resource pack gr10-term4

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GRADE 10 TERM 4

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GRADE 10, TERM 4: RESOURCE PACK

2 Grade 11 MATHEMATICS Term 2

GRADE: 10 TERM: 4

PROBABILITYRESOURCE 1

LESSON 1

EXPERIMENTAL PROBABILITY AND RELATIVE FREQUENCY INVESTIGATION

Flip a coin 30 times. Fill in the tables and complete the graph below.

Throw 1 2 3 4 5 6 7 8 9 10

Result

Throw 11 12 13 14 15 16 17 18 19 20

Result

Throw 21 22 23 24 25 26 27 28 29 30

Result

Now consider the tails in the above table.

Remember:

Relative Frequencynumber of throwsnumber of tails

=

After… 5 Throws 10 Throws

15 Throws

20 Throws

25 Throws

30 Throws

Number of Tails

RelativeFrequency

1.00

0.50

00 5

Theory line

Number of Throws

10 15 20 25

Rel

ativ

e Fr

eque

ncy



3Grade 10 MATHEMATICS Term 4


RESOURCE 2

REVISION: WEEK 1

SUM

MARY

NOT

ES –

PAP

ER 1

ALGE

BRAI

C EX

PRES

SION

S AN

D EX

PONE

NTS

The

Rea

l Num

ber s

yste

m

NAT

UR

AL

NU

MB

ER

SZ

ER

O

RE

AL

NU

MB

ER

S

RAT

ION

AL

NU

MB

ER

SIR

RAT

ION

AL

NU

MB

ER

S

FRA

CTI

ON

SIN

TEG

ER

S

NE

GAT

IVE

INTE

GE

RS

WH

OLE

NU

MB

ER

S

Prod

ucts

Gen

eral

Rul

e w

ith b

rack

ets:

All

term

s in

one

bra

cket

mus

t be

mul

tiplie

d by

all

term

s in

the

othe

r bra

cket

.

Bino

mia

l x B

inom

ial:

Use

FO

IL (fi

rst,

oute

r, in

ner,

last

)

REM

EMBE

R:

z

“Nam

es” d

on’t

chan

ge w

hen

addi

ng a

nd s

ubtra

ctin

g, o

nly

the

coeffi

cien

t doe

s (

abab

ab2

46

+=

not

a

b6

22)

z

The

sign

to th

e le

ft of

a te

rm b

elon

gs to

it

If th

ere

is m

ore

than

one

set

of b

rack

ets:

Wor

k fro

m th

e in

side

ou

t.

OR

DER

OF

OPE

RAT

ION

S is

ALW

AYS

impo

rtant

and

nee

ds to

be

follo

wed

.

B

E

DM

AS

Brack

et

Exp

onen

tDivide/Multip

ly

Addition

/Sub

trac

tion

Fact

ors

To fa

ctor

ise

an e

xpre

ssio

n is

the

oppo

site

ope

ratio

n to

find

ing

the

prod

uct.

1.

Com

mon

fact

or (i

nclu

ding

gro

upin

g an

d si

gn c

hang

ing)

z

Find

HC

F to

take

out

z

Rem

aini

ng fa

ctor

s go

into

bra

cket

z

Orig

inal

num

ber o

f ter

ms

mus

t equ

al th

e nu

mbe

r of

term

s le

ft ov

er in

the

brac

ket.

Exam

ple:

(

)x

xx

x3

93

13

23

2-

=-




2.

Gro

upin

g

z

Four

or m

ore

term

s us

ually

requ

ires

grou

ping

z

Look

at t

he ra

tios

of th

e co

effici

ents

to h

elp

deci

de w

hich

term

s gr

oup

toge

ther

z

Ther

e ne

eds

to b

e a

sign

bet

wee

n th

e br

acke

ts a

fter g

roup

ing.

If

the

sign

in fr

ont o

f a b

rack

et is

‘-‘ w

e ne

ed to

cha

nge

the

sign

in

the

brac

ket f

ollo

win

g it.

z

Onc

e gr

oupi

ng h

as b

een

done

, the

com

mon

fact

or s

houl

d be

w

hate

ver i

s in

the

brac

ket.

Ex

ampl

e p

qp

qpq

64

38

³³

²²

-+

-

(6

& 3

giv

es th

e sa

me

ratio

as

8 &

4)

(

)(

)p

pq

qq

p3

24

2²

–²

=+

+

[]

pq

qp

22

+=

+

(

)()

pq

pq

23

42

2=

+-

3.

Diff

eren

ce o

f tw

o sq

uare

s

z

Alw

ays

two

term

s se

para

ted

by a

min

us s

ign

z

Both

term

s m

ust b

e pe

rfect

squ

ares

z

Rem

embe

r not

to m

ultip

ly o

ut fi

rst i

f bra

cket

s ar

e in

volv

ed

Fo

r exa

mpl

e,

()

ab

316

2+

-

[(

)]

ab

ab

34

34–

=+

++

^h

6@

4.

Sum

and

Diff

eren

ce o

f 2 c

ubes

z

Alw

ays

2 te

rms

sepa

rate

d by

a p

lus

or m

inus

sig

n

z

Both

term

s m

ust b

e pe

rfect

cub

es

z

Fact

ors

will

alw

ays

be a

bin

omia

l and

a tr

inom

ial

z

Bino

mia

l bra

cket

: Cub

e ro

ot e

ach

term

and

kee

p sa

me

sign

z

Trin

omia

l bra

cket

:

1st

term

:

Squa

re 1

st te

rm fr

om b

inom

ial b

rack

et

2nd

term

: Fi

nd p

rodu

ct o

f 2 te

rms

from

bin

omia

l bra

cket

an

d ch

ange

the

sign

3rd

term

: Sq

uare

the

2nd te

rm fr

om th

e bi

nom

ial B

rack

et

Ex

ampl

e,

xy

2764

33

+

(

)()

xy

xxy

y3

49

1216

22

=+

-+

5.

Trin

omia

ls

z

Alw

ays

3 te

rms

and

fact

oris

es in

to tw

o fa

ctor

s (h

ence

the

two

brac

kets

) If

coeffi

cien

t of

x2 is

not

1:

z

Cho

ose

the

appr

opria

te s

igns

to m

atch

the

prod

uct o

f the

last

te

rm

z

Find

fact

ors

of th

e fir

st te

rm a

nd la

st te

rm

z

Use

cro

ss m

ultip

licat

ion

to fi

nd th

e fa

ctor

s th

at w

ork




Exam

ples

:

Wor

king

(sig

ns a

nd fa

ctor

s) a

nd s

olut

ion

xx

1037

72+

+La

st te

rm p

ositi

ve Ñ

nee

d tw

o si

gns

the

sam

e2n

d te

rm p

ositi

ve Ñ

(+)(+

)Fa

ctor

s of

term

1: 1

10#

and

25#

Fact

ors

of la

st te

rm: 1

7#

2x

1

5x

7

5x

14

x

Beca

use

the

sign

s ar

e th

e sa

me

thes

e te

rms

shou

ld a

dd u

p to

the

mid

dle

term

(37x

). Th

ey d

o no

t – th

e co

mbi

natio

n ne

eds

re-th

inki

ng:

