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Biology

BIOL-UA 6

davidjenkins

Spring

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Mathematics IGCSE notes
Index
1. Decimals and standard form
2. Accuracy and Error
3. Powers and roots
click on a topic to
4. Ratio & proportion visit the notes
5. Fractions, ratios
6. Percentages
7. Rational and irrational numbers
8. Algebra: simplifying and factorising
9. Equations: linear, quadratic, simultaneous
10. Rearranging formulae
11. Inequalities
12. Parallel lines, bearings, polygons
13. Areas and volumes, similarity
14. Trigonometry
15. Circles
16. Similar triangles, congruent triangles
17. Transformations
18. Loci and ruler and compass constructions
19. Vectors
20. Straight line graphs
21. More graphs
22. Distance, velocity graphs
23. Sequences; trial and improvement
24. Graphical transformations
25. Probability
26. Statistical calculations, diagrams, data collection
27. Functions
28. Calculus
29. Sets
{also use the intranet revision course of question papers and answers by topic } 1. Decimals and standard form top
(a) multiplying and dividing
(i) 2.5×1.36 Move the decimal points to the right until each is a whole
number, noting the total number of moves, perform the multiplication, then
move the decimal point back by the previous total:
→× 25 136 = 3400 , so the answer is 3.4
{Note in the previous example, that transferring a factor of 2, or even better,
4, from the 136 to the 25 makes it easier:
25×1=36 25×(4×34) = (25×4)×34 = 100×34 = 3400 }
(ii)0.00175÷0.042 Move both decimal points together to the right until the
divisor is a whole number, perform the calculation, and that is the answer.
→1.75÷42, but simplify the calculation by cancelling down any factors first.
In this case, both numbers share a 7, so divide this out:.25÷6 , and
0.0416▯
▯
60 .25 , so the answer i0.0416
(iii) decimal places
To round a number to n d.p., count n digits to the right of the decimal point. If
the digit following the n is ≥5 , then the n digit is raised by 1.
e.g. round 3.012678 to 3 d.p. 3.012678 → 3.012|678 so 3.013 to 3 d.p.
(iv) significant figures
To round a number to n s.f., count digits from the left starting with the first
non-zero digit, then proceed as for decimal places.
e.g. round 3109.85 to 3 s.f.3109.85 → 310|9.85 so 3110 to 3 s.f.
e.g. round 0.0030162 to 3 s.f.,0.0030162 → 0.00301|62 , so 0.00302 to 3 s.f.
(b) standard form
(iii) Convert the following to standard form: (a) 25 000 (b) 0.0000123
Move the decimal point until you have a number x where 1≤1 x < 0 , and the
number of places you moved the point will indicate the numerical value of the
power of 10. So 25000 =× 2.5 10 , and 0.0000123 =× 1.23 10 −5
(iv) multiplying in standard form: (4.4×1×0 ) (3.5×10 6) As all the
elements are multiplied, rearrange them thus:
=×(4.4 3.5) × 1()×10 6 = 15.4×10 11 = 1.54×10 12
2 3.2×10 12
(v) dividing in standard form: 3 Again, rearrange the calculation to
2.5×10
(3.2÷×2.5) (10 ÷10 ) 3 = 1.28×10 9
(vi) adding/subtracting in standard form:(2.5×1+0 ) (3.75×10 ) The
hardest of the calculations. Convert both numbers into the same denomination,
i.e. in this case 10 or 10 , then add.
7 7 7
=(×0.25 10 ) + (3.75×10 ) = 4×10
Questions
(a) 2.54×1.5
(b) 2.55÷0.015
(c) Convert into standard form and multiply: 25000000×0.00000000024
3 2 −
(d) (2.6×1÷0 ) (2×10 )
(e) (1.55× −0 ) −( )51× 0
Answers
(a) →× 254 15 = 3810 , so2.54×1.5 = 3.81
(b) 2.55÷0=.015 2550÷15 . Notice a factor of 5, so let’s cancel it first:
=÷510 3 = 170
7 1 − 0 −3
(c) =×(2.5 10 )×(2.4×10 ) = 6×10
(d) =÷(2.6 2)×(10 ÷10 ) = 1.3×10 5
(e) =×(1.55 10 )−(0.25×10 ) = 1.3×10 −3
3 2. Accuracy and Error top
To see how error can accumulate when using rounded values in a calculation,
take the worst case each way: e.g. this rectangular space is
measured as 5m by 3m, each measurement being to the nearest 3m
metre. What is the area of the rectangle? 5m
To find how small the area could be, consider the lower bounds of the two
measurements: the length could be as low as 4.5m and the width as low as
2.5m. So the smallest possible area i4.5×2.5 = 11.25 m . Now, the length
could be anything up to 5.5m but not including the value 5.5m itself (which
would be rounded up to 6m) So the best way to deal with this is to use the
(unattainable) upper bounds and get a ceiling for the area as
5.5×=.5 19.25 m , which the area could get infinitely close to, but not equal
2 2
to. Then these two facts can be expressed as 11.25m ≤ area < 19.25m .
