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INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
UNIT 2 :
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Trigonometry
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INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
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UNIT 2 :
Trigonometry
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TRIGONOMETRY : Question 1
Find the area of the following triangle to the nearest cm2.
B
40cm
A
25°
C
50cm
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TRIGONOMETRY : Question 1
Find the area of the following triangle to the nearest cm2.
B
40cm
A
25°
C
50cm
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1. For area of a
2. Always
triangle
questions
remember
to
with
an angle
round use:
your
present
answer
if the
1
Aquestions
 2 ab sinasks
C
you to.
TRIGONOMETRY : Question 1
Find the area of the following triangle to the nearest cm2.
B
40cm
A
25°
C
50cm
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Area of  = 423cm2
Question 1
1. For area of triangles questions
where an angle is present use:
Find the area of the
following triangle
Area of  = ½ bcsinA°
= 50 x 40 x sin25°  2
to the nearest cm2.
B
= 422.61…
40cm
A
2. Remember to round if asked to.
25°
C
50cm
= 423 to nearest unit
Area of  = 423cm2
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Markers Comments
Check formulae list for:
1. For area of triangles questions
where an angle is present use:
1
Area of triangle = 2 absinC
Area of  = ½ bcsinA°
= 50 x 40 x sin25°  2
(Note: 2 sides and the included
angle)
= 422.61…
Relate formula to labels being
used.
B
2. Remember to round if asked to.
40cm
c
= 423 to nearest unit
Area of  =
423cm2
A
25°
b
50cm
C
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TRIGONOMETRY : Question 1B
Find the area of the following triangle.
K
6.5m
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L
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150°
8m
M
TRIGONOMETRY : Question 1B
Find the area of the following triangle.
K
6.5m
L
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150°
8m
1. For area of a
2. Always
triangle
questions
remember
to
with
an angle
round use:
your
present
answer
if the
Aquestions
 12 ab sinasks
C
you to.
M
TRIGONOMETRY : Question 1B
Find the area of the following triangle.
K
6.5m
L
150°
8m
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Area of  = 13m2
M
Question 1B
1. For area of triangles questions
where an angle is present use:
Find the area of the
Area of  = ½ kmsinL°
= 8 x 6.5 x sin150°  2
following triangle.
K
= 13
6.5m
L
150
°
Area of  = 13m2
8m
Begin Solution
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M
Markers Comments
Check formulae list for:
1. For area of triangles questions
where an angle is present use:
Area of  = ½ kmsinL°
= 8 x 6.5 x sin150°  2
absinC
(Note: 2 sides and the included
angle)
Relate formula to labels being
used.
= 13
Area of  =
Area of triangle =
13m2
K
6.5m m
L
150
°
k
8m
M
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TRIGONOMETRY : Question 2
Two helicopters leave an air base. The first flies on a bearing of 340°
at 160km/hr. The second flies due east at 200km/hr.
How far apart will they be after 21/2 hours?
Answer
N
to the nearest 10km.
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340°
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TRIGONOMETRY : Question 2
Two helicopters leave an air base. The first flies on a bearing of 340°
at 160km/hr. The second flies due east at 200km/hr.
How far apart will they be after 21/2 hours?
Answer
N
to the nearest 10km.
What would you like to do now?
Reveal answer only
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1. Identify
what you
2.
Calculate
as
3.
Make
a
sketch
to
340°
5.need
Substitute
known
4. Identify
which
trig
to findofand
many
the
clarify
matters.
values,
remembering
to
rule to use:
the missing
information
you
angles
as
use
brackets
as
Two
sides
+
two
angles
have topossible.
help you.
appropriate.
= sine rule
Go to Comments
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Three sides + one angle
= cosine rule
TRIGONOMETRY : Question 2
Two helicopters leave an air base. The first flies on a bearing of 340°
at 160km/hr. The second flies due east at 200km/hr.
How far apart will they be after 21/2 hours?
Answer
N
to the nearest 10km.
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340°
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Distance is 740km
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Question 2
1. Identify what needs to be found.
160km/hr.
2. Need distances travelled and angle
between flight paths.
N
340°
How far apart will
they be after
21/2 hours?
20°
90°
d1 = speed1 x time = 160 x 2.5 = 400km
d2 = speed2 x time = 200 x 2.5 = 500km
340° clockwise = 20° anti-clockwise
Full angle = 20° + 90° = 110°
3. Sketch triangle.
200km/hr
b
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110°
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a
400km
A
500km
c
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Question 2
4. Apply Cosine rule.
160km/hr.
a2 = b2 + c2 – (2bccosA°)
N
340°
How far apart will
they be after
21/2 hours?
20°
90°
200km/hr
Continue Solution
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5. Substitute known values and
remember to use brackets.
= 4002 + 5002 – (2 x 400 x 500 x cos110°)
= 546808.05..
a = 546808.05..
= 739.46....
6. Remember to round answer if
asked to.
= 740
Distance is 740km
Markers Comments
1. Identify what needs to be found.
2. Need distances travelled and
angle between flight paths.
Note:
Bearings are measured
clockwise from N.
d1 = speed1 x time = 160 x 2.5 = 400km
d2 = speed2 x time = 200 x 2.5 = 500km
340° clockwise = 20° anti-clockwise
Full angle = 20° + 90° = 110°
3. Sketch triangle.
400km
110°
500km
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Markers Comments
4. Apply Cosine rule.
a2 = b2 + c2 – (2bc cosA°)
Check formulae list for the
cosine rule:
a2 = b2 + c2 – 2bc cosA
5. Substitute known values and
remember to use brackets.
(2 sides and the included
angle)
= 4002 + 5002 – (2 x 400 x 500 x cos110°)
Relate to variables used
= 546808.05..
a2 = b2 + c2 – 2bc cosA
a = 546808.05..
= 739.46....
6. Remember to round answer if
asked to.
= 740
Distance is 740km
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TRIGONOMETRY : Question 2B
Two ships sail from a port. The first travels for two hours on a
bearing of 195° at a speed of 18mph. The second travels south-east
for three hours at a speed of 15mph.
N
How far apart will they be?
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195°
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SE
TRIGONOMETRY : Question 2B
Two ships sail from a port. The first travels for two hours on a
bearing of 195° at a speed of 18mph. The second travels south-east
for three hours at a speed of 15mph.
N
How far apart will they be?
What would you like to do now?
Reveal answer only
Go to full solution
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EXIT
2. Calculate
1. Identify
what you as
3. Make a sketch
toof the
many
need
to
find
and
4. Identify
which
trig as
clarify
matters.
missing
angles
therule
information
you
to use:
possible.
have
to
help
you.
Two sides
+ two angles
5. Substitute
known
= sine remembering
rule
values,
to
use brackets as
195° + one angle
Three sides
SE
appropriate.
= cosine rule
TRIGONOMETRY : Question 2B
Two ships sail from a port. The first travels for two hours on a
bearing of 195° at a speed of 18mph. The second travels south-east
for three hours at a speed of 15mph.
N
How far apart will they be?
Distance is 41.2 miles
What would you like to do now?
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195°
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SE
Question 2B
How far apart will
they be?
1. Identify what needs to be found.
N
2. Need distances travelled and angle
between paths.
1350
3hr@15mph
2hr@18mph
195°
d1 = speed1 x time = 18 x 2 = 36miles
600
SE
d2 = speed2 x time = 15 x 3 = 45miles
NB: SE = 135°
Angle = 195° - 135° = 60°
3. Sketch triangle.
A
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60°
45miles
c
36miles
b
a
Question 2B
How far apart will
they be?
4. Apply Cosine rule.
N
a2 = b2 + c2 – (2bc cosA°)
5. Substitute known values and
remember to use brackets.
1350
3hr@15mph
2hr@18mph
= 362 + 452 – (2 x 36 x 45 x cos60°)
= 1701
195°
600
SE
a = 1701
= 41.243....
= 41.2
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Distance is 41.2 miles
Comments
1. Identify what needs to be found.
2. Need distances travelled and
angle between flight paths.
Note:
Bearings are measured
clockwise from N.
d1 = speed1 x time = 18 x 2 = 36miles
d2 = speed2 x time = 15 x 3 = 45miles
NB: SE = 135°
Angle = 195° - 135° = 60°
3. Sketch triangle.
A
60°
45miles
c
36miles
b
a
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Comments
4. Apply Cosine rule.
a2 = b2 + c2 – (2bc cosA°)
5. Substitute known values and
remember to use brackets.
= 362 + 452 – (2 x 36 x 45 x cos60°)
= 1701
Check formulae list for the
cosine rule:
a2 = b2 + c2 – 2bc cosA
(2 sides and the included
angle)
Relate to variables used
a2 = b2 + c2 – 2bc cosA
a = 1701
= 41.243....
= 41.2
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Distance is 41.2 miles
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TRIGONOMETRY : Question 3
In the kite shown below PQ = 10cm, QR = 15cm &
diagonal PR = 22cm.
(a) Find the size of angle QPR.
(b) Hence find the area of the kite to the nearest square unit.
P
10cm
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Q
S
22cm
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15cm
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R
TRIGONOMETRY : Question 3
In the kite shown below PQ = 10cm, QR = 15cm &
diagonal PR = 22cm.
(a) Find the size of angle QPR.
(b) Hence find the area of the kite to the nearest square unit.
P
1. Identify which trig
3. For10cm
area of a
rule to use:
What would you like to do now?
triangle
with an
4.
Always
Two sides + two angles
angle
present to
use:
remember
2.
Substitute
known
Q
= sine rule
Reveal answer only
round
your 22cmto
values,
remembering
1
answer
if the
use
brackets
A

sin
C as
Three sides + one
2 abangle
Go to full solution
questions
appropriate.
= cosine
rule asks
15cm
you to.
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R
S
TRIGONOMETRY : Question 3
In the kite shown below PQ = 10cm, QR = 15cm &
diagonal PR = 22cm.
(a) Find the size of angle QPR. angleQPR = 35.3°
(b) Hence find the area of the kite to the nearest square unit.
P
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= 127cm2
10cm
Q
S
22cm
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15cm
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R
Question 3
P
1. To find angle when you have 3 sides
use 2nd version of cosine rule:
10cm
Q
S
22cm
cosP
r2 + q2 - p2
= 2rq
= 102 + 222 - 152
2 x 10 x 22
15cm
= (102 + 222 - 152)  (2 x 10 x 22)
R
(a) Find the size of angle QPR
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= 0.8159…
2. Remember to use inverse function
to find angle.
angleQPR = cos-1(0.8159..) = 35.3°
Question 3
P
1. Kite = 2 identical triangles. For area
of triangles where an angle is present
use:
10cm
Q
S
22cm
(b) Area QPR = ½ qrsinP°
= 10 x 22 x sin35.3°  2
15cm
= 63.56..cm2
2. Remember to double this and
round answer.
R
(b) Hence find the area of the kite
to the nearest square unit
Continue Solution
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Area of kite = 2 x 63.56..
= 127.12..cm2
= 127cm2
Markers Comments
1. To find angle when you have 3
sides use 2nd version of cosine rule:
cosP
r2
q2
+
= p22rq
Check the formulae list for the
second form of the cosine rule:
2 + c 2 – a2
b
cosA =
2bc
( 3 sides)
= 102 + 222 - 152
2 x 10 x 22
= (102 + 222 - 152)  (2 x 10 x 22)
= 0.8159…
2. Remember to use inverse function
to find angle.
angleQPR = cos-1(0.8159..) = 35.3°
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Markers Comments
Relate to variables used:
1. To find angle when you have 3
sides use 2nd version of cosine rule:
cosP
cosP =
r 2 + q2 = p22rq
=
= 102 + 222 - 152
2 x 10 x 22
= (102 + 222 - 152)  (2 x 10 x 22)
q2 + r2 – p2
2qr
222 + 102 – 152
2x22x10
P
10cm
r
Q
q 22cm
= 0.8159…
2. Remember to use inverse function
to find angle.
angleQPR = cos-1(0.8159..) = 35.3°
15cm p
R
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Markers Comments
Note:
1. To find angle when you have 3
sides use 2nd version of cosine rule:
cosP
r 2 + q2 = p22rq
= 102 + 222 - 152
2 x 10 x 22
When keying in to calculator
work out the top line and the
bottom line before dividing or
use brackets.
= (102 + 222 - 152)  (2 x 10 x 22)
= 0.8159…
2. Remember to use inverse function
to find angle.
angleQPR = cos-1(0.8159..) = 35.3°
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TRIGONOMETRY : Question 3B
The sides of a rhombus are each 15cm while the main
diagonal is 25cm
Find the size of angle EFH and hence find the area of the
rhombus to the nearest square unit.
Get hint
E
15cm
Reveal answer only
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15cm
25cm
H
F
Go to Comments
15cm
15cm
Go to Trigonometry Menu
G
EXIT
TRIGONOMETRY : Question 3B
The sides of a rhombus are each 15cm while the main
diagonal is 25cm
Find the size of angle EFH and hence find the area of the
rhombus to the nearest square unit.
What would you like to do now?
E
Reveal answer only
15cm
1. Identify which trig 15cm
Go to full solution
H
Go to Comments
Go to Trigonometry Menu
EXIT
3. For area of a
rule to use:
triangle
with an
4. Always
Two sides + 25cm
two
angles
angle
present to
use:
2. remember
Substitute
known
= sine
rule
round
your
values,
remembering
to
1
answer
if the
use
brackets
A

