Download Maths Revision Powerpoints for level 3/4

LJR August 2011
LJR August 2011
LJR August 2011
LJR August 2011
LJR August 2011
The internal angles in a
triangle add to 180°
The angles at a point on a
straight line add to 180°
LJR August 2011
Since any quadrilateral can be split into two
triangles its internal angles add to 360°
LJR August 2011
A pentagon can be split into
three triangles so its
internal angles add to 540°
A hexagon can be split into
four triangles so its
internal angles add to 720°
LJR August 2011
Internal angles
180°  3 = 60°
60° 120°
External angles
180° - 60° = 120°
Internal angles
360°  4 = 90°
External angles
180° - 90° = 90°
90°
90°
LJR August 2011
Internal angles
540°  5 = 108°
External angles
180° - 108° = 72°
108° 72°
Internal angles
720°  6 = 120°
External angles
180° - 120° = 60°
120° 60°
LJR August 2011
Sides of
Polygon
Angle total
Internal
angles
External
angles
3
180°
60°
120°
4
360°
90°
90°
5
540°
108°
72°
6
720°
120°
60°
n
(n-2)180°
(n-2)180°
n
180°-
(n-2)180°
n
LJR August 2011
Draw a regular hexagon of side 4cm.
Sketch to
identify angles
4cm
120º
60º
60º
Use this information to accurately
draw the hexagon.
LJR August 2011
Draw a regular pentagon of side 6cm.
Sketch to
identify angles
6cm
72º
108º
72º
Use this information to accurately
draw the pentagon.
LJR August 2011
Draw a rhombus with diagonals 8cm and 6cm.
Sketch first
now draw
4 cm
3cm
3cm
8cm
3cm
3cm
8 cm
4 cm
6cm
LJR August 2011
Click here to repeat
this section.
Click here to practise
this further.
Click here to return
to the main index.
LJR August 2011
c 2  a 2  b2
c
b
a
In any right angled
triangle the square
on the hypotenuse is
equal to the sum of
the squares on the
two shorter sides.
LJR August 2011
Calculating the hypotenuse.
Examples:
c2 = a2 + b2
Calculate x in these two triangles.
9cm
x 2  82  62
 64  36
 100
x  100
 10m
8m
x
x
7cm
x 2  92  72
 81  49
 130
x  130
 11  4cm to 1 dp
6m
LJR August 2011
Pythagoras theorem can be rearranged so that a shorter
side can be calculated.
c 2  a 2  b2
a 2  b2  c 2
a 2  c 2  b2
also
b2  c 2  a 2
Write the biggest number first
Add to find the hypotenuse
Subtract to find a shorter side
LJR August 2011
Calculating a shorter side.
Examples:
c2 = a2 + b2
Calculate x in these two triangles.
x
5m
13m
x 2  132  52
 169  25
 144
x  144
x
2
2
2
x  27  24
24cm
 729 576
 153
x  153
 12 37cm to 2 dp
 12m
LJR August 2011
Find the distance
between A and B.
B
AB2  72  52
 49 25
 74
A
x  74
 11  4 cm to 1 dp
LJR August 2011
Click here to repeat
this section.
Click here to try some
Pythagoras problems.
Click here to return
to the main index.
LJR August 2011
When we talk about the speed of an object we usually mean
the average speed. A car may speed up and slow down during
a journey but if the distance covered in one hour is 50 miles,
we would say its average speed was 50mph.
When we are doing calculations using speed, distance and time,
it is important to keep the units consistent.
If distance is measured in kilometres and time is measured in
hours, then the speed is in kilometres per hour (km/h).
LJR August 2011
S =
D
T
D
T =
D
S
S T
D = S x T
LJR August 2011
Problem:
Stewart walks 15km in 3
hours.
D
Calculate his average speed.
S T
Stewart covers 15km in 3hours
So his average speed is 15  3 = 5 km/h
Speed = 5km/h
distance covered
Average speed =
time taken
LJR August 2011
D
S T
Problem
Claire cycled at a steady speed
of 11 kilometres per hour.
How far did she cycle in 3
hours?
In 1 hour she covers 11 km
So in 3 hours she covers 11 X 3 = 33km
Distance = 33km
Distance = average speed X time taken
D = S X T
LJR August 2011
D
S T
Problem
Paul drives 144 kilometres at an
average speed of 48km/h.
