Maths Revision Notes - IGCSE

Mathematics IGCSE notes
Index
1. Decimals and standard form
2. Accuracy and Error
3. Powers and roots
click on a topic to
visit the notes4. Ratio & proportion
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) 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:
2.5 1.36×
, so the answer is25 136 3400→ × = 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:
}25 136 25 (4 34) (25 4) 34 100 34 3400× = × × = × × = × =
(ii) Move both decimal points together to the right until the0.00175 0.042÷
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: , and0.25 6→ ÷
0.0416
6 0.25 , so the answer is 0.0416
(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 nth
is , then the n5≥ th
digit is raised by 1.
e.g. round 3.012678 to 3 d.p. so3.012678 3.012|678→ 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 so310|9.85→ 3110 to 3 s.f.
e.g. round 0.0030162 to 3 s.f., 0.0030162 , so0.00301|62→ 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 10x≤ < , and the
number of places you moved the point will indicate the numerical value of the
power of 10. So 4
25000 2.5 10= × , and 5
0.0000123 1.23 10−
= ×
(iv) multiplying in standard form: ( )5
(4.4 10 ) 3.5 10× × × 6
As all the
elements are multiplied, rearrange them thus:
( )5 6 11 12
(4.4 3.5) 10 10 15.4 10 1.54 10= × × × = × = ×
2

(v) dividing in standard form:
12
3
3.2 10
2.5 10
×
×
Again, rearrange the calculation to
12 3 9
(3.2 2.5) (10 10 ) 1.28 10÷ × ÷ = ×
(vi) adding/subtracting in standard form: The
hardest of the calculations. Convert both numbers into the same denomination,
i.e. in this case 10
6 7
(2.5 10 ) (3.75 10 )× + ×
6
or 107
, then add.
7 7
(0.25 10 ) (3.75 10 ) 4 10= × + × = × 7
Questions
(a) 2.54 1.5×
(b) 2.55 0÷ .015
(c) Convert into standard form and multiply: 25000000 0.00000000024×
(d) ( 3 2
2.6 10 ) (2 10 )−
× ÷ ×
1.55 10 ) 2.5 10− −
× − ×(e) ( ( )3 4
Answers
(a) → × , so254 15 3810= 2.54 1.5 3.81× =
(b) 2.55 . Notice a factor of 5, so let’s cancel it first:0.015 2550 15÷ = ÷
510 3 170= ÷ =
(c) 7 10
(2.5 10 ) (2.4 10 ) 6 10 3− −
= × × × = ×
(d) 3 2
(2.6 2) (10 10 ) 1.3 10− 5
× ÷ = ×= ÷
(e) 3 3
(1.55 10 ) (0.25 10 ) 1.3 10− −
= × − × = × 3−
3

2. Accuracy and Error top
3m
5m
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
metre. What is the area of the rectangle?
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 is 4.5 2.5 11.25× = m2
. 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
m5.5 3.5 19.25× = 2
, which the area could get infinitely close to, but not equal
to. Then these two facts can be expressed as 11.25m2
area≤ < 19.25m2
.
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) , both measured to 2 s.f. What are the possible value of
?
3.0, 2.5a b= =
a b−
Answers
(a) lower bound volume = 2.45 4.95 19.95 241.943625× × = cm3
.
upper bound volume = 2.55 5.05 20.05 258.193875× × = cm3
.
So 241.943625cm3
≤volume<258.193875cm3
(b) Since speed =
distance
time
, for the lower bound we need to take the smallest
value of distance with the biggest value of time, and vice-versa for the upper
bound.
So
99.5 100.5
speed
12.5 11.5
< < , i.e. 7.96ms-1
< speed < 8.739…ms-1
(c) for the smallest value of a b− , we need to take the smallest value of a
together with the biggest value of b, etc.
So , i.e.2.95 2.55 3.05 2.45a b− < − < − 0.4 0.6a b< − <
4

3. Powers and roots top
1) a b a b
x x x +
× =
2) a b a b
x x x −
÷ =
3) ( )a b ab
x x=
4)
1a
a
x
x
−
=
5) 0
1x =
(a) whole number powers
Note that the base numbers (x’s) have to be the same;
5
2 3× 2
5
cannot be simplified any further.
1) e.g. 3 2
x x x× = , 3 7 1
2 2 2× = 0
2
aIf in doubt, write the powers out in full: a3
× means
( ) (a a a a a)× × × × which is 5
a
2) 6 2 4
x x x÷ = , 5 58 2 6
5÷ =
2
Again, if in doubt, spell it out:
6
a a÷ means
a a a a a a
a a
× × × × ×
×
which cancels down to
4a a a a a a
a
a a
× × × × ×/ /
=
×/ /
3) ( )3 2 6
x x= ) 3=
)
, (32 4 8
To check this, ( 3 2
x means 3
( ) ( )3
x x× which is 6
x
4) 3
3
1
x
x
−
= , 3
3
1 1
10 0.001
10 1000
−
= = =
5) 10 1= .0
5

Questions
Simplify the following as far as possible:
(a) 5 3
x x×
(b) 3 5
a a a÷ × 6
(c)
4 7
5
3 3
3 3
×
×
(d)
5 10
6 3
2 4
8 4
×
÷
Answers
(a) 5 3 5 3 8
x x x x+
× = =
(b) 3 5 6 3 5 6
a a a a a− +
÷ × = = 4
(c)
4 7 4 7 11
11 6 5
5 5 1 6
3 3 3 3
3 3
3 3 3 3
+
−
+
×
= = = =
×
(d)
5 10 5 2 10 5 20 25
13
6 3 3 6 2 3 18 6 12
2 4 2 (2 ) 2 2 2
2
8 4 (2 ) (2 ) 2 2 2
× × ×
= = =
÷ ÷ ÷
=
(b) fractional powers
6)
1
nn
x x=
7) ( )
p
q qp pq
x x x= =
6) e.g.
1
33
x x= ,
1
2
9 9= = 3
7)
2
2 233
27 ( 27) (3) 9= = =
Note: if you can find the qth root of x easily then it’s better to use the ( )
q p
x version.
6

Q. Simplify the following as far as possible:
(a)
1
2
16
(b)
1
3
64
(c)
3
2
4
(d)
3
4
81
(e)
2
6 3
( )x
Answers.
(a)
1
2
16 16 4= =
(b)
1
33
64 64 4= =
(c)
3
3 32
4 ( 4) (2) 8= = =
(d)
3
3 344
81 ( 81) (3) 27= = =
(e) easier to use power law (3) above:
2 2
6
6 43 3
( )x x x
×
= =
7

4. Ratio and Proportion
(a) Using ratios
This is really a special case of proportion. If quantities are linearly related,
either directly or inversely, (like number of workers and time taken to do a job),
calculate by multiplying by a ratio:
e.g. If 8 workers can together do a job in 6 days, how long would the same job take
with 12 workers?
ans: it will take less time, so we multiply by the ratio 8
12 . So it takes
8
126 4× = days.
e.g. If a workforce of 20 can produce 12 cars in 15 days, how many workers
should be used if 15 cars are needed in 10 days?
ans: no. of workers = 15 15
12 1020× × = 1
237 , ie 38.
(b) Proportion
Where quantities are related not necessarily linearly.
(i) Direct proportion
This is when an increase in one quantity causes an increase in the other.
eg , and you are given that y is 7.2 when x is 6.2
y x∝
Rewrite as , and substitute the given values to find k:2
y kx=
2
7.2 6k= × , so
0.2k = . The relationship can now be written as
2
0.2y x= , and any problems solved.
(ii) Inverse proportion
This is when an increase in one quantity causes an decrease in the other.
e.g. If y is inversely proportional to the cube of x,
then 3
1
y
x
∝
Rewrite as 3
k
y
x
= , and proceed as usual.
(iii) Multiplier method
For the cunning, it is possible (but harder) to solve a problem without calculating k.
e.g. Radiation varies inversely as the square of the distance away from the source. In
suitable units, the radiation at 10m away from the source is 75. What is the radiation
at 50m away?
ans: as distance increases by a factor of 5, radiation must decrease by a factor of ,2
5
so the radiation is 75 = 3.25÷
8
top

Questions.
(a) Water needs to be removed from an underground chamber before work can
commence. When the water was at a depth of 3m, five suction pipes were used and
emptied the chamber in 4 hours. If the water is now at a depth of 5m (same cross-
section), and you want to empty the chamber in 10 hours time, how many pipes need
to be used?
(b) y is proportional to 2
x and when x is 5 y is 6. Find
(i) y when x is 25 (ii) x when y is 8.64
(c) The time t seconds taken for an object to travel a certain distance from rest is
inversely proportional to the square root of the acceleration a. When a is 4m/s2
, t is
2s.
What is the value of a if the time taken is 5 seconds?
Answers
(a) No. of pipes =
5 4
5 =
3 10
× × 1
33 , so it would be necessary to use 4 pipes to be sure
of emptying within 10 hours.
(b) and we know when x is 5, y is 6, so
2
2
y x
y kx
∝
=
, so2
6 k= ×5 6
25k = , and we can write the relationship as
26
25y x= .
(i) When x is 25, 26
25 25 150y = × = .
(ii) When y is 8.64, 26
258.64 , sox=
2 25 8.64
6
x
×
= no, don’t reach for the calculator yet!
2
25 1.44x = × , so .5 1.2 6x = × =
(c)
1
t ,
a
∝
So
k
t
a
= . Substituting given values:
2
4
k
= , so , ie4k =
4
t
a
= .
When t = 5,
4
5
a
= , so
4
5
a = , and
16
25
a or 0.64 m/s= 2
.
9

