Mathematics · Oct / Nov 2023 · Paper 11 · Question 9

YMA01

As printed in the paper

The question as text

Diagram NOT 9. accurately drawn Q C P O S R Figure 3 Figure 3 shows the plan view of the area being used for a ball‑throwing competition. Competitors must stand within the circle C and throw a ball as far as possible into the target area, PQRS, shown shaded in Figure 3. Given that • circle C has centre O • P and S are points on C • OPQRSO is a sector of a circle with centre O • the length of arc PS is 0.72 m • the size of angle POS is 0.6 radians

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Show the answer and mark scheme

9(a)[1]
or 0.72
0.72
=
OP
×
0.6
OP
=
1.2
0.6 =
1.2
or 0.72
=
1.2 0.6
×
OP
=
1.2
*
9(b)[3]
1
×
(
+
1.2)
2
×
0.6
x
2
1
1
×
(
x
+
1.2)
2
×
0.6
−
×
1.2
2
×
0.6
=
90
5
2
+
12
−
1500
=
0
*
x
x
2
2
9(c)[3]
−+
6
4 471
awrt 16.2
x
=
=
5
Perimeter
=
0.72
+
"16.2" "16.2" 0.6 ("16.2" 1.2)
+
+
×
+
=
awrt 43/ awrt 44
m
(see notes)
(a)
B1*:
Either
=
rθ
l =
0.72
θ=
0.6
OP =
forms an equation using l
with
and
and concludes
e.g. 0.72
0.6
1.2
= ×
r
r
=
states 0.72
0.6 =
1.2
forms the equation using 0.72
=
1.2 0.6
×
and concludes that
OP =
1.2
Condone the use of r or any other variable instead of OP. Condone lack of units or incorrect
units.
Allow the angle to be in degrees instead or attempts to find the diameter first, but
they must not round the value at any point in their working.
0.6 180
×
π
e.g.
0.72
=
π
d
d
=
2.4
r
=
1.2
scores B1 (exact angle used)
360
e.g. 0.72
=
2
π
r
r
=
1.2
scores B0 (34.4 is a rounded angle)
34.4
360
(b)
B1:
Correct expression for the area of sector OPQRSO
1
(
1.2)
2
0.6
Missing brackets may be implied by further work.
×
x
+
×
2
1
May appear within an equation e.g.
×
0.6 (
x
+
1.2)
2
−
1.2
2
=
90
2
1
Condone incorrect equations i.e.
×
(
x
+
1.2)
2
×
0.6
=
90
2
...(
x±
1.2)
2
×
0.6 ... 1.2
−
×
2
×
0.6
=
90
M1:
Proceeds from an equation of the form
or equivalent,
(i.e. both should be positive but may be missing e.g. ½ ) and attempts to multiply out the
brackets. Condone slips and invisible brackets.
A1*: Achieves the given answer with no errors seen including invisible brackets. They must have
multiplied out any brackets before proceeding to the given answer to score this mark.
54
18
3
e.g.
is acceptable as a final step to
+
x
+
x
2
−
0.432
=
90
5
x
2
+
12
x
−
1500
=
0
125
25
10
score A1*
(c)
6
4 471
−+
B1:
=
awrt 16.2
=
(ignore the negative root) May be seen on the diagram or
x
5
next to the quadratic in the question. It may also be implied in their working to find the
perimeter. It cannot be for the values embedded in the quadratic formula.
M1:
A correct method to find the perimeter using their value for x which must be positive. They
must proceed as far as a value for the perimeter to score this mark. Do not condone the
lengths 0.72 and 1.2 to be rounded, but condone for example their “16.2” to be “16”.
−+
6
4 471
12 471
2
0.72
May see exact values used in the expression e.g.
+
+
5
25
A1:
awrt 43 m or awrt 44 m, including units, following a correct method. Typically look
for 43.4…m or 43.5…m
Note some incorrect methods lead to awrt 43 which is maximum B1M0A0.
e.g. "16.2" 0.6 "16.2" "16.2" 0.72
42.7
×
+
+
+
=
which is B1M0A0 (uses 16.2 instead of
16.2+1.2)
e.g. "16.2 1.2" 0.6 "16.2" "16.2"
+
×
+
+
=
42.8
which is B1M0A0 (missing arc PS)
(
) 3
10ai
n =
ii
1080
b
(1620,
−
3)
c
1 12
3
9
k
=
−
=
e.g.
2
5
5
10
Note: answers may appear next to questions or on the diagram.
If there is a contradiction between the answer by the question/on the diagram and the main body
of the work then the answer in the main body of the work takes precedence.
(a)
(i)
(
) 3
B1:
n =
(ii)
B1:
1080 (must be in degrees). Do not accept e.g. (1080, 3)
(b)
B1:
One of the two coordinates (1620,
−
3)
Condone missing brackets. Allow x=… , y = ….
Condone 9π instead of 1620 for this mark and condone the coordinates to be the wrong way
round e.g. ( 3, 1620)
scores B1B0
−
B1:
Both coordinates(1620,
Condone missing brackets. Allow x =… , y = ….
−
3)
(c)
M1:
Attempts to find a value for k by either
1 12
3
attempting to find
k
=
−
=
...
.
2
5
5
12
1 12
3
3
1 12
3
attempting to find
k
=
−
+
=
...
or
k
= −
+
+
=
...
5
2
5
5
5
2
5
5
forming two correct simultaneous equations and eliminating sin a (may be labelled as a
different variable, condone a for sin a)
12
3
( )
2sin
+
=
,
−
2sin( )
+
= −
=
...
scores M1
a
k
a
k
k
5
5
They must have correctly usedsin(
−
a
)
= −
sin( )
a
. If they eliminate k instead, they must
proceed to find a or sina and then substitute back into one of the equations to find a
value for k
12
i.e.
4sin
=
3
=
48.59...
2sin(48.59...)
+
=
=
...
a
a
k
k
5
9
A1:
10 or 0.9 (this is not awrt 0.9 so if they achieve a rounded value e.g. 0.899… and round to 0.9
then M1A0)

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