Transcript Standard Grade - mathsrevision

Paper 2 Questions
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Paper 2 Questions
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Credit Level
S
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 at
an angle of 40°.
T
h
35°
40°
500 km
Calculate the height of the satellite
above the ground.
Find ST using the Sine Rule
332.7
105°
ASA
Sine Rule
Find  TSV = 180° - 75° = 105°
500sin40
ST
500
 ST =
= 332.7km
=
sin105
sin40 sin105
h
h =332.7 ×sin35 h =190km
Draw perpendicular from S sin35 =
332.7
Paper 2 Questions
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Find the volume of the trough,
correct to
2 significant figures.
0.6 m
0.25 m
Volume of a prism = A  l
Area of cross-section =
A = area of cross-section
1
2
0.6 × 0.25 + p × 0.3 = 0.29137cm2
2
Volume of trough =
0.29137 × 4 = 1.17 m3
Paper 2 Questions
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x
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An oil tank has a circular cross section
of radius 2.1 metres.
It is filled to a depth of 3.4 metres.
a) Calculate x , the width in metres
of the oil surface.
3.4 m
b) What other depth of oil would give
the same surface width.
(a)
1.3?m
d
2.12 =1.32 + d2
2.1 -1.3 = d
2
2.1 m
(b) 2.1 – 1.3 = 2.1 – 0.8
2
d =1.65
2
width of oil x = 2d
width of oil x = 3.30 m
Paper 2 Questions
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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 blend
which can be made.
Brazilian : Columbian
2:3
20 : 30
Insufficient Columbian
1 kg of coffee contains
400 gms Brazilian
600 gms Columbian
There are 25 000 gms Columbian – 600 gm are needed for each tin
25000 ÷ 600= 41.7
41 tins of the blend can be made
Paper 2 Questions
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The diagram shows part of
the graph of y = sin x.
The line y = 0.4 is drawn and
cuts the graph of y = sin x
at A and B.
Find the x-coordinates of A and B.
Solve equation
acute
sinx = 0.4
x =23.6°
S
A
T
C
x = sin-1 0.4
acute
x = sin-1 0.4
x =23.6° or 156.4°
A(23.6°,0.4) and B(156.4°,0.4)
Paper 2 Questions
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Esther has a new mobile phone and considers the following daily rates.
Easy Call
Green Call
25 pence per minute for
the first 3 minutes
5 pence per minute after
the first three minutes.
40 pence per minute for
the first 2 minutes
2 pence per minute after
the first two minutes.
a) For Easy Call, find the cost of ten minutes in a day.
3×25 + 7 ×5 = £1.10
b) For Easy Call, find a formula for the cost of “m” minutes in a day, m > 3
75 + (m-3) ×5 = 5m+60
c) For Green Call, find a formula for the cost of “m” minutes in a day, m > 2
80 + (m-2) ×2 = 2m+ 76
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.
2m + 76 < 60 + 5m
16 < 3m
3m > 16
m > 5.3
6 minutes is
least number of minutes
Paper 2 Questions
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A weight on the end of a string is spun
in a circle on a smooth table.
The tension, T, in the string varies directly
as the square of the speed, v, and inversely
as 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 radius
of the string is halved.
What happens to the tension in the string.
v2
T
r
kv2
T=
r
 × 32 × 2
 × 18
Paper 2 Questions
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2n =32
(a)
Solve the equation
(b)
A sequence of numbers can be grouped and added
together as shown.
The sum of 2 numbers :
(1 + 2) =
(1 + 2 + 4) =
4–1
8–1
The sum of 3 numbers :
The sum of 4 numbers :(1 + 2 + 4 + 8) = 16 – 1
Find a similar expression for the sum of 5 numbers
(c)
Find a formula for the sum of the first n number
of this sequence
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2n =32
n =5
(1 + 2) = 4 – 1
(1 + 2 + 4) = 8 – 1
(1 + 2 + 4 + 8) = 16 – 1
(1+2+ 4 +8+16) = 32-1 = 25 -1
(20 +21 +22 +23 +24 ) = 32-1 = 25 -1
(1+2+ 4 +..... +2n-1 ) = 2n -1
22 -1
23 -1
24 -1
Paper 2 Questions
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A metal beam, AB, is 6 metres long.