2x

7

5

x 1

35x

2x

Thes

e ad

d up

to 3

7x. T

he fa

ctor

s w

ill m

ake

up

the

brac

kets

Solu

tion:

x

x5

12

7+

+^

^hh

xx

810

32-

+La

st te

rm p

ositi

ve Ñ

nee

d tw

o si

gns

the

sam

e2n

d te

rm n

egat

ive Ñ

(-)(-

)Fa

ctor

s of

term

1: 1

8#

and

24#

Fact

ors

of la

st te

rm: 1

3#

4x

3

2x

7

6x

4x

Beca

use

the

sign

s ar

e th

e sa

me

thes

e te

rms

shou

ld a

dd u

p to

the

mid

dle

term

– th

ey d

o.So

lutio

n:

xx

43

21

--

^^h

h

xx

65

62+

-La

st te

rm n

egat

ive Ñ

nee

d tw

o di

ffere

nt s

igns

Fact

ors

of te

rm 1

: 16#

and

23#

Fact

ors

of la

st te

rm: 1

6#

and

23#

2x

3

3x

2

9x

4x

Beca

use

the

sign

s ar

e di

ffere

nt th

ese

term

s sh

ould

mak

e a

diffe

renc

e of

the

mid

dle

term

–

they

do.

Use

the

term

s to

dec

ide

whi

ch fa

ctor

s go

with

w

hich

sig

ns. T

o ge

t x5

+,

xx

94

+-

is re

quire

d.

The

sign

of

x4 m

oves

dire

ctly

abo

ve a

nd th

e si

gn o

f x9

is p

lace

d be

twee

n th

e to

p tw

o fa

ctor

s.So

lutio

n:

xx

23

32

+-

^^h

h




xx

1211

152-

-La

st te

rm n

egat

ive Ñ

nee

d tw

o di

ffere

nt s

igns

Fact

ors

of te

rm 1

: 112#

and

26#

and

34#

Fact

ors

of la

st te

rm: 1

15#

and

53#

3x

5

4

x 3

20x

9x

Beca

use

the

sign

s ar

e di

ffere

nt th

ese

term

s sh

ould

mak

e a

diffe

renc

e of

the

mid

dle

term

–

they

do.

Use

the

term

s to

dec

ide

whi

ch fa

ctor

s go

w

ith w

hich

sig

ns. T

o ge

t x

11-

, x

x20

9-

+ is

re

quire

d. T

he s

ign

of

x9 m

oves

dire

ctly

abo

ve

and

the

sign

of

x20

is p

lace

d be

twee

n th

e to

p tw

o fa

ctor

s.So

lutio

n:

xx

23

32

+-

^^h

h

REM

EMBE

R: A

LWAY

S lo

ok fo

r a h

ighe

st c

omm

on fa

ctor

firs

t.

ALGE

BRAI

C FR

ACTI

ONS

1.

Mul

tiplic

atio

n an

d D

ivis

ion

z

If di

visi

on, c

hang

e to

mul

tiplic

atio

n an

d re

cipr

ocat

e

z

Fact

oris

e al

l num

erat

ors

and

deno

min

ator

s fu

lly

z

Sim

plify

by

look

ing

for c

omm

on fa

ctor

s in

any

num

erat

or

and

deno

min

ator

(rem

embe

r: yo

u ca

nnot

sim

plify

‘nex

t to’

an

addi

tion

or s

ubtra

ctio

n)

Ex

ampl

e

aaa

aa

aaa

aa

aa

aa

aa

aa

aa

aa

436

943

96

22

33

33

2

2

22

2

2

22

2

2

'#

#

--+

--

=--

-+

-

=+

--

+-

+-

=+

^

^

^

^ ^

^ ^h

h

h

h h

h h

2.

Addi

tion

and

Subt

ract

ion

z

Ensu

re a

ll de

nom

inat

ors

are

fully

fact

oris

ed

z

Find

LC

D (l

owes

t com

mon

den

omin

ator

)

z

Cha

nge

num

erat

ors

acco

rdin

gly

to e

nsur

e eq

uiva

lent

frac

tions

z

Col

lect

like

term

s

Fo

r exa

mpl

e, x

xx

xx

12

23

32

12

22

-+

+-

-+

+

xx

xx

xx

xx

xx

xx

xx

xx

xx

xx

xx

xx

xx

xx

11

21

23

21

1

11

22

23

11

1

11

22

43

31

11

24

8

11

24

2

11

4

=+

-+

-+

-+

+

=+

-+

++

+-

-

=+

-+

++

+-

+

=+

-+

+

=+

-+

+

=+

-

^ ^ ^ ^^^

^

^ ^ ^ ^^^

^^

^ ^^

^

^

^

^^

^h h h hhh

h

h h h hhh

hh

h hh

h

h

h

hh

h




EXPO

NENT

S

Rul

es:

Whe

n:th

en:

mul

tiply

ing

pow

ers

that

ha

ve th

e sa

me

base

aa

a3

58

#=

keep

the

base

and

add

th

e ex

pone

nts

divi

ding

pow

ers

that

ha

ve th

e sa

me

base

aaa

2108

=ke

ep th

e ba

se a

nd

subt

ract

the

expo

nent

s

rais

ing

a po

wer

to

anot

her p

ower

aa

52

10=

^h

mul

tiply

the

expo

nent

s

mor

e th

an o

ne b

ase

is

rais

ed to

a p

ower

()

aba

b3

33

=th

e ex

pone

nt b

elon

gs to

ea

ch b

ase

a ba

se is

rais

ed to

a

nega

tive

expo

nent

aa1

22

=-

reci

proc

ate

and

chan

ge

the

sign

of t

he e

xpon

ent

a ba

se h

as a

n ex

pone

nt o

f zer

oa

10

=It

will

equa

l 1

NO

TE C

ON

CER

NIN

G A

LL R

ULE

S: 2

22

35

8#

= (n

ot 4

8)

The

rule

s re

mai

n th

e sa

me

for b

ases

of n

umer

ical

val

ues

EQUA

TION

S AN

D IN

EQUA

LITIE

S

1.

Line

ar e

quat

ions

z

If th

ere

are

brac

kets

, use

the

dist

ribut

ive

law

to re

mov

e al

l br

acke

ts z

Col

lect

like

term

s on

eac

h si

de z

Get

all

the

term

s w

ith th

e va

riabl

e in

them

on

LHS

z

Get

all

cons

tant

s on

RH

S (b

ut re

mem

ber,

wha

teve

r is

done

to o

ne s

ide

of th

e eq

uatio

n m

ust b

e do

ne to

the

othe

r sid

e to

kee

p th

e eq

uatio

n ba

lanc

ed)

z

Col

lect

like

term

s on

eac

h si

de a

gain

z

Get

the

varia

ble

on it

s ow

n us

ing

divi

sion

2.

Equa

tions

with

frac

tions

z

Find

LC

D z

Mul

tiply

ALL

term

s th

roug

hout

equ

atio

n by

LC

D in

ord

er to

re

mov

e al

l fra

ctio

ns (n

o m

ore

deno

min

ator

s) z

Ther

e sh

ould

be

NO

frac

tions

AT

ALL

in th

e ne

xt s

tep.

z

Con

tinue

the

sam

e as

for l

inea

r equ

atio

ns

3.