abangle
sin
C as
Three
sides
+
one
2
15cm
15cm
questions
appropriate.
= cosine
rule asks
you to.
G
F
TRIGONOMETRY : Question 3B
The sides of a rhombus are each 15cm while the main
diagonal is 25cm
angleEFH = 33.6°
Find the size of angle EFH and hence find the area of the
rhombus to the nearest square unit.
= 415cm2
E
What would you like to do now?
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15cm
15cm
25cm
H
F
Go to Comments
15cm
15cm
Go to Trigonometry Menu
G
EXIT
E
Question 3B
15
15
25
H
1. To find angle when you have 3 sides
use 2nd version of cosine rule:
F
e 2 + h2 - f 2
(a) cosF =
2eh
= 152 + 252 - 152
2 x 25 x 15
15
15
G
(a) Find the size of angle EFH
= (152 + 252 - 152)  (2 x 25 x 15)
= 0.8333…
2. Remember to use inverse function
to find angle.
angleEFH = cos-1(0.8333..) = 33.6°
Continue Solution
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E
Question 3B
15
15
25
H
33.60
15
15
G
(b) Hence find the area
to the nearest square unit
1. Rhombus = 2 identical triangles. For
area of triangles where an angle is
present use:
(b) Area EFH = ½ ehsinF°
F
= 25 x 15 x sin33.6°  2
= 207.52....cm2
2. Remember to double this and
round answer.
Area of kite = 2 x 207.52......
= 415.04..cm2
= 415cm2
Continue Solution
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Markers Comments
1. To find angle when you have 3
sides use 2nd version of cosine rule:
(a) cosF =
e2
h2
+
2eh
f2
Check the formulae list for the
second form of the cosine rule:
2 + c 2 – a2
b
cosA =
2bc
( 3 sides)
= 152 + 252 - 152
2 x 25 x 15
= (152 + 252 - 152)  (2 x 25 x 15)
= 0.8333…
2. Remember to use inverse function
to find angle.
angleEFH = cos-1(0.8333..) = 33.6°
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Markers Comments
Relate to variables used:
1. To find angle when you have 3
sides use 2nd version of cosine rule:
cosF =
e2 + h 2 - f 2
(a) cosF =
2eh
= 0.8333…
2. Remember to use inverse function H
to find angle.
222 + 102 – 152
2x22x10
=
= 152 + 252 - 152
2 x 25 x 15
= (152 + 252 - 152)  (2 x 25 x 15)
e 2 + h2 – f 2
2eh
E
15cm
h
f
15cm
e
25cm
F
angleEFH = cos-1(0.8333..) = 33.6°
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Markers Comments
Check formulae list for:
1. Rhombus = 2 identical triangles.
For area of triangles where an angle
is present use:
(b) Area EFH = ½ ehsinF°
= 25 x 15 x sin33.6°  2
= 207.52....cm2
2. Remember to double this and
round answer.
1
Area of triangle = 2 absinC
(Note: 2 sides and the included
angle)
Relate formula to labels being
used.
E
15cm
h
f
Area of kite = 2 x 207.52......
= 415.04..cm2
= 415cm2
15cm
e
H
25cm
F
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TRIGONOMETRY : Question 4
In triangle TUV, angle U = 35°, angle T = 105°, TV = 5.9cm
and UV = 10cm.
Find the perimeter to one decimal place.
T
Get hint
105°
5.9cm
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35°
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U
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V
10cm
TRIGONOMETRY : Question 4
In triangle TUV, angle U = 35°, angle T = 105°, TV = 5.9cm
and UV = 10cm.
Find the perimeter to one decimal place.
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35°
U
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T
3. Identify which trig
rule105°
to use:
5. Always
Two sides + two angles
5.9cm
1. For perimeter
remember
to
=2.sine
rule
Calculate
need
all three
round
your
unknown
sides.ifSo
answer
themust find
Three sides
+ one angle
angles.
TU.
questions
asks
4. Substitute
known
= cosine rule
you to.
values,
remembering to
use
brackets as
10cm
appropriate.
V
TRIGONOMETRY : Question 4
In triangle TUV, angle U = 35°, angle T = 105°, TV = 5.9cm
and UV = 10cm.
= 22.6cm
Find the perimeter to one decimal place.
T
105°
5.9cm
What would you like to do now?
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35°
U
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V
10cm
T
Question 4
1. Perimeter requires all three sides.
105°
So we need to find TU .
5.9c
m
U
35
°
2. Whether you use Sine or Cosine
rule need angle V.
400
Angle V = 180° - 35° - 105° = 40°
V
10cm
Find the perimeter to one
decimal place.
3. If we use Sine rule:
v
=
sinV°
t
sinT°
4. Substitute known values:
Continue Solution
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v
sin40°
=
10
sin105°
T
Question 4
4. Substitute known values:
105°
v
= 10
sin40°
sin105°
5.9c
m
5. Cross multiply:
U
35
°
400
V
v x sin105°
= 10 x sin40°
10cm
v = 10 x sin40°  sin105°
Find the perimeter to
= 6.654…
one decimal place.
= 6.7cm
6. Answer the question:
Continue Solution
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Perim of  = (6.7 + 10 + 5.9)cm
= 22.6cm
Markers Comments
1. Perimeter requires all three sides.
So we need to find TU .
2. Whether you use Sine or Cosine
rule need angle V.
Since we can pair off two angles
with the opposite sides
Sine Rule
Refer to the Formulae List :
a
b
c
=
=
Sine A
Sine B
Sine C
T
Angle V = 180° - 35° - 105° = 40°
3. If we use Sine rule:
v =
sinV°
105°
t
sinT°
4. Substitute known values:
v
= 10
sin40°
sin105°
5.9cm
v
U
35°
t
10cm
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V
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Markers Comments
4. Substitute known values:
v
= 10
sin40°
sin105°
Go straight to values :
10
v
=
Sine 105˚
Sine 40˚
5. Cross multiply:
v x sin105°
v = 10 x sin40°  sin105°
= 6.654…
= 6.7cm
6. Answer the question:
Perim of  = (6.7 + 10 + 5.9)cm
= 22.6cm
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TRIGONOMETRY : Question 5
Lighthouse C is on a bearing of 050° from lighthouse A
and northwest of lighthouse B. Lighthouse B is 12km
due east of lighthouse A. A ship(S) is sailing directly from
lighthouse B to lighthouse A. How close does it come to
lighthouse C?
Hints
C
N
Answer only
Full solution
Comments
50°
Trig Menu
A
EXIT
12km
B
S
TRIGONOMETRY : Question 5
Lighthouse C is on a bearing of 050° from lighthouse A
and northwest of lighthouse B. Lighthouse B is 12km
due east of lighthouse A. A ship(S) is sailing directly from
lighthouse B to lighthouse A. How close does it come to
lighthouse C?
C
What would you like to do now?
2. Identify which trig
1.4.Calculate
missing
N
rule to
use:
Closest
Answer only
Full solution
Comments
50°
Trig Menu
A
EXIT
angles.
Two sides
distance
= + two angles
(NB.
North= West
= 3150)
sine rule
perpendicular
distance.
3. Substitute known
Three
+ one angle
Create
asides
rightvalues,
remembering
to
= cosine
ruleas
angleduse
triangle
brackets
and use SOH
CAH
appropriate.
TOA.
12km
S
B
TRIGONOMETRY : Question 5
Lighthouse C is on a bearing of 050° from lighthouse A
and northwest of lighthouse B. Lighthouse B is 12km
due east of lighthouse A. A ship(S) is sailing directly from
lighthouse B to lighthouse A. How close does it come to
lighthouse C?
C
Closest distance is 5.48km
What would you like to N
do now?
Full solution
Comments
50°
Trig Menu
A
EXIT
12km
B
S
Question 5
C
1. Calculate missing angles.
NW = 3150
AngleA = 90° - 50° = 40°
N
& angleB = 315° - 270° = 45°
950
So angle C = 180° - 40° - 45° = 95°
A
5
0
°
2. Draw a sketch
400
450
C
b
12
km
B
A
95°
40°
45°
12km
S
3. If we use Sine rule:
Continue Solution
Comments
Trigonometry Menu
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b =
sinB°
c
sinC°
B
Question 5
4. Substitute known values:
C
N
b
= 12
sin45°
sin95°
950
5. Cross multiply:
A
5
0
°
400
b x sin95°
450
= 12 x sin45°
b = 12 x sin45°  sin95°
12
km
B
S
Continue Solution
Comments
Trigonometry Menu
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= 8.5176.…
= 8.52km
Question 5
6. Closest point is perpendicular so
sketch right angled triangle:
C
N
C
950
8.52km
= 8.52km
A
A
5
0
°
400
a
40°
S
450
7. Now using SOHCAH TOA:
12
km
B
sin40°
=
a
8.52
S
a = 8.52 x sin40°
Continue Solution
Comments
Trigonometry Menu
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= 5.476….
= 5.48
Closest distance is 5.48km
Markers Comments
1. Calculate missing angles.
AngleA = 90° - 50° = 40°
NW = 3150
Since we can pair off two angles
with the opposite sides
Sine Rule
& angleB = 315° - 270° = 45°
So angle C = 180° - 40° - 45° = 95°
2. Draw a sketch
Refer to the Formulae List :
a
b
c
Sine A = Sine B = Sine C
C
95°
40
45
° 12km °
A
B
3. If we use Sine rule:
b
sinB°
=
c
sinC°
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Markers Comments
6. Closest point is perpendicular so
sketch right angled triangle:
Note: the shortest distance from
a point to a line is the
perpendicular distance.
C
8.52km
A
C
a
40°
S
A
B
7. Now using SOHCAH TOA:
sin40°
=
a
8.52
a = 8.52 x sin40°
= 5.476….
= 5.48
Closest distance is 5.48km
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TRIGONOMETRY : Question 6
Two supply vessels are approaching an oil platform.
From the deck of ship A the angle of elevation of the drill tower is 2.9°
while from the deck of ship B it is 3.8°.
The ships are 400m apart.
D
How far from the platform is each vessel ?
Hints
Answer only
2.9°
Full solution
3.8°
400m
O
Comments
A
Trig Menu
EXIT
B
TRIGONOMETRY : Question 6
Two supply vessels are approaching an oil platform.
From the deck of ship A the angle of elevation of the drill tower is 2.9°
while from the deck of ship B it is 3.8°.
The ships are 400m apart.
D
How far from the platform is each vessel ?
What would you like to do now?
Answer only
2.9°
Full solution
400m
Comments
A
Trig Menu
EXIT
4. Then use SOH
2. Identify
trig
1.
Calculate
missing
CAH
TOA to which
find
rulefrom
to angles.
use
to
find
length
rig
to
3. Substitute
known
common
side to both
first
ship.
Second
values, remembering
to
triangles:
ship
is then 400
m
3.8°
use
brackets
as
Two further.
sides + two angles
appropriate.
= sine rule
O
Three sides + one angle
= cosine rule
B
TRIGONOMETRY : Question 6
Two supply vessels are approaching an oil platform.
From the deck of ship A the angle of elevation of the drill tower is 2.9°
while from the deck of ship B it is 3.8°.
The ships are 400m apart.
D
How far from the platform is each vessel ?
What would you like to do now?
Try another like this
2.9°
Full solution
3.8°
400m
O
Comments
A
Trig Menu
EXIT
B
ShipA is 1685m away
& ShipB is 1285m away
D
Question 6
1. Calculate missing angles.
Angle B = 180° - 3.8° = 176.2°
Angle D = 180° - 2.9° - 176.2° = 0.9°
0.9°
2. Draw a sketch
2.9
°
400
m
D
3.8
°
0.9°
O
A
2.9° 176.2°
400m
A
B
Continue Solution
Comments
Trigonometry Menu
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B
3. BD is common link to both
triangles so find it using sine rule.
a
=
sinA°
d
sinD°
D
Question 6
4. Substitute known values:
a
= 400
sin2.9°
sin0.9°
0.9°
5. Cross multiply:
2.9
°
400
m
3.8
°
a x sin0.9° = 400 x sin2.9°
a = 400 x sin2.9°  sin0.9°
O
= 1288m
A
B
Continue Solution
Comments
Trigonometry Menu
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to nearest metre
D
Question 6
6. Sketch right angled triangle:
D
1288m
0.9°
3.8°
B
d
2.9
°
400
m
3.8
°
7. Now using SOHCAH TOA:
cos3.8°
1285 m
A
B
Continue Solution
O
O
=
d
1288
d = 1288 x cos3.8°
= 1285m to nearest metre
Try another like this
OA is 1285m+400m = 1685m
Comments
ShipA is 1685m away
& ShipB is 1285m away
Trigonometry Menu
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Markers Comments
Since we can pair off two angles
with the opposite sides
1. Calculate missing angles.
Angle B = 180° - 3.8° = 176.2°
Angle D = 180° - 2.9° - 176.2° = 0.9°
Sine Rule
Refer to the Formulae List :
2. Draw a sketch
D
a
b
c
=
=
Sine A
Sine B
Sine C
0.9°
A
2.9° 176.2°
400m
B
3. BD is common link to both
triangles so find it using sine rule.
a =
sinA°
d
sinD°
Next Comment
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Markers Comments
6. Sketch right angled triangle:
D
Since angle BOD is right - angled
use SOHCAHTOA in triangle BOD
1288m
3.8°
B
d
O
7. Now using SOHCAH TOA:
cos3.8°
=
d
1288
d = 1288 x cos3.8°
= 1285m to nearest metre
OA is 1285m+400m = 1685m
ShipA is 1685m away
& ShipB is 1285m away
Next Comment
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TRIGONOMETRY : Question 6B
Two supply vessels are approaching an oil platform.
From the deck of ship A the angle of elevation of the drill
tower is 27° while from the deck of vessel B it is 35°.
The ships are 80m apart and the height of their decks
is 5m.
D
How high is the drill tower?
Hints
Answer only
27°
Full solution
35°
80m
O
Comments
A
Trig Menu
EXIT
B
TRIGONOMETRY : Question 6B
Two supply vessels are approaching an oil platform.
From the deck of ship A the angle of elevation of the drill
tower is 27° while from the deck of vessel B it is 35°.
The ships are 80m apart and the height of their decks
is 5m.
D
How high is the drill tower?
What would you like to do now?
Answer only
27°
Full solution
80m
Comments
A
Trig Menu
EXIT
4. Then use SOH
2. Identify
trig
1.
Calculate
missing
CAH
TOA to which
find
ruleoftorig
use
to find
angles.
Height
from
3. Substitute
known
common
side to both
deck
level.
values, remembering
to
triangles:
Remember
decks
use
brackets
as
35°
Two
sides
+
two
angles
are
5m
above
sea
appropriate.
= sine rule
level.
O
Three sides + one angle
= cosine rule
B
TRIGONOMETRY : Question 6B
Two supply vessels are approaching an oil platform.
From the deck of ship A the angle of elevation of the drill
tower is 27° while from the deck of vessel B it is 35°.
The ships are 80m apart and the height of their decks
is 5m.
D
How high is the drill tower?
What would you like to do now?
27°
Full solution
35°
80m
O
Comments
A
Trig Menu
EXIT
B
Height of tower is 123m
D
Question 6B
1. Calculate missing angles.
Angle B = 180° - 35° = 145°
Angle D = 180° - 145° - 27° = 8°
8°
2. Draw a sketch
D
27
°
80m
8°
35
°
O
A
B
Continue Solution
Comments
Trigonometry Menu
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A
27°
80m
145°
B
3. BD is common link to both
triangles so find it using sine rule.
a
=
sinA°
d
sinD°
D
Question 6B
4. Substitute known values:
a
= 80
sin27°
sin8°
8°
5. Cross multiply:
a x sin8°
27
°
80m
35
°
a = 80 x sin27°  sin8°
O
A
B
Continue Solution
Comments
Trigonometry Menu
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= 80 x sin27°
= 261m
to nearest metre
D
Question 6B
6. Sketch right angled triangle:
D
261m
8°
b
27°
B
O
7. Now using SOHCAH TOA:
27
°
80m
35
°
sin27°
O
=
b
261
b = 261 x sin27°
A
B
Continue Solution
Comments
Trigonometry Menu
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= 118m to nearest metre
Height = OD + deck height
= 118 + 5 = 123m
Height of tower is 123m
Comments
1. Calculate missing angles.
Since we can pair off two angles
with the opposite sides
Angle B = 180° - 35° = 145°
Angle D = 180° - 145° - 27° = 8°
Sine Rule
Refer to the Formulae List :
2. Draw a sketch
D
a
b
c
=
=
Sine A
Sine B
Sine C
8°
A
27°
80m
145°
B
3. BD is common link to both
triangles so find it using sine rule.
a
=
sinA°
d
sinD°
Next Comment
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Markers Comments
6. Sketch right angled triangle:
D
261m
Since angle BOD is right - angled
use SOHCAHTOA in triangle BOD
b
27°
B
O
7. Now using SOHCAH TOA:
sin27°
=
b
261
b = 261 x sin27°
= 118m to nearest metre
Height = OD + deck height
Next Comment
= 118 + 5 = 123m
Trigonometry Menu
Height of tower is 123m
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INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
You have chosen to study:
UNIT 2 :
Statistics
Please choose a question to attempt from the following:
1
EXIT
2
3
4
Back to
Unit 2 Menu
5
STATISTICS : Question 1
A taxi company was phoned each night of the week and the
response time in minutes of their cars were noted. They were ….
35
10
7
7
15
9
(a) Find the mean and standard deviation for this data.
(b) A similar experiment was conducted with a second
company. The results for this were….
mean = 12
and standard deviation = 1.88
How does the second company compare to the first?
Get hint
Reveal answer only
Go to full solution
EXIT
Go to Comments
Go to Statistics Menu
STATISTICS : Question 1
A taxi company was phoned each night of the week and the
response time in minutes of their cars were noted. They were ….
35
(a)

