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Formula List

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Formula List
NUMBERS
1. Natural number (𝑡): counting numbers.
Example: 1 , 2 , 3 , 4 , … …
2. Integer (β„€): Positive and negative whole numbers.
… … − 3 , −2 , −1 , 0 , 1 , 2 , 3 … …
Example:
β„€−
β„€+
Neutral Integer
3. Real number (ℝ):
1
−3.5 , −2 , 0 ,
2
Can be placed in number line.
Example:
, πœ‹ , √2 … …
4. Unreal number:
Cannot be placed in the number line
5. Square number:
1, 4, 9, 25, 36, 49, … …
6. Cubic number:
1, 8, 27, 64, 125, 216, 343 … …
1
0
Example: √−2 , , 𝑒𝑑𝑐
7. Prime number (𝑷): ∎ Integer starts from 2
∎ Divisible by 1 & itself
∎ Has exactly two factors. Note: 1 is not a prime number.
Example: 2 , 3 , 5 , 7 , 11 , 13 , 17, 19 … …
Factor of 6
8. Factors of πŸ”:
1, 2, 3, 6
9. Multiples of πŸ”:
6, 12 , 24 , 36 , 48 , 60, … …
10. Even numbers:
−4, −2, 0 , 2 , 4 , 6 , 8 , … …
11. Odd numbers:
−5, −3, −1, 1 , 3 , 5 , 7 , 9, … ….
1 , 2 , 3 , 6 , 12 , 18 … …
Multiple of 6
π‘Ž
12. Rational Number (𝑸): A number that can be converted into fraction of integers (𝑏 ) .
Example:
0
0
=1
2.5
=
25
10
−5.1
=−
51
10
√16
4
=1
π‘Ž
13. Irrational Number (𝑸′ ): A number that cannot be converted into fraction of integers (≠ 𝑏 ) .
14. Product of prime factors: [Example]
2 48
2 24
2 12
2 6
3
∴ 48 = 2 × 2 × 2 × 2 × 3
= 24 × 3
2 80
2 40
2 20
2 10
5
∴ 48 = 2 × 2 × 2 × 2 × 5
= 24 × 5
Formula List
SEQUENCE
∎ Type – 1 (Consecutive numbers):
π‘»πŸ
+2
π‘»πŸ
π‘»πŸ‘
4
5
π‘»πŸ
π‘»πŸ‘
−1
0
3
π‘»πŸ
−3
−2
𝑻𝒏
…..
+2
𝑛+2
𝑻𝒏
…..
−3
𝑛−3
∎ Type – 2 (Common Difference):
π‘»πŸ
π‘»πŸ
π‘»πŸ‘
π‘»πŸ’
7
9
11
13
2
2
𝑻𝒏
π‘Ž + (𝑛 − 1)𝑑
2
= 7 + (𝑛 − 1) × 2
π‘»πŸ
π‘»πŸ
π‘»πŸ‘
π‘»πŸ’
π‘»πŸ“
6
10
16
24
34
4
6
2
8
2
10
2
𝑻𝒏
π‘Ž + (𝑛 − 1)𝑑1 +
(𝑛−1)(𝑛−2)
= 6 + (𝑛 − 1) × 4 +
2
𝑑2
(𝑛−1)(𝑛−2)
2
×2
Formula List
GEOMETRY
Acute Angle
0 < πœƒ < 90°
Obtuse Angle
90° < πœƒ < 180°
Reflex Angle
180° < πœƒ < 360°
Right Angle
πœƒ = 90°
πœƒ
πœƒ
πœƒ
πœƒ
External ∠ of βˆ†
Triangle
Allied Angle
Parallelogram
Quadrilateral
𝑏
π‘Ž
𝒂=𝒃
∠𝒔 centred at a point
Vertically opposite angles
Adjacent ∠𝒔 on a straight line
Adjacent ∠𝒔 on a right angle
Corresponding angles
Alternate angles
Formula List
CIRCLE THEOREM
𝐴
𝑦°
π‘₯°
π‘₯°
2𝑦°
𝑧°
.π‘œ
.π‘œ
2π‘₯°
𝐴
𝐡
𝐢
𝐡
𝑦°
𝐴
𝐡
∠𝐴𝐢𝐡 = 90°
∠𝐴𝑂𝐡 = 2π‘₯
π‘₯° = 𝑦° = 𝑧°
Reflex ∠𝐴𝑂𝐡 = 2𝑦
𝐡
𝐴
𝐡
𝐴
𝑏°
π‘Ž°
𝑑°
π‘Ž°
𝑐°
𝐷
𝐷
.π‘œ
𝑐°
𝐢
π‘₯°
𝐴
𝐢
π‘Ž° + 𝑐° = 180°
𝐡
𝐢
∠𝐴𝐢𝑂 = ∠𝐡𝐢𝑂 = 90°
π‘Ž° = π‘₯°
𝑏° + 𝑑° = 180°
𝐴
𝐴
.𝑂
𝐡
π‘Ž°
𝑏°
𝐡
𝑋
π‘₯°
𝑦°
π‘Œ
π‘Ž° = 𝑦°
𝑏° = π‘₯°
𝐢
𝑍
𝐴𝐡 = 𝐡𝐢
Formula List
LOCUS CONSTRUCTION
1. Locus of a point 𝑷, equidistant from 𝑨 and 𝑩:
𝑷
𝑩
𝑨
← Locus
∴ Perpendicular bisector is the locus of a point 𝑃, equidistant from 𝐴 and 𝐡.
2. Locus of a point 𝑷, equidistant from 𝑨𝑩 and 𝑩π‘ͺ:
𝑨
← Locus
𝑷
𝑩
π‘ͺ
∴ Angle bisector is the locus of a point 𝑃, equidistant from 𝐴𝐡 and 𝐡𝐢.
3. Locus of a point 𝑷, so that it is πŸ’ cm from 𝑷:
𝑷′
πŸ’ cm
← Locus
𝑷
∴ Circle is the locus of a point 𝑃, 4 cm from it.
Formula List
4. Locus of a point 𝑷, so that it is always πŸ‘ cm from a line 𝑨𝑩:
𝑷
πŸ‘ cm
πŸ‘ cm
πŸ‘ cm
𝑩
𝑨
πŸ‘ cm
Locus
∴ Capsule is the locus of a point 𝑃, so that 𝑃 is always 3 cm from 𝐴𝐡.
Μ‚ 𝑩 is always πŸ—πŸŽ°:
5. Locus of a point 𝑷, so that 𝑨𝑷
𝑷
𝑨
Locus
𝑩
∴ Circle is the locus of a point 𝑃, so that 𝐴𝑃̂ 𝐡 is always 90°.
6. Locus of a point 𝑷, so that triangle 𝑨𝑷𝑩 always has an area of 𝟏𝟐 𝐜𝐦𝟐 :
Locus
𝑷
𝑷
πŸ‘ cm
πŸ‘ cm
𝑨
𝑩
πŸ– cm
𝑷
1
Locus
∴ Parallel lines πŸ‘ cm from 𝐴𝐡 are the locus of a point 𝑃, so that triangle 𝐴𝑃𝐡 is always 2 × 8 × 3 = 12 cm2.