Qua

drat

ic E

quat

ions

z

Rec

ogni

sabl

e by

the

“squ

are”

. You

sho

uld

be e

xpec

ting

two

answ

ers.

z

Get

ALL

term

s on

LH

S so

that

RH

S =

0 z

Fact

oris

e th

e LH

S fu

lly z

Find

the

2 po

ssib

le s

olut

ions

usi

ng th

e co

ncep

t tha

t tw

o fa

ctor

s m

ultip

lied

to e

qual

zer

o w

ill m

ean

that

eac

h on

e of

the

fact

ors

coul

d po

ssib

ly e

qual

zer

o.

4.

Sim

ulta

neou

s Eq

uatio

ns (G

iven

two

equa

tions

with

two

varia

bles

to

sol

ve fo

r)

Subs

titut

ion

met

hod

z

Get

ON

E of

the

varia

bles

by

itsel

f in

ON

E of

the

equa

tions

z

Use

this

info

rmat

ion

to s

ubst

itute

bac

k in

to th

e se

cond

equ

a-tio

n. Y

ou s

houl

d no

w h

ave

an e

quat

ion

with

onl

y on

e un

know

n va

riabl

e in

z

Solv

e fo

r thi

s va

riabl

e




z

Use

the

info

rmat

ion

just

foun

d to

sub

stitu

te b

ack

into

the

first

eq

uatio

n an

d so

lve

for t

he s

econ

d va

riabl

e.

Elim

inat

ion

met

hod

(wor

ks w

ell i

f one

of t

he v

aria

bles

has

the

sam

e co

-effi

cien

t in

each

of t

he tw

o eq

uatio

ns).

z

Plac

e th

e tw

o eq

uatio

ns b

elow

eac

h ot

her e

nsur

ing

all t

he li

ke

term

s ar

e un

dern

eath

eac

h ot

her

z

Subt

ract

the

one

equa

tion

from

the

othe

r (if

the

co-e

ffici

ents

ha

ve th

e sa

me

sign

) to

elim

inat

e th

at v

aria

ble

OR z

Add

the

2 eq

uatio

ns to

geth

er (i

f the

co-

effici

ents

hav

e op

posi

te

sign

s) to

elim

inat

e on

e of

the

varia

bles

z

Solv

e th

e ‘n

ew’ e

quat

ion

whi

ch h

as o

nly

one

unkn

own

z

Use

the

info

rmat

ion

just

foun

d to

sub

stitu

te b

ack

into

one

of

the

orig

inal

equ

atio

ns to

sol

ve fo

r the

mis

sing

var

iabl

e.

5.

Expo

nent

ial e

quat

ions

z

Base

s m

ust b

e th

e sa

me

to s

olve

exp

onen

tial e

quat

ions

– if

th

e ba

ses

are

the

sam

e, th

e ex

pone

nts

will

be th

e sa

me

z

If ba

ses

are

not t

he s

ame,

use

prim

e fa

ctor

s to

mak

e th

em th

e sa

me.

6.

Lite

ral E

quat

ions

z

Trea

t as

if it

is a

n or

dina

ry li

near

equ

atio

n fir

st (t

ry a

nd ig

nore

th

e fa

ct th

at th

ere

are

man

y va

riabl

es a

nd fe

w o

r no

num

bers

) z

Focu

s on

the

varia

ble

you

have

bee

n as

ked

to s

olve

for.

z

Get

all

term

s w

ith th

is v

aria

ble

in o

n on

e si

de a

nd a

ll te

rms

with

out t

his

varia

ble

in o

n th

e ot

her s

ide.

z

If th

e va

riabl

e yo

u ar

e so

lvin

g fo

r is

in m

ore

than

one

term

(and

th

ey’re

all

on o

ne s

ide

now

), fa

ctor

ise

by ta

king

this

var

iabl

e ou

t as

a co

mm

on fa

ctor

. z

Div

ide

both

sid

es b

y an

y ot

her v

aria

bles

‘in

the

way

’ and

get

th

e va

riabl

e yo

u’re

sol

ving

for o

n its

ow

n.

7.

Line

ar in

equa

litie

s z

Trea

t the

sam

e as

a li

near

equ

atio

n z

IF it

is re

quire

d to

div

ide

by a

neg

ativ

e in

tege

r to

get t

he v

ari-

able

alo

ne, t

he s

ign

(< o

r >) n

eeds

to b

e ch

ange

d. z

Thes

e so

lutio

ns m

ay n

eed

to b

e re

pres

ente

d on

a n

umbe

r lin

e.

Ineq

ualit

ies,

Inte

rval

Not

atio

n an

d R

epre

sent

atio

n on

a n

umbe

r lin

e Ineq

ualit

y si

gnw

ords

Ope

n/cl

osed

dot

>G

reat

er th

anO

pen

$G

reat

er th

an o

r eq

ual t

oC

lose

d

<Le

ss th

anO

pen

#Le

ss th

an o

r eq

ual t

oC

lose

d




> a

nd $

repr

esen

t an

arro

w th

is w

ay $

< a

nd $

repr

esen

t an

arro

w th

is w

ay #

Exam

ples

:

Ineq

ualit

yIn

terv

al

nota

tion

x2

>(

;)

x23

!-2

-10

4

x2$

[;

)x

23

!-2

-10

4

x2

6##

[;

]x

26

!1

-10

34

25

67

89

x2

6<

<(

;)

x2

6!

1-1

03

42

56

78

9

x2

6<

#[

;)

x2

6!

1-1

03

42

56

78

9

x2

6<#

(;

]x

26

!1

-10

34

25

67

89

Inte

rval

Not

atio

n is

use

d to

repr

esen

t a s

et o

f Rea

l Num

bers

as

it is

im

poss

ible

to li

st th

em.

NUM

BER

PATT

ERNS

Sequ

ence

: A s

et o

f num

bers

writ

ten

in o

rder

acc

ordi

ng to

som

e m

athe

mat

ical

rule

.

The

num

bers

in a

seq

uenc

e ar

e ca

lled

term

s.

The

term

s of

a s

eque

nce

are

indi

cate

d by

the

sym

bol T

n

Exa

mpl

e,

T 2 is

the

seco

nd te

rm o

f the

seq

uenc

e.

T

n, t

he n

th te

rm g

ives

the

rule

for t

he s

eque

nce.

A se

quen

ce th

at g

oes

up o

r dow

n in

equ

al s

teps

is c

alle

d an

arit

hmet

ic

sequ

ence

.

In a

n ar

ithm

etic

seq

uenc

e, a

con

stan

t val

ue is

eith

er a

dded

or s

ub-

tract

ed to

gen

erat

e th

e ne

xt te

rm in

the

sequ

ence

.

The

diffe

renc

e be

twee

n an

y 2

term

s in

an

arith

met

ic s

eque

nce

is

know

n as

the

com

mon

diff

eren

ce.

