Tom and Geri have a competition. Initially, each player has one attempt at hitting a target. If
one player hits the target and the other does not then the successful player wins. If both
players hit the target, or if both players miss the target, then each has another attempt, with
the same rules applying. If the probability of Tom hitting the target is always
5
4
and the
probability of Geri hitting the target is always
3
2
, what is the probability that Tom wins the
competition?
A
15
4
B
15
8
C
3
2
D
5
4
E
15
13
Solution:
C
Let the probability that Tom wins the competition be
p.
The probability that initially Tom hits and
Geri misses is
15
4
3
1
5
4
=
×
. The probability that initially they both hit is
15
8
3
2
5
4
=
×
and that they both
miss is
15
1
3
1
5
1
=
×
. So the probability that either they both hit or both miss is
5
3
15
9
15
1
15
8
=
=
+
. If they
both hit or both miss the competition is in the same position as it was initially. So Tom’s
probability of winning is then
p
. Therefore,
p
p
5
3
15
4
+
=
. So
15
4
5
2
=
p
and hence
3
2
=
p
.
24.
have length 1. It may be divided into three congruent
quadrilaterals as shown.
What is the area of the tile?
A
2
3
2
1
+
B
3
3
4
C
6
D
4
3
4
3
+
E
2
3
3
Solution:
B
The tile is made up of three congruent quadrilaterals one of which,
PQRS
shown in the diagram on the right. We let
T
be the foot of the perpendicular
from
Q
to
SR
and we let
U
be the point shown. Then
1
=
=
UR
SR
.
Also
PQ=QU.
Let their common length be
x.
Hence
x
QR
+
=
1
and
x
TR
−
=
1
. Because the three congruent quadrilaterals fit together at
Q
, we
have
0
120
=
∠
PQR
. As
PQ
and
SR
are both perpendicular to
PS
, they are
parallel and hence
0
60
=
∠
QRT
. Hence, from the rightangled
triangle
QRT
we have
0
60
sin
=
QR
TR
, that is,
2
1
1
1
=
+
−
x
x
, and therefore
.
1
2
2
x
x
+
=
−
Hence
1
3
=
x
and
so
3
1
=
x
.
Therefore
3
2
=
TR
and
3
4
=
QR
. Now
0
60
cos
=
QR
QT
and hence
0
60
cos
QR
QT
=
3
3
2
2
3
3
4
0
3
4
)
(
60
cos
=
=
=
. [We could also use Pythagoras’ Theorem applied to the triangle
QRT
to
deduce this length.]
We can now work out the area of the quadrilateral
PQRS
in two different ways.
Most
straightforwardly, area(
PQRS
) = area(
PQTS
) + area(
QRT
)
9
3
4
3
2
3
3
2
2
1
3
1
3
3
2
)
(
=
×
+
×
=
.
Alternatively,
as
PQ
is parallel to
SR
,
PQRS
is a trapezium.
We now use the fact that the area of a trapezium is its
height multiplied by the average length of its parallel sides. Hence,
PQRS
has area.
)
1
(
3
1
2
1
3
3
2
+
×
9
3
4
3
2
3
3
2
=
×
=
. The tile is made up of three copies of
PQRS.
So its area is
.
3
3
3
4
9
3
4
=
×
1
1
1
1
1
1
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View Full Document25.
How many distinct pairs
)
,
(
y
x
of real numbers satisfy the equation
)
4
)(
4
(
)
(
2
−
+
=
+
y
x
y
x
?
A
0
B
1
C
2
D
3
E
4
Solution:
B
Method 1.
The most straightforward method is to rewrite the given equation as a quadratic in
x
,
whose coefficients involve
y
, and then use the “
0
4
2
≥
−
ac
b
” condition for a quadratic,
0
2
=
+
+
c
bx
ax
, to have real number solutions.
Now ,
16
4
4
2
)
4
)(
4
(
)
(
2
2
2
−
+
−
=
+
+
⇔
−
+
=
+
y
x
xy
y
xy
x
y
x
y
x
0
16
4
4
2
2
=
+
−
+
+
+
⇔
y
y
x
xy
x
0
)
16
4
(
)
4
(
2
2
=
+
−
+
+
+
⇔
y
y
x
y
x
.
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 Spring '13
 MRR
 Math, Prime Numbers, Prime number, Square number

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