5. Fractions and ratios top
(a) Fractions
(i) Adding/subtracting: e.g.
1
3 1
6 3
2
− . Convert to vulgar form first:
19 5
6 3
− ,
then find the lowest common denominator, in this case 6. Then
19 5 19 2 5 9 1
1
6 3 6 6 2
− ×
− = = = .
(ii) Multiplying/dividing: e.g.
1 7
5
3 8
× . Convert to vulgar form:
16 7
3 8
× , and
then always cancel any factor in the numerator with a factor in the
denominator if possible, before multiplying together:
2
1
16 7 16 7 2 7 14
3 8 3 3 1 38
/ / ×
× = × = =
×/
.
To divide, turn the ÷ into a × and invert the second fraction.
(iii) Converting to and from decimals: e.g. what is
3
40
as a decimal?
0.075
40 3.000 so
3
40
is 0.075.
But what is 0.075 as a fraction? 0.075 means
75
1000
, then cancel down to
3
40
.
(b) Ratios
(iv) To divide a quantity into 3 parts in the ratio 3: 4:5, call the divisions 3
parts, 4 parts and 5 parts. There are 12 parts altogether, so find 1 part, and
hence the 3 portions.
(v) To find the ratio of several quantities, express in the same units then cancel
or multiply up until in lowest terms e.g. what is the ratio of 3.0m to 2.25m to
75cm?
Perhaps metres is the best unit to use here, so the ratio is 3 :2.25: 0.75.
Multiplying up by 4 (or 100 if you really insist) will render all numbers
integer. So the ratio is 12 : 9 : 3, and we can now cancel down to 4:3:1
10

Questions
(a) 23 5
(2 ) 1
4 11
×
(b)
1 3 1
(1 ) 2
3 5 5
− ÷
(c) What is 0.0875 as a fraction in lowest terms?(c) What is 0.0875 as a fraction in lowest terms?
(d) Split £5000 in the ratio of 1:2:5(d) Split £5000 in the ratio of 1:2:5
Answers
(a) 2
11 1
1 1
11 16 121 16 121 16
( ) 11
4 11 16 11 16 11
/ / / / //
= × = × = × =
/ / //
(b)
4 3 11 5 4 3 3 5 11 5 1
( )
3 5 5 15 11 15 11 3
× − ×
= − ÷ = × = × =
(c)
875 35 7
10000 400 80
= = =
(d) 1:2:5 means 8 parts altogether. Each part is £5000 8 £625÷ = , so the
£5000 splits into £625, £1250, and £3125.
11
Questions
23 5
) 1
4 11
×(a) (2
1 3 1
) 2
3 5 5
− ÷(b) (1
Answers
2
11 1
1 1
11 16 121 16 121 16
( ) 11
4 11 16 11 16 11
/ / / / //
= × = × = × =
/ / //
(a)
4 3 11 5 4 3 3 5 11 5 1
( )
3 5 5 15 11 15 11 3
× − ×
÷ = × = × =(b) = −
875 35 7
10000 400 80
= = =
000 8 £625
(c)
(d) 1:2:5 means 8 parts altogether. Each part is £5 ÷ = , so the
£5000 splits into £625, £1250, and £3125.
11

6. Percentages top
(i) What is 75g as a percentage of 6kg? Express as a fraction, then multiply by
100 to covert to a percentage. As a fraction, it is
75
6000
, which is
75 75 1
100% % 1 %
6000 60 4
× = = .
(ii) Find 23% of 3.2kg. This is
23 23
3.2kg 3200g 23 32g 736g
100 100
× = × = × = (or 0.736kg.)
(iii) Increase £20 by 12%. The original amount is always regarded as 100%,
and this problem wants to find 112%. The simplest method is to first find 1%,
then 112%, by dividing by 100 then multiplying by 112. This can be
accomplished in one go, however, by multiplying by
112
100
, i.e. 1.12.
So the answer is £20 .1.12 £22.40× =
(iv) Decrease £20 by 12%. This means we are trying to find 88% of the
original, so the answer is £20 0.88 £17.60× = .
(v) Reverse problems: An investment is worth £6000 after increasing by 20%
in a year: how much was it worth last year? If you are going to make a
mistake, this is where. The 20% refers to 20% of the original amount which
we don’t know, not 20% of £6000. A safe way of handling these “reverse”
problems is to call the unknown original amount £x. The information says that
so£ 1.2 £6000x× =
6000
5000
1.2
x = = .
(vi) Anything weird, and use the simple unitary method, i.e. find what is 1%.
e.g. A coke can advertises 15% extra free, and contains 368ml. How much
extra coke was there?
15%
extra
free
This can contains 115% of the original, so 1% is
368 115 3.2÷ = ml.
So the extra amount, 15%, is 15 3.2 48× = ml.
12

Questions
(a) One part of a company produces £350 000 profit, while the whole
company makes £5.6 million. What percentage of the whole company’s profits
does this part produce?
(b) How much VAT at 17½% is added to a basic price of £25?
(c) An investment earns 8% interest every year. My account has £27000 this
year. How much is contained in my account (i) next year (ii) in ten years’ time
(iii) last year?
(d) Inflation runs at 4% per year in Toyland. Big Ears can buy 24 toadstools
for £1 this year. How many will he be able to buy for £1 in 5 years’ time?
Answers
(a)
350000 1
100% 6 %
5600000 4
× =
(b) 17½% of £25 is
1
217 175
25 25
100 1000
× = × = £4.38
(c) (i) £27000 1.08 £29160× =
(ii) 10
£27000 1.08 £58290.97× =
(iii) £ 1.08 £27000x× =
27000
1.08
x = , so it was worth £25000.
(d) Inflation at 4% per year means that if you pay £100 for some goods this
year, the same goods will cost you £104 in next years’ money. So 24
toadstools will cost in 5 years’ time, and so £1 will
buy him
5
£1 1.04 £1.2166529...× =
1
24
1.2166529...
× , i.e. 19.7… or 19 whole toadstools!
13

7. Rational and irrational numbers top
A rational number is one which can be expressed as
a
b
where a and b are
integers. An irrational number is one which can’t. Fractions, integers, and
recurring decimals are rational. Examples of rationals:
2
3
, 1, 0.25, 3
8 .
Examples of irrationals: , 2, 0.1234....π (not recurring).
(i) Converting rationals to the form
a
b
(to confirm they really are rational)
A terminating decimal:
125 1
0.125
1000 8
= =
A recurring decimal: . Call the number x, so0.123 0.123123123......x =
Multiply by a suitable power of 10 so the recurring decimal appears exactly
again: 1000 123.123123..... 123 0.123123....x = = +
so 1000 , then 999123x x= + 123x = and
123 41
999 333
x = = .
(ii) rationalising a denominator:
2
3
3
has a 3 in the denominator, so multiply top and bottom by 3 (which
does not change the value of the expression, only the shape):
2 3 6
3 3 6
33 3
× = = .
(iii) a b ab=
a a
bb
=
and the same with cube roots, etc.
To simplify expressions using these:
200 100 2 100 2 10 2= × = × =
18 18
9 3
22
= = =
(iv) Finding irrational numbers in a given area:
e.g. find an irrational number between 5 and 6. Note that most square roots
are irrational (except for
4
16, ,
9
etc) are irrational, so as 5 and25=
6 3= 6 , pick a root in between, e.g. 28 . (Or say 2π + for example).
14

Questions
(a) Convert into the form
a
b
: (i) 0.375 (ii) 0.36
i i
(b) Simplify (i)
6
2
(ii)
50
2
(iii) 72 (iv) 3
250
(c) Find an irrational number between 1 and 1.1
Answers
(a) (i)
375 3
0.375
1000 8
= = (ii) 0.36363636......x = so 100 36.363636.....x =
36 0.363636 36 x= + = + . Therefore 99 36x = , so
36 4
99 11
x = =
(b) (i)
6 2 6 2
3 2
22 2
= × = =
(ii)
50
25 5
2
= = =
(iii) 72 36 2 36 2 6 2= × = × =
(iv) 3 33 3
125 2 125 2 5 2= × = × =
(c) e.g.
101
2 0.3,
10
− etc
15