How long will the journey take?
He drives 48 km in 1 hour.
144  48 = 3
(there are three 48s in 144)
So the journey takes 3 hours.
Time = 3 hours.
distance covered
Time taken =
average speed
D
T =
S
LJR August 2011
D
S T
D = S X T
Problem
A car travelled for 2 hours at an
average speed of 90km/h.
How far did it travel?
D
= 90 km X 2 hours
= 180 km
The car travelled 180 km
LJR August 2011
D
Problem
A car on a 240km journey can travel
at 60km/h.
S T
How long will the journey take?
D
T =
T
S
= 240 km  60
= 4 hours
The journey will take 4 hours
LJR August 2011
Click here to repeat
this section.
Click here to try some
harder SDT.
Click here to return
to the main index.
LJR August 2011
LJR August 2011
Diameter
LJR August 2011
Circumference
• The formula for the
circumference of a
circle is:
C = d
where C is circumference
and d is diameter
LJR August 2011
Area
• The formula for the
area of a circle is:
2
A = r
where A is area
and r is radius
LJR August 2011
Calculate the circumference
of this circle.
8cm
C = d
= 3·14  8
= 25·12cm
An
approximation
for  is 3·14
LJR August 2011
Calculate the area
of this circle.
2
10cm
A = r
= 3·14  5
2
= 3·14  25
Diameter is 10cm
= 78·5cm
2
 Radius is 5cm
LJR August 2011
LJR August 2011
• Calculate the diameter and radius of a
circle with a circumference of 157m.
C = d
157 = 3·14  d
d = 157 ÷ 3·14
d = 50m
r = 50 ÷ 2 = 25m
LJR August 2011
• Calculate the radius and diameter of a
2
circular slab with an area of 6280cm .
2
A = r
2
6280 = 3·14  r
2
r = 6280 ÷ 3·14
= 2000
r = 2000
= 44·72135955
 44·7cm  d = 89·4cm
LJR August 2011
Composite Shapes
• Calculate the Perimeter of this shape
C = d
= 3·14  9
9m
= 28·26
12m
28·26  2 = 14·13m
Perimeter = 14·13 + 12 + 12 + 9
= 47·13m
LJR August 2011
Composite Shapes
• Calculate the shaded Area.
Area of square = 28  28
= 784cm
28cm
2
2
A = r
2
= 3·14  14
28cm
= 3·14  196
= 615·44cm
2
Shaded area = 784  615·44
2
= 168·56cm
LJR August 2011
Click here to repeat
this section.
Click here to try some
more Circle work.
Click here to return
to the main index.
LJR August 2011
LJR August 2011
height
1
Area   base  height
2
1
A  bh
2
base
Example: Calculate the area of this triangle.
9cm
7cm
4cm
1
A  bh
2
1
 74
2
 14cm2
LJR August 2011
1
Area   diagonal1 diagonal2
2
1
A dd
2 1 2
d1
d1
d2
d2
LJR August 2011
Example: Calculate the area of these shapes.
3m
4m
8m
1
A dd
2 1 2
1
  11  8
2
 44m2
1
A dd
2 1 2
1
 96
2
6m
9m
 27m2
LJR August 2011
height
base base
This shows that the area of a parallelogram is
similar to the rectangle.
A re a  ba se  h e i g h t
LJR August 2011
Area  base  height
height
A  bh
base
Example: Calculate the area of this parallelogram.
A  bh
5cm
8cm
 85
 40cm2
LJR August 2011
A composite shape can be split into parts so that the area
can be calculated.
Examples: Calculate the area of the following shapes.
11cm
10cm
A
B
6cm
Area A = 10 × 6 = 60
Area B =
6 × 5 = 30
90cm2
6cm
LJR August 2011
12m
Area A = 12 × 11 = 132
11m
1
Area B = × 6 × 11 = 33
2
A
B
165m2
18m
A shape can be split into as many parts as necessary.
LJR August 2011
Click here to repeat
this section.
Click here to practise
more like this.
Click here to return
to the main index.