LINE

AR P

ATTE

RNS

All t

hese

pat

tern

s ha

ve a

com

mon

diff

eren

ce b

etw

een

each

term

. In

othe

r wor

ds,

TT

TT

21

32

-=

-

The

gene

ral t

erm

for a

line

ar p

atte

rn c

an b

e w

ritte

n as

Tbn

cn

=+

or

Tan

qn

=+

This

form

is li

ke th

e st

anda

rd fo

rm o

f the

stra

ight

-line

gra

ph w

hich

sh

ows

it is

a li

near

pat

tern

. A p

atte

rn h

owev

er w

ould

be

repr

esen

ted

by d

iscr

ete

poin

ts a

nd n

ot a

con

tinuo

us li

ne.




To fi

nd th

e ge

nera

l pat

tern

(als

o kn

own

as th

e nth

term

): z

Find

the

com

mon

diff

eren

ce z

Subs

titut

e in

to b

in th

e ge

nera

l for

mat

Subs

titut

e ‘n

’ with

1 to

repr

esen

t th

e 1s

t ter

m a

nd fi

nd w

hat c

(o

r q) s

houl

d be

to e

qual

the

first

te

rm.

Exam

ple:

Find

the

gene

ral t

erm

for t

he

patte

rn 4

7

1

0 1

3 ..

.C

omm

on d

iffer

ence

: 3

( 74

3-

= a

nd 1

07

3-

=)

Tbn

c

Tn

c

c c

3

31

4 1

n n

=+

=+

+= =

^h T

n31

n`

=+

Giv

en th

e po

sitio

n, lo

okin

g fo

r th

e te

rm: s

ubst

itute

n w

ith

posi

tion

give

n an

d fin

d T

n

Exam

ple.

Fin

d th

e 20

th te

rm o

f th

e ab

ove

patte

rn

Tn

T

31

320

161

n 20

=+

=+

=^h

Giv

en th

e te

rm, l

ooki

ng fo

r the

po

sitio

n: m

ake

an e

quat

ion

and

solv

e fo

r n

(sub

stitu

te in

Tn)

Exam

ple:

In w

hich

pos

ition

will

the

term

151

be

in th

e ab

ove

patte

rn?

Tn n n n3

1

151

31

150

3

50

n=

+

=+

= =

151

is th

e 50

th te

rm

FINA

NCE

AND

GROW

TH

Sim

ple

Inte

rest

()

AP

in1

=+

A =

Fin

al a

mou

ntP

= P

rinci

pal a

mou

nti

= in

tere

st ra

ten

= n

umbe

r of t

imes

inte

rest

is

calc

ulat

ed*

*In s

impl

e in

tere

st it

is a

lway

s an

nual

ly.*In

Gra

de 1

0, in

com

poun

d in

tere

st

it is

alw

ays

annu

ally.

Com

poun

d In

tere

st

()

AP

i1

n=

+

Hire

Pur

chas

e(b

uyin

g an

item

from

a s

hop

on c

redi

t - y

ou a

re o

ffici

ally

hi

ring

the

item

unt

il th

e fin

al

paym

ent w

hen

you

have

fin

ally

pur

chas

ed it

)

z

Alw

ays

use

sim

ple

inte

rest

fo

rmul

a (

.)

AP

in

1=

+ z

If in

sura

nce

is re

quire

d, it

is

alw

ays

on th

e to

tal p

urch

ase

pric

e re

gard

less

of d

epos

its p

aid

z

Dep

osits

are

sub

tract

ed fr

om

purc

hase

pric

e to

find

am

ount

ne

eded

to b

e ‘b

orro

wed

’.

Infla

tion

(The

incr

ease

in a

n ite

m

over

the

cour

se o

f tim

e)

z

Alw

ays

use

com

poun

d in

tere

st

form

ula

z

Whe

n w

orki

ng to

war

ds a

pr

evio

us ti

me

perio

d th

en y

ou

usua

lly h

ave

‘A’ a

nd a

re lo

okin

g fo

r ‘P’

.




Exch

ange

rate

sTh

e ra

te o

f one

cou

ntry

’s m

oney

ag

ains

t ano

ther

cou

ntry

’s m

oney

.U

se ra

tios

to c

onve

rt be

twee

n on

e cu

rrenc

y an

d an

othe

r.

Perc

enta

ge in

crea

se/

decr

ease

old

new

old

100

#-

If yo

ur a

nsw

er h

as a

‘min

us’ s

ign

then

it is

a d

ecre

ase

- do

not p

ut th

e m

inus

in y

our a

nsw

er.

FUNC

TION

S

Ther

e ar

e tw

o ty

pes

of fu

nctio

ns:

ON

E-TO

-ON

EM

ANY-

TO-O

NE

A si

ngle

x-v

alue

for a

par

ticul

ar

y-va

lue

Mor

e th

an o

ne x

-val

ue fo

r a

parti

cula

r y-v

alue

THER

E C

AN O

NLY

BE

ON

E y-

VALU

E

The

stra

ight

-line

gra

ph (L

inea

r fun

ctio

n)

Stan

dard

form

: y

axq

=+

G

radi

ent

V

ertic

al s

hift

(up/

dow

n)

:a

0>

:

a0

<

y-

inte

rcep

t

To d

raw

Find

the

x-in

terc

ept (

mak

e y

= 0

)Fi

nd th

e y-

inte

rcep

t (m

ake

x =

0)

To fi

nd th

e eq

uatio

n z

Giv

en th

e y-

inte

rcep

t and

ano

ther

poi

nt:

Subs

titut

e y-

inte

rcep

t for

‘q’

Subs

titut

e ot

her p

oint

(x; y)

to fi

nd ‘a

’ z

Giv

en tw

o po

ints

U

se tw

o po

ints

to fi

nd g

radi

ent (

‘a’)

Use

any

poi

nt to

sub

stitu

te a

nd fi

nd ‘q

’N

ote:

che

ck th

e va

lues

‘a’ o

f ‘q’

acc

ordi

ng to

w

hat t

hey

repr

esen

t (fo

r exa

mpl

e, if

you

hav

e fo

und

that

a0

<, c

heck

that

the

line

has

a ne

gativ

e sl

ope)

Dom

ain

and

Ran

geD

omai

n (a

ll po

ssib

le x

-val

ues

on th

e fu

nctio

n):

xR!

Ran

ge (a

ll po

ssib

le y

-val

ues

on th

e fu

nctio

n):

yR!

