2.tips maths

8/6/2019 2.Tips Maths

1/17

T I P S

PROBLEM SOLVING IN MATHEMATICS

Problem solving is the main focus in the teaching and learning of mathematics. Therefore the teaching

and learning process must include problem solving skills which are comprehensive and cover the whole

curriculum. The development of problem solving skills need to be emphasized so that students are able to

solve various problems effectively.

The skills involved are:

Understanding the problem

Devising a plan

Carrying out the plan; and

Looking back at the solutions

UNDERSTAND THE KEYWORDS

NO. KEYWORD WHAT IS ITS IMPLICATION ?

1 Hence Usually you have to use the answer obtained from the previous

section in your calculation.

2 Express No numerical answer is required. Answers are usually given in terms

of variables.

3 Prove, Show The answer is usually given. You are required to show clearly the

steps how you arrive at the answer. In this type of question, you must

be familiar with the mathematical formulas, rules and laws.

4 Write, State The answer can be worked out mentally. Hence, you can write downthe answer without showing the working.

5 Solve Find the root(s) of a given equation

6 Estimate For example, estimate the median from an ogive. A certain range of

answers is acceptable.

7 Calculate,fnd,determine Obtain the answer by showing proper steps of working.

8 Draw ( a graph ) Prepare a table of values, plot the points on a graph paper using a

suitable scale and then join the points using a straight line ( or a

smooth curve )

9 Sketch ( a graph ) A graph paper is not required. The accuracy of the graph isnegligible. The important factors are the shape of the graph and the

position of the graph with respect to the axes.

MATHEMATICS


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T I P S

IMPORTANT FORMULAE

RELATION SHAPES AND SPACE

ac

bd

bcadA

11

P(A) = SA

P(A / ) = 1 P(A)

Distance = 2122

12 yyxx

Midpoint (x,y) =

2,

2

2121 yyxx

Average speed = distance traveled

Time taken

Mean = sum of data

Number of data

Pythagoras / Theorem

c 2 = a 2 + b 2

gradient , m =12

12

xx

yy

erceptx

ercety

m int

int

Scale factor , k =PA

PA/

Area of image = k2 x area of object

Sum of interior angles of a polygon

= ( n 2 ) x 180 0

Volume of right pyramid

=3

1x base area x height

Volume of sphere = 3

3

4r

Volume of cone = hr

2

3

1

Volume of cylinder = hr2Volume of right prism

= cross sectional area x length

Surface area of sphere = 24 rCurved surface area of cylinder = rh2

Area of circle = 2rCircumference of circle = rd 2Area of trapezium

=2

1x sum of parallel sides x height


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T I P S

1. Sentence of the truth is known is a statement.

2.

)(1,,

)(

/ APAPS

BABandAP

S

BABorAP

3. Range difference between the smallest value and the largest value.

4. m = tan , where is the angle made by the line with the horizontal and

measured in an anticlockwise direction.

5. Equation of the straight line: y = mx + c where m = gradient and c = y-intercept

at y intercept , x = 0 and x intercept , y = 0

The x-intercept and the y-intercept are not written in the form of coordinates.

6. A compound statement containing `` if and only if`` can be written as twoimplications.

Implication 1 : If p then q

Implication 2 : If q then p

7. The number of subsets = 2 n , where n is the number of elements in the set.

8. Standard form , A x 10 n where 1 < A < 10 and n is an integer.

9. To factorise ax 2 + bx = x ( ax + b ) , To factorise a 2 x 2 b 2 = (ax b )( ax + b )

10. In a unit circle, the value of sin = y coordinate , cos = x coordinate ,

tan =coordinatex

coordinatey

11. Cumulative frequency the total of the frequencies of the class interval and all the

classes before it.

12. The intersection of two straight lines is found by solving simultaneous equations.

13. If the lines are parallel then m1 =m 2 else m 1 m 2

14. A statement is a sentence which is either true or false. A sentence is not a statement if

we cannot determine whether it is true or false.

15. The mode and mean of grouped data:

(a) The modal class is the class with the highest frequency

(b) The midpoint of a class =2

1( upper class limit + lower class limit )

(c) Mean = Sum of ( midpoint of class x frequency )

Sum of frequency

SMART NOTES


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T I P S

MISCONCEPTION CORNER

QUESTIONS INCORRECT CORRECT

1. Factorise 16k2 - 25 16k2 25

= ( 4k 5 ) 2(4k) 2 5 2

= ( 4k 5)(4k + 5 )

2. Solve y 2 2y = 3 y ( y 2 ) = 3

y = 3 0r y 2 = 3y = 5

y 2 2y 3 = 0

( y 3)( y +1 ) = 0y = 3 , y = - 1

3. Round off 7951 correct

to two significant figures.

