What is
f
We have
t
z
Fo t Foo
as
Zoo
Zo
decimal
a
Fo
Foto t
Foot
222500
4441000
0.444
the point where all
the centre of the
four squares meet In
diagrams
third of the circle
In each diagram below
Which
exactly
at
of the
shaded
one
the
In A
it is 90
In C
it
is
while the angle
the
In
90
The angle in B
so
of
centre
the
circle
should be
120
360 3
E
is
is less than the
in
answer
D
is
greater
must
be 13
Consider the squares mud
digit
it
180
1350
45
How many squares have 7
units
is
is
one
The angle
C
one
circle
as
10
angle
C
than the angle in
their
to
in
units
examine
digit
their
n
O
I
2
3
4
5
n
O
I
4
9
6
5
answer
is
So the
6
6
of the Following is not the
E 13
A S
D 11
B 7 c 9
5 21 3
We have
So the
8
9
9
4
1
0
Which
7
7
9
5
21 7
13
21 11
sum
of
2 primes
2
is D
answer
The diagram shows two circles with the same centre
The radius of the outer circle is twice the radius of
the inner circle The region between the inner
circle and
outer circle is divided into six
equal segments as shown
what Fraction of the area of the outer circle is shaded
radius
Suppose the
inner circle has
outer circle has radius 2r
The
area
The
area
of the inner circle
of the
So the
area
shaded
area
outer
between them
is
Iz
31T
r
Then the
Tr
is
4Th
31172 and
circle is
is
IT
2
therefore the
Hence the answer is
72
Tr
38
Fitr
The angles of a quadrilateral are
ratio
in the
3 4 S b What is the difference between the largest
angle and the smallest angle
The angle sum of a quadrilateral is 3600
The four angles are
3k 4k Sx 6k So
3kt 4
18K
t
360
6x
360
20
x
So the
t Sx
difference
between
angles is Gu 3x
Four different positive
the
3x
smallest
have
a
What is the smallest possible
of
range
mean
ntltnt2tn
4
Lint 6
4
20 6
So the
answer
is
4
the
nt2
2017
minimum
nt3
2017
2017
which is impossible since 6 is not
It is possible to Find a set of
range 4 satisfying the condition
2015
of
the chosen
integers
Since the Four integers are all different
possible range is 3 from the list n ntl
In that case we would need
n
largest
600
3 20
integers
and
2018
multiple of
integers with
a
2019
4
of the Following numbers is the largest
A 1.3542
B 1.3542 c l 354 2 D 1.35 42 E 1.3542T
Consider the Fifth decimal place of each value
Which
A O
B 2
4
C
D
is D
So the
answer
The number tu is the 2
digit t and units digit u
distinct and
value of
We have
non
ab
ba
ab
zero
n
digit
The
What
is
ba
that b
9A
E
S
3
number
digits
with tens
and b
a
the largest possible
Wb ta
94
9 a b
So
we
want to
maximise
a
b
choose
a
9
9 8
72
919 l
The diagram shows three rectangles Find
x
We have
angles
Then
ba
ab
at 90
a
and
pst 90
29
180
supplementary
61
43
180
p
are
47
b
I
1
Finally
By angle
triangle
8
Otp
I
jp
X t 90
t
180
x
72
a
180
t
90
of
47
180 61
8
T
sum
72
t
360
8
1080
The diagram shows four equilateral triangles with sides
of length 42,3 4 The area of the shaded region is
equal to n times the area of the unshaded
with side length I Find n
The area
of the unshaded triangle
1 x Sth 60
x1
is
triangle
E
4
the shaded
Asia
Aside
is
area
Aside
77
Asia
9
3
Aside
14124 5
By 26
44
so
3
n
26
The combined age of Alice and Bob is 39
The combined age of Bob and Clare is 40
The combined age of Clare and
Dan is 38
The combined age of Dan and Eve is
44
The total of all Five
ages is 105
Which of the Five is
youngest
We have
ATB
Btc
39
CTD
38
40
44
Dt E
At Btc 1 Dt E
C A
I
c
B D
2
B
Dt2
E
6
E
Ctb
C
and
From 20
At 40
From
ios
21
we
So the answer is
The diagram
shows
At 7
have
105
t 44
B
Atl
39
A
B
18
D
16
21
c
22
E
28
Dan
made From two
similar right angled triangles PQR and PRS The length of
PQ is 3 the length of QR is 4 and
PRQ
L PSR
a
quadrilateral PQRS
What is the perimeter of
PQRs
By
Pythagoras
PR
5
By similar triangles
and
Ig
So the
Ps
Psy
perimeter
64
235
3 t 4 t
is
For what value of
We have
RS
RST
is
x
5125
If
t
5125
64
2915
2b
Gx
_LRS Q
also Now LSPQ
we have Lps
Since
SQ
2LRSQ
2x
QR
2x
and
QR
we
have
and
45
7.5
452
In the diagram shown PQ SQ
LSPQ 2 LRSQ What is LQRS
x
5
9
x
et
22
235
LSRQ
since
PQ
x
SQ
By
of
angle sum
triangle
a
2n t x
Zx
Gx
I
So
27
t
180
x
x
L QRS
180
x
30
300
two positive integers is equal to
twice their sum This product is also equal to
six times the difference between the
integers
