Mathematics · Oct / Nov 2025 · Paper 11 · Question 7
YMA01
As printed in the paper
The question as text
7. Q P O T S R Figure 2 In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. Figure 2 shows the plan view of a design for a swimming pool. The design consists of a sector POS of a circle centre O joined to a major sector QORTQ of a different circle, also with centre O. Given that • angle POS is 1.65 radians 2 • the area of sector POS is 30 m • PQ = 2.8 m
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
7(a)[3]
1
2
2
2 r
×
1.65 =30
r
=
... or
r
=
...
r =
6.03
(
)6.03 2.8
OQ =
+
=
8.83(m)
*
(a) Alternative
1
(
)
(
)
(
)
−
2
×
=
−
2
=
−
=
2 OQ
2.8
1.65
30
OQ
2.8
... or
OQ
2.8
...
(
)
OQ−
=
2.8
6.030...
(
)6.03 2.8
OQ =
+
=
8.83(m)*
Notes
(a)
1
2
2 r ×
= in an attempt to find a value for r2 or r (where r is OP)
M1: Uses
1.65
400
11 or 20
11 or 20 11
r =
A1: '' ''
6.03...
or exact
or
36.36...
11
A1*: (
)
20 11
OQ
=
6.03 ore.g.
or
36.36...
+
2.8
=
8.83(m)
11
Must be 8.83 not awrt 8.83
The “OQ =” is not required.
20 11
There must be some indication that they are adding
6.03 ore.g.
to 2.8
11
There is no requirement for units but if any are given they must be correct.
180
Allow equivalent correct work in degrees e.g.
1.65rads
=
1.65
×
° etc.
π
Alternative:
1
M1: Uses (
)
2
in an attempt to find a value for (
)
2
or (
2 OQ −
2.8
×
1.65
=
30
OQ −
2.8
OQ−
400
11 or 20
11 or 20 11
A1: (
)
OQ−
2.8
=
6.03...
or exact
or
36.36...
11
20 11
=
+
=
A1*:
OQ
6.03 ore.g.
2.8
8.83(m)
11
Must be 8.83 not awrt 8.83
No requirement for units but if any are given they must be correct.
Note that any attempts to verify the result in (a) would need a complete argument e.g. attempts to find
the area for 2 suitable values of OQ such as 8.835 and 8.825 for M1, A1 for both areas evaluated
correctly and A1 for fully reasoning that OQ is 8.83 to 2dp.
7(b)[3]
2
(
) (
)
Attempts
1
2 8.83
×
×
2
π
−
1.65 = 180.__
2
(
)
Total area =
1
2 8.83
×
×
2
π
−
1.65 + 30=...
awrt 210 (m2) or 211 (m2)
(
)
(
) (
)
7(c)[3]
×
×
π
−
''6.03'' 1.65= 9.95
or
8.83
2
1.65 = 40.91
(
)
'6.03'' 1.65 8.83
×
+
×
2
π
−
1.65
+
2.8 2
×
Full method for perimeter =
= awrt 56 (m)
Notes
Allow equivalent correct work in degrees.
(b)
(
)
2
2
×
2
×
π
−
π×
−×
×
M1: Attempts
1
2 8.83
2
1.65
or equivalent e.g.
8.83
1
8.83
1.65
2
May be implied by awrt 180 or awrt 181 provided no incorrect working is seen.
2
(
)
×
×
π
−
dM1: Attempts
1
2 8.83
2
1.65 + 30 =...
Adds 30 to a correct attempt at the area of the major sector of the larger circle.
A1:
awrt 210 (m2) or awrt 211 (m2)
No requirement for units but if any are given they must be correct.
Correct answer only scores no marks.
(c)
×
M1: Attempts ''6.03'' 1.65
which may be implied by awrt 9.95 or 3sf accuracy for their 6.03
provided no incorrect working is seen.
OR
(
)
Attempts
8.83
×
2
π
−
1.65
which may be implied by awrt 40.9 provided no incorrect working
is seen.
(
)
"6.03'' 1.65 8.83
×
+
×
2
π
−
1.65
+
2.8 2
×
dM1: Full method for perimeter =
A1:
awrt 56 (m) . No requirement for units but if any are given they must be correct.
Correct answer only scores no marks.
180,181
2.8
8.83
40.9
1.65
9.95
4.63
6.03
30
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