SD  (b)



EXIT
10
15
9
Draw a7table 7
n
data deviation for this data.
2 comparing
Find the mean
and
standard
mean 
x  x  to mean.
 Then square
A similar experiment
was conducted with a second
n
Thevalues.
company.
results for this were….
n 1

When comparing

data sets= always
mean = 12 and standard deviation
1.88
make comment on
How does the second companythe
compare
to(which
the first?
average
is
bigger etc.) and the
spread of the data.



Reveal answer only
Go to Comments
Go to full solution
Go to Statistics Menu
What would you like to do now?
STATISTICS : Question 1
A taxi company was phoned each night of the week and the
response time in minutes of their cars were noted. They were ….
35
10
7
7
15
9
(a) Find the mean and standard deviation for this data.
(b) A similar experiment was conducted with a second
company. The results for this were….
mean = 12
and standard deviation = 1.88
How does the second company compare to the first?
(a) Mean = 8
SD  3.87
(b)The second company has a longer average response time.
The smaller standard deviation means their arrival time is
more predictable.
Go to Comments
What would you like to do now?
EXIT
Go to full solution
Go to Statistics Menu
Question 1
A taxi company was phoned each
night of the week and the
response time in minutes of their
cars were noted.
They were ….
3
5
10
7
7
15
9
(a) Find the mean and standard
deviation for this data.
Begin Solution
Continue Solution
Comments
Statistics Menu
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1. Calculate mean.
(a)
Mean = (3+5+10+7+7+15+9)7 = 8
So x = 8
and no.pieces of data = n = 7
Question 1
A taxi company was phoned each
night of the week and the
response time in minutes of their
cars were noted.
They were ….
3
5
10
7
7
15
9
(a) Find the mean and standard
deviation for this data.
x 8
Begin Solution
Continue Solution
Comments
Statistics Menu
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2. Draw table comparing data to mean.
x

3
5
10
7
7
15
9
-5
-3
2
-1
-1
7
1
25
9
4
1
1
49
1
 (x - x )2 =
90
xx


xx

2
Question 1
A taxi company was phoned each
night of the week and the
response time in minutes of their
cars were noted.
They were ….
3
5
10
7
7
15
9
(a) Find the mean and standard
deviation for this data.
Begin Solution
Continue Solution
Comments
Statistics Menu
Back to Home
3. Use formula to calculate standard
deviation.


xx


SD 
 n 1


 90 
SD   
 6 
SD  3.87

2





Just found!!
Question 1
(b) A similar experiment was
conducted with a second
company. The results for this
were….
mean = 12 and
standard deviation = 1.88
How does the second company
compare to the first?
x 8
1. Always compare mean and standard
deviation.
(b)The second company has a
longer average response time.
The smaller standard deviation
means their arrival time is more
predictable.
SD  3.87
What would you like to do now?
Begin Solution
Continue Solution
Comments
Statistics Menu
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Comments
3. Use formula to calculate standard
deviation.