Oth

er z

If 2

lines

are

par

alle

l, th

en m

m1

2=

z

If 2

lines

are

per

pend

icul

ar, t

hen

mm

11

2#

=-

z

A lin

e pe

rpen

dicu

lar t

o th

e x-

axis

and

pa

ralle

l to

the

y-ax

is (a

ver

tical

line

): th

e eq

uatio

n w

ill be

in th

e fo

rm x

c=

z

A lin

e pe

rpen

dicu

lar t

o th

e y-

axis

and

pa

ralle

l to

the

x-ax

is (a

hor

izon

tal l

ine)

: the

eq

uatio

n w

ill be

in th

e fo

rm y

c=




The

para

bola

(Qua

drat

ic fu

nctio

n)

Stan

dard

form

: y

axq

2=

+

:

a0

>

:

a0

<

Verti

cal s

hift

(up/

dow

n)

y-in

terc

ept

open

s up

war

dop

ens

dow

nwar

dop

ens

upw

ard

o

pens

dow

nard

To d

raw

Find

the

x-in

terc

ept (

mak

e y

0=

)Fi

nd th

e y-

inte

rcep

t (m

ake

x0

=)

Find

the

axis

of s

ymm

etry

: x

0=

(In

Gr 1

0 th

is

is a

lway

s th

e ca

se)

Find

the

turn

ing

poin

t: ;q0^h

(In

Gr 1

0 th

is is

al

way

s th

e ca

se)

To fi

nd th

e eq

uatio

n z

Giv

en th

e y-

inte

rcep

t and

ano

ther

poi

nt:

Subs

titut

e y-

inte

rcep

t for

‘q’

Subs

titut

e ot

her p

oint

(x; y)

to fi

nd ‘q

’ z

Giv

en tw

o po

ints

Su

bstit

ute

each

poi

nt to

form

two

equa

tions

an

d so

lve

sim

ulta

neou

sly

Not

e: c

heck

the

valu

es o

f ‘a’ &

‘q’ a

ccor

ding

to

wha

t the

y re

pres

ent (

for e

xam

ple,

if y

ou h

ave

foun

d th

at a

0<

, ch

eck

that

the

para

bola

op

ens

dow

nwar

ds/is

ups

ide

dow

n)

Dom

ain

and

Ran

geD

omai

n (a

ll x-

poss

ible

val

ues

on th

e fu

nctio

n):

xR!

Ran

ge (a

ll y-

poss

ible

val

ues

on th

e fu

nctio

n):

If :

[;

)a

yq

0>

3!

If

:(

;]q

ay

0<

3!

-

Oth

erPa

rabo

las

can

have

a m

inim

um o

r a m

axim

um

valu

e. z

If a

0>

, the

re is

a m

inim

um v

alue

Th

e m

inim

um v

alue

is y

q=

z

If a

0<

, the

re is

a m

axim

um v

alue

Th

e m

axim

um v

alue

is y

q=

The

hype

rbol

a (H

yper

bolic

func

tion)

Stan

dard

form

: y

xaq

=+

:a

0>

:

a0

<

V

ertic

al s

hift

(up/

dow

n)H

oriz

onta

l as

ympt

ote

( y

q=

)




To d

raw

Dra

w in

the

horiz

onta

l asy

mpt

ote:

yq

=

Find

the

x-in

terc

ept (

mak

e y

0=

)Fi

nd a

few

mor

e po

ints

if n

eces

sary

with

po

ssib

le x

-val

ues.

R

emem

ber t

hat t

he v

ertic

al a

sym

ptot

e is

x0

=

(the

y-ax

is)

To fi

nd th

e eq

uatio

n z

Giv

en th

e ho

rizon

tal a

sym

ptot

e an

d an

othe

r po

int:

Subs

titut

e th

e va

lue

of th

e as

ympt

ote

for ‘

q’

Subs

titut

e ot

her p

oint

(x; y)

to fi

nd ‘a

’ z

Giv

en tw

o po

ints

Su

bstit

ute

each

poi

nt to

form

two

equa

tions

an

d so

lve

sim

ulta

neou

sly

Not

e: c

heck

the

valu

es o

f ‘a’ &

‘q’ a

ccor

ding

to

wha

t the

y re

pres

ent (

for e

xam

ple,

if y

ou h

ave

foun

d th

at a

0<

, che

ck th

at th

e hy

perb

ola

is in

th

e co

rrect

qua

dran

ts)

Dom

ain

and

Ran

geD

omai

n (a

ll po

ssib

le x

-val

ues

on th

e fu

nctio

n):

xR!

; x

0!

Ran

ge (a

ll po

ssib

le y

-val

ues

on th

e fu

nctio

n):

yR!

; y

q!

Oth

erA

hype

rbol

a ha

s tw

o ax

es o

f sym

met

ry.

z

One

has

a g

radi

ent o

f ‘1’

and

the

othe

r has

a

grad

ient

of ‘

-1’

z

They

bot

h pa

ss th

roug

h th

e po

int w

here

the

asym

ptot

es m

eet.

(;

)q0

(In

Gr 1

0 th

is is

al

way

s th

e ca

se)

The

expo

nent

ial g

raph

(Exp

onen

tial f

unct

ion)

Stan

dard

form

: .

ya

bq

x=

+

(;

)a

b0

0!

!

V

ertic

al s

hift

(up/

dow

n)

Hor

izon

tal a

sym

ptot

e

( yq

=)

b1

>b

01

<<

a0

>

a0

<

To d

raw

Dra

w in

the

horiz

onta

l asy

mpt

ote:

yq

=

Find

the

y-in

terc

ept (

mak

e x

0=

)Fi

nd a

few

mor

e po

ints

if n

eces

sary

with

po

ssib

le x

-val

ues.




To fi

nd th

e eq

uatio

n z

Giv

en th

e as

ympt

ote

and

anot

her p

oint

: Su

bstit

ute

the

valu

e of

the

asym

ptot

e fo

r ‘q’

Su

bstit

ute

othe

r poi

nt (x

; y)

to fi

nd ‘a

’ z

Giv

en tw

o po

ints

Su

bstit

ute

each

poi

nt to

form

two

equa

tions

an

d so

lve

sim

ulta

neou

sly

z

Not

e: c

heck

the

valu

es o

f ‘a’ o

r ‘b’

& ‘q

’ ac

cord

ing

to w

hat t

hey

repr

esen

t (O

nly

2 va

lues

will

be re

quire

d)

Dom

ain

and

Ran

geD

omai

n (a

ll po

ssib

le x

-val

ues

on th

e fu

nctio

n):

xR!

Ran

ge (a

ll po

ssib

le y

-val

ues

on th

e fu

nctio

n):

If :

a0

>

[;

)y

q3

!

If :

a0

<

(;

]qy

3!

-

Oth

erA

hype

rbol

a ha

s tw

o ax

es o

f sym

met

ry.

z

One

has

a g

radi

ent o

f ‘1’

and

the

othe

r has

a

grad

ient

of ‘

-1’

z

They

bot

h pa

ss th

roug

h th

e po

int w

here

the

asym

ptot

es m

eet.

(;

)q0

(In

Gr 1

0 th

is is

al

way

s th

e ca

se)

GENE

RAL

INFO

RMAT

ION

REGA

RDIN

G FU

NCTI

ONS

1.

Find

the

valu

es o

f fo

r whi

ch:

()

fx

gx

=^h

M

ake

the

equa

tions

equ

al a

nd s

olve

for

x.

If th

e co

ordi

nate

s ar

e as

ked

for,

subs

titut

e th

e x-

valu

e(s)

into

any

func

tion

and

solv

e fo

r y

.

For e

ach

of th

e qu

estio

ns b

elow

:Fi

rst fi

nd th

e pa

rt of

the

grap

h th

at a

nsw

ers

the

ques

tion

(hig

hlig

ht it

if

poss

ible

)Fi

nd th

e x-

valu

es th

at c

orre

spon

d to

the

part

of th

e gr

aph

that

sa

tisfie

s th

e st

atem

ent.