8. Algebra: top
(a) Simplifying
Multiply out brackets and gather up like terms: e.g.
2 2
3 ( 2 ) 2 (3 ) 3 6 6 2 3 22 2
x x y y x y x xy xy y x y+ − − = + − + = +
(b) Factorising
(i) extracting the highest common factor: 2
6 3 3 (2 )x xy x x y− = −
(multiply out the answer to check)
(ii) quadratics:
(a) 2
2x x− (no number term): ( 2x x )= −
(b) (no x term): if it is difference of two squares as in this case:2
16x −
( 4)( 4x x= − + )
)(c) ( a full quadratic): start with ( ) form, then look for
two numbers which multiply to give
2
3 4x x− − (x x
4− and add to give 3− . These are
and +1. So is the answer. (multiply out the answer to check!)
4−
( 4)( 1x x− + )
4
)
+ +
)+
2
ab ac b bc+ − −
)
(d) 2 9 (full quadratic with more than one x2
x x+ + 2
): multiply the 2 by the 4
to get 8, and repeat the previous process i.e. look for two numbers which
multiply to 8 and add up to 9. These are +8 and+1. Now split the middle term
accordingly and group into 2 pairs:
2 2 2
2 9 4 2 8 4 (2 8 ) ( 4x x x x x x x x+ + = + + + = + + + Then factorise each
group, = + , and notice the bracket factor which you now
extract: = + .
2 ( 4) ( 4)x x x
( 4)(2 1x x
(iii) grouping: unusual, but reminiscent of part of (d) above, expressions like
may be able to be factorised even though there are
apparently no common factors. 2
( ) ( ) ( ) (ab ac b bc a b c b b c= + − + = + − + ,
and there just happens to be a big factor: ( )(b c a b)= + −
16

Questions
(a) Simplify ( ) ( ) (a b c b c a c a b− + − + − )
(b) Factorise (i) 2
4 6p q p− q 6(ii) 2 2
(iii) 4 2
x x+ 1x −
(iv) (v) 3 12
10 21x x+ + 2
1 6x x+ + (vi) 2 6 3ab ac b c− + −
Answers
(a) 0ab ac bc ab ac bc= − + − + − =
(b) (i) (ii)2 (2 3)pq p= − 2 ( 3)x x= + (iii) (2 1)(2 1)x x= − +
)(iv) (v) 3( 7)( 3x x= + + 6× gives 18, so 9 and 2 are the required numbers:
and finally
.
2 2
3 9 2 6 (3 9 ) (2 6) 3 ( 3) 2( 3)x x x x x x x x x+ + + = + + + = + + +
(3 2)( 3)x x= + +
(vi) 2 ( 3 ) ( 3 ) (2 1)( 3 )a b c b c a b c= − + − = + −
17

9. Equations: linear, quadratic, simultaneous top
(a) Linear
Perform the same operation on both sides to isolate x:
2 1
3 2
x
x+ = [ ]6×
4 6x x+ = 3
[ ]10 3x = 10÷
3
10
x = .
(b) Quadratic
(i) rearrange the equation if necessary to get 0 on the right.
If it can be factorised, do so (see 8. Algebra). Then:
(2 1)( 3) 0x x− + = means one of the brackets must be 0, so
2 1 0x − = or , which can be solved to give3 0x + =
1
, 3
2
x = −
If not, use
2
4
2
b b a
x
a
− ± −
=
c
5
4
and round the answers suitably.
(c) Simultaneous
2 linear equations:
(i) elimination
Multiply both equations until either the x’s or the y’s are the same then
add/subtract so that they disappear.
2 7
3 2
x y
x y
− =
+ =
multiply equation 1 by 2, then add:
4 2 1
3 2 5
x y
x y
− =
+ =
7 1x 9=
solve and substitute back in to equation 1 to find y.
(ii) substitution
isolate x or y from one equation and substitute its value into the other:
2 3
5 2
x y
y x
− =
= −
5
5
Use the expression for y in equation 2 and substitute it into equation 1:
, and proceed.2 3(5 2)x x− − =
18

one linear, one quadratic:
2 2
25
0.8
x y
x y
+ =
+ =
Rearrange the linear equation and substitute into the quadratic:
, so . Multiply out, and solve the quadratic in x.0.8y x= − 2 2
(0.8 ) 25x x+ − =
Note that each x answer will then produce a y answer, and this gives two pairs,
as it should because the equations represent the intersection of :
Questions
(a) Solve
1
1
3 2
x x−
− =
(b) Solve 2
2 1x x+ = 5
0(c) Solve 2
2 6x x+ − =
(d) Solve
1
2x
x
− =
(e) Solve the simultaneous equations 2 2
2 5
3
x y
x y
+ =
− = −
Answers
(a)
1
1
3 2
x x−
− = [× ]6
2 3(1 )x x− − = 6 6
9
2 3 3x x→ − + =
5x = , so
9
5
x = .
(b) 2
2 15x x+ − = 0
0
2
0
( 5)( 3)x x+ − =
5, 3x = − .
(c) , so look for two numbers which multiply to –12 and add to 1.
These are 4, -3.
2 6 1×− = −
So 2
2 4 3 6x x x+ − − =
2
(2 4 ) (3 6) 0x x x+ − + =
2 ( 2) 3( 2) 0x x x+ − + =
(2 3)( 2) 0x x− + = , which gives
3
2,
2
x = − .
19

20
1
2x
x
− = [× ]x(d)
2
1 2x x− =
2
2 1 0x x− − =
2
2 ( 2) 4 1 1
2 1
x
± − − × × −
=
×
2
2
±
=
8
{Note that 8 4 2 2 2= × =
and so 2 can be cancelled}: 1= ± 2 , so x = -0.41, 2.41 to 2 d. p.
(e) rearrange equation1 : 52 2
2 5
3
x y
x y
+ =
− = −
2x y= − , and substitute:
2 2
(5 2 ) 3y y∴ − − = −
2 2
25 20 4 3y y y∴ − + − = −
2
3 20 28y y 0∴ − + =
(3 14)( 2) 0y y∴ − − =
14
, 2
3
y∴ = . These lead to
13
, 1
3
x = − , so the two answers are
13 14
( , ) ( , ), (1,2)
3 3
x y = − .

10. Rearranging formulae top
(i) with a variable which only appears once, treat like an equation and isolate
the variable: e.g. make x the subject of
ax b
d
c
+
= : [ c× ] gives ,ax b cd+ =
[ ] gives , and finally [b− ax cd b= − a÷ ] gives
cd b
x
a
−
= .
(ii) with a variable which appears more than once, gather together and
factorise: e.g. [ax bx c= + bx− ] gives ax bx c− = , factorising
gives( ) , then [a b x c− = (a b)÷ − ] gives
c
x
a b
=
−
.
Questions
(a) The Centigrade and Fahrenheit scales are related linearly by
9
( 32
5
C F= − ) . Rearrange it to make F the subject.
(b) Make x the subject of
x a
b
x
−
=
Answers
(a)
9
( 32
5
)= −
)
C F [× ]5
5 9( 32C F∴ = − [ ÷ ]9
5
32
9
C
F∴ = − [ + ]32
5
32
9
F C∴ = +
(there are different ways to approach this, but all (correct) answers
will be equivalent even though they may look different)
(b)
x a
b
x
−
= [× ]x
x a bx∴ − = [ − ],bx a+
x bx a∴ − = [factorise]
(1 )x b a∴ − = [ ÷ − ](1 )b
1
a
x
b
∴ =
−
.
{ Note that
1
a
x
b
−
=
−
would also be correct, as top and bottom
are multiplied by –1}
21

11. Inequalities top
(a) linear
treat exactly like an equation, except if you multiply/divide by a negative
number, the inequality sign must be reversed. Avoid it!
(b) quadratic
e.g. First treat like an equation and factorise if possible (formula
otherwise): , then
2
4x <
2
4 0x − < ( 2)( 2)x x 0− + < . This gives “critical values” of
–2 and +2. Draw a number line, and a sketch of the function (in this case a
“happy” parabola) which reveals the region in which 2
4 0x − < :
y
x1 2 3– 1– 2– 3
2 2x− < < .
Had the question been , the answer would be2
4x > 2x < or 2x > .
(c) 2 variable linear inequalities
e.g. 3 2 . Plot the boundary line6 6x y− ≥ 3 2x y− = , then take a trial point
(e.g. the origin) to determine which side of the line to accept.
y
x1 2 3– 1– 2
1
2
– 1
– 2
– 3
– 4
The origin’s coordinates make 3 0 2 0× − × which is not ≥ , so that side is
rejected:
6
y
x1 2 3– 1– 2
1
2
– 1
– 2
– 3
– 4
22

y
x1 2 3 4 5– 1– 2– 3– 4
1
2
3
4
5
– 1
– 2
– 3
– 4
y
x1 2 3 4– 1– 2– 3– 4– 5
5
10
– 5
– 10
– 15
Questions
(a) Solve 2(1 ) 6x− <
(b) Solve 2
12 x x− ≤
(c) Find the 3 inequalities which identify this region:
Answers
(a) [ ]2(1 ) 6x− < 2÷
1 3x∴ − < [ ], 3x+ −
2x > −
(b) 2
12 x x− ≤ [rearrange]
2
12 0x x+ − ≥
( 4)( 3)x x 0∴ + − ≥ , giving critical values of –4 and +3.
so 4x ≤ − or 3x ≥
(c) The three line equations are
1
2 1, 2, 4
2
y x y x x y= + = − + = .
By considering a point (e.g. origin) in the shaded region, the inequalities are
2 1y x< + ,
1
2
2
y x> − , and 4x y+ < .
23