Formula List
SIMILARITY
1. If three angles equal, triangles are similar.
2. If two angles equal, 3rd angle automatically is equal.
3. Opposite sides of equal angles are corresponding sides, 𝑙1 and 𝑙2 .
4.
𝑙1
𝑙2
5.
𝐴1
𝐴2
6.
π‘š
= π‘š1
2
𝑙 2
𝑙2
𝑉1
𝑙1 3 𝑉1
π‘š
=
(
) , 𝑉 = π‘š1
𝑉2
𝑙2
2
2
𝐴 3
𝑉 2
= ( 1)
1
1
2
2
𝑀
= 𝑀1
2
7. (𝐴 ) = (𝑉 )
CONGRUENCY
POLYGON
A regular polygon has all sides equal and all interior angles equal
For a 𝒏-sided polygon:
Sum of all exterior angles (Regular + Irregular polygon)
360°
Sum of all interior angles (Regular + Irregular polygon)
(𝑛 − 2) × 180°
Each exterior angle (Regular polygon only)
Each interior angle (Regular polygon only)
Sum of an exterior angle and an interior angle
360°
𝑛
(𝑛 − 2) × 180°
𝑛
(𝑛 − 2) × 180° 360°
+
= 180°
𝑛
𝑛
Formula List
Naming of polygons:
3-sided: Triangle
4-sided: Quadrilateral
5-sided: Pentagon
6-sided: Hexagon
7-sided: Heptagon
8-sided: Octagon
9-sided: Nonagon
10-sided: Decagon
TRIGONOMETRY
SOH CAH TOA
sin πœƒ =
π‘‚π‘π‘π‘œπ‘ π‘–π‘‘π‘’ 𝑠𝑖𝑑𝑒 𝑂
=
π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’
𝐻
𝐴
cos πœƒ =
π΄π‘‘π‘—π‘Žπ‘π‘’π‘›π‘‘ 𝑠𝑖𝑑𝑒 𝐴
=
π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’
𝐻
Opposite
For a right-angled triangle, 𝑨𝑩π‘ͺ:
πœƒ
π‘‚π‘π‘π‘œπ‘ π‘–π‘‘π‘’ 𝑠𝑖𝑑𝑒 𝑂
tan πœƒ =
=
π΄π‘‘π‘—π‘Žπ‘π‘’π‘›π‘‘ 𝑠𝑖𝑑𝑒 𝐴
𝐡
𝐢
Adjacent
SINE RULE
𝐴
π‘Ž
𝑏
𝑐
=
=
sin 𝐴 sin 𝐡 sin 𝐢
𝑐
𝐡
𝑏
π‘Ž
𝐢
COSINE RULE
To find the length of a side
π‘Ž2 = 𝑏 2 + 𝑐 2 − 2𝑏𝑐 cos 𝐴
𝑏 2 = π‘Ž2 + 𝑐 2 − 2π‘Žπ‘ cos 𝐡
To find an angle when all three sides are given
cos 𝐴 =
𝑏 2 + 𝑐 2 − π‘Ž2
2𝑏𝑐
cos 𝐡 =
π‘Ž2 + 𝑐 2 − 𝑏 2
2π‘Žπ‘
cos 𝐢 =
π‘Ž2 + 𝑏 2 − 𝑐 2
2π‘Žπ‘
𝑐 2 = π‘Ž2 + 𝑏 2 − 2π‘Žπ‘ cos 𝐢
Formula List
BEARING
The bearing of a point 𝐡 from another point 𝐴 is:
•
an angle measured from the north at 𝐴,
•
in a clockwise direction.
𝑁
𝐡
𝑁
𝐡
050°
230°
Example: The bearing of 𝐡 from 𝐴 is 050°.
The bearing of 𝐴 from 𝐡 is 230°.
𝐴
𝐴
ELEVATION – DEPRESSION
1. Angle of Elevation → line of sight above horizontal line.
2. Angle of Depression → line of sight below horizontal line.
Angle of elevation of 𝑻 from 𝑴 = 𝜽°
Angle of depression of 𝑴 from 𝑻 = 𝜽°
Formula List
ARITHMETIC
∎ Ratio:
If the amount 𝐴 is divided between 𝑃, 𝑄 and 𝑅 in the ratio 𝑝: π‘ž: π‘Ÿ
Share of 𝑃 =
𝑃
𝑝+π‘ž+π‘Ÿ
× π΄,
Share of 𝑄 =
π‘ž
𝑝+π‘ž+π‘Ÿ
×𝐴
Share of 𝑅 =
π‘Ÿ
𝑝+π‘ž+π‘Ÿ
×𝐴
∎ Percentage:
loss
1. Profit = sale price – cost price
4. Percentage loss = cost price × 100
2. Loss = cost price – sale price
increase
5. Percentage increase = original value × 100
profit
3. Percentage profit = cost price × 100
𝐴
6. % of 𝐴 compared with 𝐡 = 𝐡 × 100
∎ Simple Interest:
Here,
𝑃𝑅𝑇
𝐼=
100
𝐼 = Interest
𝑃 = Principle
∎ Compound Interest:
𝑇 = Time (Year)
𝐴 = 𝑃 (1 +
𝑅
)
100
𝑇
𝑅 = Rate of Interest
𝐴 = Final Amount
∎ Speed, Distance, Time:
1.
•
𝑑 =𝑠×𝑑
•
𝑠=
•
𝑑=
𝑑
𝑑
𝑑
𝑠
2.
𝑑
𝑠
𝑑
𝟏𝐬𝐭 part
• 𝑠 = π‘₯ km/h
• 𝑑 = 𝑑1 h
• 𝑑 = π‘₯𝑑1
𝟐𝐧𝐝 part
• 𝑠 = 𝑦 km/h
• 𝑑 = 𝑑2 h
• 𝑑 = 𝑦𝑑2
Avg. speed =
=
Total distance
Total time
π‘₯𝑑1 +𝑦𝑑2
𝑑1 +𝑑2
∎ Map Scale:
1. Scale of a map = (map distance): (actual distance)
2. Scale of a map is written in the form 1: 𝑛, where 1 is map distance ratio and 𝑛 is actual distance ratio.
𝑑
1
3. 𝑑1 = 𝑛
where 𝑑1 = map distance and 𝑑2 = actual distance.
2
𝐴
1 2
4. 𝐴1 = (𝑛)
2
𝑉
1 3
5. 𝑉1 = (𝑛)
2
where 𝐴1 = map area and 𝐴2 = actual area.
where 𝑉1 = model volume and 𝑉2 = actual volume.