xx


SD 
 n 1


 90 
SD   
 6 
SD  3.87

2





Check the list of Formulae for the
Standard Deviation Formula:
 ( x  x)
n 1
2
or
2
(
x
)
 x 2  n
n 1
The second formula can be used
in the calculator paper.
The calculator must be in
Stats. Mode to allow the data
to be entered.
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STATISTICS : Question 2
Weeks of training
A TV personality takes part in a 20 week training schedule to copy a 100m
sprinter. Her times are recorded every 2 weeks and plotted in the graph
then a line of best fit is drawn.
EXIT
Continue question
STATISTICS : Question 2
Get hint
Reveal answer only
Go to full solution
Go to Comments
Go to Stats Menu
Weeks of training
(i) Find the equation of the line in terms of T and W.
(ii) Use answer (i) to predict
(a) her time after 12 weeks of training
(b) the week when her time was 11.5secs
EXIT
STATISTICS : Question 2
Find
gradient
Use
Use
your
your and
note
intercept
equation
equation
and
and of
T axis.
substitute
substitute
WT= =12.
11.5
Reveal answer only
Go to full solution
Go to Comments
Go to Stats Menu
Weeks of training
(i) Find the equation of the line in terms of T and W.
(ii) Use answer (i) to predict
(a) her time after 12 weeks of training
(b) the week when her time was 11.5secs
EXIT
What would you like to do now?
STATISTICS : Question 2
What would you like to do now?
Go to full solution
Go to Comments
Go to Stats Menu
Weeks of training
(i) Find the equation of the line in terms of T and W.
(ii) Use answer (i) to predict
EXIT
T = -¼ W + 15
(a) her time after 12 weeks of training
12secs.
(b) the week when her time was 11.5secs
14 weeks
Question 2
(i) Find the equation of the line
in terms of T and W.
1. Find gradient and note intercept of
T axis.
(a) m = (y2 – y1) 10 - 15
= -¼
=
(x2 – x1)
20 - 0
Intercept at 15
Equation is
Go to graph
Begin Solution
Continue Solution
Comments
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T = -¼ W + 15
Question 2
(ii) Use answer (i) to predict
(a) her time after 12 weeks of
training
(b) the week when her time was
11.5secs
2. Use your equation and substitute
W = 12.
(b)(i) If w = 12 then T = -¼ W + 15
becomes
T = (-¼ x 12) + 15
= -3 + 15
= 12
Time at 12 weeks is 12secs.
Begin Solution
Continue Solution
Comments
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Question 2
(ii) Use answer (i) to predict
(a) her time after 12 weeks of
training
(b) the week when her time was
11.5secs
3. Use your equation and substitute
T = 11.5
(b)(i) If t = 11.5 then
becomes
T = -¼ W + 15
-¼ W + 15 = 11.5
(-15) (-15)
-¼ W = -3.5 x (–4)
W = 14
Reach a time of 11.5sec after 14 weeks.
What would you like to do now?
Begin Solution
Continue Solution
Comments
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Comments
1. Find gradient and note intercept
of T axis.
(a)
m = (y2 – y1) 10 - 15
= -¼
=
(x2 – x1)
20 - 0
To find the equation of a
line from the graph:
Must Learn:
y = mx + c
Intercept at 15
Equation is
T = -¼ W + 15
gradient
intercept
So you need to find these!!
Next Comment
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Comments
1. Find gradient and note intercept
of T axis.
(a)
m = (y2 – y1) 10 - 15
= -¼
=
(x2 – x1)
20 - 0
vertical
m =
horizontal
Note: Always draw the
horizontal before the vertical:
Intercept at 15
horizontal (+ve)
Equation is
T = -¼ W + 15
vertical (-ve)
Next Comment
Statistics Menu
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STATISTICS : Question 3
A sample of 180 teenagers were asked their opinions on the TV series the
“Simpsons” & the movie “Shrek” and their responses were displayed in
the following table…
Like
Don’t like
Simpsons
Like
Shrek
Don’t like
Shrek
Simpsons
126
?
15
21
(a) What percentage liked Shrek but not the Simpsons?
(b) If someone is picked at random what is the probability that
(i) they liked the Simpsons but not Shrek?
(ii) they liked neither?
Get hint
EXIT
Reveal
Full solution
Comments
Statistics Menu
STATISTICS : Question 3
A sample of 180 teenagers were asked their opinions on the TV series the
“Simpsons” & the movie “Shrek” and their responses were displayed in
the following table…
Like
Don’t like
Simpsons
Like
Shrek
Don’t like
Shrek
126
15
Simpsons
Remember
To find probabilities
all entries in
use:
? P = no oftable must
add to total
favourable / no of
sample size
data
21
(a) What percentage liked Shrek but not the Simpsons?
(b) If someone is picked at random what is the probability that
(i) they liked the Simpsons but not Shrek?
(ii) they liked neither?
EXIT
What would you like to do now?
Full solution
Statistics Menu
Comments
Reveal
STATISTICS : Question 3
A sample of 180 teenagers were asked their opinions on the TV series the
“Simpsons” & the movie “Shrek” and their responses were displayed in
the following table…
Like
Don’t like
Simpsons
Like
Shrek
Don’t like
Shrek
Simpsons
126
?
15
21
(a) What percentage liked Shrek but not the Simpsons?
10%
(b) If someone is picked at random what is the probability that
(i) they liked the Simpsons but not Shrek?
(ii) they liked neither?
Full solution
EXIT
Comments
7/
60
Statistics Menu
What now?
1/
12
Question 3
1. Use P = no of favourable / no of data
Like
Simpsons
Like
Shrek
Don’t like
Shrek
Don’t like
Simpsons
126
?
15
21
(a) What percentage liked Shrek
but not the Simpsons?
Begin Solution
Continue Solution
Comments
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NB: There are 180 in survey!!
(a)
Like Shrek but not Simpsons
= 180 – 126 – 15 – 21
= 18
% = 18/180 =
1/
10
= 10%
Question 3
1. Use P = no of favourable / no of data
Like
Simpsons
Like
Shrek
Don’t like
Shrek
Don’t like
Simpsons
126
?
15
21
what is the probability that
(i) they liked the Simpsons but
not Shrek?
Begin Solution
Continue Solution
Comments
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(b)(i) Prob =
15/
180
=
1/
12
Question 3
1. Use P = no of favourable / no of data
Like
Simpsons
Like
Shrek
Don’t like
Shrek
Don’t like
Simpsons
126
?
15
21
(b)(ii) Prob =
21/
180
=
7/
60
What would you like to do now?
(ii) they liked neither?
Begin Solution
Continue Solution
Comments
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Comments
1. Use P = no of favourable / no of data
NB: There are 180 in survey!!
(a)
Like Shrek but not Simpsons
Note: To change a fraction to
a % multiply by 100%
18
180
=
18
180
x 100%
= 180 – 126 – 15 – 21
= 18
% = 18/180 =
1/
10
=
10%
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Comments
To calculate simple probabilities:
1. Use P = no of favourable / no of data
Probability =
(b)(i) Prob =
15/
180
=
1/
12
Number of favourable outcomes
Number of possible outcomes
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STATISTICS : Question 4
Get hint
On a college course you have to pick a language plus a
Reveal ans
leisure activity from the following lists
LANGUAGE
FRENCH
GERMAN
SPANISH
LEISURE
MUSIC
VIDEO
PRODUCTION
BASKETBALL
SWIMMING
Full solution
Comments
Stats Menu
(a) Make a list of all possible combinations of courses.
(b) If a combination is selected at random what is the
probability that it is ……
(i)
Includes Spanish?
(ii) Includes swimming?
EXIT
(iii) Doesn’t include French but includes music?
What would you like to do now?
STATISTICS : Question 4
On a college course you have to pick a language plus a
Reveal ans
leisure activity from the following lists
LANGUAGE
FRENCH
GERMAN
SPANISH
LEISURE
MUSIC
VIDEO
PRODUCTION
BASKETBALL
SWIMMING
Full solution
Comments
Stats Menu
Use a tree
(a) Make a list of all possible combinations of courses. diagram &
“branch out”
(b) If a combination is selected at random what is the with
Noweach
list
language.
probability that it is ……
the pairs
Use your
From list
eachof
of
language
“branch
(i) Includes Spanish?
possible
subjects.
out” with each
combinations
to find
(ii) Includes swimming?
leisure activity.
probabilities.
EXIT
(iii) Doesn’t include French but includes music?
STATISTICS : Question 4
On a college course you have to pick a language plus a
leisure activity from the following lists
LANGUAGE
FRENCH
GERMAN
SPANISH
Full solution
LEISURE
MUSIC
VIDEO
PRODUCTION
BASKETBALL
SWIMMING
(a) Make a list of all possible combinations of courses.
Comments
Stats Menu
CLICK
(b) If a combination is selected at random what is the
probability that it is ……
(i)
Includes Spanish?
(ii) Includes swimming?
EXIT
1/
3
1/
4
1/
(iii) Doesn’t include French but includes music?
6
(a)
French/Music
French/Video
French/Bsktbll
French/Swim
German/Music
GERMAN
German/Video
German/Bsktbll
German/Swim
Spanish/Music
Spanish/Video
Spanish/Bsktbll
Spanish/Swim
Hints
Use a tree diagram &
“branch out” with
each language.
From each language
“branch out” with
each leisure activity.
Now list the pairs of
subjects.
Hints
French/Music
Have list of combinations
handy.
French/Video
French/Bsktbll
(b)(i) Prob = 4/12 =
1/
3
Remember to simplify.
French/Swim
What now?
German/Music
German/Video
Comments
(b)(ii) Prob = 3/12 =
1/
4
Stats Menu
German/Bsktbll
What is probability:
German/Swim
(i)
Includes Spanish?
Spanish/Music
Spanish/Video
Spanish/Bsktbll
Spanish/Swim
(ii) Includes swimming?
(b)(iii) Prob = 2/12 = 1/6
(iii)Doesn’t include French
but includes music?
Comments
(a)
French/Music
French/Video
No. of possible
outcomes
French/Bsktbll
French/Swim
GERMAN
= 3 x 4 = 12
German/Music
To calculate simple
probabilities:
German/Video
Probability =
German/Bsktbll
German/Swim
Number of favourable
Number of possible
Spanish/Music
Spanish/Video
Spanish/Bsktbll
Next Comment
Spanish/Swim
Statistics Menu
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STATISTICS : Question 5
The delivery times for a fast food company are shown in the
following cumulative frequency table.
Time
No. Deliveries
up to 4 mins
up to 8 mins
up to 12 mins
5
13
37
up to 16 mins
up to 20 mins
53
60
Get hint
Reveal ans
Full solution
Comments
Stats Menu
(a)
How many deliveries took longer than 16 mins?
(b)
Use the data to construct a cumulative frequency graph.
(c)
Use the graph to find the median and semi-interquartile
range for this data.
STATISTICS : Question 5
The delivery times for a fast food company are shown in the
following cumulative frequency table.
What would you like to do now?
For how many
greater Time
than 16
Median is
findup
difference
to 4 mins
middle
between end
valueso 8we
upofto
mins
value
values
SIQR
=
½(Q3
– Q1)
want
up
to
16.
up to 12 mins
halfwayQ1 – 25% point &
point
on
up–to
16point
mins
Q3
75%
frequency
up
to 20 mins
axis.
No. Deliveries
5
13
37
53
60
Reveal ans
Full solution
Comments
Stats Menu
(a)
How many deliveries took longer than 16 mins?
(b)
Use the data to construct a cumulative frequency graph.
(c)
Use the graph to find the median and semi-interquartile
range for this data.
STATISTICS : Question 5
The delivery times for a fast food company are shown in the
following cumulative frequency table.
Time
No. Deliveries
up to 4 mins
up to 8 mins
up to 12 mins
5
13
37
up to 16 mins
up to 20 mins
53
60
Full solution
Comments
Stats Menu
=7
(a)
How many deliveries took longer than 16 mins?
(b)
Use the data to construct a cumulative frequency graph.
(c)
Use the graph to find the median and semi-interquartile
range for this data. Median = 11 mins SIQR = 2.75 mins
(a) Deliveries taking longer than 16 mins = 60 – 53 = 7
Time Deliveries
Cum freq
4
5
¾ of 60 = 45
Q3
Q2
8
13
12
37
Comments
½ of 60 = 30
16
53
20
60
Stats Menu
¼ of 60 = 15
What would you like to do now?
Q1
8.5
(C) Median = 11mins.
11
14
Delivery time(mins)
(c) SIQR = ½(Q3 – Q1) = (14 – 8.5)  2 = 2.75mins
Comments
Note:
In a Cumulative Frequency
Diagram:
Cumulative
Frequency
On y-axis (cumulative
frequency)
60
Q1 at 25%,
50%
Q2 at 50%,
Q3 at 75%
11
20
Delivery
Time
And read values from the
delivery time scale (x-axis).
End of Statistics
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INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
You have chosen to study:
UNIT 2 :
Graphs, Charts
& Tables
Please choose a question to attempt from the following:
1
Stem
& Leaf
EXIT
2
Dot
Plot
3
Cum
Freq
Table
4
Dot to
boxplot
Back to
Unit 2 Menu
5
Stem to
boxplot
6
Piechart
GRAPHS, CHARTS, TABLES : Question 1
The following stem & leaf diagram shows the distribution of wages for
employees in a small factory …..
16
2
3
6
9
17
1
1
1
8
8
9
18
2
3
3
5
6
7
19
1
2
8
20
1
5
5
21
8
6
7
n = 25
17 4 = £174
(a) Use this information to find the
(i) median
(ii) lower & upper quartiles
(iii) the semi-interquartile range
(b)What is the probability that someone chosen at random earns
less than £180?
Go to full solution
Get hint
EXIT
Reveal answer
Go to Comments
GRAPHS, CHARTS, TABLES : Question 1
The following stem & leaf diagram shows the distribution of wages for
employees in a small factory …..
16
2
3
6
17
1
1
1
18
2
3
3
19
1
2
8
20
1
5
5
21
8
(a) Use this information to find the
9
Use median Q1 is midpoint
position
8
8= (n+1)9/ from start to
2 to find median
median
5
6
7
7
6
Q3 is midpoint
nfrom
= 25median to
end
17 4 = £174
(i) median
(ii) lower & upper quartiles
What would you like to do now?
(iii) the semi-interquartile range
(b)What is the probability that someone chosen at random earns
less than £180?
Graphs etc Menu
Go to full solution
EXIT
Reveal answer
Go to Comments
GRAPHS, CHARTS, TABLES : Question 1
The following stem & leaf diagram shows the distribution of wages for
employees in a small factory …..
16
2
3
6
9
17
1
1
1
8
8
9
18
2
3
3
5
6
7
19
1
2
8
20
1
5
5
21
8
6
7
n = 25
17 4 = £174
(a) Use this information to find the
What would you like to do now?
median = £183
(ii) lower & upper quartilesQ1 = £171
(i) median
Q3 = £195
(iii) the semi-interquartile range
= £12
(b)What is the probability that someone chosen at random earns
less than £180? = 2/5
Go to full solution
EXIT
Graphs etc Menu
Go to Comments
Question 1
16
17
18
19
20
21
2