Questions
(a) A gold block in the shape of as cuboid measures 2.5cm by 5.0cm by
20.0cm, each to the nearest 0.1cm. What is the volume of the block?
(b) A runner runs 100m, measured to the nearest metre, in 12s, measured to
the nearest second. What is the speed of the runner?
(c) a = 3.0, b = 2.5, both measured to 2 s.f. What are the possible value of
a−b ?
Answers
3
(a) lower bound volume = 2.45×4.95×1 =9.95 241.943625 cm .
upper bound volume = 2.55×5.05×2 =0.05 258.193875 cm .
3 3
S o 241.943625cm ≤volume<258.193875cm
distance
(b) Since speed = , for the lower bound we need to take the smallest
time
value of distance with the biggest value of time, and vice-versa for the upper
bound.
99.5 100.5
So < 4 , the answer would be x < 2 or x > 2.
(c) 2 variable linear inequalities
e.g.3x2− ≥ 6. Plot the boundary line32− = 6 , then take a trial point
(e.g. the origin) to determine which side of the line to accept.
y
2
1
–2 –1 1 2 3 x
–2
–3
–4
The origin’s coordinates make 30 −× 2 0 which is not ≥ 6, so that side is
rejected:
y
2
1
–2 –1 1 2 3 x
–2
–3
–4
22 Questions y
5
3
(a) Solve2(1− x) < 6 2
1
2
(b) Solve12−≤ x x –4–3–2–1 1 2 3 4 5 x
–2
(c) Find the 3 inequalities which identify this region: –3
–4
Answers
(a)2(1− x) < 6 [ ÷2]
∴1−3− 2
(b)12−≤ x x2 [rearrange]
2
xx+− 12 ≥ 0
∴(x4+−)( 3) ≥ 0 giving critical values of –4 and +3.
y
10
5
x
–5–4–3–2––5 1 2 3 4
– 10
– 15
so x ≤− 4 or x ≥ 3
1
(c) The three line equations ayx 21 +=, y x 2, x + y= 4.
2
By considering a point (e.g. origin) in the shaded region, the inequalities are
1
y <21 ,y > x 2 , and x+ < 4.
2
23 12. Parallel lines, bearings, polygons top
(a) Parallel lines
correspondingangles equal allied or interior add up to
alternate angles equal 180°
N
(b) bearings
A
Bearings are measured clockwise from North:
bearing of B from A is 135º 45°
B
(c) polygons
for a polygon with n sides, exterior angle
sum of interior angles = (2− )180 º interior angle
sum of exteriors = 360º
24 Questions
a
(a) In the diagram opposite, find A
E
the value ofθ in terms of a and b. D
θ C
(b) The bearing of B from A is 090º,
B b
and the bearing of C from B is 120º.
Given also that AB = BC, find the
bearing of C from A.
(c) A pentagon has exactly one line of symmetry, and angles all of which are
either 100º or 120º. Make a sketch of the pentagon, marking in the angles.
Answers
(a) DAE = a (opposite), DEC = b (corresponding), so
▯ ▯
AED =180−b (angles on a straight line). ADE =θ (opposite). We
now have the three angles in triangle ADE, sab(180−+) θ =180 .
A rearrangement gives θ = b−a .
N N
(b) Angle at point B means B 120°
A
ABC =−360 90−120 =150º
▯ C
Triangle ABC is isosceles, so BAC =15º.
The bearing of C from A is therefore 105º.
(c) Sum of internal angles (n− 2)180 = 540 º
for any pentagon. A line of symmetry means c
the set up is like this: b b
The only way of allocating 100º and 120º to a, b, c
and make a total of 540º is to have three 100º’s a a
and two 120º’s. So there are two possible pentagons:
100° 100°
100 ° 100°
120°12 0°
120°120°
100°100 °
25 13. Areas and volumes, similarity top
(a) Areas of plane figures
CIRCLE
TRIANGLE
B
a
r h
b C A
b
π r2 1 1
bh or absinC
2 2
TRAPEZIUM
PARALLELOGRAM
b
h h
b a
1
bh (a + b)h
2
(b) Surface area and volume
Shape surface area volume
PRISM
Prism p×l Al× A
p
r l
Cylinder 2πrh πr h
h CYLINDER
1 2 CONE
Cone πrl πr h h l
3
r
2 4 3 SPHERE r
Sphere 4πr πr
3
1
Pyramid ×base area× h PYRAMID h
3
3
Pipe flow: number of m /s flowing through (or out of) a pipe
= cross-sectional are× speed
v
26 (b) Similarity
Enlargement scale factor =
k
Area scale factor =
2
k
Volume scale factor =
k 3
Questions
3
(a) A cylinder has volume 100cm , and height 5cm.