7951 = 80 7951 = 8000

4. Determine if the

following mathematical

sentence is a statement or

not. Give a reason for your

answer.

`` 7 + 3 = 21 `

7 + 3 = 21 is not a

statement.

7 + 3 = 21 is a statement

because it is false.

Find the gradient of thestraight line y 3x = 7 y 3x = 7Therefore, the gradient

is - 3

y 3x = 7y = 3x + 7

Compare this equation with

y = mx + c

Therefore, the gradient

is 3 .

Given that L = and M = , find

5,4,3,2,1

6,5 ML

725

25)(

ML

MandL 6

6,5,4,3,2,1

LK

LK

Complete the following

argument.Premise 1 : ___________

Premise 2 : p - 3 10Conclusion : p 13

If p - 3 = 10 , then p = 13 If p = 13 , then p - 3 = 10

Express

5

3

2

14

as a single matrix.

12

16

3

44

52

314

5

3

2

14

3

7

58

34

5

3

8

4

5

3

2

14


5/17

T I P S

MISCONCEPTION CORNER

QUESTIONS INCORRECT CORRECT

Find the values of

(a) sin 160 0 (b) tan 290 0(a) sin 160 0

= sin ( 160 0 900 )

= sin 70 0

= 0.9397

(b) tan 290 0

= - tan ( 290 0 270 0 )

= - tan 20 0

= - 0.3640

(a) sin 160 0

= + sin (180 0 160 0 )

= sin 20 0

= 0.3420

(b) tan 290 0

= - tan ( 360 0 290 0 )

= - tan 70 0

= - 2.7474

6. Find the gradient of a

straight line passing

through P(2,3) and

Q ( 4,6 )

Gradient PQ

2

3

42

36

Gradient PQ

2

3

24

36

Find the difference in

longitudes between points

P ( 45 0 N, 87 0 E ) and

Q ( 45 0 N, 33 0 W )

Difference in longitudes

= 87 0 33 0

= 54 0

Difference in longitudes

= 87 0 + 33 0

= 120 0

A box contains 3 yellow

and 5 white balls. Two

balls are selected at

random, one after another,

with replacement. Find the

probability that the selected

balls are of differentcolours.

Let Y and W be the events of

selecting a yellow ball and a

white ball respectively.

P ( different colours )

= P ( yellow and white )

= P ( Y W )

=85

83

=64

15

Let Y and W be the events

of selecting a yellow ball

and a white ball

respectively.

P ( different colours )

= P ( Y W ) + P (WY)

32

15

8

3

8

5

8

5

8

3

7. A bag contains 3 black

and 2 green marbles.Find the sample

space S

S = GB, S= 21321 ,,,, GGBBB


6/17

T I P S

CONSTRUCTION REQUIREMENT

KNOWLEDGE SKILL

1. Express ( 3x - 1)( 2x + 5 ) in the simplest

form.

1. Solve the equation

( 2y 1)( y + 3) = 2 (y + 1 )

2. Factorise each of the following:

(a) p q 2 p (b) 3x 2 5x 2

2. Using matrices, find the values

of x and y that satisfy thefollowing simultaneous equations.

3x + 4y = 5

4x 3y =2

5

3. Explain why the mathematical sentence

` 2 3 + 1 = 7 ` is a statement.

3. Make a general conclusion by

Induction for the number

sequence 5 , 9 , 13, 17, .

5 = 4(2) 3

9 = 4(3) 3

13 = 4(4) 317 = 4(5) - 3


7/17

T I P S

CLONE SPM QUESTIONS ( PAPER 1 )

ALGEBRAIC EXPRESSIONS

1. 2 ( h 3k ) 2 + 3 hk =

A - 2h 2 + 15hk + 18k2 C - 2h 2 + 15hk 18k2

B - 2h 2 + 9hk 18k2 D - 2h 2 9hk + 18 k2

Answer : C

Solution :

- 2 ( h 3k )( h 3k ) + 3hk

= - 2 ( h 2 6hk + 9k2 ) + 3hk

= - 2h 2 + 12hk 18k2 + 3hk

= - 2h 2 + 15hk 18k2

LINEAR EQUATIONS

2. Given that5

1

3

3

kk, find the value of k .