1
2
1
1
8
3
1
3
2
5
6
1
3
8
5
1. Use median = (n+1) / 2 to find
median
9
8
5
6
(a)(i) Since n = 25 then the median is
8
6
9
7
7
n = 25
17 4 = £174
(i) Median
(ii) lower & upper quartiles
(iii) the semi-interquartile range
Begin Solution
Continue Solution
Comments
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13th value
ie median = £183
(NOT 3!!!)
2. There are 12 values before median
so Q1 position = 13 - (12 + 1) / 2
(ii) Both 6th & 7th values are £171
so Q1 = £171
3. There are 12 values after median so
Q3 position = 13 + (12 + 1) / 2
19th is £192 & 20th is £198
so Q3 = £195
What would you like to do now?
Question 1
16
17
18
19
20
21
2
1
2
1
1
8
3
1
3
2
5
6
1
3
8
5
4. Use SIQR = ½ (Q3 – Q1 ) / 2
9
8
5
6
8
6
9
7
(iii) SIQR = ½(Q3 – Q1)
= (£195 - £171)  2
7
n = 25
17 4 = £174
(i) Median
(ii) lower & upper quartiles
(iii) the semi-interquartile range
Begin Solution
Continue Solution
Comments
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= £12
Question 1
16
17
18
19
20
21
2
1
2
1
1
8
3
1
3
2
5
6
1
3
8
5
5. Use P = no of favourable / no of data
9
8
5
6
No of favourable ( under £180) = 10
8
6
9
7
7
n = 25
17 4 = £174
(b)What is the probability that
someone chosen at random
earns less than £180?
Begin Solution
Continue Solution
Comments
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No of data = n = 25
(b) Prob(under £180) =
10/
25
= 2/ 5 .
Comments
1. Use median = (n+1) / 2 to find
median
(a)(i) Since n = 25 then the median is
13th value
ie median = £183
2. There are 12 values before median
so Q1 position = 13 - (12 + 1) / 2
(ii) Both 6th & 7th values are £171
so Q1 = £171
Median:
the middle number in the
ordered list.
25 numbers in the list.
1 – 12
13
14 - 25
12 numbers on either side of the
median
median is the
13th number in order.
3. There are 12 values after median
so Q3 = 13 + (12 + 1) / 2
19th is £192 & 20th is £198
so Q3 = £195
Next Comment
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Comments
1. Use median = (n+1) / 2 to find
median
To find the upper and lower
quartiles deal with the numbers
on either side of the median
separately.
(a)(i) Since n = 25 then the median is
13th value
ie median = £183
2. There are 12 values before median
so Q1 position = 13 - (12 + 1) / 2
Q1
12 numbers before median.
6 numbers either side of Q1
is midway between
the 6th and 7th number.
(ii) Both 6th & 7th values are £171
so Q1 = £171
3. There are 12 values after median
so Q3 = 13 + (12 + 1) / 2
19th is £192 & 20th is £198
so Q3 = £195
Next Comment
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Comments
1. Use median = (n+1) / 2 to find
median
To find the upper and lower
quartiles deal with the numbers
on either side of the median
separately.
(a)(i) Since n = 25 then the median is
13th value
ie median = £183
2. There are 12 values before median
so Q1 position = 13 - (12 + 1) / 2
Q3
12 numbers after median.
6 numbers either side of Q3
is midway between
the 19th and 20th number.
(ii) Both 6th & 7th values are £171
so Q1 = £171
3. There are 12 values after median
so Q3 = 13 + (12 + 1) / 2
19th is £192 & 20th is £198
so Q3 = £195
Next Comment
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Charts, Graphs & Tables : Question 2
The weights in grams of 20 bags of crisps were as follows
28 29 29 30 31 30 28 30 29 28
29 30 30 28 28 29 29 29 29 28
a) Illustrate this using a dot plot.
b) What type of distribution does this show?
c) If a bag is chosen at random what is the probability it
will be heavier than the modal weight?
Get hint
EXIT
Reveal answer
Go to full solution
Go to Comments
Charts, Graphs & Tables : Question 2
The weights in grams of 20 bags of crisps were as follows
Establish lowest &
28 29 29 30 31 30 28 30highest
29 28
values and
Plot a dot for
draw
line with use:
scale.
For
probability
each piece of
29 30 30 28 28 29 29 29 29 28
data and label
P = no of favourable
diagram. /
no of data
a) Illustrate this using a dot plot.
b) What type of distribution does this show?
c) If a bag is chosen at random what is the probability it
will be heavier than the modal weight?
What would you like to do now?
EXIT
Graphs etc Menu
Go to full solution
Reveal answer
Go to Comments
Charts, Graphs & Tables : Question 2
The weights in grams of 20 bags of crisps were as follows
28 29 29 30 31 30 28 30 29 28
29 30 30 28 28 29 29 29 29 28
a) Illustrate this using a dot plot.
CLICK
b) What type of distribution does this show?
Tightly clustered
c) If a bag is chosen at random what is the probability it
3/10
will be heavier than the modal weight?
Graphs etc Menu
EXIT
Go to full solution
Go to Comments
Question 2
28
30
29
29
29
28
30
29
29
30
30
29
30
29
28
29
31
28
28
28
1. Establish lowest & highest values
and draw line with scale.
(a) Lowest = 28 & highest = 31.
Weights in g
Illustrate this using a dot plot.
26
Begin Solution
Continue Solution
Comments
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28
30
32
2. Plot a dot for each piece of data and
label diagram.
Question 2
28
30
29
29
29
28
30
29
29
30
30
29
30
29
28
29
31
28
28
28
3. Make sure you know the possible
descriptions of data.
Weights in g
What type of distribution does
this show?
26
28
30
Begin Solution
Continue Solution
Comments
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(b) Tightly clustered distribution.
32
Question 2
28
30
29
29
29
28
30
29
29
30
30
29
4. Use P = no of favourable / no of data
30
29
28
29
31
28
28
28
Mode!
Weights in g
If a bag is chosen at random
what is the probability it will
be heavier than the modal
weight?
26
Begin Solution
28
30
No of favourable ( bigger than 29) = 6
Continue Solution
Comments
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32
No of data = n = 20
(c) Prob(W > mode) =
6/
3/
=
20
10 .
What would you like to do now?
Comments
Other types of distribution:
3. Make sure you know the possible
descriptions of data.
Weights in g
26
28
30
(b) Tightly clustered distribution.
32
Next Comment
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Comments
Other types of distribution:
3. Make sure you know the possible
descriptions of data.
Weights in g
26
28
30
(b) Tightly clustered distribution.
32
Next Comment
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Comments
Other types of distribution:
3. Make sure you know the possible
descriptions of data.
Weights in g
26
28
30
(b) Tightly clustered distribution.
32
Next Comment
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Comments
4. Use P = no of favourable / no of data
Mode!
To calculate simple probabilities:
Probability =
Weights in g
Number of favourable outcomes
Number of possible outcomes
2
2
3
3
6
8
0
2
No of favourable ( bigger than 29) = 6
Next Comment
No of data = n = 20
Menu
(c) Prob(W > mode) =
6/
20
= 3/10 .
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Charts, Graphs & Tables : Question 3
The results for a class test were
18 14 16 17 14 16 13 11 13 13 16 14 13 18 15
10 14 17 13 15 15 18 14 17 13 16 10 14 13 17
(a) Construct a cumulative frequency table for this data.
(b) What is the median for this data?
(c) What is the probability that a pupil selected at random
scored under 14?
Get hint
Graphs etc Menu
EXIT
Reveal answer
Go to full solution
Go to Comments
Charts, Graphs & Tables : Question 3
The results for a class test were
lowest &
18 14 16 17 14 16 13 11 13 13 16 14Establish
13 18
Use15median =
highest values and
(n+1)
/ 2 to
10 14 17 13 15 15 18 14 17 13 16 10 14draw
13
17
Complete
each
table.
For probability
use:
establish
row
1 step in
at a
which
row
time,
P = no of favourable
lies./
(a) Construct a cumulative frequency table formedian
this
data.
calculating
no of data
running total as
(b) What is the median for this data?
you go.
(c) What is the probability that a pupil selected at random
scored under 14?
What would you like to do now?
EXIT
Graphs etc Menu
Go to full solution
Reveal answer
Go to Comments
Charts, Graphs & Tables : Question 3
The results for a class test were
18 14 16 17 14 16 13 11 13 13 16 14 13 18 15
10 14 17 13 15 15 18 14 17 13 16 10 14 13 17
(a) Construct a cumulative frequency table for this data. CLICK
(b) What is the median for this data?
Median = 14
(c) What is the probability that a pupil selected at random
scored under 14?
1/3
Graphs etc Menu
EXIT
Go to full solution
Go to Comments
Question 3
1. Establish lowest & highest values and
draw a table.
18 14 16 17 14 16 13 11 13
(a) Lowest = 10 & highest = 18
13 16 14 13 18 15 10 14 17
Mark
13 15 15 18 14 17 13 16 10
14 13 17
(a) Construct a cumulative
frequency table for this
data.
Begin Solution
Continue Solution
10
11
12
13
14
15
16
17
18
Frequency Cum Frequency
2
1
0
7
6
3
4
4
3
2
3
3
10
16
19
23
27
30
Comments
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2. Complete each row 1 step at a time,
calculating running total as you go.
Question 3
3. Use median = (n+1) / 2 to establish in
which row median lies.
18 14 16 17 14 16 13 11 13
Mark
Frequency Cum Frequency
13 16 14 13 18 15 10 14 17
13 15 15 18 14 17 13 16 10
14 13 17
(b) What is the median for
this data?
What would you like to do now?
Begin Solution
Continue Solution
Comments
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10
11
12
13
14
15
16
17
18
2
1
0
7
6
3
4
4
3
2
3
3
10
16
19
23
27
30
For 30 values median is between
15th & 16th both of which are in row 14.
Median Mark = 14
Question 3
4. Use P = no of favourable / no of data
18 14 16 17 14 16 13 11 13
Mark
Frequency Cum Frequency
13 16 14 13 18 15 10 14 17
13 15 15 18 14 17 13 16 10
14 13 17
(c) What is the probability that
a pupil selected at random
scored under 14?
What would you like to do now?
Begin Solution
Continue Solution
Comments
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10
11
12
13
14
15
16
17
18
2
1
0
7
6
3
4
4
3
2
3
3
10
16
19
23
27
30
No of favourable ( under 14) = 10
No of data = n = 30
(c) Prob(mark<14) =
10/
30
= 1/ 3 .
Comments
Median:
Mark
10
11
12
13
14
15
16
17
18
Freq
2
1
0
7
6
3
4
4
3
Cum Freq
2
3
3
10
16
19
23
27
30
For 30 values median is between
15th & 16th both of which are in row 14.
Median = 14
1 – 15 Q2 16 - 30
Median = 14
Find the mark at which the
cumulative frequency first
reaches between 15th and
16th number.
Next Comment
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Comments
To calculate simple probabilities:
Mark
Freq
Cum Freq
Probability =
10
11
12
13
14
15
16
17
18
2
1
0
7
6
3
4
4
3
2
3
3
10
16
19
23
27
30
Number of favourable outcomes
Number of possible outcomes
No of favourable ( under 14) = 10
No of data = n = 30
(c) Prob(mark<14) =
10/
30
=
1/
3
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.
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Charts, Graphs & Tables : Question 4
The dot plot below shows the number of matches per box in a
sample of 23 boxes.
48
(a) Find the
50
52
54
56
58
(i) median (ii) lower quartile (iii) upper quartile
(b) Construct a boxplot using this data.
(c) In a second sample the semi-interquartile range was 2.5.
How does this distribution compare to the above sample?
Get hint
EXIT
Graphs etc Menu
Reveal answer
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Charts, Graphs & Tables : Question 4
The dot plot below shows the number of matches per box in a
sample of 23 boxes.
median
Q1Use
is midpoint
position
= (n+1)
from start
to /
2 remember
tomedian
find median
bigger SIQR
means more
Q3 is midpoint
variation
from median to
(spread) in
end
48
50
52
54
56
58
data.
(a) Find the
(i) median (ii) lower quartile (iii) upper quartile
(b) Construct a boxplot using this data.
(c) In a second sample the semi-interquartile range was 2.5.
How does this distribution compare to the above sample?
What would you like to do now?
Graphs etc Menu
EXIT
Reveal answer
Go to full solution
Go to Comments
Charts, Graphs & Tables : Question 4
The dot plot below shows the number of matches per box in a
sample of 23 boxes.
Median = 50
So Q1 = 49
So Q3 = 52
48
(a) Find the
50
52
54
56
58
(i) median (ii) lower quartile (iii) upper quartile
(b) Construct a boxplot using this data.
CLICK
(c) In a second sample the semi-interquartile range was 2.5.
How does this distribution compare to the above sample?
the data is distributed more widely than (or not as clustered as)
the above data
EXIT
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Full solution
Comments
1. Use median = (n+1) / 2 to find
median
Question 4
(a) (i) Sample size = 23
so median position is 12.
ie (23+1)2
48
50
52
(a) Find the
54
56
(i) median
(ii) lower quartile
(iii) upper quartile
58
Median = 50
2. There are 11 values before median
so Q1 position = 12 - (11 + 1) / 2
(ii) Middle of 1st 11 is position 6.
So Q1 = 49
Begin Solution
Continue Solution
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3. There are 11 values after median so
Q3 position = 12 + (11 + 1) / 2
(iii) Middle of 2nd 11 is position 18.
So Q3 = 52
Question 4
4. Draw number line with scale.
Make sure you note highest & lowest
as well as Q1, Q2, Q3.
(b)Lowest = 48, Q1 = 49, Q2 = 50,
Q3 = 52 & Highest = 58.
48
50
52
54
56
58
(b) Construct a boxplot using
this data.
Begin Solution
Continue Solution
Comments
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48
50
52
54
56
58
Question 4
5. Calculate SIQR then compare
remember bigger SIQR means more
variation (spread) in data.
(c) For above sample
SIQR = (52 - 49)  2 = 1.5
48
50
52
54
56
58
(c) In a second sample the
semi-interquartile range was
In a sample where the SIQR is 2.5 the
data is distributed more widely than
(or not as clustered as) the above data
2.5. How does this compare?
What would you like to do now?
Begin Solution
Continue Solution
Comments
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Comments
1. Use median = (n+1) / 2 to find
median
The median:
23 numbers in the list:
(a) (i) Sample size = 23