12. Parallel lines, bearings, polygons top
(a) Parallel lines
allied or interior add up to180°corresponding angles equalalternate angles equal
45°
B
A
N
exterior angle
interior angle
(b) bearings
Bearings are measured clockwise from North:
bearing of B from A is 135º
(c) polygons
for a polygon with n sides,
sum of interior angles = ( 2)180n − º
sum of exteriors = 360º
24

25
θ
b
a
A
B
C
D
E
A
B
C
N N
120°
aa
b
c
b
100°
120°
100°
100°
100° 100°
100°
120°
120° 120°
Questions
(a) In the diagram opposite, find
the value of θ in terms of a and b.
(b) The bearing of B from A is 090º,
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
180AED b= − (angles on a straight line). ADE θ= (opposite). We
now have the three angles in triangle ADE, so (180 ) 180a b θ+ − + = .
A rearrangement gives b aθ = − .
(b) Angle at point B means
360 90 120 150ABC = − − = º
Triangle ABC is isosceles, so 15BAC = º.
The bearing of C from A is therefore 105º.
(c) Sum of internal angles is ( 2)180 540n − = º
for any pentagon. A line of symmetry means
the set up is like this:
The only way of allocating 100º and 120º to a, b, c
and make a total of 540º is to have three 100º’s
and two 120º’s. So there are two possible pentagons:

13. Areas and volumes, similarity top
(a) Areas of plane figures
CIRCLE
A
p
PRISM
lr
h CYLINDER
r
h
lCONE
rSPHERE
hPYRAMID
v
r
π r
2
b
h
TRIANGLE
1
or
C A
B
a
b
1
2
bh
2
absinC
h
b
PARALLELOGRAM
TRAPEZIUM
h
b
a
bh 1
2
(a + b)h
(b) Surface area and volume
Shape surface area volume
Prism p l× A l×
Cylinder 2 rhπ 2
r hπ
Cone rlπ 21
3
r hπ
Sphere 2
4 rπ 34
3
rπ
Pyramid
1
3
× base area× h
Pipe flow: number of m3
/s flowing through (or out of) a pipe
= cross-sectional area × speed
26

(b) Similarity
Enlargement scale factor =
k
Area scale factor =
k
2
Volume scale factor =
k
3
1m
3m
10m
25m
Questions
(a) A cylinder has volume 100cm3
, 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)
(c) The empty swimming pool shown opposite
is to be filled with water. The speed of flow of
water in the pipe is 2m/s, and the radius of the
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) 2
5 100rπ =
2 100
5
r
π
∴ = , so
20
2.52r
π
= = cm. Whoops! Diameter asked for!
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
the base is therefore 2 21 1
10 20 5 10
3 3
π π× − × which factorises to
1
1750
3
π =
1830cm3
to 3sf
(c) Pool is a prism with cross section the side, which is a trapezium.
So volume of pool =
1
(1 3)25 10
2
+ × = 500 m3
.
Rate of egress of water is c.s.a. × speed = 2
5 200 5000π π× = cm3
, which
is 5000 m6
10π ÷ 3
/s. (Units!!) So time taken =
5
6 10
500 (5000 10 ) 31831π
π
÷ ÷ = = s, i.e. approx 8hrs 51 mins.
(d) assuming same density material, weight is directly proportional to volume.
The volume factor is 20, i.e. 3
20k = . 3
20k∴ = , and so the ratio of surface
areas,1 : , is2
k 1 : 7.37
28

θ
O
A
H
14. Trigonometry top
sin
O
H
θ =
cos
A
H
θ =
tan
O
A
θ =
Sine rule:
sin sin sin
a b c
A B C
= =
Cosine rule: 2 2 2
2 cosa b c bc A= + −
x
y
Two opposite pairs: use sine rule
yx
z
Three sides and one angle: use cosine rule
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:
y
x90 180 270 360
1
– 1
y
x90 180 270 360
1
– 1
y
x90 180 270 360
5
– 5
sinx cosx tanx
29

A B
C
D
40° θ
6cm
Questions
(a) ADB is a straight line of length 20cm. Find θ .
(b) In triangle ABC, AB = 5cm, BC = 8cm, and 30BCA = º. 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 sin 0.5x = −
30

B
CA
5 cm 8 cm
30°
N
N
5
6
O
A B
45°
y
x90 180 270 360– 90
1
– 1
-30
-0.5
y
x90 180 270 360– 90
1
– 1
-30
-0.5
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
sin 40
6
CD
= , º
6sin 40 3.85....CD∴ = =
º
cos 40
6
AD
= , º
6cos40 4.59.....AD∴ = = , so BD = 20 4.59... 15.4...− =
Then
3.85...
tan 0.250...
15.4...
θ = = , so º
14.1θ = to 3 s.f.
(b) If we target angle A then we have 2 opposite
side/angle pairs, so use the sine rule:
º
sin sin30
8 5
A
= , so , andºˆ 53.1..A =
ˆ 180 53.1... 30B = − − = 96.9º
(c) Using cosine rule,
2 2 2
5 6 2 5 6 cos135OB = + − × × × º
so OB = 10.2 miles (to 3 s.f.)
Now using the sine rule,
ºˆsin sin135
6 10.16...
AOB
= , which gives ºˆ 24.7AOB =
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.
So
1
tan
2
θ = . an d θ = 35.3º to 3 s.f.
(e) First find the principal value from
the calculator: -30º. Where are there
other angles in our window with the
same sine?
Clearly at 30º beyond 180º and 30º back from 360º.
So x = 210º, 330º
31

15. Circles top
(a) arcs, sectors, segments
θ
r
arc
sector
segment
Arc length = 2
360
r
θ
π×
Sector area = 2
360
r
θ
π×
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

33
θ
120°
35°
θ 40°
θ
25 cm
25 cmOIL
50 cm 25 cm
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) Find θ in the following diagrams:
(a) (b) (c)
Answers
(a) Arc length = 2 20
360
θ
π× and this is given as 10cm. Rearranging
gives
90
θ
π
= . Therefore sector area = 2
20
360
θ
π× = 290
20
360
π
π
×
which simplifies nicely to 100cm2
.
{Would you have reached for the calculator at
90
θ
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
split it down the line of symmetry:
This gives an angle of = 60º, and a base1
cos 0.5−
of 43.3…cm by Pythagoras. So the angle at the centre
of the sector is 120º. Therefore the area of the segment
is 2120
= 1535cm50 43.3... 25
360
π× − × 2
. The oil is in the
shape of a prism with volume 1535 200 307092× = cm3
= 0.307m3
.

θ
35°
θ
A
B
C
40°
θ
T
AB
C
(c) The angle subtended at the centre is
360º – 120º = 240º, so 120θ = º by the
angle at the centre theorem.
ˆ 35C = º (angles in the same segment)
ˆ 90CAB = º (angle in a semicircle)
so 180 35 90 55θ = − − = º (angle sum
of a triangle)
ˆ 40ABT = º (alternate segment theorem)
Isosceles triangle gives ˆ 70BAT = º, and
so Cˆ 70= º (angles in same segment)
34

16. Similar triangles, congruent triangles top
(a) Similar triangles
same shape, different size; related by enlargement (may be
different orientation)
SSS
SAS
AAS
ASA
RHS
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.
to prove congruent:
but not ASS – there are sometimes two different triangles with the same
ASS
35

6
8
6
4A
B
C
DE
A B
CD
X
Questions
(a) (i) Prove that triangles BCD and
ACE are similar. (ii) Hence find the lengths BD and
DE. (iii)If the
area of triangle BCD is 12 what is the area of the trapezium
ABDE?
(b) Use congruent triangles to prove that the diagonals of a
parallelogram bisect each other.
Answers
(a) (i) andˆ ˆEAC DBC= ˆ ˆAEC BDC= (corresponding). The third angle is
shared, so AAA is established and they are similar.
(ii) scale factor of enlargement is
12 3
8 2
= . So BD =
3
6
2
÷ , or
2
6
3
× = 4. CE is
3
6 , so DE is 9 – 6 = 39
2
× =
(iii) Area of triangle ACE = 23
12 {note area scale factor = k( )
2
× 2
}
= 27. So the trapezium has area 27 – 12 = 15.
(b) The two pairs of marked angles are equal
(alternate), and the top and bottom sides are
equal (parallelogram). So we have two
congruent triangles ABX and DCX by
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
1
k
(i) translation by vector shifts a to the right and b up.
a
b
⎛ ⎞
⎜ ⎟
⎝ ⎠
(ii) rotation about P through θ . [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 another pair, then where
the two perpendicular bisectors meet is the centre of rotation.
Alternatively, if it’s a 90º rotation, find the centre by trial and error then
confirm by using L shapes.
(iii) reflection through a line l.
To find the mirror line of a given
reflection, join a point to its image and
mark the mid-point . Repeat this with another
pair of points, and join the two mid-points to
form the mirror line.
(iv) enlargement from P with a scale factor k.
Note distance from centre of enlargement is multiplied by k.
37