Formula List
SETS
1. Notation:
Number of elements in set 𝐴
𝑛(𝐴)
𝐴 = {𝑝, π‘ž, π‘Ÿ, 𝑠}
∴ 𝑛(𝐴) = 4
Intersection
∩
𝐴 = {𝑓, 𝑔, β„Ž, 𝑖}, 𝐡 = {𝑓, β„Ž, 𝑗, π‘˜}
∴ 𝐴 ∩ 𝐡 = {𝑓, β„Ž}
Union
∪
𝐴 = {𝑓, 𝑔, β„Ž, 𝑖}, 𝐡 = {𝑓, β„Ž, 𝑗, π‘˜}
∴ 𝐴 ∪ 𝐡 = {𝑓, 𝑔, β„Ž, 𝑖, 𝑗, π‘˜}
Universal set
πœ€
Complement of set 𝐴
𝐴′
𝐴 ∪ 𝐴′ = πœ€
πœ€ = {1, 2, 3, 4}, 𝐴 = {2, 3}
∴ 𝐴′ = {1, 4}
Null set/Empty set
πœ™
𝐴 = {π‘₯, 𝑦, 𝑧}, 𝐡 = {2, 3, 4}
∴ 𝐴∩𝐡 = πœ™
∴ 𝑛(𝐴 ∩ 𝐡) = 0
Is an element of
∈
𝐴 = {π‘₯, 𝑦, 𝑧}
π‘₯ ∈ 𝐴, 2 ∉ 𝐴
Is a subset of
⊂
𝐴 = {π‘₯, 𝑦, 𝑧}, 𝐡 = {𝑦, π‘₯}, 𝐢 = {π‘₯, 𝑝}
∴ 𝐡 ⊂ 𝐴 and 𝐢 ⊄ 𝐴
πœ€
2. All possible subsets:
•
•
If 𝑨 = {𝒙, π’š, 𝒛}, subsets of 𝑨 are: {𝒙}, {π’š}, {𝒛}, {𝒙, π’š}, {𝒙, 𝒛}, {π’š, 𝒛}, {𝒙, π’š, 𝒛}, 𝝓.
Both 𝝓 and 𝑨 are subsets of 𝑨.
3. De Morgan’s rules:
𝐴′ ∩ 𝐡′ = (𝐴 ∪ 𝐡)′
𝐴′ ∩ 𝐡′ ∩ 𝐢 ′ = (𝐴 ∪ 𝐡 ∪ 𝐢)′
𝐴′ ∪ 𝐡′ = (𝐴 ∩ 𝐡)′
𝐴′ ∪ 𝐡′ ∪ 𝐢 ′ = (𝐴 ∩ 𝐡 ∩ 𝐢)′
4. Subset:
Isosceles
Rightangled
Acute
Equilateral
Scalene
Obtuse
Formula List
πœ€ = π‘ƒπ‘œπ‘™π‘¦π‘”π‘œπ‘›π‘ 
•
•
•
•
•
•
Rectangle
Rhombus
Parallelogram
Square
Kite
Quadrilateral
Trapezium
Rhombus → all sides equal.
Rectangle → all angles equal.
Square → all angles 90°, all sides equal.
Parallelogram → opposite sides parallel.
Kite → one line symmetry.
Trapezium → one pair of parallel sides.
Triangle
πœ€ = π‘π‘’π‘šπ‘π‘’π‘Ÿπ‘ 
Real, ℝ
Rational, β„š
Integer, β„€
•
Natural, β„•
•
1, 2, 3, …
•
•
… , −1, 0, 1, …
β„• is the set of natural numbers or
positive integers {1, 2, 3, 4, … … }.
β„€ is the set of integers
{… , −2, −1, 0, 1, 2, … }.
β„š is the set of rational numbers.
ℝ is the set of real numbers.
½, ¾, 2.5
πœ‹, ln 2, √3
5. Description of a Venn diagram:
•
•
•
•
•
•
•
•
•
•
•
•
•
•
1 → 𝑂𝑛𝑙𝑦 𝐴.
2 → 𝑂𝑛𝑙𝑦 𝐡.
3 → 𝑂𝑛𝑙𝑦 𝐢.
4 → 𝑂𝑛𝑙𝑦 𝐴 π‘Žπ‘›π‘‘ 𝐡 π‘œπ‘Ÿ 𝑗𝑒𝑠𝑑 𝐴 π‘Žπ‘›π‘‘ 𝐡 π‘œπ‘Ÿ 𝐴 π‘Žπ‘›π‘‘ 𝐡 𝑏𝑒𝑑 π‘›π‘œπ‘‘ 𝐢.
5 → 𝑂𝑛𝑙𝑦 𝐴 π‘Žπ‘›π‘‘ 𝐢 π‘œπ‘Ÿ 𝑗𝑒𝑠𝑑 𝐴 π‘Žπ‘›π‘‘ 𝐢 π‘œπ‘Ÿ 𝐴 π‘Žπ‘›π‘‘ 𝐢 𝑏𝑒𝑑 π‘›π‘œπ‘‘ 𝐡.
6 → 𝑂𝑛𝑙𝑦 𝐡 π‘Žπ‘›π‘‘ 𝐢 π‘œπ‘Ÿ 𝑗𝑒𝑠𝑑 𝐡 π‘Žπ‘›π‘‘ 𝐢 π‘œπ‘Ÿ 𝐡 π‘Žπ‘›π‘‘ 𝐢 𝑏𝑒𝑑 π‘›π‘œπ‘‘ 𝐴.
7 → 𝐴𝑙𝑙 𝐴, 𝐡 π‘Žπ‘›π‘‘ 𝐢.
8 → π‘π‘œπ‘‘ 𝐴, 𝐡 π‘Žπ‘›π‘‘ 𝐢.
4, 7 → π΅π‘œπ‘‘β„Ž 𝐴 π‘Žπ‘›π‘‘ 𝐡 π‘œπ‘Ÿ π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ 𝐴 π‘Žπ‘›π‘‘ 𝐡.
5, 7 → π΅π‘œπ‘‘β„Ž 𝐴 π‘Žπ‘›π‘‘ 𝐢 π‘œπ‘Ÿ π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ 𝐴 π‘Žπ‘›π‘‘ 𝐢.
6, 7 → π΅π‘œπ‘‘β„Ž 𝐡 π‘Žπ‘›π‘‘ 𝐢 π‘œπ‘Ÿ π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ 𝐡 π‘Žπ‘›π‘‘ 𝐢.
1, 4, 5, 7 → 𝐴.
2, 4, 6, 7 → 𝐡.
3, 5, 6, 7 → 𝐢.
