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If you can do these questions without looking at the answers then you are well on the way
to passing the exam.(Use solution line by line if needed)
(Click to start)
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A TV signal is sent from a transmitter T, via a satellite S, to a village V, as shown in the diagram.
The village is 500 km from the transmitter.The signal is sent out at an angle of 35° and is received in the village atan angle of 40°.
Calculate the height of the satellite above the ground.
ST 500=
sin40 sin105
ASA Sine Rule
35° 40°
S
T V
500 km
Find ST using the Sine Rule
Draw perpendicular from S
h
Find TSV = 180° - 75° = 105°
105°
500sin40ST = =332.7km
sin105
332.7
35° 40°
hsin35 =
332.7h=332.7×sin35 h=190km
0.6 m
0.25 m
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0.6×0.25 21+ p×0.3
230.29137×4 =1.17m
Area of cross-section = 2cm=0.29137
Find the volume of the trough,correct to
2 significant figures.
Volume of a prism = A l A = area of cross-section
Volume of trough =
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An oil tank has a circular cross sectionof radius 2.1 metres.
It is filled to a depth of 3.4 metres.
x
2.1 m
3.4 m
a) Calculate x , the width in metres of the oil surface.
b) What other depth of oil would give the same surface width.
2.1 m
?1.3 m
d2 2 22.1 =1.3 +d2 2 22.1 -1.3 =d
d=1.65 width of oil x = 3.30 m
width of oil x = 2d
(a)
(b)2.1 – 1.3 =2.1 – 0.8
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A coffee shop blends its own coffee and sells it in one-kilogram tins.One blend consists of two kinds of coffee, Brazilian and Columbian,in the ratio 2 : 3.
The shop has 20 kilograms of Brazilian and 25 kilograms of Columbian in stock. What is the maximum number of one-kilogram tins of this blendwhich can be made. Brazilian : Columbian
2 : 3
20 : 30 Insufficient Columbian
1 kg of coffee contains
400 gms Brazilian600 gms Columbian
There are 25 000 gms Columbian – 600 gm are needed for each tin
25000÷600=41.741 tins of the blend can be made
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The diagram shows part of the graph of y = sin x.
The line y = 0.4 is drawn andcuts the graph of y = sin xat A and B.
Find the x-coordinates of A and B.
Solve equation sinx =0.4 -1x =sin 0.4 -1acute x =sin 0.4
acute x =23.6°AS
CT
x =23.6° or 156.4°
A(23.6°,0.4) and B(156.4°,0.4)
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25 pence per minute forthe first 3 minutes
5 pence per minute afterthe first three minutes.
Green Call40 pence per minute for
the first 2 minutes2 pence per minute after
the first two minutes.
Esther has a new mobile phone and considers the following daily rates.
a) For Easy Call, find the cost of ten minutes in a day.
b) For Easy Call, find a formula for the cost of “m” minutes in a day, m > 3
c) For Green Call, find a formula for the cost of “m” minutes in a day, m > 2
d) Green Call claims that its system is cheaper.Find algebraically, the least number of minutes (to the nearest minute)which must be used each day for this claim to be true.
3×25 +7×5 =£1.10
75 +(m-3)×5 =5m+60
80 +(m-2)×2 =2m+76
2m+76 <60+5m
16 <3m3m>16m>5.3
6 minutes isleast number of minutes
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A weight on the end of a string is spunin a circle on a smooth table.
The tension, T, in the string varies directlyas the square of the speed, v, and inverselyas the radius, r, of the circle.
a) Write down a formula for T in terms of v and r.
b) The speed of the weight is multiplied by 3 and the radiusof the string is halved.
What happens to the tension in the string.
2vT
r
2kvT =
r2×3 ×2 ×18
(b) A sequence of numbers can be grouped and addedtogether as shown.
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n2 =32
(1 + 2) = 4 – 1
(1 + 2 + 4) = 8 – 1
(1 + 2 + 4 + 8) = 16 – 1
(a) Solve the equation
The sum of 2 numbers :
The sum of 3 numbers :
The sum of 4 numbers :
Find a similar expression for the sum of 5 numbers
(c) Find a formula for the sum of the first n numberof this sequence
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n2 =32
(1 + 2) = 4 – 1
(1 + 2 + 4) = 8 – 1
(1 + 2 + 4 + 8) = 16 – 1
22 -1
n=5
32 -142 -1
5(1+2+4+8+16) =32-1=2 -1
n-1 n(1+2+4+.....+2 ) = 2 -1
0 1 2 3 4 5(2 +2 +2 +2 +2 ) =32-1=2 -1
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A metal beam, AB, is 6 metres long.It is hinged at the top, P, of a vertical post1 metre high.
When B touches the ground, A is 1.5 metresabove the ground, as shown In Figure 1.
When A comes down to the ground, B rises,as shown in Figure 2.By calculating the length of AP, or otherwise,find the height of B above the ground.Do not use a scale drawing.
Similar Triangles BP 1=
BA 1.5
BP 1= BP=4
6 1.5
So, AP = 2 m
AP 1=
AB height of B2 1
=6 height of B
metresheight of B=3
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In 1999, a house was valued at £90,000
and the contents were valued at £60,000.