LJR August 2011
Organise this data in a Stem & Leaf chart
27
24
31
28
33
42
50
29
30
26
32
45
48
51
45
34
26
51
33
41
44
37
22
52
35
2
7 6 4 8 2 9 6
2
2 4 6 6 7 8 9
3
3 2 1 7 3 4 5 0
3
0 1 2 3 3 4 5 7
4
1 5 4 8 2 5
4
1 2 4 5 5 8
5
1 2 0 1
5
0 1 1 2
Leaf
n = 25
Stem
4|2 means 42
LJR August 2011
Click here to repeat
this section.
Click here to try some
more Stem & Leaf.
Click here to return
to the main index.
LJR August 2011
Calculations must be carried out in a certain order.
Brackets first and any ‘of’ questions,
then multiply and divide before add and subtract.
rackets
f
ivide
ultiply
dd
ubtract
eg: ¼ of 20
LJR August 2011
Examples
3 5  7  2
1
of (8  7)
3
7  5 - 12  3
 15  14
 35 - 4
 29
 31
1

of 15
3
 5
7  (2  1) - 8  2
 7 3 - 8  2
 21 - 4

17
78
7  2 4
15


3
5
8-3
5
Evaluate top and bottom
separately first then divide.
LJR August 2011
Given a = 4, b = 5 and c = 3, find the values of:
2a  3b  2  4  3  5
 8  15
2abc  2  4  5  3
 120
 23
7(b - c)  2a  7(5 - 3)  2  4
 7 2  8
 14  8
 22
ab  6c 4  5  6  3

4  3-5
a  c b
20  18

2
38

2
 19
LJR August 2011
Click here to repeat
this section.
Click here to try some
more problems.
Click here to return
to the main index.
LJR August 2011
Very large or very small numbers can be written in
scientific notation (also known as standard form) to
ease calculations and allow the use of a calculator.
300 can be written as 3 x 100 and we know that 100 can
be written as 102 , so 300 can be written as 3 x 102
300 = 3 x 102 This is scientific notation.
In general a x 10n where 1  a  10 and n is an integer.
Positive or negative
whole number.
LJR August 2011
Normal numbers to scientific notation
Examples
7000000000 7 1000000000
 7109
530000000 5  3100000000
 5  3  108
4710000 4  71 1000000
 4  71106
7000000000 
decimal point moves
9 places left
530000000 
decimal point moves
8 places left
4710000 
decimal point moves
6 places left
LJR August 2011
3
3
0  03  3  100 
 2  3  10-2
100 10
You do not need to remember this but it is the reason why
we can write small numbers as follows.
Examples
0  0000000008 8 10-10
0 0000000008
decimal point moves
10 places right
0  000000692 6  9210-7
0 000000692
decimal point moves
7 places right
LJR August 2011
Scientific notation to normal numbers.
Examples
5 109  5 1000000000
 5000000000
4  7 103  4  7 1000
 4700
3  89105  3  89100000
 389000
5000000000 
decimal point moves
9 places right
4700 
decimal point moves
3 places right
389000 
decimal point moves
5 places right
LJR August 2011
Scientific notation to normal numbers.
Examples
-7
4 10
-4
1  6 10
 0  0000004
 0  00016
-5
2  5410
 0  0000254
0 0000004
decimal point moves
7 places left
0 00016
decimal point moves
4 places left
00000254
decimal point moves
5 places left
LJR August 2011
You must be able to enter and understand scientific
notation on a calculator.
To enter 4 107
To enter 3  1  10-4
4
7
EXP
•
3
1
EXP
+/-
4
On a calculator display
41007
3·110-04
represents 4  107
represents 3  1  10-4
LJR August 2011
Click here to repeat
this section.
Click here to try some
more problems.
Click here to return
to the main index.
LJR August 2011
Everything in the bracket is multiplied by what is outside the bracket.
4 ( x  3 )  4  x  4  3  4 x  12
3 (2 p  5 q)  6 p  15 q
2
3t (2 t  7 )  6 t  21t
2 a(b  5 a)  2 ab 10 a2
LJR August 2011
The reverse process is called factorising.
To factorise
• look for factors which are common to all terms.
• identify the highest common factor.
Factorise 4 x  12
factors of 4 x are 1 and 4 x, 2 and 2 x, 4 and x.
factors of 12 are 1 and 12 , 2 and 6 , 3 and 4.
 4 x  12  4 ( x  3 )
LJR August 2011
Try to work out the answer to each question before
pressing the space-bar.