πœ€
𝐴
4
1
5
8
7
3
𝐢
2
6
𝐡
Formula List
MATRIX
1. Multiplication by a scalar
π‘Ž
(
𝑐
π‘Ž
(
𝑐
π‘Ž
(
𝑐
2. Addition of matrix
3. Subtraction of matrix
4. Multiplication of matrix
π‘Ž 𝑏
π‘˜π‘Ž π‘˜π‘
π‘˜(
)=(
)
𝑐 𝑑
π‘˜π‘ π‘˜π‘‘
𝑀 π‘₯
π‘Ž+𝑀 𝑏+π‘₯
𝑏
) + ( 𝑦 𝑧) = (
)
𝑐+𝑦 𝑑+𝑧
𝑑
𝑀 π‘₯
π‘Ž−𝑀 𝑏−π‘₯
𝑏
) − ( 𝑦 𝑧) = (
)
𝑐−𝑦 𝑑−𝑧
𝑑
π‘Žπ‘€ + 𝑏𝑦 π‘Žπ‘₯ + 𝑏𝑧
𝑏 𝑀 π‘₯
)(
)=(
)
𝑐𝑀 + 𝑑𝑦 𝑐π‘₯ + 𝑑𝑧
𝑑 𝑦 𝑧
5. If two matrices have order (𝑝 × π‘ž) and (π‘Ÿ × π‘ ), they can be multiplied if π‘ž = π‘Ÿ and the order of
the resultant matrix will be (𝑝 × π‘ ).
𝑀 π‘₯
π‘Ž 𝑏
(
) = ( 𝑦 𝑧)
𝑐 𝑑
∴ π‘Ž = 𝑀, 𝑏 = π‘₯, 𝑐 = 𝑦, 𝑑 = 𝑧.
6. Equal matrix
π‘Ž 𝑏
7. Determinant of (
) is π‘Žπ‘‘ − 𝑏𝑐.
𝑐 𝑑
1
π‘Ž 𝑏
𝑑 −𝑏
8. Inverse of (
) is
(
). If π‘Žπ‘‘ − 𝑏𝑐 = 0, the matrix has no inverse.
π‘Žπ‘‘−𝑏𝑐
𝑐 𝑑
−𝑐 π‘Ž
9. Identity matrix:
(1 × 1)
(2 × 2)
(3 × 3)
1 0 0
𝐈 = (0 1 0)
0 0 1
10. If any matrix is multiplied by identity matrix, the matrix remains same, i.e., 𝟏 × π€ = 𝐀.
The product of inverse matrix and original matrix is identity matrix i.e., 𝐀−1 × π€ = 𝐈,
1 0
where 𝐈 = (
).
0 1
𝐈 = (1)
1 0
𝐈=(
)
0 1
MATRIX TRANSFORMATION
Formula List
Formula List
5. Identification of transformation and their fully description:
Transformation
Identification
Need to describe fully
Enlargement
Change of area or change of side-length
Centre and scale factor
Translation
Change of position without rotation
Column vector
Reflection
180° rotation
Corresponding object – image joining
lines are parallel
Corresponding object – image joining
lines intersects at the same point, which
is the centre of 180° rotation
Other rotation
-
6. πŸπŸ–πŸŽ° rotation is same as enlargement, 𝑺. 𝑭. = −𝟏.
7. How to find unknown matrix?
Equation of reflection line
Centre
Angle, direction and centre
Formula List
Case – 2: For known transformation, use the following matrices.
Transformation
Rotation 90°, anticlockwise, centre (0, 0)
Rotation 90°, clockwise, centre (0, 0)
Rotation 180°,centre (0, 0)
Reflection in π‘₯ −axis
Reflection in 𝑦 −axis
Reflection in 𝑦 = π‘₯
Reflection in 𝑦 = −π‘₯
Enlargement, scale factor π‘˜, centre (0, 0)
Matrix
0 −1
(
)
1 0
0 1
(
)
−1 0
−1 0
(
)
0 −1
1 0
(
)
0 −1
−1 0
(
)
0 1
0 1
(
)
1 0
0 −1
(
)
−1 0
π‘˜
(
0
0
)
π‘˜
8.
π‘€π‘Žπ‘‘π‘Ÿπ‘–π‘₯ × π‘‚π‘π‘—π‘’π‘π‘‘ = πΌπ‘šπ‘Žπ‘”π‘’
9. 𝐑𝐓𝐄(β„Ž, π‘˜) means (β„Ž, π‘˜) is transformed by 𝐄, followed by 𝐓, followed by 𝐑.
Case – 3:
Formula List
VECTOR
1
2
Column Vector:
π‘₯1
• πšβƒ— = (𝑦 ) or π‘₯1 𝐒 + 𝑦1 𝐣
1
6
βƒ—βƒ—βƒ—βƒ—βƒ—
𝐴𝐡
−−−/−−−
Step 1: 𝑃𝑇
= −−−/−−− = π‘˜ (constant)
βƒ—βƒ—βƒ—βƒ—βƒ—
•
|πšβƒ—| = √π‘₯1 2 + 𝑦1 2
•
Unit Vector of πšβƒ—: |πšβƒ—|
•
∠ between πšβƒ— and π‘₯-axis: tan πœƒ =
•
π‘₯1
π‘₯2
π‘₯ +π‘₯
πšβƒ— + 𝐛 = (𝑦 ) + (𝑦 ) = (𝑦1 + 𝑦2 ).
•
π‘₯1
π‘˜π‘₯
π‘˜πšβƒ— = π‘˜ (𝑦 ) = ( 1 ).
π‘˜π‘¦1
1
Step 2: ∴ 𝐴𝐡 βˆ₯ 𝑃𝑇
βƒ—
𝐚
1
Show that 𝑨𝑩 βˆ₯ 𝑷𝑻:
Use the following steps:
2
1
𝑦1
π‘₯1
7
2
Show 𝑨, 𝑩 and π‘ͺ lie on a straight line
(collinear).
βƒ—βƒ—βƒ—βƒ—βƒ—
𝐴𝐡
−−−/−−−
Step 1: βƒ—βƒ—βƒ—βƒ—βƒ— = −−−/−−− = π‘˜ (constant)
𝐡𝐢
Step 2: ∴ 𝐴𝐡 βˆ₯ 𝐡𝐢
Step 3: Since 𝐡 is common, therefore 𝐴, 𝐡
and 𝐢 lie on a straight line.
3
•
•
π’™πŸ
π’™πŸ
If (π’š ) = (π’š ),
𝟏
8
𝟐
o π‘₯1 = π‘₯2 ------------ (i)
o 𝑦1 = 𝑦2 ------------ (ii)
π’™πŸ
π’™πŸ
If (π’š ) || (π’š ),
𝟏
𝟐
π‘₯1
π‘₯2
o (𝑦 ) = π‘˜ (𝑦 )
1
2
o π‘₯1 = π‘˜π‘₯2 ------------ (i)
o 𝑦1 = π‘˜π‘¦2 ------------ (ii)
o
π‘₯
•
5
•
•
•
2
9
β„Ž
𝐴(β„Ž, π‘˜) means βƒ—βƒ—βƒ—βƒ—βƒ—
𝑂𝐴 = ( ).
π‘˜
In terms of position vector:
βƒ—βƒ—βƒ—βƒ—βƒ— = 𝑂𝐡
βƒ—βƒ—βƒ—βƒ—βƒ— − 𝑂𝐴
βƒ—βƒ—βƒ—βƒ—βƒ—
o 𝐴𝐡
o
∴ 𝐴𝐡𝐢𝐷 is a parallelogram.