A 6 B 9 C4

1D

4

3

Answer: B

Solution :

5k 15 = 3k + 3

5k 3k = 3 + 15

2k = 18 , k = 9

NUMBER BASES

3. Given that x 5 = 44 10 , then x =

A 34 B 43 C 134 D 431

Answer : C

5 44 Remainder

5 8 4

5 1 3

0 1


8/17

T I P S

ALGEBRAIC FRACTIONS

4. Express212

52

6

1

p

p

p

as a single fraction in its simplest form.

2222 12

13

12

5

12

3

3

1

p

pD

pC

p

pB

pA

Answer : C

Solution :

22

2

2

2

12

5

12

522

12

522

12

52

62

21

12

52

6

1

pp

pp

ppp

p

p

pp

p

p

p

p

EARTH AS A SPHERE

5. P ( 30 0 N, 70 0 E ) and Q are two points on the surface of the earth such that PQ is a

diameter of the earth. State the location of Q.

A ( 30

0

N, 70

0

W ) B ( 30

0

N,110

0

B) C ( 30

0

S, 110

0

E)D ( 30 0 S , 110 0 W )

Answer: D

Location of P P ( 30 0 N, 70 0 E )

Location of Q Q ( 30 0 S , 110 0 W )

FORMULAE

7 Given that 75

3

k

n

, then k =

57

35

7

357549

22

22

nD

nCnBnA

Answer: C


9/17

T I P S

TYPE OF QUESTION

LEVEL OF DIFFUCULTY

LOW

1. Express 8 2 + 5 as a number in base 8.

2. Given that1

2

h

hk , express h in terms of k .

3. Express 3.71 x 10 5 as a single number.

4. Find the difference in longitude between each pair of meridians given.

(i) 37 0E and 78 0E (ii) 43 0 E and 59 0 W

5. If K = , determine the number of possible subsets.7,5,4,3,2,1

MEDIUM

1. Calculate the values of m and n that satisfy the following simultaneous equations:

- m + n = 11 , 2m + 3n = 8

2. List all the integers y that satisfy the inequalities 4y 11 < y 2 < 3y + 2 .

3. There are 180 red, yellow and green marbles in a bag. 60 of them are red.

If a marble is picked at random from the bag, the probability of picking a

green marble is4

1. How many yellow marbles are there in the bag ?

HIGH

1. Express210

42

5

3

m

m

m

as a single fraction in its simplest form.

2. Simplify 3m 0 x m 5 x ( 2m ) 3

3.

82

73

01

64

4. Given thatk

k

32

12

9

19

, find the value of k .


10/17

T I P S

CRITICAL AND CREATIVE THINKING SKILLS (CCTS )

1. Express 4107 8 as a number in base two.

( Contexts Number Base , Difficulty level - Low )

2. Factorise k ( h 3 ) + 4 ( 1 h ) completely.( Contexts Algebraic Expressions, Difficulty level Low )

3. P and Q are two points on the equator. The difference in longitude between

P and Q is 26 0 . Find the distance, in nautical miles, between P and Q

measured along the equator.

( Contexts Earth as a Sphere , Difficulty level Medium )

4. There are two lorries with 2.4 x 10 6 kg and 8.3 x 10 5 kg of flour respectively.

Find the difference of mass, in kg, between the two lorries.

( Contexts Standard Form , Difficulty level - Medium )

5. Calculate the values of h and k that satisfy both of the following equations:

2h + k = - 2 , - 6h 4k = 2

( Contexts Simultaneous Equations , Difficulty level - High )

EXAMPLES AND MARKING SCHEME OF PAPER 2

1. Solve the quadratic equation2

1

3

52 2

k

k( 4 marks )

Answer: 4k2 3k 10 = 0 ( 1 mark )

(4k + 5)(k 2 ) = 0 ( 1 mark )

k = 2,4

5 ( 1 mark)( 1 mark )

2. Find the value of v and of w that satisfy the simultaneous linear equation below.

3 v 4 w = - 2 dan 1322

1 wv ( 4 marks )

Answer:

v + 4w = 26 or equivalent ( 1 mark )

4v = 24 or equivalent ( 1 mark )

v = 6 ( 1 mark )w = 5 ( 1 mark )

3. A straight line PQ is parallel to line y = - 2x + 5 and passes through point ( 4,-2 )

Find

(a) the gradient of the straight line PQ

(b) the equation of the straight line PQ and hence, state its y intercept.