Examples:
Factorise
8 y  6  2 (4 y  3 )
4 p2  8 p  4 p( p  2 )
18 x  14 x  2 x(9 7 x)
6 t 3  21t  3t (2 t 2  7 )
24 g  18 g 2  6 g (4  3 g )
25 xy  10 x2  5 x(5 y  2 x)
2
LJR August 2011
Click here to repeat
this section.
Click here to try some
more problems.
Click here to return
to the main index.
LJR August 2011
Letters are used to represent missing numbers.
Expressions
An expression contains letters and numbers.
x + 3, 2t – 5, 7 + 4y etc are all expressions.
The value of an expression depends on the value
given to the letters in the expression.
If x = 4, give the value of (i) x + 3
(i) x + 3
=4+3
=7
(ii) 5x – 7
(ii) 5x – 7
= 20 – 7
= 13
LJR August 2011
Find an expression for the number of matches in design x.
5
9 13
4
An expression for the no. of matches in design x is 4x + 1
If a = 3, b = 0, c = 5 and d = 7 find the value of the following
expressions (i) 4b + 2d
(ii) 3c – 2a + 5d
(i) 4b + 2d
(ii) 3c – 2a + 5d
= 0 + 14
= 15 – 6 + 35
= 14
= 44
LJR August 2011
Click here to repeat
this section.
Click here to try some
more problems.
Click here to return
to the main index.
LJR August 2011
Example:
Solve
2 (4 x  3 )  5  13
8 x  6  5  13
8 x  11  13
8 x  24
x 3
Example:
Solve
7 x  8  3 x  24
4 x  8  24
4 x  32
x 8
LJR August 2011
Example:
Solve
3 (2 x  5 )  8  17
6 x  15  8  17
6 x  23  17
6 x  6
x  1
Example:
Solve
5 x  4  3 x  22
2 x  4  22
2 x  22
x  11
LJR August 2011
Click here to repeat
this section.
Click here to try more
Equations/Inequations.
Click here to return
to the main index.
LJR August 2011
Here is some information (or data) – imagine it is a set of test
marks belonging to a group of children
19
21
20
17
24
20
16
20
18
This data can be organised and used in different ways.
LJR August 2011
The mode (or modal value) is the value in the data that occurs most
frequently.
19
21
20
17
24
20
16
20
18
First of all rearrange the data in order -
16
17
18
19 20 20 20 21 24
The mode is 20 as it occurs most often.
LJR August 2011
19
21
20
17
24
20
16
20
18
The median is the value in the middle of the data when it is
arranged in order.
16
17
18
19 20 20 20 21 24
The median is 20 as this is the value which is in the middle.
The range is a measure of spread: it tells us how the data is
spread out. The range = the highest value – lowest value.
The range is 24 – 16 = 8.
The value of the range is 8.
LJR August 2011
Temperatures in ºC
13
13 11 14 17 19 18
11
13
13
14
17
18
19
The sum of the values is 105.
The number of values is 7.
The mean is
105
7
= 15
LJR August 2011
This set of data shows shoe sizes. Find the mean, median, mode and range.
3
4
6
3
7
5
4
5
4
3
3
4
4
4
5
5
6
7
The mode is 4 as it occurs most often.
The median is the middle value 4.
The range is 7 – 3 = 4
The sum of the values is 41.
The number of values is 9.
The mean is
41
5
4
9
9
LJR August 2011
Here is another set of data. Find the mean, median, mode and range.
19
21
20
17
22
18
28 27
17
18
19
20
21
22 27 28
There is no mode as each value occurs just once.
The median is the middle value. As there is an even number of
data, the median is half way between 20 and 21.
The median is 20•5
The range is 28 – 17 = 11.
The value of the range is 11.
The sum of the values is 172.
The number of values is 8.
The mean is
172
 21  5
8
LJR August 2011
Number of
goals scored
Frequency
0
13
1
21
2
11
11 X 2 goals = 22
3
8
8 X 3 goals = 24
4
6
6 X 4 goals = 24
The sum of the values is:
13 X 0 goals = 0
21 X 1 goal = 21
So the sum of the values is: 0 +21 + 22 + 24 + 24 = 91
The number of values is the total frequency:
13 + 21 + 11 + 8 + 6 = 59
The mean of the goals scored is
91
59
= 1•54 to 2dp
LJR August 2011
Marks out of
10
Frequency of
marks
5
2
6
6
7
9
8
10
9
7
The mode is 8 because this test mark has the highest frequency.