Or prove βƒ—βƒ—βƒ—βƒ—βƒ—
𝐴𝐷 = βƒ—βƒ—βƒ—βƒ—βƒ—
𝐡𝐢
∴ 𝐴𝐡𝐢𝐷 is a parallelogram.
⇒ π‘₯πšβƒ— + 𝑦𝐛 = πœ†(π‘πšβƒ— + π‘žπ›)
∴ π‘₯ = πœ†π‘ ------------ (i)
∴ 𝑦 = πœ†π‘ž ------------ (ii)
βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—
For two geometrical properties of 𝑨𝑩
and βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—
𝑷𝑸:
Use the following steps:
Step 1: Ratio •
βƒ—βƒ—βƒ—βƒ—βƒ—
𝑃𝑄 = βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—
𝑂𝑄 − βƒ—βƒ—βƒ—βƒ—βƒ—
𝑂𝑃
βƒ—βƒ—βƒ—βƒ—βƒ— = (5).
Example: Given 𝐴(3, 2) and 𝐴𝐡
2
βƒ—βƒ—βƒ—βƒ—βƒ—
Find 𝑂𝐡.
5
• ⇒ βƒ—βƒ—βƒ—βƒ—βƒ—
𝑂𝐡 − βƒ—βƒ—βƒ—βƒ—βƒ—
𝑂𝐴 = ( ).
2
3
8
5
βƒ—βƒ—βƒ—βƒ—βƒ—
• ∴ 𝑂𝐡 = ( ) + ( ) = ( ).
2
4
2
Show that 𝑨𝑩π‘ͺ𝑫 is a parallelogram:
βƒ—βƒ—βƒ—βƒ—βƒ— = 𝐷𝐢
βƒ—βƒ—βƒ—βƒ—βƒ—
Either prove 𝐴𝐡
𝑦
π‘ž
βƒ—βƒ—βƒ—βƒ—βƒ— = πœ† × π΄πΆ
βƒ—βƒ—βƒ—βƒ—βƒ—
Or otherwise, 𝐴𝐡
π‘₯
Position Vector:
•
π‘₯
𝑝
Step 2: Use shortcut = to find πœ†.
Shortcut: 𝑦1 = 𝑦2
1
4
Given 𝑨, 𝑩 and π‘ͺ are collinear (or
parallel), find 𝝀.
Step 1:
Find βƒ—βƒ—βƒ—βƒ—βƒ—
𝐴𝐡 = π‘₯πšβƒ— + 𝑦𝐛 and βƒ—βƒ—βƒ—βƒ—βƒ—
𝐴𝐢 = π‘πšβƒ— + π‘žπ›.
βƒ—βƒ—βƒ—βƒ—βƒ—
𝐴𝐡
βƒ—βƒ—βƒ—βƒ—βƒ—
𝑃𝑄
=
π‘š
𝑛
(constant)
Step 2: Properties • βƒ—βƒ—βƒ—βƒ—βƒ—
𝐴𝐡 βˆ₯ βƒ—βƒ—βƒ—βƒ—βƒ—
𝑃𝑄 .
•
10
(ii) 𝐴𝐡: 𝑃𝑄 = π‘š: 𝑛.
Ratio example:
βƒ—βƒ—βƒ—βƒ—βƒ— = 𝐩
βƒ— + 2πͺ
βƒ—.
Given 𝐴𝐡: 𝐡𝐢 = 5: 3 and 𝐴𝐡
Find βƒ—βƒ—βƒ—βƒ—βƒ—
𝐴𝐢 .
βƒ—βƒ—βƒ—βƒ—βƒ—
𝐴𝐡
5
•
⇒ βƒ—βƒ—βƒ—βƒ—βƒ— = 5+3
•
⇒
•
βƒ—βƒ—βƒ—βƒ—βƒ— = 𝐩 +
∴ 𝐴𝐢
5
𝐴𝐢
𝐩+2πͺ
𝐴𝐢
=
8
5
8
16
πͺ
5
Formula List
11
Area Ratio:
1
βˆ† 𝑨𝑩𝑫 2 × 2 × β„Ž
=
βˆ† 𝑩𝑫π‘ͺ 1 × 3 × β„Ž
2
1
βˆ† 𝑨𝑷𝑸 2 × 3 × 1 × sin πœƒ
=
βˆ† 𝑨𝑩π‘ͺ 1 × 4 × 2 × sin πœƒ
2
βˆ† 𝑷𝑺𝑻
3 2
=( )
βˆ† 𝑷𝑸𝑹
5
PROBABILITY
𝑠 = number of 𝐴, 𝑑 = number of 𝐡, 𝑒 = number of 𝐢, 𝑛 = total number.
𝑠
𝑑
𝑒
1. 𝑃(𝐴) = 𝑛, 𝑃(𝐡) = 𝑛, 𝑃(𝐢) = 𝑛
𝑠
𝑛
𝑠
𝑛
2. (i) 𝑃(𝐴𝐴) = × , [If replaced]
𝑠
𝑠
𝑠
3. (i) 𝑃(𝐴𝐴𝐴) = 𝑛 × π‘› × π‘› [If replaced]
𝑠
𝑑
4. (i) 𝑃(𝐴𝐡) = 𝑛 × π‘› × 2 [If replaced]
𝑠
𝑛
(ii) 𝑃(𝐴𝐴) = ×
𝑠
𝑠−1
𝑛−1
[If not replaced]
𝑠−1
𝑠−2
(ii) 𝑃(𝐴𝐴𝐴) = 𝑛 × π‘›−1 × π‘›−2 [If not replaced]
𝑠
𝑑
(ii) 𝑃(𝐴𝐡) = 𝑛 × π‘›−1 × 2 [If not replaced]
5. 𝐴, 𝐡 and 𝐢 are arranged in the following 6 ways:
𝐴𝐡𝐢, 𝐴𝐢𝐡, 𝐡𝐴𝐢, 𝐡𝐢𝐴, 𝐢𝐴𝐡, 𝐢𝐡𝐴.
𝑠
𝑑
𝑒
6. (i) 𝑃(𝐴𝐡𝐢) = 𝑛 × π‘› × π‘› × 6 [If replaced]
𝑠
𝑑
𝑒
(ii) 𝑃(𝐴𝐡𝐢) = 𝑛 × π‘›−1 × π‘›−2 × 6 [If not replaced]
7. For 2 draws, 𝑃(diferent color) = 1 − 𝑃(same color).
8. AND rule: 𝑃(𝐴 and 𝐡) = 𝑃(𝐴) × π‘ƒ(𝐡).
9. OR rule: 𝑃(𝐴 or 𝐡) = 𝑃(𝐴) + 𝑃(𝐡).
10. 𝑃(𝐴) + 𝑃(𝐡) + 𝑃(𝐢) = 1.
11. The probabilities of disc, ball, bead, sweet etc. may change or not according to replacement.
But probabilities of die, coin etc. always remain unchanged.
