clockwise around a third semicircle with centre
)
0
,
3
(
.
Where does Peri end this expedition?
A
)
0
,
0
(
B
)
0
,
1
(
C
)
0
,
2
(
D
)
0
,
4
(
E
)
0
,
6
(
Solution:
C
As may be seen from the diagram, Peri first moves along the
semicircle with centre
)
0
,
4
(
from the point
)
0
,
0
(
to the point
)
0
,
8
(
, then along the semicircle with centre
)
0
,
6
(
to the point
)
0
,
4
(
, and finally along the semicircle with centre
)
0
,
3
(
to end
up at the point
)
0
,
2
(
.
19.
The shaded region shown in the diagram is bounded by four arcs, each of
the same radius as that of the surrounding circle. What fraction of the
surrounding circle is shaded?
A
1
4
B
4
1
C
2
1
D
3
1
E it depends on the radius of the circle
Solution:
A
Suppose that the surrounding circle has radius
r
. In the diagram we have
drawn the square with side length 2
r
which touches the circle at the points
where it meets the arcs. The square has area
2
2
4
)
2
(
r
r
. The unshaded area
inside the square is made up of four quarter circles with radius
r
, and thus has
area
2
r
. Hence the shaded area is
2
2
2
)
4
(
4
r
r
r
. The circle has area
2
r
.
So the fraction of the circle that is shaded is
1
4
4
)
4
(
2
2
r
r
.
20.
A rectangle with area 125 cm
2
has sides in the ratio 4:5. What is the perimeter of the rectangle?
A
18 cm
B 22.5 cm
C 36 cm
D 45 cm
E 54 cm
Solution:
D
Since the side lengths of the rectangle are in the ratio 4:5, they are 4
a
cm and 5
a
cm, for some positive
number
a
. This means that the rectangle has area
2
20
5
4
a
a
a
cm
2
. Hence
125
20
2
a
. So
2
a
4
25
20
125
, and hence
2
5
a
. Hence the rectangle has perimeter
2
5
18
18
)
5
4
(
2
a
a
a
45
cm.
3
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21.
The parallelogram
PQRS
is formed by joining together four
equilateral triangles of side 1 unit, as shown.
What is the length of the diagonal
SQ
?
A
7
B
8
C 3
D
6
E
5
Solution:
A
Let
T
be the foot of the perpendicular from
Q
to the line
SR
extended. Now
RQT
is half of an equilateral triangle with side
length 1. Hence the length of
RT
is
2
1
and hence
ST
has length
2
5
2
1
1
1
. By Pythagoras’ Theorem applied to the right angled triangle
RQT
,
2
2
2
1
2
)
(
1
QT
.
Therefore
4
3
4
1
2
2
1
2
2
1
)
(
)
1
(
QT
. Hence, by Pythagoras’ Theorem applied to the right angled
triangle
SQT
,
7
)
(
4
3
4
25
4
3
2
2
5
2
2
2
QT
ST
SQ
. Therefore,
7
SQ
.
22.
What is the maximum possible value of the median number of cups of coffee bought per
customer on a day when Sundollars Coffee Shop sells 477 cups of coffee to 190 customers,
and every customer buys at least one cup of coffee?
A
1.5
B 2
C 2.5
D 3
E 3.5
Solution:
E
Put the set of numbers of cups of coffee drunk by the individual customers into numerical order with
the smallest first. This gives an increasing sequence of positive integers with sum 477. Because 190 is
even, the median of these numbers is the mean of the 95th and 96th numbers in this list. Suppose
these are
a
and
b
, respectively. Then the median is
)
(
2
1
b
a
.
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 Spring '13
 MRR
 Math, Prime number, triangle, Divisor, perfect number

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