What is its diameter?
(b) A cone of base radius 10cm and height 20cm is sliced parallel to the base
half way up into two pieces. What is the volume of the base part? (frustum)
25m
(c) The empty swimming pool shown opposite 10m
is to be filled with water. The speed of flow of
water in the pipe is 2m/s, and the radius of the 1m 3m
pipe is 5cm. How long will the pool take to fill?
(d) Two blocks are geometrically similar,
and the big blocks weighs 20 times the
small block. What is
the ratio of surface areas of the two
blocks?
Answers
(a)πr 51 = 00
∴ r2 =100 , so r = 20 = 2.52 cm. Whoops! Diameter asked for!
5π π
diameter = 5.05cm to 3sf
{Note the pre-corrected value was doubled resulting in 5.05 when itself
rounded, not 5.04}
27 (b) The upper small cone has base radius 5cm and height 10cm. The volume of
1 2 1 2 1
the base is thereforeπ 10 × 20 − π 5 × 10 which factorises to π1750 =
3 3 3 3
1830cm to 3sf
(c) Pool is a prism with cross section the side, which is a trapezium.
1 3
So volume of pool = (1 3)25 10 = 500 m .
2
Rate of egress of water is c.s.a.peed = π5 ×2 =00 5000π cm , which
6 3
is5000π ÷10 m /s. (Units!!) So time taken =
105
500 (5000 π 10 ) = = 31831 s, i.e. approx 8hrs 51 mins.
π
(d) assuming same density material, weight is directly proportional to volume.
The volume factor is 20, i.ek = 20 . ∴k = 20 , and so the ratio of surface
areas,1 :k2, is 1 : 7.37
28 14. Trigonometry top
O
sinθ =
H H
O A
θ cosθ = H
A
tanθ =O
A
a b c
Sine rule: = =
sin A sin B C sin
2 2 2
Cosine rule:ab=+ c − 2cc os A
y
x
Two opposite pairs: use sine rule
x y Three sides and one angle: use cosine rule
z
Angle between line and plane is the angle between the line and its
projection on the plane: e.g. for the angle between this diagonal and
the base, draw the projection, and the angle is shown here:
Trigonometric functions for all angles:
1y 1 y 5y
90 180 270 360 x 90 180 270 360 x 90 180 270 360 x
–1 –1 –5
tanx
sinx cosx
29 C
Questions
6cm
(a) ADB is a straight line of length 20cm. .ind 40° θ
A D B
(b) In triangle ABC, AB = 5cm, BC = 8cm, andCA= 30º. Find ABC
(c) A yacht sails 5 miles at 045º then 6 miles at 090º. How far and at what
bearing is it from its original point?
(d) Is an internal diagonal of a cube at 45º elevation from the base?
(e) Find two values of x in the range 0º to 360º for which− 0.5
30 Answers
(a) draw a diagram. No, a decent diagram!
θ lies in the triangle on the right, and all the information
we have is in the left triangle. To connect with the triangle BCD
it would be helpful to calculate CD and AD.
º CD º
sin 40 = , ∴CD = 6 =sin40 3.85....
6
º AD º
cos40 = 6 , ∴AD = 6cos40 = 4.59....., so BD = 20− 4.59... = 15.4...
Then tanθ = =.85... 0.250..., soθ =14.1 º to 3 s.f.
15.4...
B
(b) If we target angle A then we have 2 opposite
side/angle pairs, so use the sine rule: 5 cm 8 cm
º 30 °
sin A = sin30 , so A= 53.1.. , and A C
8 5
ˆ
B =−80 53.1... −0 = 96.9º
N
(c) Using cosine rule, N A 6 B
OB =+ 25 6 − 25×6 × ×cos135 º 5
so OB = 10.2 miles (to 3 s.f.)
45°
Now using the sine rule, O
sin AOB sin135 º º
= , which gives AOB = 24.7
6 10.16...
The bearing of B from O is therefore 069.7º
(d) Assume the length of side of the cube is 1.
(Enlargement won’t make any difference to the
angles). Pythagoras gives the projection on the
base as 2, and the opposite side is 1.
1
So tanθ = . an dθ = 35.3º to 3 s.f.