Formula List
MENSURATION
1. Triangles
𝐡
𝐡
𝐴
𝐢
𝑏
𝐴
𝐢
𝑏
𝑏
𝑐
1
1
Area of βˆ†π΄π΅πΆ = × π‘π‘ × sin πœƒ
Area of βˆ†π΄π΅πΆ = × π‘ × β„Ž
2
2
2.
πœƒ
π‘Ž
β„Ž
β„Ž
SINE RULE
𝐴
π‘Ž
sin 𝐴
𝑏
𝑐
𝐡
𝐢
π‘Ž
=
𝑏
sin 𝐡
=
𝑐
sin 𝐢
COSINE RULE
For sides: π‘Ž2 = 𝑏 2 + 𝑐 2 − 2𝑏𝑐 cos 𝐴
For angles: cos 𝐡 =
π‘Ž 2 +𝑐 2 −𝑏2
2π‘Žπ‘
3. Area of Quadrilateral
𝑏
β„Ž2
πœƒ
β„Ž1
𝑏
β„Ž
π‘Ž
Parallelogram
Rhombus
i) π΅π‘Žπ‘ π‘’ × β„Žπ‘’π‘–π‘”β„Žπ‘‘ 𝑖. 𝑒 (π‘Ž × β„Ž1 ) π‘œπ‘Ÿ (𝑏 × β„Ž2 )
i) π‘π‘Žπ‘ π‘’ × β„Žπ‘’π‘–π‘”β„Žπ‘‘
ii) π‘Žπ‘ sin πœƒ
ii) π‘Ž2 sin πœƒ
Kite
Trapezium
1
1
π΄π‘Ÿπ‘’π‘Ž = (π‘Ž + 𝑏) × β„Ž
π΄π‘Ÿπ‘’π‘Ž = × π‘‘1 × π‘‘2
2
2
1
iii) 𝑑1 𝑑2
2
4. Circle
Length
Area
r
𝑙 = 2πœ‹π‘Ÿ
Major arc
Minor arc
r
𝑙=
2πœ‹π‘Ÿ
2
𝑙=
2πœ‹π‘Ÿ
4
𝐴 = πœ‹π‘Ÿ 2
𝐴=
πœ‹π‘Ÿ 2
2
𝐴=
πœ‹π‘Ÿ 2
4
Formula List
5. Sector and Arc
π‘€π‘Žπ‘—π‘œπ‘Ÿ π‘Žπ‘Ÿπ‘ =
π‘₯
360
π‘€π‘Žπ‘—π‘œπ‘Ÿ π‘ π‘’π‘π‘‘π‘œπ‘Ÿ =
× 2πœ‹π‘Ÿ
π‘₯
360
× πœ‹π‘Ÿ 2
π‘€π‘–π‘›π‘œπ‘Ÿ π‘Žπ‘Ÿπ‘ =
π‘₯
πœƒ
πœƒ
360
π‘€π‘–π‘›π‘œπ‘Ÿ π‘ π‘’π‘π‘‘π‘œπ‘Ÿ =
× 2πœ‹π‘Ÿ
πœƒ
360
× πœ‹π‘Ÿ 2
6. Area of polygon:
Step 1: πœƒ =
360
𝑛
1
Step 2: π‘ƒπ‘œπ‘™π‘¦π‘”π‘œπ‘› π‘Žπ‘Ÿπ‘’π‘Ž = [2 × π‘Ÿ × π‘Ÿ × sin πœƒ] × π‘›
𝑻. 𝑺. 𝑨 = 𝑻𝒐𝒕𝒂𝒍 𝒔𝒖𝒓𝒇𝒂𝒄𝒆 𝒂𝒓𝒆𝒂
7. Cylinder (solid)
π‘ͺ. 𝑺. 𝑨 = π‘ͺ𝒖𝒓𝒗𝒆𝒅 𝒔𝒖𝒓𝒇𝒂𝒄𝒆 𝒂𝒓𝒆𝒂
Cone (solid)
Sphere
Hemisphere (solid)
π‘Ÿ
𝑙
β„Ž
β„Ž
𝑂
π‘Ÿ
π‘Ÿ
π‘Ÿ
π‘‰π‘œπ‘™π‘’π‘šπ‘’ = πœ‹π‘Ÿ 2 β„Ž
𝑙 = √β„Ž2 + π‘Ÿ 2
𝐢. 𝑆. 𝐴 = 2πœ‹π‘Ÿβ„Ž
𝑇. 𝑆. 𝐴 = 2πœ‹π‘Ÿβ„Ž + 2πœ‹π‘Ÿ 2
4
π‘£π‘œπ‘™π‘’π‘šπ‘’ = πœ‹π‘Ÿ 3
π‘‰π‘œπ‘™π‘’π‘šπ‘’ = πœ‹π‘Ÿ 2 β„Ž
𝐢. 𝑆. 𝐴 = 4πœ‹π‘Ÿ 2
𝐢. 𝑆. 𝐴 = 2πœ‹π‘Ÿ 2
𝐢. 𝑆. 𝐴 = πœ‹π‘Ÿπ‘™
𝑇. 𝑆. 𝐴 = πœ‹π‘Ÿπ‘™ + πœ‹π‘Ÿ 2
𝑇. 𝑆. 𝐴 = 4πœ‹π‘Ÿ 2
𝑇. 𝑆. 𝐴 = 2πœ‹π‘Ÿ 2 + πœ‹π‘Ÿ 2
3
1
Pyramid
2
π‘‰π‘œπ‘™π‘’π‘šπ‘’ = πœ‹π‘Ÿ 3
3
3
Cuboid
Cube
𝑉
𝐷
𝐡
𝐢
𝑂
𝐴
𝐴
𝐡
(Tetrahedron)
π‘‰π‘œπ‘™π‘’π‘šπ‘’ =
1
3
× π΅π‘Žπ‘ π‘’ π‘Žπ‘Ÿπ‘’π‘Ž × β„Ž
𝑆. 𝐴 = π΅π‘Žπ‘ π‘’ + π‘†π‘™π‘Žπ‘›π‘‘ π‘†π‘’π‘Ÿπ‘“π‘Žπ‘π‘’
π‘‰π‘œπ‘™π‘’π‘šπ‘’ = 𝑙 × π‘ × β„Ž
𝑇. 𝑆. 𝐴 (πΆπ‘™π‘œπ‘ π‘’π‘‘) = 2𝑙𝑏 + 2π‘™β„Ž + 2π‘β„Ž
π‘‰π‘œπ‘™π‘’π‘šπ‘’ = 𝑙 3
𝑇. 𝑆. 𝐴(π‘π‘™π‘œπ‘ π‘’π‘‘) = 6𝑙 2
Formula List
Prism
𝑙
𝑙
Cross Sections:
π‘½π’π’π’–π’Žπ’† 𝒐𝒇 π’‘π’“π’Šπ’”π’Ž = (𝒂𝒓𝒆𝒂 𝒐𝒇 𝒄𝒓𝒐𝒔𝒔 π’”π’†π’„π’•π’Šπ’π’) × π’π’†π’π’ˆπ’•π’‰
8. Volume of Liquid issued (speed in m/s)
𝑖𝑛 1 𝑠𝑒𝑐 = 𝐢. 𝑆 × π‘ π‘π‘’π‘’π‘‘
𝑖𝑛 1 π‘šπ‘–π‘›π‘’π‘‘π‘’ = 𝐢. 𝑆 × π‘ π‘π‘’π‘’π‘‘ × 60
COORDINATE GEOMETRY
(i) Length 𝐴𝐡 = √(π‘₯2 − π‘₯1 )2 + (𝑦2 − 𝑦1 )2