so median position is 12.
1 - 11 12
13 - 23
ie (23+1)2
Median = 50
2. There are 11 values before median
so Q1 position = 12 - (11 + 1) / 2
Q2
11 numbers on either side
of the median
(ii) Middle of 1st 11 is position 6.
So Q1 = 49
3. There are 11 values after median
so Q3 position = 12 + (11 + 1) / 2
(iii) Middle of 2nd 11 is position 18.
So Q3 = 52
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Comments
1. Use median = (n+1) / 2 to find
median
(a) (i) Sample size = 23
For quartiles:
1 - 5
so median position is 12.
2. There are 11 values before median
so Q1 position = 12 - (11 + 1) / 2
(ii) Middle of 1st 11 is position 6.
So Q1 = 49
3. There are 11 values after median
so Q3 position = 12 + (11 + 1) / 2
(iii) Middle of 2nd 11 is position 18.
So Q3 = 52
7
- 11
12
Q2
12
Q2
Q1
ie (23+1)2
Median = 50
6
13 - 17 18
19 - 23
Q3
Now count through the list until you
reach the 6th, 12th,and 18th number
in the list.
Next Comment
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Comments
5. Calculate SIQR then compare
remember bigger SIQR means
more variation (spread) in data.
(c) For above sample
SIQR = (52 - 49)  2 = 1.5
In a sample where the SIQR is 2.5
the data is distributed more widely
than or not as clustered as the
above data
The semi-interquartile range is a
measure of the range of the
“middle” 50%.
S.I.R. = 1 (Q3 - Q1)
2
It is a measure of how
spread-out and so how
“consistent” or “reliable”
the data is.
Remember: when asked to
compare data always consider
average and spread.
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Charts, Graphs & Tables : Question 5
The stem & leaf diagram below shows the weight distribution of 26
people when they joined a slimming club.
Get hint
6
7
8
9
10
11
12
0
1
2
4
5
2
3
2
4
7
5 7 7
2 5 6 6 8 9
6 9 9
7
Reveal answer
Full solution
11 4 = 114kg
1 1 3
Comments
(a) Find the median, lower & upper quartiles for this data.
(b) Use the data to construct a boxplot.
(c) The boxplot below shows the weight distribution for these
people after several months. Compare the two & comment
on the results.
EXIT
60
70
80
90
100
110
120
Charts, Graphs & Tables : Question 5
What now?
The stem & leaf diagram below shows the weight distribution of 26
people when they joined a slimming club.
Menu
median
Q1Use
is midpoint
6 0 2
position
= (n+1)
from start
to /
7 1 3 5 7 7
When
Reveal answer
2 tomedian
find median
8 2 2 2 5 6 6 8 9comparing
position
9 4 4 6 9 9
two data sets
Q3 is midpoint
Full solution
10 5 7 7
comment on
from11median
to
= 114kg
11
spread4 and
end
12 1 1 3
Comments
average
(a) Find the median, lower & upper quartiles for this data.
(b) Use the data to construct a boxplot.
(c) The boxplot below shows the weight distribution for these
people after several months. Compare the two & comment
on the results.
EXIT
60
70
80
90
100
110
120
Charts, Graphs & Tables : Question 5
The stem & leaf diagram below shows the weight distribution of 26
people when they joined a slimming club.
6
7
8
9
10
11
12
0
1
2
4
5
2
3
2
4
7
median = 87
5 7 7
2 5 6 6 8 9
6 9 9
7
Menu
Q1 = 77
Q3 = 99
Full solution
11 4 = 114kg
1 1 3
Comments
(a) Find the median, lower & upper quartiles for this data.
(b) Use the data to construct a boxplot.
CLICK
(c) The boxplot below shows the weight distribution for these
people after several months. Compare the two & comment
on the results.
EXIT
60
70
80
90
100
110
120
CLICK
Question 5
6
7
8
9
10
11
12
0
1
2
4
5
2
3
2
4
7
5 7 7
2 5 6 6 8 9
6 9 9
7
11 4 = 114kg
1 1 3
1. Use median = (n+1) / 2 to find
median
(a)(i) Since n = 26 then the median is
between 13th & 14th value
ie median = 87
2. There are 13 values before median
so Q1 position is 6th value
(a) Find the median, lower &
upper quartiles for this data. (ii)
so
Q1 = 77
3. There are 13 values after median so
Q3 position is 20th position
Begin Solution
Continue Solution
Comments
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so
Q3 = 99
Question 5
6
7
8
9
10
11
12
0
1
2
4
5
2
3
2
4
7
5 7 7
2 5 6 6 8 9
6 9 9
7
11 4 = 114kg
1 1 3
4. Draw number line with scale.
Make sure you note highest & lowest
as well as Q1, Q2, Q3.
(b)Lowest = 60, Q1 = 77, Q2 = 87,
Q3 = 99 & Highest = 123.
(b) Use the data to construct
a boxplot.
Begin Solution
Continue Solution
Comments
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60
70
80
90
100
110
120
5. Compare spread and relevant
average.
Question 5
(c) The boxplot below shows
the weight distribution for
(c) Lightest has put on weight –
these people after several
lowest now 65,
months.
heaviest 3 have lost weight –
Compare the two &
highest now 115,
comment on the results.
median same but overall
spread of weights has decreased
as Q3-Q1 was 22
60
70
80
90
100
110
120
but is now only 15.
Begin Solution
Continue Solution
Comments
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What would you like to do now?
Comments
Remember:
4. Draw number line with scale.
Make sure you note highest & lowest
as well as Q1, Q2, Q3.
(b)Lowest = 60, Q1 = 77, Q2 = 87,
To draw a boxplot you need
a “five-figure summary”:
Box Plot :
Q3 = 99 & Highest = 123.
Lowest
60
70
80
90
100
110
Q1
Q2 Q3 Highest
120
five-figure summary
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Charts, Graphs & Tables : Question 6
The pie chart below shows the breakdown of how a sample of
630 people spent their Saturday nights.
(a) How many people
Watching TV
went clubbing?
144°
(b) If 84 people went to the
cinema
x° theatre
clubbing
theatre then how big is x°?
Get hint
EXIT
Graphs etc Menu
Go to full solution
Reveal answer
Go to Comments
Charts, Graphs & Tables : Question 6
The pie chart below showsangle
the breakdown
of how a sample of
amount
=
630 people spent their Saturday
360° nights. 630
(a) How many people
Watching TV
went clubbing?
144°
(b) If 84 people went to the
cinema
x° theatre
clubbing
theatre then how big is x°?
What would you like to do now?
EXIT
Graphs etc Menu
Go to full solution
Reveal answer
Go to Comments
Charts, Graphs & Tables : Question 6
The pie chart below shows the breakdown of how a sample of
630 people spent their Saturday nights.
(a) How many people
Watching TV
cinema
= 252
went clubbing?
144°
(b) If 84 people went to the
x° theatre
clubbing
theatre then how big is x°?
= 48°
What would you like to do now?
Graphs etc Menu
EXIT
Go to full solution
Go to Comments
Question 6
Watching
TV
144°
1. Set up ratio of angles and sectors
and cross multiply.
cinema
x° theatre
clubbin
g
How many people went
clubbing?
Begin Solution
Continue Solution
Comments
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(a) The angle is 144° so …..
angle = amount
360°
630
144° = amount
360°
630
360 x amount = 144 x 630
amount = 144 x 630  360
= 252
Question 6
Watching
TV
144°
2. Set up ratio of angles and sectors
and cross multiply.
cinema
x° theatre
clubbin
g
(b) If 84 people went to the
theatre then how big is x°?
Begin Solution
Continue Solution
Comments
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(b) The amount is 84 so …..
angle = amount
360°
630
angle =
84
360°
630
630 x angle = 360° x 84
angle = 360° x 84  630
= 48°
Comments
1. Set up ratio of angles and sectors
and cross multiply.
(a) The angle is 144° so …..
angle = amount
360°
630
144° = amount
360°
630
360 x amount = 144 x 630
Can also be tackled by using
proportion:
amount 
angle at centre
 total sample
360
Amount =
144
360
x 630
amount = 144 x 630  360
= 252
Next Comment
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Comments
2. Set up ratio of angles and sectors
and cross multiply.
(b) The amount is 84 so …..
angle = amount
360°
630
angle =
84
360°
630
Can also be tackled by using
proportion:
amount 
angle at centre
 total sample
360
84 =
x
360
x 630
630 x = 84 x 360
630 x angle = 360° x 84
angle = 360° x 84  630
= 48°
x = 84 x 360
630
End of graphs, charts etc.
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INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
You have chosen to study:
UNIT 2 :
Simultaneous
Equations
Please choose a question to attempt from the following:
1
EXIT
2
3
Back to
Unit 2 Menu
4
Simultaneous Equations : Question 1
The diagram shows the graph of
3x + 2y = 17.
Copy the diagram and on your
diagram draw the graph of
y = 2x – 2 ,
hence solve
3x + 2y = 17
y = 2x – 2
3x + 2y = 17.
EXIT
Reveal answer only
Go to Comments
Go to full solution
Go to Sim Eq Menu
Get hint
Simultaneous Equations : Question 1
The diagram shows the graph of
3x + 2y = 17.
Copy the diagram and on your
diagram draw the graph of
To draw
graph:
Construct
a join
Plot and
table ofSolution
points.
values with at
is where lines
least 2 xcross
y = 2x – 2 ,
hence solve
3x + 2y = 17
y = 2x – 2
coordinates.
3x + 2y
= 17.would you like to do now?
What
EXIT
Reveal answer only
Go to Comments
Go to full solution
Go to Sim Eq Menu
Simultaneous Equations : Question 1
The diagram shows the graph of
3x + 2y = 17.
Copy the diagram and on your
diagram draw the graph of
y = 2x – 2 ,
hence solve
3x + 2y = 17
y = 2x – 2
Solution is
x=3 & y=4
3x + What
2y = 17.
would you like to do now?
Try another like this
EXIT
Go to full solution
Go to Comments
Go to Sim Eq Menu
Question 1
The diagram shows the graph
of 3x + 2y = 17.
y = 2x - 2
Copy the diagram and on your
diagram draw the graph of
y = 2x – 2 ,
hence solve
3x + 2y = 17
y = 2x – 2
Begin Solution
Continue Solution
Comments
Sim Eq Menu
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1. Construct a table of values with at
least 2 x coordinates.
x
y
0
-2
5
8
Question 1
The diagram shows the graph
2. Plot and join points. Solution is
where lines cross.
of 3x + 2y = 17.
Copy the diagram and on your
diagram draw the graph of
y = 2x – 2 ,
hence solve
3x + 2y = 17
y = 2x – 2
Begin Solution
Try another like this
Comments
Solution is
x=3 & y=4
Sim Eq Menu
What would you like to do now?
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Markers Comments
1. Construct a table of values with
at least 2 x coordinates.
y = 2x - 2
x
y
0
-2
5
8
There are two ways of drawing
the line y = 2x - 1
Method 1
Finding two points on the line:
x=0
y
= 2 x 0 – 1 = -1
x=2
y
=2x2-1 = 3
First Point (0,-1)
Second Point (-1,3)
Plot and join (0,-1), and (-1,3).
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Markers Comments
2. Plot and join points. Solution is
where lines cross.
Method 2
Using y = mx + c form:
y = mx + c
gradient
y - intercept
y = 2x - 1
gradient
m=2
y - intercept
c = -1
Plot C(0, -1) and draw line
with m = 2
Next Comment
Solution is
x=3 & y=4
Sim Eqs Menu
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Simultaneous Equations : Question 1B
The diagram shows the graph of
2x + y = 8.
Copy the diagram and on your
diagram draw the graph of
y = 1/2x + 3 ,
hence solve
2x + y = 8
y = 1/2x + 3
EXIT
Reveal answer only
Go to Comments
Go to full solution
Go to Sim Eq Menu
Get hint
Simultaneous Equations : Question 1B
The diagram shows the graph of
2x + y = 8.
Copy the diagram and on your
To draw
graph:
Construct
a join
Plot and
table ofSolution
points.
values with at
is where lines
least 2 xcross
coordinates.
diagram draw the graph of
y = 1/2x + 3 ,
hence solve
2x + y = 8
y = 1/2x + 3
What would you like to do now?
EXIT
Reveal answer only
Go to Comments
Go to full solution
Go to Sim Eq Menu
Simultaneous Equations : Question 1B
The diagram shows the graph of
2x + y = 8.
Copy the diagram and on your
diagram draw the graph of
y = 1/2x + 3 ,
hence solve
2x + y = 8
y = 1/2x + 3
Solution is
x=2 & y=4
What would you like to do now?
Go to Comments
EXIT
Go to full solution
Go to Sim Eq Menu
Question 1B
1. Construct a table of values with at
The diagram shows the graph of
least 2 x coordinates.
2x + y = 8.
y = 1/2x + 3
Copy the diagram and on your
diagram draw the graph of
y = 1/2x + 3 ,
hence solve
2x + y = 8
y = 1/2x + 3
Begin Solution
Continue Solution
Comments
Sim Eq Menu
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x
y
0
3
6
6
Question 1B
The diagram shows the graph
of 2x + y = 8.
2. Plot and join points. Solution is
where lines cross.
y = 1/2x + 3
Copy the diagram and on your
x
y
0
3
6
6
diagram draw the graph of
y = 1/2x + 3 ,
What would you like to do now?
hence solve
2x + y = 8
y = 1/2x + 3
Begin Solution
Continue Solution
Comments
Sim Eq Menu
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Solution is
x=2 & y=4
Markers Comments
1. Construct a table of values with at
least 2 x coordinates.
y = 1/2x + 3
x
y
0
3
6
6
There are two ways of drawing
the line y = ½ x + 3
Method 1
Finding two points on the line:
x=0
y
=½ x0+3= 3
x=2
y
= ½ x 2 +3 = 4
First Point
(0, 3)
Second Point (2, 4)
Plot and join (0, 3), and (2, 4).
Next Comment
Sim Eqs Menu
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Markers Comments
2. Plot and join points. Solution is
where lines cross.
y=
1/ x
2
+3
x
y
0
3
6
6
Method 2
Using y = mx + c form:
y = mx + c
gradient
y - intercept
y=½x+3
gradient
m=½
y - intercept
c = +3
Plot C(0, 3) and draw line
with m = ½
Next Comment
Sim Eqs Menu
Solution is
x=2 & y=4
Back to Home
Simultaneous Equations : Question 2
Solve
3u - 2v = 4
2u + 5v = 9
Get hint
Reveal answer only
Go to full solution
Go to Comments
Go to Sim Eq Menu
EXIT
Simultaneous Equations : Question 2
Solve
3u - 2v = 4
2u + 5v = 9
What would you like to do now?
Reveal answer only
Go to full solution
Go to Comments
Go to Sim Eq Menu
EXIT
Eliminate
either
variable by
making
coefficient
same.
Substitute
found value
into either of
original
equations.
Simultaneous Equations : Question 2
Solve
3u - 2v = 4
2u + 5v = 9
Solution is
u=2 & v=1
What would you like to do now?
Try another like this
Go to full solution
Go to Comments
Go to Sim Eq Menu
EXIT
Question 2
Solve
3u - 2v = 4
2u + 5v = 9
1. Eliminate either u’s or v’s by making
coefficient same.
3u - 2v = 4
2u + 5v = 9
(x5)
(x2)
1
2
Now get:
15u  10v  20