The range is 9 – 5 = 4.
The value of the range is 4.
The total frequency is 34.
The median is ½(n +1) value so ½(34 +1) = ½(35) = 17•5
The 17th value is 7
and the 18th value is 8.
The median is 7•5
LJR August 2011
Click here to repeat
this section.
Click here to try some
more problems.
Click here to return
to the main index.
LJR August 2011
Scatter graphs are used to identify any correlation
between two measures.
Graph the following data taken from a class of S3 students.
Height (cm)
Shoe size
125 130 135 140 140 145 150 150 155 160 165 175
2
4
3
5
6
7
6
7
8
10
9
11
Does this show a connection between height and shoe size?
LJR August 2011
Height and shoe size in S3
13
12
11
10
This graph shows
a strong positive
correlation.
Shoe size
9
8
7
6
5
4
3
2
1
120
130
140
150
160
170
180
Height (cm)
LJR August 2011
28
Exam Marks & Attendance
26
24
22
Exam mark
20
This graph shows
a strong negative
correlation.
18
16
14
12
10
8
6
4
2
0
2
4
6
8
Absence (in days)
10
12
14
LJR August 2011
Exam marks & Height
28
26
24
22
Exam mark
20
This graph shows
no correlation.
18
16
14
12
10
8
6
4
2
0
120
130
140
150
160
Height (cm)
170
180
LJR August 2011
Click here to repeat
this section.
Click here to practise
more of this.
Click here to return
to the main index.
LJR August 2011
A factor is a number that divides another number exactly.
Find all the factors of 24
1 x 24
2 x 12
3 x 8
4 x 6
Factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24
LJR August 2011
Find the prime factors of 24
24 = 2 x 12
=2x3x4
=2x3x2x2
2 x 2 x 2 = 23
= 23 x 3
LJR August 2011
Find the prime factors of 72
72 = 2 x 36
= 2 x 3 x 12
=2x3x3x4
=2x3x3x2x2
3
2
=2 x3
2 x 2 x 2 = 23
3 x 3 = 32
LJR August 2011
Click here to repeat
this section.
Click here to investigate
further.
Click here to return
to the main index.
LJR August 2011
A ratio compares quantities
You must be able to
• Simplify ratios
• Find one quantity given the other
• Share a quantity in a given ratio
LJR August 2011
The ratio of cars to buses is 4 to 3
also written as 4:3
It is essential to write ratios in the
correct order
cars to buses is 4:3
LJR August 2011
The ratio of eggs to bunnies is 3 to 5
also written as 3:5
eggs to bunnies is 3:5
Remember order is important
LJR August 2011
Simplify the ratio 28:21
both numbers divide by 7  28:21
4:3
Simplify the ratio 32:56
both numbers divide by 8  32:56
4:7
You can take
as many steps
as you need
32:56
16:28
8:14
4:7
LJR August 2011
The ratio of boys to girls in a class is 5:4
If there are 12 girls how many boys are there?
Boys : Girls
5 x 3 = 15
3
5 : 4
15 :12
3
The ratio of oranges to apples in a fruit bowl is 2:3
If there are 8 oranges how many apples are there?
Oranges : Apple
4
2 : 3
8 : 12
4
3 x 4 = 12
LJR August 2011
Share £800 between 2 partners in a business in
the ratio 3:5
3 + 5 = 8 shares
 £800  8 = £100
3 x £100 = £300
5 x £100 = £500
£300 + £500 = £800
The partners receive £300
and £500 respectively.
LJR August 2011
Share 45 sweets between 2 friends in the ratio 5:4
5 + 4 = 9 shares
 45  9 = 5 sweets
5 x 5 = 25 sweets
4 x 5 = 20 sweets
25 + 20 = 45
The friends receive 25 and
20 sweets each.
LJR August 2011
Click here to repeat
this section.
Click here to try some
more problems.
Click here to return
to the main index.
LJR August 2011
Remember :
A right angle is 90°
A straight angle is 180°
There are 360° round a point
An acute angle is less than 90°
An obtuse angle is more than 90°
and less than 180°
A reflex angle is more than 180°
and less than 360°
LJR August 2011
More Angle terms
A reflex angle is greater than 180°
but less than 360°
A
Reflex ABC = 280°
280°
B
80°
C
LJR August 2011
Two lines are perpendicular if
they intersect at right angles.