π‘ž
π‘₯1 +π‘₯2 𝑦1 +𝑦2
,
)
2
2
(ii) Midpoint = (
(iii) Gradient =
(iv) Co-ordinates of 𝑃 = (
(v) Equation of 𝐴𝐡,
•
𝑝
𝑦2 −𝑦1
π‘₯2 −π‘₯1
𝑝π‘₯2 +π‘žπ‘₯1 𝑝𝑦2 +π‘žπ‘¦1
, 𝑝+π‘ž )
𝑝+π‘ž
𝐡 (π‘₯2 , 𝑦2 )
𝑃
𝐴 (π‘₯1 , 𝑦1 )
𝑦 − 𝑦1 = π‘š(π‘₯ − π‘₯1 )
∎ To find gradient:
Case-1:
Case-2:
Case-3:
𝐡 (π‘₯2 , 𝑦2 )
2
•
𝐴
3
𝐴 (π‘₯1 , 𝑦1 )
𝑦 −𝑦
Gradient, π‘š = π‘₯2 −π‘₯1
2
m=
1
(vi) For parallel lines,
Rise
Run
2
𝑑𝑦
=3
mat 𝐴 = 𝑑π‘₯
(vii) For perpendicular lines,
π‘š1 = π‘š2
π‘š1 × π‘š2 = −1
π‘š1
π‘š2
π‘š2
1
∴ π‘š2 = − π‘š
1
π‘š1
Formula List
∎ To find equation:
Case-1:
Case-2:
𝑦
𝑦
(3, 5)
•
•
5
π’š = π’Žπ’™ + 𝒄
𝑙1
𝑦=5
For 𝑙1 ⟢ 𝑐 = 1, π‘š =
1
π‘₯=3
π‘₯ = −2
1
2
1
∴ 𝑦 = 2π‘₯ + 1
π‘₯
0
For 𝑙2 ⟢ 𝑐 = 1, π‘š = −1
−2
3
0
π‘₯
∴ 𝑦 = −π‘₯ + 1
𝑙2
β†ͺ How to know if a point lies on the line or not
Case-3:
𝑦
• (4, 8) does not lie on the line
3π‘₯ + 7𝑦 = 18
π‘š
•
(π‘₯1 , 𝑦1 )
6
as when π‘₯ = 4, 𝑦 = 7 (≠ 8)
• (6, 0) lies on the line
as when π‘₯ = 6, 𝑦 = 0
π‘₯
𝑦 − 𝑦1 = π‘š(π‘₯ − π‘₯1 )
(viii)
𝐢
𝐡 (π‘₯1 , 𝑦1 )
Area
=
•
𝐴 (π‘₯2 , 𝑦2 )
=
•
1
|π‘₯
2 3
𝑦3
1
|( β†˜ )
2
𝐡
π‘₯2
𝑦2
𝐴
π‘₯1
𝑦1
𝐢
π‘₯3 |
𝑦3
− ( ↙ )|
•
𝐢 (π‘₯1 , 𝑦1 )
𝑦
(x)
𝑄
π‘Žπ‘₯ + 𝑏𝑦 = 𝑐
0
𝑃
π‘₯
For 𝑷
For 𝑸
𝑦=0
π‘Žπ‘₯ + (𝑏 × 0) = 𝑐
𝑐
π‘₯=π‘Ž
π‘₯=0
(π‘Ž × 0) + 𝑏π‘₯ = 𝑐
𝑐
𝑦=𝑏
Formula List
(xi)
− − − −(i)
For intersecting point 𝑃,
•
Solve equations (i) and (ii)
𝑃
− − − −(ii)
(xii)
Tangent,
𝑑𝑦
π‘š 𝑇 = 𝑑π‘₯
(π‘₯1 , 𝑦1 )
Normal,
1
π‘šπ‘ = − π‘š
𝑇
Equation of tangent: 𝑦 − 𝑦1 = π‘š 𝑇 (π‘₯ − π‘₯1 )
Equation of normal: 𝑦 − 𝑦1 = π‘šπ‘ (π‘₯ − π‘₯1 )
FUNCTION
1.
π‘Ž
𝑝
π‘Ž
𝑏
π‘ž
𝑐
π‘Ÿ
𝑏
𝑐
𝑝
π‘ž
𝑏
π‘ž
π‘Ÿ
One to Many
Many to One
One to One
𝑝
π‘Ž
2. Vertical line test:
𝑦
π‘œ
𝑦
𝑦
π‘₯
π‘₯
π‘œ
π‘₯
π‘œ
One to one
Many to one
One to many
Function
Function
Not Function
Inverse
Inverse does not exist
𝑁/𝐴
Formula List
3. 𝒇(𝒙) to 𝒇−𝟏 (𝒙)
4. 𝑓(π‘₯) = π‘Ž
Step – 1:
Let 𝑓(π‘₯) = 𝑦
Step – 2:
𝑦 → π‘₯, π‘₯ → 𝑦
Step – 3:
Make 𝑦 subject
Step – 4:
𝑦 = 𝑓 −1 (π‘₯)
π‘₯ = 𝑓 −1 (π‘₯)
5. 𝐷 of 𝑓(π‘₯) = 𝑅 of 𝑓 −1 (π‘₯)
𝑅 of 𝑓(π‘₯) = 𝐷 of 𝑓 −1 (π‘₯)
6.
7. 𝑓(π‘₯) = 2π‘₯ + 2
π‘“π‘”β„Ž(π‘₯)
𝑔(π‘₯) = 4π‘₯
= 𝑓𝑔(π‘₯ 3 )
β„Ž(π‘₯) = π‘₯ 3
= 𝑓(4π‘₯ 3 )
= 2(4π‘₯ 3 ) + 2
STATISTICS
∎ Histogram:
Q1
50 < 𝑀 ≤ 60
60 < 𝑀 ≤ 75
75 < 𝑀 ≤ 80
𝒇
80
90
60
80 < 𝑀 ≤ 100
80
Weight (in kg)
Step – 1:
𝐹
𝑓. 𝑑. =
𝑐. 𝑀.