2
y
1
(e) First find the principal value from
the calculator: -30º. Where are there -30
other angles in our window with the – 90 90 180 270 360 x
same sine? -0.5
–1
y
1
-30
– 90 90 180 270 360 x
-0.5
–1
Clearly at 30º beyond 180º and 30º back from 360º.
So x = 210º, 330º
31 15. Circles top
(a) arcs, sectors, segments
arc
Arc length = θ ×2πr θ sector
360
r
Sector area = θ ×π r 2
360
segment
Segment area = Sector – Triangle
(b) circle theorems
(i) Angle subtended at the centre =2× angle subtended
at the circumference by the same arc
(ii) Angles in the same segment are equal
(iii) Angle in a semicircle = 90º
(iv) Opposite angles of a cyclic quadrilateral
add up to 180º
(v) Exterior angle of a cyclic quadrilateral
= interior opposite
(vi) Alternate segment theorem: the angle between
tangent and side = interior opposite
32 Questions
(a) The arc of a sector of a circle of radius 20cm has length 10cm.
Find the area of the sector.
(b) A cylindrical tank, radius 50cm and length 2m with horizontal axis is
partially filled with oil to a maximum depth of 25cm. How much oil is
contained in the cylinder?
(c) Findθ in the following diagrams:
θ
(a) 120° (b) (c)
35°
θ θ 40 °
Answers
θ
(a) Arc length = 360 ×2π2 0 and this is given as 10cm. Rearranging
gives θ = 90 . Therefore sector area =θ ×π 202 = 90 ×π 20 2
π 360 360π
2
which simplifies nicely to 100cm .
90
{Would you have reached for the calculator atθ = π , and missed the
beautiful cancellation later?}
(b) We need to find the area of the segment
comprising the cross-section of the oil.
Above the oil is an isosceles triangle, so 25 cm
split it down the line of symmetry: OIL 25 cm
50 cm
25 cm
−1
This gives an angle ofcos 0.5 = 60º, and a base
of 43.3…cm by Pythagoras. So the angle at the centre
of the sector is 120º. Therefore the area of the segment
is120 ×− 502 43.3... 25 = 1535cm . The oil is in the
360
shape of a prism with volume 1535×200= 307092 cm 3
3
= 0.307m .
33 (c) The angle subtended at the centre is
360º – 120º = 240º, so =120 º by the
angle at the centre theorem.
θ
C = 35º (angles in the same segment)
ˆ C
CAB = 90º (angle in a semicircle)
so θ =−80 35−90= 55 º (angle sum
of a triangle) 35°
A
θ B
ABT = 40º (alternate segment theorem) C
θ
Isosceles triangle givesT = 70 º, and B A
so C = 70 º (angles in same segment)
T 40°
34 16. Similar triangles, congruent triangles top
(a) Similar triangles
same shape, different size; related by enlargement (may be
different orientation)
to prove similar: AAA
(each pair of angles equal)
to solve problems use either (a) scale factor or (b) ratio of sides equal
(b) Congruent triangles
same shape and size, i.e. identical though usually in different positions.
SSS
to prove congruent: SAS
AAS
ASA
RHS
but not ASS – there are sometimes two different triangles with the same
ASS
35 Questions
E D 6
(a) (i) Prove that triangles BCD and C
ACE are similar. (ii) Hence find the lengths BD and
8
DE. (iii)If the B
area of triangle BCD is 12 what is the A 4 area of the trapezium
ABDE?
(b) Use congruent triangles to prove that the diagonals of a
parallelogram bisect each other.
Answers
(a) (i)AC = DBC and AEC = BDC (corresponding). The third angle is
shared, so AAA is established and they are similar.
12 3 3
(ii) scale factor of enlargement =s . So BD =6 ÷ , or
82 2
2 3
6 × = 4. CE is6 ×= 9, so DE is 9 – 6 = 3
3 2
3 2 2
(iii) Area of triangle AC12 ( 2 {note area scale factor = k }
= 27. So the trapezium has area 27 – 12 = 15
A B
(b) The two pairs of marked angles are equal
(alternate), and the top and bottom sides are
X
equal (parallelogram). So we have two
congruent triangles ABX and DCX by D C
ASA. (Note each would have to be rotated
180º about X to transform onto the other).
So AX = XC and DX = XB, i.e. the diagonals bisect each other.
36 17. Transformations top
a
(i) translation by vector⎜ ⎟ shifts a to the right and b up.
⎝⎠
(ii) rotation about P throughθ . [Note e.g. +90º means 90º anticlockwise]
perform a rotation using compasses,
or if a multiple of 90º, use the L shape:
To find the centre of a rotation already performed, perpendicularly bisect
a line joining any point with its image. Repeat with

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