A line parallel to the earth’s horizon is horizontal.
A line perpendicular to a horizontal is called vertical.
LJR August 2011
A
D
If two angles make a right angle they
are said to be complementary.
B
ABD and DBC are complementary
C
D
If two angles make a straight angle
they are said to be supplementary.
ABD and DBC are supplementary
A
B
C
LJR August 2011
A
B
D
60°
ABD = 90° – 60° = 30°
D
C
DBC = 180° – 40° = 140°
A
B
40°
C
Vertically opposite angles are equal.
B
A
•
D
*
E
*
•
C
AED = BEC
AEB = DEC
LJR August 2011
Angles and parallel Lines
When two parallel lines are involved F and Z shapes can be
used to calculate angles.
K
A
B
C
E
80°
L
D
G
F
H
Parallel lines (F shape)
so HFG = 80°
M
P
N
65°
R
Q
S
Parallel lines (Z shape)
so PQM = 65°
LJR August 2011
Fill in all the missing angles.
117°
63°
63°
117°
117°
63° 63°
117°
180° - 63° = 117°
LJR August 2011
Click here to repeat
this section.
Click here to try some
more problems.
Click here to return
to the main index.
LJR August 2011
1
S cale factor
2
1
S cale factor
2
1
S cale factor
4
S c a le f a c to r 2
S c a le f a c to r 2
S c a le f a c to r 4
LJR August 2011
Identify the scale factor
scale
factor 3
scale
factor 2
scale
factor ½
scale
factor 6
scale
factor ¼
What other scale factors can you identify?
LJR August 2011
scale
factor 3
scale
factor 4
scale
factor 2
scale
factor 1½
scale
factor ¾
scale
factor ½
LJR August 2011
Click here to repeat
this section.
Click here to try some
more problems.
Click here to return
to the main index.
LJR August 2011
To draw a pie chart we need to work out the different
fractions for each group.
One evening the first 800 people to enter a Cinema
complex are asked which film they plan to see.
The results are as follows:
Lord of the Rings
160
Calendar Girls
150
Pirates of the Caribbean
190
The Last Samurai
170
Touching the Void
130
Use these results to draw a pie chart.
LJR August 2011
Lord of the Rings
160
Calendar Girls
144
Pirates of the Caribbean
192
The Last Samurai
176
Touching the Void
128
Touching
the Void
16%
22%
The Last
Samurai
Lord of
the Rings
20%
18%
Calendar
24%
Girls
Pirates of
the Caribbean
16 0
 10 0  2 0 %
800
14 4
 10 0  18 %
800
19 2
 10 0  2 4 %
800
17 6
 10 0  2 2 %
800
12 8
 10 0  16 %
800
100%
LJR August 2011
Lord of the Rings
160
Calendar Girls
144
Pirates of the Caribbean
192
The Last Samurai
176
Touching the Void
128
Touching
the Void
58°
79°
The Last
Samurai
Lord of
the Rings
72°
65°
Calendar
86°
Girls
Pirates of
the Caribbean
16 0
 3 6 0  7 2
800
14 4
 3 6 0  6 4  8   6 5
800
19 2
 3 6 0  8 6  4   8 6
800
17 6
 3 6 0  7 9  2   7 9
800
12 8
 3 6 0  5 7  6   5 8
800
360°
LJR August 2011
Click here to repeat
this section.
Click here to try some
more problems.
Click here to return
to the main index.
LJR August 2011
Probability is a measure of chance between 0 and 1.
Probability of an impossible event is 0.
Probability of a certain event is 1.
½
0
Impossible
Unlikely
Even chance
1
Most likely
Certain
N u mber o f f a v o u ra ble o u tc o me
P ro ba bi li ty 
To ta l n u mber o f o u tc o mes
LJR August 2011
The probability of throwing a 3
1
is 1 out of 6
Pr(3)
6
The probability of throwing an
3 1
even number is 3 out of 6 Pr(3)  
6
The probability of choosing a 5 of diamonds
from a pack of cards is 1 out of 52
5
5
2
1
Pr(5 of diamonds) 
52
13 1
Pr(diamond) 

52 4
LJR August 2011
Click here to repeat
this section.
Click here to try some
more problems.
Click here to return
to the main index.
LJR August 2011