What is the sum of these two
integers
Let the two integers be ma d n with m
n
The product of
We have
2 Mtn
Mn
6cm
so
3
Mtn
n
m
3m
Hence
2n
2n
m
212N
n
2n
3N
3
n
m
Mtn
tn
213N
n
So
3h
Lin
2n
n
since
are
disregard the
6
61 3
both
9
positives we
o
n
solution
The diagram shows two rectangles and a regular pentagon
One side of each rectangle has been extended to meet at X
Find
x
is
The green angle
1
90
108
108
18
The red angle is
180 18 108
u
angle
L
sum
of
54
a
triangle
The yellow angle is also
540
vertically opposite
r7
The purple angle is
By angle sun of
360 108
1620
90
quadrilateral
a
360
90
162
54
540
A water tank is 5 6 Full When 30L of water
is 415 Full
are removed From the tank the tank
How much water does the tank hold when Full
Suppose
the
tank holds
5
6
30
25k
900
x
900
Then
litres
Is
24
a
so
Pars
is
a
square
Point T
I 2 Point U lies
PT TQ
The perimeter of PTUS
is
of PTUS
P
x
T
2x
on
lies
SR
on
PQ
s.t
SU UR
s
Draw
I 2
the
40cm what is
d
t
area
diagram
a
The perimeter of PTUS is
3kt act 3k 1
In
s
3k
3h
x
So
x
2k
40
5
Hence the
is
x 3 1
U
82C
x
area
5
of PTUS
15
75cm
R
circular
The diagram shows 7
arcs
and a heptagon
with equal sides but unequal angles The sides of the
heptagon have length 4 The centre of each arc is a
vertex of the heptagon and the ends of the arc
midpoints of the two adjacent sides What is the
shaded area
Let the angles in the heptagon be
Where
4
t
t
dy
7 2
180
a
900
92
are
the
total
97
The circles have radius 2
So the shaded area is
it
b
22
I
34ft 7 360
IT
077
900
2520
To
coat
181T
Brachycephalus Frogs
and have 3 toes on
are
tiny
less than 1cm long
each foot and two Fingers on
each hand whereas the common Frog has five toes on
each Foot and four Fingers on each hand Some
Brachycephalus and common Frogs are in a bucket Each
Frog has all its Fingers and toes Between them they
have 122 toes and 92 Fingers How many frogs are
in the bucket
Let
the number of Brachycephaly frogs and
be the number of common frogs
13 be
We have
613 t 10C
413 t 8C
2CBtc
122
C
122
92
92
Btc
30
15
arc
The diagram shows an
PQ of a circle with
centre 0 and radius 8 Angle LOOP is a right
angle
the point M is the midpoint of OP and N
lies
on
the
arc
PQ
so
that
to OP Which of the following
MN
is
is
perpendicular
closest to the
length of
the perimeter of triangle PNM
B 18
A 17
D 20
C 19
E
21
Since the
OM
ON
Construct line
a radius
it
8
8
and
8
00
has
circle
has
Consider triangles
DMNP
4
MN
4
Mp
LOMN
are
by
SAS
2 sides and included
congruent
By Pythagoras
The perimeter of PNM
4
t
8
t
548
Since it is
length
4
DONM and
LPMN
the
angle
FF
MN
4
MP
common
on
So
radius 8
Fb
triangles
548
is
I
121
7
19
Since
49
is
the
closest
square
to
48
Two brothers and three sisters form a single line for
next to
a photograph The two boys refuse to stand
each other How
different line ups are possible
Ignoring the restriction the
of 5
people
is
total
of arrangement
number
120
5
calculate the number of those line ups in
which the boys are adjacent
We take the two boys as a single group and
arrange this group plus the 3 girls in a line
Now
we
4
life
so the
Ide
ups
is
answer
48
2
x
their
of boys in
48
120
block
72
sequence is calculated by
where K
multiplying together all the numbers
takes all the integer values from 2
to ntl
inclusive
The
nth term
in
a
1ft
ey 3rd
term is
ftp
FI
which is the smallest value of
the nth term of the sequence is
We have
So
the
2nd term
3rd term
KI
K
it
First term is
FE
FI
fz.FI
fZ
FI
FE
if
for which
h
an
integer
4th term
FE FE
B
Sth term
53
FE
FE
6th term
FE JE
So the
answer
ru
2
6
is
radius 2 and a
the
square with sides of length 2 The centre of
circle lies on the perpendicular bisector of a side of
the square at a distance k from the side The
The diagram shows
shaded
a
region has
circle
with
2
Find
area
X
Let 0 be the centre of
B
FA
D
g
d
F
E
the circle and label the
as
points
consider
BF
2
shown
SBOF
We
have
side length of
square and OB of
radii of the circle
the
So
D BOF is equilateral
and therefore
B OF
GO
2
The
area
of sector BOF is
36
The
area
of
x
DOBF
and so
the
Consider
DOAB
AD
The area
B
C
22
0A
x
Fs
of rectangle
f
D
T
do
F
E
2
22
231T
Fs
2 x Fez
53
is
area
green
By Pythagoras
So
is
IT
OA
i
BEEF
is
2hr3
D
x