To find a centre of enlargement, join a point to its image and extend
the line back. Repeat, and the centre is where the lines intersect.
y
x1 2 3 4– 1– 2– 3– 4
1
2
3
4
– 1
– 2
– 3
– 4
A
B
C
D
y
x1 2 3 4– 1– 2– 3– 4
1
2
3
4
– 1
– 2
– 3
– 4
E
F
y
x1 2 3 4– 1– 2– 3– 4
1
2
3
4
– 1
– 2
– 3
– 4
A
D A'
Questions
(a) What single transformation will carry
triangle A onto (i) B (ii) C?
(b) A “glide reflection” is a reflection
followed by a translation. A is transformed
onto D by a glide reflection, in which the
mirror line is . What is the vector1y x= −
of the subsequent translation?
(c) E is transformed onto F: state the
single transformation which accomplishes
this.
Answers
(a) (i) -90º rotation about (-1,2). (Check with L shapes)
(ii) reflection through the line 1y x= − −
(b) The diagram shows A reflected to
A’. The vector of translation necessary
to take A’ onto D is .
5
1
−⎛ ⎞
⎜ ⎟
⎝ ⎠
(c) Draw lines joining points with
their images, and extend them downwards.
They all meet at the centre of enlargement.
So it’s an enlargement, centre (-2, -4) with scale factor ½.
38

18. Loci and ruler and compass constructions top
(a) In 2-D: locus of points equidistant from:
1 fixed point is a circle
2 fixed points A and B is the perpendicular bisector of AB
A
B
C
Circumcentre
3 fixed points A, B and C the circumcentre of ABC
(b) locus of points equidistant from:
2 fixed lines is the angle bisector
incentre
3 fixed lines is the incentre of the
triangle
AB
C
Questions
(a) Construct the triangle ABC where AB = 8cm, BC = 5cm and
CA = 6cm. Construct the region of points within ABC which
are closer to AB than AC, and also closer to A than B.
(b) In 3-D, describe the locus of points exactly 1cm away from the
nearest point on a line segment AB.
Answers
(a) Using compasses, construct ABC
accurately. Then note that the boundary
lines for the two requirements are the
angle bisector of and the perpendicularˆBAC
bisector of AB, and the intersection of the
two regions must be selected.
(b) This is a cylinder of radius 1cm and axis AB,
and also two hemispheres of radius 1cm and centres
A and B. (or a hollow sausage, as we say in the trade).
39

19. Vectors top
Vectors are most easily seen as journeys for a particular distance
in a particular direction. location is not relevant, e.g. the opposite sides
of a parallelogram can both be represented by a .
a b+ : join the arrows nose to tail:
a
b
a+b
{Note in books and exam papers vectors
will be bold lower case letters without bars.
You write bars underneath –okay?}
a− : is a reversed
a
3a
ka where k is a scalar is a vector parallel to a , k times as long.
To get from A to B via given vectors, the route chosen
doesn’t matter - the expressions will all simplify down to
the same answer.
40

X
AB
C D
a
bc
O
B
A
K
a
b X
AB
C D
a
bc c
aY
Questions
(a) ABCD is a trapezium with AB parallel
to CD and CD = 2AB. The vectors
a , b , and c are defined as shown.
X is a point ¼ of the way along CD.
(i) Find two different expressions for in terms ofAX
→
a , b , and c .
(ii) Are the previous expressions really different? Explain.
(b) In triangle OAB, OA a
→
= and OB b
→
= .
The line OK strikes AB one third of the way
up, and OK is 1½ times as long as OX.
Find in terms of a and b :
(i) (ii) (iii) OK (iv) AKAB
→
OX
→ → →
What does the final answer tell you geometrically?
Answers
(a) (i) Going via B and C we get 1 1
4 2(2 )AX a c a a
→
c= − + + = − +
Via D, however, we get 3 3
4 2(2 )AX b a b
→
= − − = − − a .
(ii) These two expressions must be the same. So 31
2 2a c b a− + = − − ,
which simplifies to 0a b c+ + = . This means that each can be expressed
in terms of the others, so one is superfluous. This relationship can be seen
easily if we join A to the mid-point of CD and observe that there is a closed
triangle illustrating that 0a b c+ + = :
(b) (i) AB a b
→
= − + (ii) 1
3 ( )a a b
→
OX , which simplifies to= + − +
2 1
3 3a + b (iii) 3 2 1 1
2 3 3 2( )OK = + . (iv)a b a b
→
= +
1 1
2 2( )AK a a b
→
= − + + = b .
That AK is parallel to OB and half as long.
41

y
x
∆x
∆y
20. Straight line graphs top
gradient m =
y
x
∆
∆
y
x
y = mx
Equation of a straight line through the origin,
gradient m, is y mx=
y
x
c y = mx + cEquation of a straight line gradient m and
y-intercept c is y mx c= +
Equation of a straight line gradient m and
passing through (x1,y1) is 1 1( )y y m x x− = −
y
x
Intersecting lines: solve their equations
simultaneously to find the intersection.
Parallel lines have 1 2m m=
Perpendicular lines have 1 2 1m m = − , or 2
1
1
m
m
= −
y
x1 2 3– 1
1
2
3
– 1
Questions
(a) What is the gradient, and y- intercept, of 2 6 12x y 0+ + = ?
(b) Find the equation of this line in the form
ax by c+ = where the coefficients are integers.
(c) Where do the lines 3y x 5= − and
3 2x y+ = 6 intersect?
(d) A is (2,3), B is (5,6) and C is (4,0). Find the equation
of the line through C perpendicular to AB.
42

Answers
(a) Rearrange: [-2x, -12](a) Rearrange: [-2x, -12]2 6 12x y+ + = 0
2 [6 2 1y x= − − 6÷ ]
1
3 2y x= − −
so the gradient is 1
3− and the y-intercept is –2.
(b) gradient = 2
3− and y-intercept = 2.
The equation is 2
3 2y x= − + [ 2
3 x+ ]
2
3 2x y+ = [ 3× ]
2 3x y+ = 6
(c) Solve simultaneously. In this form, substitution would be easier:(c) Solve simultaneously. In this form, substitution would be easier:
sub (1) into (2): 3 2 which gives(3 5)x x+ − = 6
16
9
x = . Sub back into
(1) gives
1
3
y = . So the intersection is
16 1
( , )
9 3
.
(d) gradient AB is
6 3
1
5 2
−
=
−
. Perpendicular gradient is
1
1
1
− = − .
So the required equation is y x c= − + but what is c? Get this by
substituting the coordinates of a point on the line, i.e. C.substituting the coordinates of a point on the line, i.e. C.
0 4 c→ = − + , giving , and the equation is4c = 4y x= − +
Answers
2 6 12 0x y+ + =
6 2 12y x= − − 6[ ÷ ]
1
3 2y x= − −
so the gradient is 1
3− and the y-intercept is –2.
2
3−(b) gradient = and y-intercept = 2.
2
3 2y x= − + 2
3 xThe equation is [ + ]
2
3 2x y+ = 3[× ]
2 3 6x y+ =
sub (1) into (2): 3 2 which gives(3 5) 6x x+ − =
16
9
x = . Sub back into
1
3
y =
16 1
)
9 3
(1) gives . So the intersection is ( , .
6 3
1
5 2
−
=
−
(d) gradient AB is . Perpendicular gradient is
1
1
1
− = −
y x c
.
So the required equation is = − +
0 4 c→ = − + 4c = 4y x
but what is c? Get this by
, giving , and the equation is = − +
43

21. More graphs top
graphs of 2 3 1
, , , , x
x x x k
x
and 2 2 2
x y r+ =
y
x
y = x
2
– x + 1
y
x
y = x
3
– x
(a) x as above, linear
(b) parabolae2
x
(c) cubics(!)3
x
y
x
y =
2
x
1
x
(d) hyperbolae
y
x1 2– 1– 2
1
2
3
4
y = 2
x
(e) , where and x is an integer0k >x
k
y
x
x
2
+ y
2
= 1
y
x1 2– 1– 2
1
2
3
4
y = x
2
y = x + 2
y
x1 2– 1– 2
1
2
3
4
y = x
2
y = 3
(f) 2 2 2
x y r+ = circle radius r, centre origin
Solving equations using graphs:
(i) Draw the graph of and on the same grid2
y x=
y = 3. What equation do the intersections solve?
Solving simultaneously, 2
3x = , i.e. the x-values
at the intersections are solutions of 2
3x = , i.e.
they are 3± .
If is drawn, the x-values at the2y x= +
intersections are solutions to 2
2x x= + ,
i.e. , which could be factorised2
2 0x x− − =
to , giving( 2)( 1)x x− + = 0 1, 2x = − , which
can be seen on the graph.
(ii) What other graph should be drawn on the
44