()

fx

gx

>^h

W

here

is th

e fu

nctio

n f

x^h

gre

ater

than

(in

othe

r w

ords

abo

ve) t

he fu

nctio

n (

)g

x.

()

fx

gx

<^h

W

here

is th

e fu

nctio

n f

x^h

less

than

(in

othe

r w

ords

bel

ow) t

he fu

nctio

n (

)g

x.

()

fx

gx

$^h

W

here

is th

e fu

nctio

n f

x^h

gre

ater

than

(in

othe

r w

ords

abo

ve) o

r equ

al to

the

func

tion

()

gx

.

()

fx

gx

#^h

W

here

is th

e fu

nctio

n f

x^h

less

than

(in

othe

r w

ords

bel

ow) o

r equ

al to

the

func

tion

()

gx

.

2.

Tran

sfor

mat

ions

of f

unct

ions

z

Refl

ectio

ns

Refl

ectio

n in

the

x-ax

is (

)y

0=

Rul

e: (

;)

(;

)x

yx

y"

-

In o

ther

wor

ds –

leav

e th

e x-

valu

e th

e sa

me

and

chan

ge th

e y-

valu

e to

neg

ativ

e

Refl

ectio

n in

the

y-ax

is (

)x

0=

Rul

e: (

;)

(;

)x

yx

y"

-

In o

ther

wor

ds –

leav

e th

e y-

valu

e th

e sa

me

and

chan

ge th

e x-

valu

e to

neg

ativ

e




PROB

ABIL

ITY

Prob

abilit

y is

the

likel

ihoo

d of

som

ethi

ng h

appe

ning

or b

eing

true

.

Prob

abilit

y is

ass

igne

d a

valu

e fro

m 0

(im

poss

ible

) to

1 (c

erta

in).

The

prob

abilit

ies

of th

e po

ssib

le o

utco

mes

in a

sam

ple

spac

e m

ust

sum

up

to 1

.

Prob

abili

ty o

f an

even

t occ

urrin

g an

d sa

mpl

e sp

ace

The

prob

abilit

y of

Eve

nt A

occ

urrin

g is

: (

)P

An

Sn

A=^ ^hh

In g

ener

al, A

is th

e to

tal n

umbe

r of w

ays

a sp

ecifi

c ev

ent c

an o

ccur

. S

is th

e to

tal n

umbe

r of p

ossi

ble

outc

omes

for t

he e

vent

.

Theo

retic

al a

nd R

elat

ive

freq

uenc

y

Rel

ativ

e fre

quen

cy is

the

prob

abilit

y fo

und

from

per

form

ing

an a

ctua

l tri

al. F

or e

xam

ple,

toss

ing

a co

in 1

00 ti

mes

and

find

ing

that

tails

ap

pear

ed 4

3 tim

es. T

here

fore

, the

pro

babi

lity

of g

ettin

g ta

ils, a

ccor

d-in

g to

the

expe

rimen

t, is

100

43.

In m

ost c

ases

, it t

akes

man

y tri

als

befo

re e

xper

imen

tal a

nd th

eore

tical

pr

obab

ilitie

s ap

proa

ch th

e sa

me

valu

e.

Not

atio

n

Prob

abilit

y of

an

even

t lie

s fro

m 0

to 1

. ;

;;

;;

;;

;;

A1

23

45

67

89

10="

,

how

ever

mea

ns th

e el

emen

ts th

at a

re p

art o

f Eve

nt A

.

z

()

PA"

The

pro

babi

lity

of E

vent

A o

ccur

ring

z

()

PA"l

The

pro

babi

lity

of E

vent

A N

OT

occu

rring

. It i

s al

so

know

n as

the

com

plem

ent o

f A

z

()

PA

orB

PA

B"

,=

^h

The

pro

babi

lity

of A

or B

occ

urrin

g.

, is

the

sym

bol f

or ‘o

r’ an

d is

als

o kn

own

as u

nion

.

z

()

PA

and

BP

AB"

+=

^h

The

pro

babi

lity

of A

and

B

occu

rring

. +

is th

e sy

mbo

l for

‘and

’ and

is a

lso

know

n as

in

ters

ectio

n.

z

nA"

^h

the

num

ber o

f ite

ms

in s

et A

.

Incl

usiv

e ev

ents

Two

even

ts th

at c

an o

ccur

at t

he s

ame

time

are

incl

usiv

e.

()

PA

B0

+!

(

)

PA

orB

PA

PB

PA

and

B=

+-

^^

^h

hh

Mut

ually

Exc

lusi

ve e

vent

s

Two

even

ts th

at a

re m

utua

lly e

xclu

sive

can

not o

ccur

at t

he s

ame

time.

Th

ere

is n

o in

ters

ectio

n.

PA

B0

+=

^h

()

PA

orB

PA

PB

=+

^^

hh




Exha

ustiv

e ev

ents

Two

even

ts A

and

B a

re e

xhau

stiv

e if

toge

ther

they

cov

er a

ll th

e el

emen

ts o

f the

sam

ple

spac

e.

PA

orB

1=

^h

Com

plem

enta

ry e

vent

s

Mut

ually

exc

lusi

ve, e

xhau

stiv

e ev

ents

are

com

plem

enta

ry e

vent

s.

They

are

the

only

two

poss

ible

out

com

es. I

f one

eve

nt d

oes

not o

ccur

, th

e ot

her e

vent

mus

t occ

ur

(

)P

not

AP

AP

A1

==

-l

^^

hh

The

addi

tion

rule

()

PA

BP

AP

BP

AB

,+

=+

-^

^^

hh

h

If th

e ev

ents

are

mut

ually

exc

lusi

ve th

en:

AB

PA

PB

,=

+^

^^

hh

h ,

as

PA

and

B0

=^

h.

Venn

dia

gram

s

Venn

dia

gram

s ar

e a

grap

hica

l way

of r

epre

sent

ing

a sa

mpl

e sp

ace

and

its e

vent

s. If

two

even

ts c

an b

oth

happ

en a

t the

sam

e tim

e, th

en a

Ve

nn d

iagr

am is

a g

ood

way

to re

pres

ent t

he s

ituat

ion.

Tree

dia

gram

s

Whe

n th

ere

are

2 or

mor

e co

nsec

utiv

e ev

ents

taki

ng p

lace

, it i

s of

ten

usef

ul to

repr

esen

t the

pos

sibl

e so

lutio

ns o

n a

tree

diag

ram

. Tre

e

diag

ram

s ar

e co

nstru

cted

by

show

ing

all p

ossi

ble

even

ts. T

hey

can

be u

sed

for d

epen

dent

or i

ndep

ende

nt e

vent

s. W

hen

deal

ing

with

tree

di

agra

ms

alw

ays

mul

tiply

alo

ng th

e br

anch

es (h

oriz

onta

l) an

d ad

d pr

obab

ilitie

s m

ovin

g do

wn

bran

ches

(ver

tical

) at t

he e

nd. W

rite

the

prob

abilit

y of

an

even

t occ

urrin

g at

the

top

of th

e br

anch

es a

nd th

e ac

tual

eve

nt a

t the

end

of t

he b

ranc

h.