The value of the house appreciates by 5% each year.
The value of the contents depreciates by 8% each year.
What will be the total value of the house and contents in 2002 ?
390000×1.05 =£ 104,186.25
House :
360000×0.92 =£ 46,721.28
Contents :
increases by 5% multiplier = 1.05
decreases by 8% multiplier = 0.92
£ 104,186.25 + £ 46,721.28 =£ 150,907.53TOTAL :
a) Using the information in the diagram, show that the equation of the line AB is
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3y- x =6
A water pipe runs between two buildings, represented by the points A and B in the diagram.
b) An emergency outlet pipe has to be built across the main pipe.
The line representing this outlet pipe has equation 4y+5x =46
Calculate the coordinates of the point on the diagram at which the outlet pipe will cut across the main water pipe.
6-2m =
12- 0
2
(12, 6)
A
B
x
y
4 1= =
12 31
y= x +23
3y=x+6 3y- x =6
3y- x =6
4y+5x =46
×5 15y-5x =30
4y+5x =46
add 19y=76 y=4
subst. 3(4)- x =6 x =6 Coordinates: (6, 4)
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A cylindrical soft drinks can is 15 cms in height and 6.5 cms in
diameter.
A new cylindrical can holds the same volume but has a reduced
height of 12 cms. What is the diameter of the new can ?
Give your answer to 1 decimal place.
radius of can = 3.25 cms
Volume = 2r h 2=π ×(3.25) ×15 3cm=497.75
New can = same volume 2497.75 =πr ×12
2 497.75r =
12π2r =13.203 r =3.6336...
diameter cms (1d.p.)=7.3
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The radio masts, Kangaroo (K), Wallaby (W)and Possum (P) are situated in the Australian outback.
Kangaroo is 250 kilometres due south of Wallaby.Wallaby is 410 kilometres from PossumPossum is on a bearing of 130° from Kangaroo.
Calculate the bearing of Possum from Wallaby.
Do not use a scale drawing.
N
W
K
P
410 km
250 km
130°
Sine or Cosine RuleNot right angled triangle
Must be Sine RuleNot Cosine ruleNeed to find WPK
sinP sin130=
250 410250sin130
sinP=410
sinP=0.467-1P=sin 0.467
WPK =27.8° WPK =28°
22°
PWK =22° Bearing is: 158°
158°
28°
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Solve algebraically the equation :tan40°=2sinx°+1 0 x 360
tan 40° = 0.839099…
Re-arrange 2sinx°=tan40°-1
tan40°-1sinx°=
2
sinx°=-0.08045..
acute x =4.61°
AS
CT
x =180+4.61°
x =360- 4.61°
x =185° or 356°to nearest degree
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The uniform cross-section is as shown in Figure 2:
Find the volume of metal required to make the doorstop.
figure 1
figure 2
Volume of prism = A l Area of cross-section =½ a b Sin C
Area of cross-section1
= ×8×14×sin1002
=55.15
Volume of prism =55.15 ×5 =275.75 3cm 3sf=276
100°8 cm 14 cm5 cm
8 cm
14 cm
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The electrical resistance, R, of copper wire varies directly as its length,L metres, and inversely as the square of its diameter, d millimetres .
Two lengths of copper wire, A and B, have the same resistance.
Wire A has a diameter of 2 millimetres and a length of 3 metres.Wire B has a diameter of 3 millimetres
What is the length of wire B.
2
kLR=
dWire A A 2
k×3R =
2 A
3kR =
4
Wire B B 2
k×LR =
3B
kLR =
9
B AR =R 3k k=
4L
9
27L=
4L =6.75 m
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Each leg of a folding table is prevented from opening too far by a metal bar. The metal bar is 21 cms long.It is fixed to the table top 14 cms from the hinge and to the table leg12 cms from the hinge.
a) Calculate the size of the obtuse angle which the table top makes with the leg.
b) Given that the table leg is 70 cms long, calculate the height of the table.
SSS - Cosine Rule
2 2 214 +12 -21cosA =
2×14×12A =-0.3006
acute A = 72.5° A = 107.5°
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Each leg of a folding table is prevented from opening too far by a metal bar.The metal bar is 21 cms long.It is fixed to the table top 14 cms from the hinge and to the table leg 12 cms from the hinge.
a) Calculate the size of the obtuse angle which the table top makes with the leg.
b) Given that the table leg is 70 cms long, calculate the height of the table.
107.5°72.5°
72.5°
70 cmh
SOH-CAH-TOA
hsin72.5 =
70
h=70×sin72.5
h=66.76height of table = 67 centimetres (2 sf)
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2A =x +50x+600
c) The area of the new vent must be at least 40% more than the original area. Find the minimum dimensions to the nearest centimeter, of the new vent.
New Area = 30+x 20+x2A =600+20x+30x+x 2=x +50x+600
New length = 30 + x
Area of old vent = 20 x 30 = 600cm2
Area of new vent = 600 1.4 = 840 cm2
2x +50x+600 >840 2x +50x-240 >02
Solve : x +50x-240 =0
New dimensions: 34cms 24 cms
Use formula: a = 1, b = 50, c = -240
x= 4.41 or -54.41 cms (nearest cm)