( 5 marks )

Answer:

(a) - 2 ( 1 mark )

(b) - 2 = - 2(4) + c ( 1 mark )

c = 6 ( 1 mark )

y = - 2x + 6 ( 1 mark )

y intercept = 6 ( 1 mark )


11/17

T I P S

Answer:

(a) - 2 ( 1 mark )

(b) - 2 = - 2(4) + c ( 1 mark )

c = 6 ( 1 mark )

y = - 2x + 6 ( 1 mark )

y intercept = 6 ( 1 mark )

4. (a) Determine whether the following is a statement or not. Give a reason to

you answer.

0121121 (b) Rewrite the following statement by inserting the word `` not `` into the

original statement. State the true value of your new statement.

`` 6 is the factor of 72 ``

(c) Construct a false statement using a suitable quantifier for the given object and

the property. Object : Triangles

Property : Have a right angle

( 5 marks )

Answer:

(a) Statement. ( 1 mark )

False statement. ( 1 mark )

(b) 6 is not a factor of 72. ( 1 mark )

False. ( 1 mark )

(c) All triangles have a right angle ( 1 mark )

5 . Eleven cards bearing the letters of the word `` MATHEMATICS`` are

well shuffled and placed in a bag. Two cards are picked at random from

the bag, one after the other, and are not replaced. Calculate the probability that

(a) the first card bears letter A but the second card does not bear the

same letter A

(b) both cards bear the same letter ( 5 marks )

Answer :

(a) P ( first A and second not A )

mark

mark

155

9

1109

112

(b) P ( both of the same letter )

= P ( AA or MM or TT )

mark

marks

1

55

3

210

1

11

2

10

1

11

2

10

1

11

2


12/17

T I P S

6. P ( 50 0 N, 40 0 E) , Q ( 50 0 N, 100 0 E ) and R are three points on the

surface of the earth such that QR is the diameter of the parallel of

latitude 50 0 N.

(a) State the longitude of point R

(b) Calculate, nautical miles,

(i) the shortest distance from R to Q via the North Pole(ii) the distance from P to R due west along the parallel

of latitude 50 0 N

(c) Given that the point W is 5100 nautical miles due south of P,

calculate the latitude of W . (12 marks)

Answer : (a) Longitude of R = ( 180 100 ) W = 80 0 W (2 marks)

(b) (i) Distance from R to Q via the North Pole

= 60 x 80 (2 marks)

= 4800 n.m (1 marks)

(ii) Distance from P to R along latitude 50 0 N

= 60 x 120 x cos 500

(3 marks)= 4628 n.m (1 mark)

(c) 8560

5100orPOW (1 mark)

Latitude of W = ( 85 50 ) S (1 mark)

= 35 0 S (1 mark)

7. (a) The inverse matrix of is

41

83

31

41 k

m

Find the values of m and k .

(b) Using matrices, calculate the values of x and y that satisfy thefollowing simultaneous linear equations.

3x + 8y = 3

x + 4y = - 1 (7 marks)

Answer:

(a) k = - 8 ( 1 mark ) , m = 4 (2 marks)

(b) marky

x1

1

3

41

83

marky

markx

y

x

marky

x

12

3

15

2

35

11

3

31

84

41


13/17

T I P S

8.

BOX NUMBER OF PENS NUMBER OF PENCILS

R 2 1

S 4 2

T 5 4

The table shows the number of pens and pencils in three boxes, R , S and T . The

probability of Hafiz choosing any of the boxes is the same. Calculate the probability that

Hafiz

(i) choose R or S (ii) chooses a pen from box T

(iii) chooses a pencil

(6 marks)

Answer :

27

10

29

4

3

1

6

2

3

1

3

1

3

1

127

5

195

31

13

2

3

1

3

1

marksiii

mark

markii

marki

9. Transformation U represents the translation . Transformation V represents

2

3

the represents the reflection in the line x = 2. State the coordinates of image of point

( -1,2) under the following transformations.

(i) V (ii) U 2 (iii) VU (5 marks)

Answer: (i ) ( 5,2 ) (ii) ( 5,-2 ) (iii) ( 2,0 )


14/17

T I P S

11.