4u  10v  18 
Begin Solution
Continue Solution
Comments
Sim Eq Menu
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19u
 38
u
 2
3
=
1
(x5)
4
=
2
(x2)
3
+
4
Question 2
Solve
3u - 2v = 4
2u + 5v = 9
2. Substitute found value into either of
original equations.
3u - 2v = 4
2u + 5v = 9
(x5)
(x2)
1
2
Substitute 2 for u in equation
2
4 + 5v = 9
5v = 5
v=1
Begin Solution
Try another like this
Comments
Sim Eq Menu
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Solution is u = 2 & v = 1
What would you like to do now?
Markers Comments
Note:
1. Eliminate either u’s or v’s by
making coefficient same.
3u - 2v = 4
2u + 5v = 9
(x5)
(x2)
When the “signs” are the
same subtract to eliminate.
1
2
When the “signs” are
different add to eliminate.
Now get:
15u  10v  20

4u  10v  18 
19u
 38
u
 2
3
=
1
(x5)
4
=
2
(x2)
e.g.
2x + 3y = 4
6x + 3y = 4
Subtract the equations
3
+
4
Next Comment
Sim Eqs Menu
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Markers Comments
Note:
1. Eliminate either u’s or v’s by
making coefficient same.
3u - 2v = 4
2u + 5v = 9
(x5)
(x2)
When the “signs” are the
same subtract to eliminate.
1
2
When the “signs” are
different add to eliminate.
Now get:
15u  10v  20

4u  10v  18 
19u
 38
u
 2
3
=
1
(x5)
4
=
2
(x2)
e.g.
2x + 3y = 4
6x - 3y = 4
Add the equations
3
+
4
Next Comment
Sim Eqs Menu
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Simultaneous Equations : Question 2B
Solve
5p + 3q = 0
4p + 5q = -2.6
Get hint
Reveal answer only
Go to full solution
Go to Comments
Go to Sim Eq Menu
EXIT
Simultaneous Equations : Question 2B
Solve
5p + 3q = 0
4p + 5q = -2.6
What would you like to do now?
Reveal answer only
Go to full solution
Go to Comments
Go to Sim Eq Menu
EXIT
Eliminate
either
variable by
making
coefficient
same.
Substitute
found value
into either of
original
equations.
Simultaneous Equations : Question 2B
Solve
5p + 3q = 0
4p + 5q = -2.6
Solution is q = -1 & q = 0.6
What would you like to do now?
Go to full solution
Go to Comments
Go to Sim Eq Menu
EXIT
Question 2B
Solve
5p + 3q = 0
4p + 5q = -2.6
1. Eliminate either p’s or q’s by making
coefficient same.
(x4)
5p + 3q = 0
4p + 5q = -2.6 (x5)
1
2
Now get:
20 p  12q  0 

20 p  25q  13
13q  13
Begin Solution
Continue Solution
Comments
Sim Eq Menu
Back to Home
q  1
3
=
1
(x4)
4
=
2
(x5)
4
-
3
Question 2B
Solve
5p + 3q = 0
4p + 5q = -2.6
2. Substitute found value into either of
original equations.
(x4)
5p + 3q = 0
4p + 5q = -2.6 (x5)
1
2
Substitute -1 for q in equation 1
5p + (- 3) = 0
5p = 3
p =3/5 = 0.6
Begin Solution
Continue Solution
Comments
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Solution is q = -1 & p = 0.6
What would you like to do now?
Markers Comments
1. Eliminate either p’s or q’s by
making coefficient same.
(x4)
5p + 3q = 0
4p + 5q = -2.6 (x5)
13q  13
q  1
When the “signs” are the
same subtract to eliminate.
1
2
When the “signs” are
different add to eliminate.
Now get:
20 p  12q  0 

20 p  25q  13
Note:
3
=
1
(x4)
4
=
2
(x5)
e.g.
2x + 3y = 4
6x + 3y = 4
Subtract the equations
4
-
3
Next Comment
Sim Eqs Menu
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Markers Comments
1. Eliminate either p’s or q’s by
making coefficient same.
(x4)
5p + 3q = 0
4p + 5q = -2.6 (x5)
13q  13
q  1
When the “signs” are the
same subtract to eliminate.
1
2
When the “signs” are
different add to eliminate.
Now get:
20 p  12q  0 

20 p  25q  13
Note:
3
=
1
(x4)
4
=
2
(x5)
4
-
3
e.g.
2x + 3y = 4
6x - 3y = 4
Add the equations
Next Comment
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Simultaneous Equations : Question 3
If two coffees & three doughnuts cost £2.90
while three coffees & one doughnut cost £2.60
then find the cost of two coffees & five doughnuts.
Get hint
Reveal answer only
Go to full solution
Go to Comments
Go to Sim Eq Menu
EXIT
Simultaneous Equations : Question 3
If two coffees & three doughnuts cost £2.90
while three coffees & one doughnut cost £2.60
then find the cost of two coffees & five doughnuts.
What would you like to do now?
Reveal answer only
Go to full solution
Go to Comments
Go to Sim Eq Menu
EXIT
Form two
Eliminate
equations,
either
c’s orcosts
keeping
Substitute
d’s
by
Remember
to to
inmaking
pence
found
value
coefficient
answer
the of
avoid
decimals.
into
either
same.
question!!!
original
equations.
Simultaneous Equations : Question 3
If two coffees & three doughnuts cost £2.90
while three coffees & one doughnut cost £2.60
then find the cost of two coffees & five doughnuts.
Two coffees & five doughnuts
= £3.90
Try another like this
Go to full solution
Go to Comments
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EXIT
Question 3
If two coffees &
1. Form two equations, keeping costs
in pence to avoid decimals.
three doughnuts cost £2.90
Let coffees cost c pence
while three coffees
& doughnuts d pence then we have
& one doughnut cost £2.60
then find the cost of
two coffees & five doughnuts.
Begin Solution
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2c + 3d = 290
3c + 1d = 260
(x1)
(x3)
1
2
2. Eliminate either c’s or d’s by making
coefficient same.
2c  3d  290 

9c  3d  780 
7c  490
c  70
3
=
1
(x1)
4
=
2
(x3)
4
-
3
Question 3
If two coffees &
3. Substitute found value into either of
original equations.
three doughnuts cost £2.90
while three coffees
& one doughnut cost £2.60
then find the cost of
two coffees & five doughnuts.
Let coffees cost c pence
& doughnuts d pence then we have
2c + 3d = 290
3c + 1d = 260
(x1)
(x3)
Substitute 70 for c in equation
What would you like to do now?
1
2
2
210 + d = 260
d = 50
Comments
Two coffees & five doughnuts
= (2 x 70p) + (5 x 50p)
= £1.40 + £2.50
Sim Eq Menu
= £3.90
Begin Solution
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Comments
1. Form two equations, keeping
costs in pence to avoid decimals.
Let coffees cost c pence
& doughnuts d pence then we have
2c + 3d = 290
3c + 1d = 260
(x1)
(x3)
1
2
i.e.
2. Eliminate either c’s or d’s by
making coefficient same.
2c  3d  290 

9c  3d  780 
7c  490
c  70
Step 1
Form the two simultaneous
equations first by introducing a
letter to represent the cost of a
coffee ( c) and a different letter
to represent the cost of a
doughnut (d).
3
=
1
(x1)
4
=
2
(x3)
4
-
3
2c + 3d
1d + 2c
=
=
290
260
Note change to
pence eases working.
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Comments
1. Form two equations, keeping
costs in pence to avoid decimals.
Let coffees cost c pence
& doughnuts d pence then we have
2c + 3d = 290
3c + 1d = 260
(x1)
(x3)
Step 2
Solve by elimination.
Choose whichever variable it is
easier to make have the same
coefficient in both equations.
1
2
2. Eliminate either c’s or d’s by
making coefficient same.
2c  3d  290 

9c  3d  780 
7c  490
c  70
3
=
1
(x1)
4
=
2
(x3)
4
-
3
Next Comment
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Comments
3. Substitute found value into either
of original equations.
Let coffees cost c pence
& doughnuts d pence then we have
2c + 3d = 290
3c + 1d = 260
Step 3
Once you have a value for
one variable you can substitute
this value into any of the
equations to find the value
of the other variable.
It is usually best to choose an
equation that you were given
in question.
Substitute 70 for c in equation
210 + d = 260
d = 50
Two coffees & five doughnuts
= (2 x 70p) + (5 x 50p)
= £1.40 + £2.50
Step 4
Remember to answer the
question!!!
Next Comment
Sim Eqs Menu
= £3.90
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Simultaneous Equations : Question 3B
A company plan to introduce a new blend of tropical fruit drink
made from bananas & kiwis.
They produce three blends of the drink for market research
purposes.
Blend 1 uses 70% banana syrup, 30% kiwi syrup
and the cost per litre is 74p.
Blend 2 has 55% banana syrup, 45% kiwi syrup
and costs 71p per litre to produce.
A third blend is made using 75% banana syrup and 25% kiwi.
How does its cost compare to the other two blends?
Get hint
EXIT
Reveal answer
Go to Comments
Go to full solution
Go to Sim Eq Menu
Simultaneous Equations : Question 3B
A company plan to introduce a new blend of tropical fruit drink
made from bananas & kiwis.
They produce three blends of the drink for market research
purposes.
Blend 1 uses 70% banana syrup, 30% kiwi syrup
and the cost per litre is 74p.
Blend 2 has 55% banana syrup, 45% kiwiEliminate
syrup
Form two
either c’s or
and costs 71p per litre to produce.
Substitute
equations,
d’s
by making
Remember
to
found
value
eliminating
coefficient
answer
the of
into
either
decimals
A third blend is made using 75% banana syrupquestion!!!
and
25% kiwi.
same.
original
wherever
How does its cost compare to the other two blends?
equations.
possible.
What would you like to do now?
EXIT
Reveal answer
Go to Comments
Go to full solution
Go to Sim Eq Menu
Simultaneous Equations : Question 3B
A company plan to introduce a new blend of tropical fruit drink
made from bananas & kiwis.
They produce three blends of the drink for market research
purposes.
Blend 1 uses 70% banana syrup, 30% kiwi syrup
and the cost per litre is 74p.
Blend 2 has 55% banana syrup, 45% kiwi syrup
and costs 71p per litre to produce.
A third blend is made using 75% banana syrup and 25% kiwi.
How does its cost compare to the other two blends?
So this blend is more expensive
than the other two.
EXIT
Go to full solution
Go to Comments
Go to Sim Eq Menu
Question 3B
Blend 1 uses 70% banana syrup,
30% kiwi syrup and the cost per
litre is 74p.
Blend 2 has 55% banana syrup,
45% kiwi syrup and costs 71p
per litre to produce.
A third blend is made using 75%
banana syrup and 25% kiwi. How
does its cost compare to the
other two blends?
Begin Solution
Continue Solution
Comments
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1. Form two equations, keeping costs
in pence to avoid decimals.
Let one litre of banana syrup cost B
pence & one litre of kiwi syrup cost K
pence.
0.70B + 0.30K = 74 (x10)
0.55B + 0.45K = 71 (x100)
1
2
2. Get rid of decimals
7 B  3K  740 