same grid as to see the solutions of2
y x=
2
3 1 0x x+ − = ? Unravel this to 2
3 1x x= − +
and so we need to draw the line 3 1y x= − + .
y
x1 2– 1– 2
1
2
3
4
y = x
2
y
x1 2– 1– 2
1
2
3
4
5
– 1
y = x
2
y = 4 – x
2
number of bacteria
x1 2 3 4 5
1000
2000
3000
Questions
(a) Plot 2
y x= and on the same grid and find the x-values2
4y = − x
5
of their intersections. To what equation are these the solutions?
(b) What are the x-coordinates at the intersections
of these two graphs? What equation is being solved
approximately by these two numbers?
(c) A colony of bacteria double in number daily,
after starting with 100 individuals. State the number
of bacteria after (i) 1 day (ii) 2 days (ii) 3 days (iv) 4 days (v) x days
Sketch the graph of the number of bacteria against x, the number of days
after the start, for . Estimate (a) when the colony has grown to0 x≤ ≤
2500 (b) the rate of growth when x = 3.
Answers
(a) Intersection x-values are approx. –1.4 and 1.4.
At intersection, and2
y x= 2
4y x= − . Solving
simultaneously, 2 2
4x x= − , 2
2x 4∴ = , so the
equation is . (which means these two values2
2x =
of x are actually 2± )
(b) At the two intersections, x = -0.6 and x = 1.6 (approx).
For the line, m = 1 and c = 1, so its equation is 1y x= + .
Therefore the equation representing x-values at intersection
is 2
1x x= + , i.e. .2
1 0x x− − =
(c) (i) 200 (ii) 400 (iii) 800 (iv) 1600 (v) 100 2x
×
(a) 2500 are attained after about 4.7 days, (b) the gradient of the tangent
at x = 3 shows the rate of growth at that moment, and is about 550
bacteria/day.
45

22. Distance, velocity graphs top
{We are really dealing with displacement, i.e. how far along a certain route,
usually a straight line, from an origin. e.g. going round a complete circle
would represent 0 displacement, but 2πr of distance}
(a) Displacement – time
s
t
A
B
v
t
A
B
y
x
d
y0
y1
y2
yn – 1 yn
Gradient measured between A and B
= average velocity
Gradient of a tangent
= velocity at that point.
(b) Velocity – time
Gradient measured between A and B
= average acceleration
Gradient of a tangent
= acceleration at that point
Area between the curve and x-axis = displacement
{note: under the x-axis, area counts negative}
(c) Trapezium rule
an approximate method for counting
area under a curve:
Area { 0 1 12 ....... 2 }
2
n n
d
y y y y−≈ + + + + .
This replaces each strip with a trapezium, i.e.
the top becomes a straight line segment, and will
under- or overestimate the true area.
46

v
t
V
T 3T
2V
v
t0.5 1
0.5
1
Questions
(a) An object moves in a straight line so that its velocity
after time t seconds is given by 2
v t= . Find
(i) its average acceleration over the first second
(ii) its instantaneous acceleration at 1
2t =
(iii) the distance it covers using the trapezium rule with 4 strips.
(b) In the journey represented in the
diagram, the total distance covered was
60m, and the acceleration over the first
part was 5 ms-2
. Find the values of V and T.
Answers
(a) (i) average acceleration = change in
velocity over time taken = 1 m/s per s
= 1ms-2
.
(ii) acceleration at t = ½ is gradient of
tangent there, i.e. 1ms-2
.
(iii) Using trapezium rule,
distance
{ }2 2 20.25
0 2 0.25 2 0.5 2 0.75 1
2
≈ + × + × + × + 2
= 0.34375, or
0.34 m to 2 s.f. {Note that 0.34375 is an overestimate due to the
concave curve}
(b) Splitting into two trapezia and a triangle, area under curve
= 1
2
1 1
which = 3 . So
1
( 2 ) (2 )
2 2 2 2
V
V V T V V T T+ + + + 0VT 3 6VT =
Acceleration on first part =
V
T
which = 5. Substituting gives
which leads to T = 2, and V = 10.2
5 2T = 0
47

23. Sequences; trial and improvement top
(a) Sequences
Numbers in a sequence u receive the names 1 2 3, , ,.... ,.....nu u u u
A sequence may be defined directly: 3 1nu n= + (that is 4, 7, 10, ….)
or recursively: and12 3n nu u −= − 1 4u = (that is 4, 5, 7, 11, 19, 35,…)
special sequences:
(i) Triangle numbers 1, 3, 6, 10, 15, 21, 28, 35, ……
where 1
2 ( 1n )nu n= +
(ii) Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, …..
where u u1 21, 1, and 1 2n n nu u u− −= += =
(b) trial and improvement
to solve an equation or maximise/minimise a quantity
e.g. Solve x x correct to 1.d.p.3
1 0− − =
x 3
1x x− −
0
1
2
1.5
1.3
1.4
1.35
-
-
+
+
-
+
+
We’ve established there is a zero between 1.3 and 1.4, but which
figure do we quote? Must go halfway, i.e. 1.35 to indicate. Answer is
between 1.3 and 1.35, so when rounded it will definitely be 1.3x =
48

Questions
(a) . What is (i) the 103 7nu n= − th
term (ii) the first term over 1000?
(b) Suggest a formula for –5, 2, 9, 14, 21, 28, ….
(c) Find the number of straight lines joining n dots, and prove your formula.
(d) In a game, a counter can move
either one or two spaces on each turn. How many different ways are there
for the counter to get from the 1st
square to the 10th
square?
(e) Find, to 1 d.p. the value of x which minimises the function 2
2x
x +
Answers
(a) (i) .10 3 10 7 23u = × − =
(ii) Need , i.e. 31000nu > 7 1000n − > which solves to 2
3335n >
so the first term is the 336th
and is 336 1001u =
(b) u n , (or, not so impressive,7 12n = − 1 7n nu u −= + and u )1 5= −
)−
(c) Each dot radiates n -1 lines to the other
dots, and as there are n dots to radiate lines
from, there are n n lines: except that we( 1
have counted every line exactly twice over.
So the number of lines between n dots is 1
2 ( 1n n )−
(d) We are having to advance the counter 9 places. Let the number of ways
of advancing it n places be called u , (and we need to find .)n 9u
The first move is either a 1 or a 2, after which the number of ways
remaining to get to the end is u 1n− or 2nu − respectively. So u u 1 2n n nu− −= +
and the sequence is our old friend the Fibonacci. Noting that 1 1u =
and 2 2u = , the sequence must go 1, 2, 3, 5, 8, 13, 21, 34, 55, …. and
9u is 55
49

y
x1– 1
1
2
3
y = x
2
+ 2
x
(e) To get an idea where to look see sketch:
The minimum is around x = -0.3
x 2
2x
x +
-0.5
-0.4
-0.3
-0.2
0.957..
0.917..
0.902..
0.910..
So far, we are assuming there is a simple minimum, but all we know
is that it’s somewhere in the vicinity of –0.3 – it may well not be closest
to that value at 1 d.p., so we need to go to a finer division:
x 2
2x
x +
-0.29
-0.28
-0.27
0.902..
0.901..
0.902..
We now know it’s between –0.27 and –0.29, so rounded to 1 d.p.
the value of x is indeed –0.3
{Provided the function is a straightforward one with no funny business}
50

24. Graphical transformations top
For any graph ,( )y f x=
y
x180 360
1
– 1
y
x180 360
1
– 1
y
x180 360
1
– 1
y = sinx y = sin
x
2
y = sin 2x
stretch by 2 in x stretch by 1/2 in x
(y f x a= − ) a translation of a steps in the + x directiona→ +
( )y a f x− = a translation of a steps in the + y directiona↑ +
{i.e. }( )y f x a= +
( )
x
y f
a
= a stretch by factor a in the + x directiona↔ ×
( )
y
f x
a
= a stretch by factor a in the + y directiona×
{i.e. }( )y af x=
e.g.
multiple transformations:
this is truly tricky. It’s different from techniques in rearranging formulae:
you must do a sequence of steps in the formula, each time replacing x (or y)
with something new, eventually to get the required equation
e.g. what transformation must be performed on the curve 2
y x= to obtain the
following: (a) (b)2
( 1)y x= − + 3 21
2( )
3
x
y
−
=
(a) starting with , replace x by2
y x= 1x − { }, then replace2
( 1)y x→ = −
y by { }. So the original parabola must be moved 1 step
to the right then 3 steps up.
3y − 2
3 ( 1)y x→ − = −
51