RESOURCE 3

REVISION: PAPER 1 2017




RESOURCE 4

REVISION

SUM

MARY

NOT

ES –

PAP

ER 2

ANAL

YTIC

AL G

EOM

ETRY

All t

hree

form

ulae

requ

ire tw

o po

ints

: ;

xy

11

^h

and

;

xy

22

^h

Gra

dien

t

mx

xy

y2

1

21

=--

Dis

tanc

e be

twee

n tw

o po

ints

(the

leng

th o

f a li

ne s

egm

ent)

dx

xy

y2

12

21

2=

-+

-^

^h

h

Mid

poin

t (th

e m

iddl

e co

ordi

nate

of a

line

seg

men

t)

;x

xy

y2

21

21

2+

+a

k

USEF

UL IN

FORM

ATIO

N:

1.

Col

linea

r poi

nts:

poi

nts

that

lie

on a

stra

ight

line

To

pro

ve th

ree

poin

ts (A

, B &

C) c

ollin

ear,

prov

e A

BB

CA

Cm

mm

==

(onl

y tw

o pa

irs re

quire

d)

2.

Two

lines

are

par

alle

l if t

heir

grad

ient

s ar

e eq

ual.

3.

Two

lines

are

per

pend

icul

ar if

the

prod

uct o

f the

ir gr

adie

nts

equa

ls -

1.

4.

To fi

nd th

e y-

inte

rcep

t of a

ny g

raph

, let

x=

0 To

find

the

x-in

terc

ept o

f any

gra

ph, l

et y

=0

5.

To s

how

that

two

lines

bis

ect e

ach

othe

r: Fi

nd th

e m

idpo

ints

of

eac

h lin

e an

d if

they

are

the

sam

e, th

en th

e lin

es b

isec

t ea

ch o

ther

.

6.

To s

how

that

a p

oint

lies

on

a gr

aph:

sub

stitu

te th

e po

int a

nd

com

pare

the

LHS

and

RH

S. I

f the

y eq

ual e

ach

othe

r, th

en

the

poin

t lie

s on

the

grap

h.

7.

To fi

nd w

here

two

grap

hs in

ters

ect,

get b

oth

into

sta

ndar

d fo

rm (y

=…

), th

en le

t the

2 y

-val

ues

equa

l eac

h ot

her a

nd

solv

e fo

r x.

To fi

nd th

e co

rresp

ondi

ng y

-val

ues,

sub

stitu

te b

ack

into

any

eq

uatio

n.

8.

Prop

ertie

s of

qua

drila

tera

ls (o

ften

need

ed):

z

Dia

gona

ls o

f rho

mbu

s bi

sect

eac

h ot

her a

t 90°

z

Dia

gona

ls o

f a re

ctan

gle

are

equa

l in

leng

th.

9.

To p

rove

a q

uadr

ilate

ral i

s a

para

llelo

gram

, pro

ve o

ne o

f the

fo

llow

ing:

z

diag

onal

s bi

sect

(sam

e m

id-p

oint

)

z

both

pai

rs o

f opp

osite

sid

es p

aral

lel (

equa

l gra

dien

ts)

z

both

pai

rs o

f opp

osite

sid

es e

qual

(equ

al le

ngth

s)

z

one

pair

of o

ppos

ite s

ides

par

alle

l and

equ

al (e

qual

le

ngth

s &

equa

l gra

dien

ts)




Find

ing

the

equa

tion

of a

str

aigh

t lin

e

Exam

ples

:

Det

erm

ine

the

equa

tion

of a

stra

ight

whi

ch is

:

a)

Para

llel t

o th

e lin

e x3

4=

-+

; pas

sing

thro

ugh

the

poin

t (

;)

A4

7.

line

isy

xc

3"

<=

-+

Su

b ;

A4

7^h

c

73

4`

=-

+^h

c

19`

=

y

x317

`=

-+

b)

Perp

endi

cula

r to

the

line

x32

2=

-+

; with

a y

-inte

rcep

t of

3-

.

line

isto

yx

cy

x23

322

"=

=+

=-

+

Su

b c

3=

-

y

x23

3`

=-

c)

Para

llel t

o th

e x-

axis

and

pas

ses

thro

ugh

the

poin

t ;43

-^

h.

y

3=

(A li

ne p

aral

lel t

o th

e x-

axis

is a

hor

izon

tal l

ine.

)

d)

Para

llel t

o th

e y-

axis

and

pas

ses

thro

ugh

the

poin

t (;

)4

3-

.

x

4=

- (A

line

par

alle

l to

the

y-ax

is is

a v

ertic

al li

ne.)

STAT

ISTI

CSU

ngro

uped

dat

aR

epre

sent

ing

ungr

oupe

d da

ta g

raph

ical

ly:

Bar

Gra

ph

Dow

ntow

n

Mal

l

45 40 35 30 25 20 15 10 5 0

Cup

s of

Cof

fee

Sol

d

Jan

Feb

Mar

chA

pril

May

Bar g

raph

s: z

are

draw

n to

dis

play

dis

cret

e da

ta.

z

have

“bin

s’’ t

hat c

an b

e fig

ured

out

eas

ily b

y lo

okin

g at

the

data

(in

this

cas

e, m

onth

s).

z

usua

lly h

as s

pace

s be

twee

n th

e ba

rs z

can

be u

sed

to c

ompa

re a

sm

all n

umbe

r of v

alue

s (a

llow

ance

pe

r per

son,

or i

ncom

e pe

r com

pany

for e

xam

ple)

.




Bro

ken

line

grap

hs

Temperature (°C)

Day

Tem

per

atur

e at

noo

n

Sun

day

Mon

day

Tues

day

Wed

nesd

ayTh

ursd

ayFr

iday

15 14 13 12 11 10 9 8 7 6 5 0

Brok

en li

ne g

raph

s re

pres

ent c

hang

es in

dat

a ov

er ti

me,

suc

h as

cha

nges

in s

tock

mar

ket p

rices

thro

ugho

ut a

day

or c

hang

es

in d

aily

tem

pera

ture

s. In

suc

h gr

aphs

, con

vent

ion

dict

ates

the

inde

pend

ent v

aria

ble

is re

pres

ente

d ho

rizon

tally

on

the

x-ax

is a

nd

the

depe

ndan

t var

iabl

e is

repr

esen

ted

verti

cally

on

the

y-ax

is.

Box

-and

-whi

sker

plo

t

1516

1718

1920

2122

2324

2526

2728

29

This

type

of g

raph

is u

sed

to s

how

the

shap

e of

the

dist

ribut

ion,

its

cent

ral v

alue

, and

its

varia

bilit

y. In

a b

ox a

nd w

hisk

er p

lot:

the

ends

of

the

box

are

the

uppe

r and

low

er q

uarti

les,

so

the

box

span

s th

e in

terq

uarti

le ra

nge.

the

med

ian

is m

arke

d by

a v

ertic

al li

ne in

side

th

e bo

x.

