
25
log0.
9 987
9
8
t =
2692.49...
so item is 2700 years old
Notes:
(a)
M1: See scheme.
A1: See scheme.
(b)
M1: Uses the information to set up a correct equation and rearranges to form
λ
a
=
b
where a and b
1
are constants. Allow
λ
5700
=
to be written down directly.
2
A1*: Evidence of taking fractional root seen, leading to the given answer. Alternatively, may see
1
−
log
2
−
log2
logs used,
Look for at least one correct
5700
lo λ
g
=
log
log
λ
=
λ
=
10
5700
=
..
.
2
570
0
intermediate step (and no incorrect ones). Allow greater accuracy e.g. 0.9998784026…
Note that the use of 5699 instead of 5700 scores M1A0
Alt:
M1: Substitutes values 0.9998775 and 0.9998785 or a tighter range containing the root (e.g.
0.999878 and 0.9998785 will do) to calculate N or just
λ at t = 5700.
t
A1*: Correct values, with suitable conclusion that as half value occurs between these λ=0.999878
to 6 d.p.
(c)
M1: Substitutes given values (or more accurate λ) into the equation and evaluates. Implied by a
correct answer, but an incorrect answer with no evidence is M0. Allow 3249 for 3250.
Condone the use of a less accurate λ e.g. 0.9999 if the intention is clear.
A1: Awrt 10.1. Units not required. Answer only can score both marks.
(d)
B1: Sets up the correct equation from the information given. It must be clear they are using the
given λ or better accuracy so clear use of e.g. 0.9999 scores B0 but allow the M1 below.
This may be implied if they write e.g. 18
and then go on to use the given value of λ.
=
25
×
λ
t
Allow this mark if a different letter is used for t as long as the equation is correct.
M1: Solves an equation of the form
a
=
bλ
t
for t to obtain a value
a
a
Must be correct log work so look for e.g.
a
=
b
λ
t
=
λ
t
=
t
log
b
λ
b
a
log
a
a
b
Or e.g.
a
=
b
λ
t
=
λ
t
=
t
log
=
log
λ
t
=
t
log
λ
=
t
b
b
log
λ
Allow a miscopied λ or a less accurate λ for this mark.
A1: Correct answer of 2700 (not awrt). Accept “27 hundred” or even “2692.49… so 27”
1
−
FYI use of
2
gives 2701.4 years.
λ
=
5700
(
3)
2
(
5)
2
2
y
=
2
x
+
k
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