(b) Instinct says replace x by x-1 first, but it don’t work! Starting with 2
y x=
Replace x by
3
x
{ 2
( )
3
x
y→ = } then replace x by 1x − { 21
(
3
x
y
−
→ = ) } and
finally replace byy
2
y
{ 21
(
2 3
y x
)
−
→ = }. Thus the transformations are: a
stretch in the x direction by factor 3, then a translation by +1 in the x direction,
and finally a stretch by factor 2 in the y direction, illustrated here:
y
x1 2 3– 1– 2
1
2
y
x1 2 3– 1– 2– 3
1
2 y
x1 2 3– 1– 2
1
2 y
x1 2 3– 1– 2
1
2
y = x
2
y =
⎝
⎛
⎜
x
3 ⎠
⎞
⎟
2
y =
⎝
⎛
⎜
x – 1
3 ⎠
⎞
⎟
2
y = 2
⎝
⎛
⎜
x – 1
3 ⎠
⎞
⎟
2
y
x180 360
1
2
– 1
– 2
y
x180 360
1
2
– 1
– 2
y
x180 360
1
2
– 1
– 2
y
x180 360
1
2
– 1
– 2
y = cos 2x y = 2cosx y = cos
x
2
– 1
y
x1 2 3 4– 1– 2
1
2
3
4
5
– 1
y
x1 2 3 4– 1– 2
1
2
3
4
5
– 1
Questions
(a) The graph of is shown. On the empty grids, sketch the graphs of
(i) (ii) (iii)
cosy = x
cos 2y x= 2cosy x= cos 1
2
x
y = −
(b) On the left grid is the graph of
1y x= + . (i) Perform a stretch on this
by a factor of 2 in the y direction,
drawing the result on the empty grid.
(ii) Show algebraically what effect
this stretch has on the equation of
the line
(c) Describe the transformations of the curve 2
y x= which result in a
curve with equation
2
1
1
2 3
x
y
⎛ ⎞
= +⎜ ⎟
⎝ ⎠
52

y
x180 360
1
2
– 1
– 2
y
x180 360
1
2
– 1
– 2
y
x180 360
1
2
– 1
– 2
y
x180 360
1
2
– 1
– 2
x
2
– 1
y
x1 2 3 4– 1– 2
1
2
3
4
5
– 1
y
x1 2 3 4– 1– 2
1
2
3
4
5
– 1
Answers
(a)(a)
(b) How do you stretch? Pick a point, measure its distance from the(b) How do you stretch? Pick a point, measure its distance from the
invariant x-axis, then double it.invariant x-axis, then double it.
The ensuing line will have equation 1
2
y
x= + , i.e. 2y x 2= + , and this
is confirmed by the diagram.is confirmed by the diagram.
(c) a fiendish trap. Suppress the urge to divide x by 3 first (as you would do in
a calculation):
(c) a fiendish trap. Suppress the urge to divide x by 3 first (as you would do in
a calculation):
replace x by x +1: . Next, replace x by2
( 1)y x→ = +
3
x
: 2
( 1)
3
x
y→ = ,+
finally replace y by 2y: 2
2 ( 1)
3
x
y→ = + which is it. So the transformations are:
translate by –1 in x direction, then stretch by factor 3 in the x direction, then
stretch by factor ½ in the y direction.
53
y
x180 360
1
2
– 1
– 2
y
x180 360
1
2
– 1
– 2
y
x180 360
1
2
– 1
– 2
y
x180 360
1
2
– 1
– 2
x
2
– 1
y
x1 2 3 4– 1– 2
1
2
3
4
5
– 1
y
x1 2 3 4– 1– 2
1
2
3
4
5
– 1
Answers
1
2
y
x= + 2 2y xThe ensuing line will have equation , i.e. , and this= +
2
( 1)y x→ = +
3
x 2
( 1)
3
x
y +replace x by x +1: . Next, replace x by : → = ,
2
2 ( 1)
3
x
y→ = +finally replace y by 2y: which is it. So the transformations are:
translate by –1 in x direction, then stretch by factor 3 in the x direction, then
stretch by factor ½ in the y direction.
53

25. Probability top
One definition of the probability of an event is the limit to which
the relative frequency (no. of successes ÷ no. of trials) tends as the
no. of trials .→ ∞
So 7 tails from 10 flips of a fair coin gives 0.7 relative frequency, as does
700 tails out of 1000 flips. However, one would expect by 1000n = to have
converged more closely to 0.5. That would cast doubt on the fairness of the
coin.
(i) For mutually exclusive events A and B, ( ) ( ) (P Aor B p A P B)= +
red
blue
first
pick
second
pick
1
3
2
3
9
14
5
14
10
14
4
14
(ii) For independent events A and B, ( ) ( ) (P Aand B P A P B)= ×
(iii) Probability trees can illustrate combined events well:
e.g. a bag contains 5 blue balls and 10 red balls. Pick one ball at random, keep
it out, then pick another ball. What is the probability of one of each colour?
The two nodes corresponding to one of each colour are marked.
The probability is
2 5 1 10
3 14 3 14
× + × which is
10
.
21
Alternatively to a tree, just spell out the sequence of events:
P(one of each colour) = P(R1 and B2 or B1 and R2) and use the
addition and multiplication laws to get the same result.
54

12
3
4
5
1 2 3 4 5 6
1
2
3
4
5
6
A
B
Questions
(a) Draw a table of results for the rolling of two dice.
What is the probability that (i) the difference is 2 (ii) the total is 6
(iii) the difference is 2 or the total is 6 ?
(b) A teacher picks 2 pupils at random to be class representatives out of a class
with 10 boys and 12 girls. What is the probability that
(i) they are both boys (ii) there is at least 1 girl ?
(c) A game consists of three turns of an arrow which lands randomly
between 1 and 5, with the scores are added together.
A prize is given for a final score of 14 or more
What is the probability that (i) a player scores the same
number on each turn (ii) a player wins a prize ?
Answers
(a) The difference being 2 is shown with dots
while the total being 6 is shown with rings.
(i) P(difference = 2) =
8 2
36 9
= .
(ii) P(total = 6) =
5
36
(iii) P(difference = 2 or total = 6)? Cannot use the addition law directly
here because they are not exclusive. (there’s an overlap). Just counting
gives
11
36
= .
(b) (i) P(B1 and B2) =
10 9 15
22 21 77
× = (ii) P(at least 1 girl) =
1 – P(both boys) =
62
77
.
(c) (i) P(111 or 222 ……..) = 3 31 1
( ) ( ) ......
5 5 2
+ + =
1
5
(ii) P(prize) = P(554 or 545 or 455 or 555) = 3 31 1 4
( ) ( ) ......
5 5 12
+ + =
5
.
55

26. Statistical calculations, diagrams, data collection top
(a) calculations
(i) averages:
mean = ix
n
∑
median = value of the middle item when listed in order
mode = most commonly occurring value
(ii) measures of spread:
range = max – min
Interquartile range = Upper quartile – lower quartile
Quartiles in small data sets: fiddly and pointless, but here we go. Median is
found. If the number of data was even, split the data into two sets; if the
number of data was odd, ignore the median and consider the remaining
values as two sets. Then the quartiles are the medians of the two remaining
sets.
(b) diagrams
oilgas
nuclear
others
(i) pie chart for categoric data (non-numerical) e.g. modes of
transport used to school
no of runs5 152535455565758595
frequency
2
4
6
8
(ii) frequency diagram
(iii) moving average in a time sequence, the mean of the last 10
(say) values is calculated then plotted. This
smooths out short term fluctuations so that a
long term trend may be seen.
56 Maths %
60 65 70 75 80 85 90 95
Latin%
60
65
70
75
80
85
90
(iv) scatter graphs to see correlation between two variables

age(years)60
cum freq
50
1
4
up
3
4
up
medianlower
quartile
upper
quartile
age(years)0 60
min
l.q.
median
u.q.
max
(v) stem and leaf diagrams the data is transcribed straight from a
table onto the stems: this is a back-to-
back stem and leaf.
Maths
9
8
7
6
56
1
458
0222568
025
088
0022558
0258
0
3 6key:
means 36%
Latin
(vi) cumulative frequency curve: grouped data with frequencies is turned
into cumulative frequencies thus:
x freq. x cum.freq
0<x 10≤
10<x≤20
20<x≤30
…….
5
8
12
……
→
0<x≤10
0<x≤20
0<x≤30
……
5
13
25
……
and the cum freq’s plotted at the right end of the interval.
(vii) box plot
57

lass(vii) histogram: no gaps allowed. If the data is integer valued, the c
boundaries will be between integers.
Height of block is not frequency, but
frequency density = freq ÷ width
tion
(i) sampling: random sampling – the population is in a numbered list,
ample
(c) data collec
then adapt random numbers to select members
repeatedly without inherent bias. e.g. a random s
of size 50 to be selected from the school population of
830: take a Ran# from calculator, 1000× , discard if
over 830, otherwise choose that me of the list.
Repeat 50 times.
mber
stratified sampling – when the population is divided into
ize 50.
strata and you wish each stratum to be represented in
the sample proportionately to its size. Calculate the
sizes and then sample randomly within each stratum.
e.g. in a prison in the age groups 18 - 25, 26 - 40, 41 –
60, and 61- 100 there are 100, 300, 250 and 150
inmates. You wish to take a stratified sample of s
From the 18 – 25 group you must take a random sample
of size
100
50
800
× , i.e. 6 (nearest integer), and so on.
(ii) questionnaires : no vague or leading questions. No
a
questions which could have a variety of possible
responses, better yes/no or tick boxes, or score on
scale of 1 to 5, say.
58