It is hinged at the top, P, of a vertical post
1 metre high.
When B touches the ground, A is 1.5 metres
above 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 heightofB
heightofB =3 metres
2
1
=
6 heightofB
Paper 2 Questions
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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 ?
House :
increases by 5%
multiplier =
1.05
90000 ×1.053 =£ 104,186.25
Contents :
decreases by 8%
multiplier =
0.92
60000 ×0.923 = £46,721.28
TOTAL :
£ 104,186.25 + £46,721.28 = £150,907.53
Paper 2 Questions
Credit Level
y
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A water pipe runs between two buildings,
represented by the points A and B in the diagram.
a) Using the information in the diagram, show that
the equation of the line AB is 3y - x = 6
A
(12, 6)
B
2
x
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.
m =
6 -2
12 - 0
3y - x = 6
4
1
=
12
3
×5
4y +5x = 46
subst.
=
1
y = x +2
3
15y -5x = 30
add
3y = x + 6
19y = 76
4y + 5x = 46
3(4) - x = 6
x =6
Coordinates:
(6, 4)
3y - x = 6
y=4
Paper 2 Questions
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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.
Volume =
 r 2h = π ×(3.25)2 ×15 = 497.75 cm3
New can = same volume
r2 =
radius of can = 3.25 cms
497.75
12π
497.75 = πr2 ×12
r2 =13.203 r = 3.6336...
diameter = 7.3 cms (1d.p.)
Paper 2 Questions
Credit Level
N
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The radio masts, Kangaroo (K), Wallaby (W)
and Possum (P) are situated in the Australian outback. W
Kangaroo is 250 kilometres due south of Wallaby.
Wallaby is 410 kilometres from Possum
Possum is on a bearing of 130° from Kangaroo.
22°
250 km
410 km
Calculate the bearing of Possum from Wallaby.
Do not use a scale drawing.
K
Not right angled triangle
Sine or Cosine Rule
Need
to find
Not Cosine
rule WPK
Must be Sine Rule
sinP sin130
=
250
410
WPK =27.8°
sinP =
250sin130
410
WPK =28°
158°
sinP = 0.467
PWK =22°
130°
28°
P
P = sin-1 0.467
Bearing is: 158°
Paper 2 Questions
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Solve algebraically the equation :
tan40° =2sinx° +1
0  x  360
tan 40° = 0.839099…
Re-arrange
2sinx° = tan40° -1
tan40° -1
sinx° =
2
sinx° = -0.08045..
acute
x = 4.61°
S
A
T
C
x =180 + 4.61°
x =360 - 4.61°
x =185° or 356°
to nearest degree
Paper 2 Questions
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5 cm
14 cm
8 cm
8 cm
100°
14 cm
figure 1
The uniform cross-section is as shown in Figure 2:
figure 2
Find the volume of metal required to make the doorstop.
Area of cross-section = ½ a b Sin C
Volume of prism = A  l
Area of cross-section
=
1
×8 ×14 × sin100
2
Volume of prism = 55.15 × 5
= 55.15
= 275.75 = 276 cm3 3sf
Paper 2 Questions
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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.
kL
R= 2
d
Wire A
Wire B
k ×L
RB = 2
3
RB = RA
k ×3
RA = 2
2
kL
RB =
9
3k k L
=
4
9
3k
RA =
4
27
L = 6.75 m
L=
4
Paper 2 Questions
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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.
SSS - Cosine Rule
A
142 +122 -212
cosA =
2 ×14 ×12
acute
= -0.3006
A = 72.5° A = 107.5°
Paper 2 Questions
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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.
72.5°
72.5°
107.5°
h
70 cm
SOH-CAH-TOA
sin72.5 =
h
70
h = 70 × sin72.5
h = 66.76
height of table = 67 centimetres (2 sf)
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A = x2 +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 length = 30 + x
New Area =
30+ x20+ x
A = 600 +20x +30x + x2 = x2 +50x + 600
Area of old vent = 20 x 30 = 600cm2
Area of new vent = 600  1.4 = 840 cm2
x2 +50x + 600 > 840
x2 +50x -240 > 0
Use formula: a = 1, b = 50, c = -240
x= 4.41 or -54.41 cms
Solve :
x2 +50x -240 = 0
New dimensions: 34cms  24 cms
(nearest cm)