55B  45K  7100
3
=
1
(x10)
4
=
2
(x100)
Question 3B
Blend 1 uses 70% banana syrup,
30% kiwi syrup and the cost per
litre is 74p.
Blend 2 has 55% banana syrup,
45% kiwi syrup and costs 71p
per litre to produce.
A third blend is made using 75%
banana syrup and 25% kiwi. How
does its cost compare to the
other two blends?
Begin Solution
Continue Solution
Comments
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3. Eliminate either B’s or K’s by
making coefficient same.
7 B  3K  740 

55B  45K  7100
105B  45K  11100

55B  45K  7100 
50B  4000
B  80
(x15)
3
4
5
=
(x15)
3
4
5
-
4
Question 3B
Blend 1 uses 70% banana syrup,
30% kiwi syrup and the cost per
litre is 74p.
Blend 2 has 55% banana syrup,
45% kiwi syrup and costs 71p
per litre to produce.
4. Substitute found value into an
equation without decimals.
7 B  3K  740 

55B  45K  7100
3
(x15)
4
Substitute 80 for B in equation
3
560 + 3K = 740
A third blend is made using 75%
banana syrup and 25% kiwi. How
does its cost compare to the
other two blends?
What would you like to do now?
Begin Solution
Continue Solution
Comments
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3K = 180
K = 60
B  80
5. Use these values to answer question.
75%B+25%K = (0.75 x 80p)+(0.25 x 60p)
= 60p + 15p
= 75p
So this blend is more expensive than
the other two.
Markers Comments
1. Form two equations, keeping costs
in pence to avoid decimals.
Let one litre of banana syrup cost B
pence & one litre of kiwi syrup cost K
pence.
0.70B + 0.30K = 74 (x10)
0.55B + 0.45K = 71 (x100)
1
2
2. Get rid of decimals
7 B  3K  740 

55B  45K  7100 
3
=
1
4
=
2
Step 1
Form the two simultaneous
equations first by introducing a
letter to represent the cost per litre
of banana syrup ( B) and a
different letter to represent the
cost per litre of kiwis fruit (K).
i.e.
0.70 B + 0.30K =
0.55B + 0.45K =
74
71
(x10)
Multiply all terms Note change to
(x100) by 100 to remove
pence eases
decimals.
working.
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Markers Comments
Step 2
Solve by elimination.
3. Eliminate either B’s or K’s by
making coefficient same.
7 B  3K  740 

55B  45K  7100
105B  45K  11100

55B  45K  7100 
50B  4000
Choose whichever variable it is
easier to make have the same
coefficient in both equations.
(x15)
3
4
5
=
(x15)
3
4
5
-
4
B  80
Next Comment
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Markers Comments
Step 3
Once you have a value for
one variable you can substitute
this value into any of the
equations to find the value
of the other variable.
4. Substitute found value into an
equation without decimals.
7 B  3K  740 

55B  45K  7100 
3
(x15)
4
Substitute 80 for B in equation
3
560 + 3K = 740
3K = 180
K = 60
B  80
5. Use these values to answer question.
It is usually best to choose an
equation that you were given
in question.
Step 4
Remember to answer the
question!!!
75%B+25%K = (0.75 x 80p)+(0.25 x 60p)
= 60p + 15p
= 75p
So this blend is more expensive than
the other two.
Next Comment
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Simultaneous Equations : Question 4
A Fibonacci Sequence is formed as follows…….
Start with two terms
add first two to obtain 3rd
add 2nd & 3rd to obtain 4th
add 3rd & 4th to obtain 5th
etc
eg starting with 3 & 7, next four terms are 10, 17, 27 & 44.
(a) If the first two terms of such a sequence are P and Q then find
expressions for the next 4 terms in their simplest form.
(b) If the 5th & 6th terms are 8 and 11 respectively then write down
two equations in P and Q and hence find the values of P and Q.
Get hint
EXIT
Reveal answer
Go to Comments
Go to full solution
Go to Sim Eq Menu
Simultaneous Equations : Question 4
A Fibonacci Sequence is formed as follows…….
Start with two terms
Write down
Solveexpressions
and
two
To find Establish
values
ofusing
P
&found
Qprevious
:
rd
substitute
add firstMatch
two
to
obtain
3
equations.
term fromEliminate
(a) with
value into
either
two
to form
values
given
in
question.
P’s or Q’s
by
rd to obtain
add 2nd & 3either
4th next.
of original
making coefficient
equations.
rd
th
same.5th
add 3 & 4 to obtain
etc
eg starting with 3 & 7, next four terms are 10, 17, 27 & 44.
(a) If the first two terms of such a sequence are P and Q then find
expressions for the next 4 terms in their simplest form.
(b) If the 5th & 6th terms are 8 and 11 respectively then write down
two equations in P and Q and hence find the values of P and Q.
What would you like to do now?
EXIT
Reveal answer
Go to Comments
Go to full solution
Go to Sim Eq Menu
Simultaneous Equations : Question 4
A Fibonacci Sequence is formed as follows…….
Start with two terms
add first two to obtain 3rd
add 2nd & 3rd to obtain 4th
add 3rd & 4th to obtain 5th
etc
eg starting with 3 & 7, next four terms are 10, 17, 27 & 44.
(a) If the first two terms of such a sequence are P and Q then find
expressions for the next 4 terms in their simplest form.
P + Q,
P + 2Q
2P + 3Q
3P + 5Q
P=7
Q  2
(b) If the 5th & 6th terms are 8 and 11 respectively then write down
two equations in P and Q and hence find the values of P and Q.
Try another like this
EXIT
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Question 4
(a) If the first two terms of such a
sequence are P and Q then find
1. Write down expressions using
previous two to form next.
(a)
First term = P
& second term = Q
expressions for the next 4 terms
3rd term = P + Q
in their simplest form.
4th term = Q + (P + Q) = P + 2Q
5th term = (P + Q) + (P + 2Q)
= 2P + 3Q
6th term = (P + 2Q) + (2P + 3Q)
= 3P + 5Q
Begin Solution
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Question 4
(b) If the 5th & 6th terms are 8
and 11 respectively then write
down two equations in P and Q
and hence find the values of
P and Q.
1. Match term from (a) with values
given in question.
2P  3Q  8

3P  5Q  11
Q  2
Try another like this
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2
(x3)
(x2)
2. Eliminate either P’s or Q’s by
making coefficient same.
6 P  9Q  24 

6 P  10Q  22
Begin Solution
1
3
=
1
(x3)
4
=
2
(x2)
4
-
3
Question 4
(b) If the 5th & 6th terms are 8
and 11 respectively then write
down two equations in P and Q
and hence find the values of
P and Q.
3. Substitute found value into either of
original equations.
2P  3Q  8

3P  5Q  11
1
2
(x3)
(x2)
Substitute -2 for Q in equation
2
2P + (-6) = 8
2P = 14
P=7
Begin Solution
First two terms are 7 and –2
respectively.
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What would you like to do now?
Comments
1. Write down expressions using
previous two to form next.
(a)
First term = P
& second term = Q
3rd term = P + Q
For problems in context it is
often useful to do a simple
numerical example before
attempting the algebraic
problem.
Fibonacci Sequence:
4th term = Q + (P + Q) = P + 2Q
3, 7, 10, 17, 27, 44, ……
5th term = (P + Q) + (P + 2Q)
P
= 2P + 3Q
6th term = (P + 2Q) + (2P + 3Q)
= 3P + 5Q
Q P + Q P + 2Q
4, 6, 10, 16, 26, 42, ……
Then introduce the variables:
P, Q,
P + Q, P + 2Q, 2P + 3Q, …
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Simultaneous Equations : Question 4B
In a “Number Pyramid” the value in each block is the
sum of the values in the two blocks underneath …..
1
7
8
5
-6
-1
3
-5
-4
-1
The two number pyramids below have the middle two
rows missing. Find the values of v and w.
(B)
-18
11
(A)
3v
2w
w – 2v v + w
v + w v – 3w
6w
v-w
Get hint
EXIT
Reveal answer
Go to Comments
Go to full solution
Go to Sim Eq Menu
Simultaneous Equations : Question 4B
In a “Number Pyramid” the value in each block is the
Form 2 underneath
equations …..
sum of the values in the two blocks
8
5
andSubstitute
eliminate found
either
Work
your way to
1
v’svalue
or w’sinto
by making
either of
ansame.
expression for
7
-6
coefficient
original equations.
the top row by
-1
-5
filling in the
3
-4
-1
middle rows.
The two number pyramids below have the middle two
rows missing. Find the values of v and w.
-18
(A)
3v
EXIT
2w
w – 2v v + w
(B)
11
v + w v – 3w
6w
v-w
Reveal answer
Go to Comments
Go to full solution
Go to Sim Eq Menu
Simultaneous Equations : Question 4B
In a “Number Pyramid” the value in each block is the
sum of the values in the two blocks underneath …..
1
7
8
5
-6
-1
3
-5
-4
-1
The two number pyramids below have the middle two
rows missing. Find the values of v and w.
-18
(A)
3v
W  1
EXIT
2w
w – 2v v + w
V=4
Go to full solution
(B)
11
v + w v – 3w
6w
v-w
Go to Comments
Go to Sim Eq Menu
Question 4B
1. Work your way to an expression for
the top row by filling in the middle
rows.
-18
(A)
3v
w – 2v v + w
2w
Pyramid (A)
2nd row 3v + 2w, -2v + 3w, -v + 2w
(B)
3rd row v + 5w, -3v + 5w
11
Top row
v + w v – 3w
6w
-2v + 10w
= - 18
v-w
Pyramid (B)
2nd row 2v - 2w, v + 3w, v + 5w
Begin Solution
Continue Solution
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3rd row 3v + w, 2v + 8w
Top row
5v + 9w
= 11
2. Form 2 equations and eliminate
either v’s or w’s by making
coefficient same.
Question 4B
-18
(A)
3v
w – 2v v + w
2w
(B)
6w
v-w
10V  50W  90 

10V  18W  22 
68W  68
Begin Solution
Continue Solution
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1
(x5)
(x2)
2
Now get:
11
v + w v – 3w
2V  10W  18

5V  9W  11 
3
=
1
(x5)
4
=
2
(x2)
4
+
3
W  1
What would you like to do now?
3. Substitute found value into either of
original equations.
Question 4B
-18
(A)
3v
w – 2v v + w
2w
2V  10W  18

5V  9W  11 
1
2
(x5)
(x2)
Substitute -1 for W in equation
(B)
11
5V + (-9) = 11
5V = 20
v + w v – 3w
6w
Begin Solution
Continue Solution
Comments
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v-w
V=4
2
Comments
1. Work your way to an expression
for the top row by filling in the
middle rows.
Use diagrams given to
organise working:
-18
(A)
Pyramid (A)
2nd
5w + v
row 3v + 2w, -2v + 3w, -v + 2w
3v + 2w
3v
5w - 3v
3w - 2v
2w
2w - v
w – 2v
v+w
3rd row v + 5w, -3v + 5w
Top row
-2v + 10w
= - 18
(B)
11
3v + w
Pyramid (B)
2nd row 2v - 2w, v + 3w, v + 5w
3rd row 3v + w, 2v + 8w
Top row
5v + 9w
= 11
8w + 2v
2v - 2w 3w + v
5w + v
v+w
v – 3w
6w
v-w
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Markers Comments
1. Work your way to an expression
for the top row by filling in the
middle rows.
Hence equations:
10w - 2v = -18
9w + 5v = 11
Pyramid (A)
2nd row 3v + 2w, -2v + 3w, -v + 2w
3rd row v + 5w, -3v + 5w
Top row
-2v + 10w
Solve by the method
of elimination.
= - 18
End of simultaneous Equations
Pyramid (B)
2nd row 2v - 2w, v + 3w, v + 5w
3rd row 3v + w, 2v + 8w
Top row
5v + 9w
= 11
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