27. Functions top
Functions are rules which require an input, x, and give a single output, ( )f x , (also
called y). So for example, pressing a calculator button performs a function.
Domain
This is the set of input values. This may be given in a question, or you
may have to find the natural domain, that is the set of all possible
input values.
Tha natural domain of ( ) 3f x x= − is , since any values of x
below 3 do not give a real output.
3x ≥
x-1 1 2 3 4 5 6 7 8 9 10
y
-1
1
2
3
Range This is the set of output values.
For ( ) 3f x x= − , the range is all the numbers between 0 and infinity,
i.e. (or ).( ) 0f x ≥ 0y ≥
{Note that the function only gives the positive square root}
Composing functions
If the output from one function f is used as the input fro another
function g, giving the composite function (said as “ g of f
of x”).
( ( ))g f x
For example, if ( ) 2 1f x x= + and ( ) 3 2g x x= − ,
then g f is which = 7.( (1) (3)g)
For this pair of functions, more generally, ( ( )) 3(2 1) 2g f x x= + − ,
which can be simplified to ( ( )) 6 1g f x x= + .
Inverse
The inverse of a function, called 1
( )f x−
, reverses the action of the
function. e.g. with ( ) 2 1f x x= − , (3) 5f = , so should = 3.1
(5)f −
To find a formula for the inverse of ( )f x , call this y, and rearrange the
formula so that x is the subject.
2 1
1 2
1
2
y x
y x
y
x
= −
∴ + =
+
∴ =
So 1 1
( )
2
y
f y− +
= , but this is normally rewritten as 1 1
( )
2
x
f x− +
= ,
since the input number to any function is usually called x. Check with
the above example,
59

1 5 1
(5) 3
2
f − +
= = , which is correct!
Questions
(a) Find the natural domains of :
(i) (ii)( ) 3 2f x x= −
1
( )f x
x
= (iii) ( ) 5f x x= −
(iv)
1
( )
2 2
f x
x
=
−
(v) ( ) (4 )f x x x= −
(b) Find the ranges of all the functions in q.1
(c) If and( ) 5 3f x x= −
3
( )
2
x
g x
x
−
=
+
, find a simplified expression for
.( ( ))g f x
(d) Find the inverse of (i)
2 3
( )
5
x
f x
+
= (ii)
3
( )
1
x
h x
x
−
=
+
60

Answers
(a) (i) the entire real line
(ii) all real numbers except 0
(iii) 5x ≤
(iv) 1x >
x-1 1 2 3 4
y
(v) the inside bit needs to be . This means0≥ (4 ) 0x x− ≥ , which is a
quadratic inequality:
it requires .0 4x≤ ≤
(b) (i) the entire real line
(ii) all real numbers except 0
(iii) 0y ≥
(iv) 0y >
(v) 0 2y≤ ≤
(c)
(5 3) 3
( ( ))
(5 3) 2
x
g f x
x
− −
=
− +
=
5 6
5 1
x
x
−
−
(d) (i)
2 3
5
x
y
+
=
5 2y x= + 3
5 3 2y x− =
5 3
2
y
x
−
= , so 1 5 3
( )
2
x
f x− −
= .
(ii)
3
1
x
y
x
−
=
+
( 1) 3x y x+ = −
3xy y x+ = −
3xy x y− = − −
( 1) 3x y y− = − −
3
1
y
x
y
− −
=
−
. Why all these minuses? Let’s multiply top and bottom by -1, and
1 3
( )
1
x
h x
x
− +
=
−
.
61

top
28. Calculus
Differentiation If , then( )y f x=
dy
dx
(or ) is the name of the gradient
function of y.
'( )f x
How to differentiate: If n
y x= then 1ndy
nx
dx
−
=
Constants multiplying a power of x remain:
If then2
10y x= 10 2 20
dy
x x
dx
= × =
Rate of change
dy
dx
is the rate of change of y with respect to x. So, for
example:
(i) on the graph of y against x,
dy
dx
represents the gradient.
(ii) if h metres is the vertical height of a ball after t seconds,
then
dh
dt
is the vertical velocity of the ball in m/s.
(iii) if P is the price of a share,
dP
dt
is the rate of change of the
share price.
Max/Min Put 0
dy
dx
= and solve.
At a max or min the gradient will be 0.
Is the stationary point ( 0
dy
dx
= ) you have located is a max or a
min? To determine this, factorise the gradient function if
possible, and calculate the gradients at the stationary point, and
also nudging a little to the left and a little to the right:
e.g. On 3
3y x x= − show that there is a staionary point at
and determine its nature.
1x =
2
3 3
dy
x
dx
= −
when ,1x = 2
3 1 3 0
dy
dx
= × − = , so there is a stationary point.
What type is it?
dy
dx
can be factorised to 3( 1)( 1)
dy
x x
dx
= − + .
1-
1+
1x
dy
dx
0- +
62

The diagram shows that we have a minimum.
{For 1-
, imagine substituting a number just less than 1 for x in
the expression 3( 1)( 1)
dy
x x
dx
= − + , and it’s clearly going to be
negative, etc}
Questions
(a) Differentiate the following functions:
(i) 3
x (ii) 17.5 (iii) 2
5x (iv) 3 ( 1)x x +
(v)
2
x
(vi)
1x
x
+
(b)Show that the curve 3
y x x= + has no stationary points.
(c) Find the coordinates of the stationary points on
, and determine whether the right hand one
is a max or a min.
3 2
3 9y x x x= + − + 5
(d)The displacement of a toy car during the first 10 seconds after
release is given by
3
2
15
t
s t= − . Find (i) the car’s speed after 2
seconds (ii) the maximum speed of the car.
63

Answers
(a) (i) 3 2
x (ii) 0 (iii) 10 (iv) ( )x 2
3f x x x= + so
'( ) 6 1f x x= + (v) ( ) 1
2f x x−
= so 2
'( ) 2 1f x x−
= ×− , i.e. 2
2
x
−
.
(vi)
1x
x
+
needs to be simplified to
1
1 , that is 1 1
x−
+
x
+ . So
2
'( ) 1f x x−
= − , i.e. 2
1
x
−
.
(b) 2
3
dy
x
dx
1= + . Now 2
3x is always at least 0, so
dy
dx
is always at least 1.
So there are no stationary points.
(c) 2
3 6
dy
9x x
dx
= + − . For stationary points, 3 62
9 0x x+ − = . Therefore
2
2 3 0x x+ − =
( 3)( 1) 0x x+ − =
3x∴ = − or 1.
Plugging back into the original equation gives coordinates as (-3, 32) and
(1,0).
Is (1,0) a max or min?
Using 3( 3)( 1)
dy
x x
dx
= + − ,
64
1-
1+
1x
dy
dx
0- +
so we have a minimum.
(d)
ds
v
dt
= , so
2
2
5
t
v t .= −
(i) After 2 seconds,
2
2
2 2
5
v = 3.2 m/s.= × −
(ii) To find the maximum value of v, we need to differentiate the
expression for v.
2
2
5
dv t
dt
= − .
Putting this = 0 solves to 5t = . Back into the formula for v gives the
maximum value of v as
2
5
2 5
5
× − = 5 m/s.

top29. Sets
Sets are groups of elements.
A small set may be shown on a Venn diagram, e.g. given the universal set is positive
integers less than 12, and A is the set of primes:
ξ 1
2
3 4
5
6
7
8
9
10
11
A
A B∩ : the intersection of A and B
A B∪ : the union of A and B
'A : the complement of A
A BA B⊂ : A is a subset of B
6 A∈ : 6 is a member of the set A
∅: the empty set
Intersections are overlaps, unions are all elements in one or the other or both.
65

Questions
(a) Shade the set 'A B∩ :
(b) A is the set of animals, B is the set of black objects, and C is the set of
cats.
(i) Translate into normal English: B C∩ ≠ ∅
(ii) Describe the set 'B A∩
(iii) Is a white mouse a member of the set '( )A B C∩ ∪
)'B∩ ∪
?
(c). In a class of 25, 12 play football, 15 play water polo, but 10 do neither
sport. How many play both football and water polo?
(d) ξ is the set of all employed people in England. A is the set of those with a
bank account. B is the set of those with a building society account. C is the set
of people who work in the catering industry.
(i) Shade the set of those in catering with a bank account but no building
society account, and describe this in set notation.
(ii) Shade the set C A , and describe the members of this set.(
A B
66

Answers
(a)
(b) (i) there do exist black cats
(ii) black inanimate objects
(iii) the white mouse is certainly not a member of black objects or cats, and it is an
animal, so yes!
(c) let’s call the number in the intersection x. Then the numbers in the other
compartments can be calculated:
Now using the fact that they represent 25 people altogether:
(12 ) (15 ) 10 25x x x− + + − + = , so 12x = ,
i.e. there are 12 people who play both football and water polo.
(Note also that there are none in the left hand compartment, i.e. ( ')n F W 0∩ = , which
means that in this case . That is, all who play footballF W⊂ also play water polo.)
(d) (i) this is 'A C B∩ ∩
(ii)
The members of '( )C A B∩ ∪ are those in the catering industry without a bank or
building society account.
A B
12-x x 15-x
10
F W
A B
C
A B
C
67

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