Time ( s )

22

16

u

0 8 12 18

Speed (ms -1 )The diagram shows the speed-time graph

of a particle over a period of 18 seconds.

Calculate

(i) its accelerations during the first

8 seconds.

(ii) the value of u , if the distance

traveled during the last 10 seconds

is 132 m .

( 5 marks )

Answer :

(i) 4

3

or - 0.75 ( 2 marks )

(ii)2

1x4x(16+u) +

2

1x u x 6 = 132 ( 2 marks )

u = 20 ( 1 mark )

10. On the graph, OPQR is a parallelogram. O is the origin.

Find

(a) the equation of the straight

line PQ

(b) the x intercept of the straightline RQ .

( 5 marks )

0

yP (2,6)

Q

R (4,2)

x

Answer: (a)40

20

04

02

ormOR ( 1 mark )

c 22

16 or c = 5 ( 2 mark )

y =2

1x + 5 ( 1 mark )

(b) - 10 ( 1 mark )


15/17

T I P S

12 cm

P

Q

R12. The diagram shows a solid cylinder with a

hemispherical PQR hole removed from one

circular end. Both the cylinder and

hemisphere have the same diameter of 10 cm.

Using 7

22

, calculate

the volume of the remaining solid.

(4 marks)

Answer :

mark

markI

markjI

marktjI

SOLID

CALHEMISPHERI

CYLINDER

196.680

157

22

3

2125

7

22

157

22

3

2

3

4

2

1

11257

22

32

33

22

13.

142 145 147 148 158 153 155 157 159 152

157 160 164 148 161 169 150 153 158 143

159 165 163 156 167 162 156 151 154 149

Table 1

Table 1 shows the distribution of heights of 30 plants.

(a) State the size of the class interval. (1 mark)

(b) Copy and complete Table 1. (4 marks)

(c) From the frequency table

(i) state the modal class

(ii) calculate the mean height (4 marks)

Answer :

(a) 5 (1 mark)

(b)

Column 1 Column 2 Column 3

Height (cm ) Frequency Midpoint

140 144 2 142145 149 5 147

150 154 6 152

155 159 9 157

160 164 5 162

165 - 169 3 167

Table 1

Column 1 (1 mark), Column 2 (2 marks), Column 3 (1 mark)

(c) (i) 155 159 ( 1 mark )

(ii) marks230

4655

= 155.17 ( 1 mark )


16/17

T I P S

14. In diagram below, OHKM is a quadrant with the centre O and OMKL is a

semicircle with the centre M.

Given that OH= 14 cm. Using7

22 , calculate

(a) the area, in cm 2, of the shaded region

(b) the perimeter, in cm, of the whole diagram. ( 7 marks )K

60 0

O H

LM

N

Answer:

(a) Area of sector OHK = mark114147

22

360

90

Area of sector ONM = mark1777

22

360

60

14147

22

360

90 mark177

7

22

360

60

= mark131128

(b) Arc HK = markKLOArcor 177

222

360

18014

7

222

360

90

= mark11477

222

360

18014

7

222

360

90

= 58 ( 1 mark )


17/17

T I P S

15. (a) Copy and complete the following table 2 of values for y = - 2x 2 + 4x + 7

x - 3 - 2.5 - 1.5 - 1 0 1 2 3 4

y - 23 - 3.5 1 7 9 7 - 9

Table 2 ( 2 marks )

(b) Using a scale of 2 cm to represent 1 unit on the x-axis and 2 cm to

Represent 5 units on the y-axis, Draw the graph of y = - 2x 2 + 4x + 7

for .43 x

(3 marks)

(c) Draw a suitable straight line on the graph to find the values of x where

which satisfy the equation 2 x 2x.43 x2 = 0 .

State these values of x.

(4 marks)

(d) On the same axes, draw the graph of 3y = - 7x + 21 and y = - 2 .Hence, shade the region defined by the following inequalities:

.22173,0 ydanxyx

(3 marks)

Answer :

(a)

x - 2.5 3

y - 15.5 1

(2 marks)

(b) Draw a graph (3 marks)

(c) 2 x 2x 2 = 0 .

- 2x 2 x + 5x + 2 + 5 = 0 + 5x + 5

y = 5x + 5 (1 mark)

In this case, the suitable straight line that should

be drawn is y = 5x + 5 (1 mark)

From the graph, x = .., x = . (2 marks)

(d) (3 marks)

2.tips maths

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