Step – 2:
Bar 1: 𝑓. 𝑑. =
Bar 2: 𝑓. 𝑑. =
Bar 3: 𝑓. 𝑑. =
Bar 4: 𝑓. 𝑑. =
80
10
90
15
60
5
80
20
=8
=6
= 12
=5
Formula List
∎ Median:
(a) Median of 𝒑, 𝒒, 𝒙, π’š, 𝒛 = π‘₯
(b) Median of 𝟎, 𝟎, 𝟐, πŸ‘, πŸ“, πŸ“ =
(c)
= 2.5
π‘΄π’‚π’“π’Œπ’”
0
1
2
3
𝒇
3
10
12
15
𝒄. 𝒇.
3
13
25
40
4
5
8
2
48
50 = 𝑁
π‘΄π’‚π’“π’Œπ’”
0
1
2
3
𝒇
3
10
12
15
𝒄. 𝒇.
3
13
25
40
4
5
8
3
48
51 = 𝑁
𝒙
0 < π‘₯ ≤ 10
10 < π‘₯ ≤ 12
12 < π‘₯ ≤ 20
𝒇
5
8
10
𝒄. 𝒇.
5
13
23 = 𝑁
(d)
(e)
2+3
2
𝑁
2
=
50
2
= 25
Median =
25π‘‘β„Ž π‘‘π‘’π‘Ÿπ‘š + 26π‘‘β„Ž π‘‘π‘’π‘Ÿπ‘š
2
∴ Median =
𝑁+1
2
=
51+1
2
2+3
2
= 2.5.
= 26
∴ Median = 26π‘‘β„Ž π‘‘π‘’π‘Ÿπ‘š = 3.
𝑁
2
=
23
2
= 11.5.
∴ Median group = 10 < π‘₯ ≤ 12.
∎ Mean:
(a) Mean of 𝒙, π’š, 𝒛, π’˜ =
(b)
(c)
π‘₯+𝑦+𝑧+𝑀
4
𝒙
0
1
𝒇
4
19
𝒇𝒙
0
19
2
3
4
25
29
23
50
87
92
𝒙
0 < π‘₯ ≤ 10
10 < π‘₯ ≤ 12
𝒙
5
11
𝒇
2
5
𝒇𝒙
10
55
12 < π‘₯ ≤ 20
16
4
64
∑ 𝑓 = 100, ∑ 𝑓π‘₯ = 248
∴ Mean =
∑ 𝑓π‘₯
∑𝑓
= 2.48
∑ 𝑓 = 11, ∑ 𝑓π‘₯ = 129
∴ Mean =
∑ 𝑓π‘₯
∑𝑓
= 11.73
Formula List
∎ Mode:
(a) π‘Ž, 𝑏, 𝑐, 𝑐, 𝑑, 𝑑, 𝑒 → Mode = 𝑦 and 𝑑.
(b) π‘Ž, 𝑏, 𝑐, 𝑐, 𝑑, 𝑑, 𝑒 → Mode = 𝑐 and 𝑑.
(c)
𝒙
𝒇
(d)
0 < π‘₯ ≤ 10
10 < π‘₯ ≤ 12
12 < π‘₯ ≤ 20
0
2
1
20
2
25
3
25
15
10
2
4
16
5
3
Modal group → 0 < π‘₯ ≤ 10.
∎ Bar Chart:
1. Aadil observed the number of people in each of 20 cars entering a car park.
The results are shown in the table below.
Number of people
Frequency
1
2
2
7
3
6
(a) Draw a bar chart to represent this information.
(b) Write down the mode.
(c) Calculate the mean number of people in each car.
Answer:
(a)
(b) Mode = 2.
(c) Mean =
(1×2)+(2×7)+(3×6)+(4×4)+(5×1)
2+7+6+4+1
= 2.75.
4
4
5
1
Mode = 2 and 3.
Formula List
∎ Pie Chart:
1. The proportion of drinks sold at the refreshment tent at a summer fair are represented by the pie
chart in figure. Only four categories of drink shown in the pie chart were sold and the number of
coffees sold was 135.
Coffee
(a) Calculate the total number of drinks sold.
(b) Calculate the number of orange juice drinks sold.
75°
100°
180 lemonade drinks were sold.
Lemonade
(c) Calculate the angle for lemonade drink.
Tea
45°
140°
Orange
(d) Calculate the number of teas sold.
Answer:
(a) 75° ≡ 135
135
⇒ 360° ≡
× 360
75
∴ Total drinks = 648.
(c) 135 ≡ 75°
75
⇒ 180 ≡
× 180°
135
∴ Lemonade ∠ = 100°.
(b) 75° ≡ 135
⇒ 140° ≡ 252
∴ Orange juice = 252.
(d) Tea = 360° − 75° − 100° − 140°
∴ Tea ∠ = 45°.
75° ≡ 135
135
⇒ 45° ≡
× 45°
75
∴ Tea = 81.
2. (a) Three times as many students said Blue as said Green.
Calculate the angle of the sector which represents the number
of students who said Green.
(b) There are 8 more students who said Red than Yellow.
How many students said Red?
Answer:
(a) Let Green be π‘₯°
∴ Blue = 3π‘₯°
∴ 90° + 110° + π‘₯° + 3π‘₯° = 360°
⇒ π‘₯ = 40°
∴ Angle for green = 40°.
Red
Yellow
(b) 20° ≡ 8 students
1° ≡
8
20
110° ≡
students
8
20
× 110° = 44 students
∴ Red = 44 students.
Green
110°
Blue
Formula List
EQUATION GRAPH
∎ Distance-Time Graph and Speed-Time Graph:
Constant acceleration
At rest
Constant speed
Constant speed (Returning)
Constant deceleration
Formula List
•
•
•
•
•
# Distance time:
• Gradient = Speed
• Vertical difference = Distance
•
Average speed =
∎ Set of values of 𝒙:
Total distance
Total time
0 to 𝑇1 = constant acceleration
𝑇1 to 𝑇2 = constant speed
𝑇2 to 𝑇3 = constant deceleration
𝑇3 to 𝑇4 = at rest
𝑇4 to 𝑇5 = returning
# Speed time:
• Gradient = Acceleration
• Area under the graph = Distance
•
Average speed =
Total distance
Total time
Formula List
∎ Gradient at given point:
To find gradient at 𝒙 = 𝟏. πŸ“:
Step 1: Draw the tangent line at π‘₯ = 1.5
Step 2: Take two suitable values from the tangent
𝑦 −𝑦
Step 3: π‘šπ‘₯=1.5 = π‘₯2 −π‘₯1
2
=
1
0.3−(−4)
3−0.75
= 1.91
∎ Solve Graphically:
To solve π’™πŸ − 𝒙 − 𝟏 = 𝟎:
Graph Equation = Line
⟹ 2π‘₯ 2 − 2π‘₯ − 2 = 0 [× 2]
⟹ 2π‘₯ 2 − 2π‘₯ − (2π‘₯) − 2 + (1) = 0 − (2π‘₯) + (1)
⟹ 2π‘₯ 2 − 4π‘₯ − 1 = −2π‘₯ + 1
⟹ 𝑦 = −2π‘₯ + 1
𝒙
π’š
0
1
2
−3
∴ Solution: π‘₯ = 2, −0.6
∎ Common Sketching:
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