Gro

uped

dat

a

Rep

rese

ntin

g gr

oupe

d da

ta g

raph

ical

ly:

His

togr

am

1.2

05101520253035

1.3

1.4

1.5

1.6

1.7

1.8

1.9

Hei

ght

(m)

A F

req

uenc

y-D

ensi

ty h

isto

gram

of t

he h

eigh

ts o

f a s

amp

leof

peo

ple

Frequency Density (m-1)

His

togr

ams

are

draw

n to

dis

play

con

tinuo

us d

ata.

Th

eref

ore

the

bars

are

touc

hing

.

Dis

cret

e da

ta h

as c

lear

sep

arat

ion

betw

een

the

diffe

rent

pos

sibl

e va

lues

, whi

le c

ontin

uous

dat

a do

esn’

t. W

e us

e ba

r gra

phs

for d

ispl

ay-

ing

disc

rete

dat

a, a

nd h

isto

gram

s fo

r dis

play

ing

cont

inuo

us d

ata.




MEA

SURE

S OF

CEN

TRAL

TEN

DENC

Y

Ung

roup

ed d

ata

Mea

nEx

ampl

e:

List

of s

hoe

size

s: 7

,9,1

2,9,

8,6,

9,12

,13,

17M

ost c

omm

only

use

d m

easu

re o

f cen

tral

tend

ency

Add

all d

ata

and

divi

de

by n

umbe

r of i

tem

s in

da

ta s

et.

The

mea

n is

dis

torte

d by

out

liers

107

912

98

69

1213

17+

++

++

++

++

10102

=

,10

2=

Med

ian

Mid

dlem

ost s

core

(o

dd n

umbe

r of d

ata)

or

ave

rage

of t

he tw

o m

iddl

e sc

ores

(eve

n nu

mbe

r of d

ata)

.N

umbe

rs n

eed

to b

e or

dere

d

6

7

8

9

9

9 1

2 1

2 1

3 1

7

29

9218

9+

==

Mod

eTh

e m

ost f

requ

ently

oc

curri

ng s

core

Can

hav

e m

ore

than

on

e m

ode

9

Gro

uped

dat

a

Estim

ate

of th

e m

ean:

z

Cal

cula

te th

e m

idpo

int o

f eac

h cl

ass

z

Mul

tiply

eac

h m

idpo

int b

y th

e fre

quen

cy fo

r tha

t int

erva

l z

Add

up a

nd d

ivid

e by

the

tota

l num

ber o

f sco

res

The

mod

al c

lass

: z

This

is th

e in

terv

al in

whi

ch th

e da

ta o

ccur

s m

ost f

requ

ently

The

med

ian:

z

The

best

way

to c

alcu

late

the

med

ian

is b

y dr

awin

g a

cum

ulat

ive

frequ

ency

cur

ve (O

give

z

A w

ay o

f rep

rese

ntin

g gr

oupe

d da

ta z

Nev

er g

oes

dow

n an

d sh

ould

form

an

S-sh

ape

z

Can

als

o be

use

d to

est

imat

e m

edia

n, q

uarti

les

and

perc

entil

es

Mea

sure

s of

Dis

pers

ion

(spr

ead

of d

ata)

1.

Ran

ge:

The

diffe

renc

e in

the

larg

est a

nd th

e sm

alle

st v

alue

in th

e da

ta s

et.

The

bigg

er th

e ra

nge

the

mor

e sp

read

out

the

data

is.

2.

Qua

rtile

s:M

easu

res

of d

ispe

rsio

n ar

ound

the

med

ian.

The

med

ian

divi

des

the

data

into

two

halv

es. T

he lo

wer

and

upp

er q

uarti

les

divi

de th

e da

ta

furth

er in

to q

uarte

rs.

To fi

nd:

Low

er q

uarti

le –

:

()

Qn

411

1+

Med

ian

–

:

()

Qn

211

2+

Upp

er q

uarti

le –

:

()

Qn

431

3+




Rem

embe

r: Th

is g

ives

the

posi

tion!

3.

Inte

r-qua

rtile

rang

e (IQ

R)

The

diffe

renc

e be

twee

n th

e up

per q

uarti

le a

nd lo

wer

qua

rtile

()

QQ

31

-

4.

Five

num

ber s

umm

ary:

z

Min

imum

: The

sm

alle

st v

alue

in th

e se

t of d

ata

z

Low

er q

uarti

le: T

he m

edia

n of

the

low

er h

alf o

f the

val

ues

z

Med

ian:

The

val

ue th

at d

ivid

es th

e da

ta in

to h

alve

s z

Upp

er q

uarti

le: T

he m

edia

n of

the

uppe

r hal

f of t

he v

alue

s z

Max

imum

: The

larg

est v

alue

in th

e da

ta.

The

box

and

whi

sker

plo

t is

a gr

aphi

cal r

epre

sent

atio

n of

the

five

num

ber s

umm

ary.

Min

imum

Low

er Q

uart

ileM

edia

nU

pp

er Q

uart

ileM

axim

um

Whi

sker

Whi

sker

Box

5060

7080

90

25%

25%

25%

25%

TRIG

ONOM

ETRY

Rig

ht-a

ngle

d tr

iang

les

SO

H

inyp

oten

euse

ppos

ite

SHO

i=

CAH

os

ypot

eneu

sedj

acen

tC

HA

i=

Oppos

ite

Hyp

oten

use

Adjace

nt

TO

A an

djac

ent

ppos

ite

TAO

i=

si

nry

i=

cos

rxi

=

ta

nxy

i=

Rec

ipro

cal f

unct

ions

sin

csc1

ii

=co

ssi

n1i

i=

cos

sec1

ii

=se

cco

s1i

i=

tan

cot1

ii

=co

tta

n1i

i=

Qua

dran

ts

Quad

rant

II

sin +

csc

+

Quad

rant

I

All p

os

tan +

cot

+

Quad

rant

III

cos

+

sec

+

Quad

rant

IV




Spec

ial a

ngle

s 90°

0°

(√3

; 1)

(√2

; √2)

(1 ;

√3)

r =

2

60°

40°

30°

(2 ;

0)

(0 ;

2)

2-di

men

sion

al p

robl

ems

horiz

onta

l

horiz

onta

l

angl

e of

dep

ress

ion

angl

e of

ele

vatio

n

Exam

ple

Find

the

heig

ht o

f the

tree

38º

4,2

m

,,

,

tan

tan

tree

tree

mtr

ee

384

24

238

328

o o#

= = =

Solv

ing

trig

onom

etric

equ

atio

ns (fi

ndin

g th

e si

ze o

f an

angl

e)

Step

s:

z

Get

the

trig

func

tion

of th

e an

gle

on it

s ow

n on

one

sid

e

z

Use

the

2nd

func

tion

on th

e ca

lcul

ator

: shi

ft ; t

rig fu

nctio

n ; r

atio

Exam

ples

:

,co

s0

85i

=

(shi

ft ; c

os ;

0,85

),

3179

o`i

=

, ,si

n

sin

x x

220

153

202

153

o o

+=

+=

^ ^

h h

(shi

ft ; s

in ;

, 21

53)

,x

2049

91o

o`

+=

,x

2991

o`

=




Usi

ng d

iagr

ams

to d

eter

min

e nu

mer

ical

val

ues

of ra

tios

(Pyt

hago

ras

ques

tions

)

Step

s:

z

Usi

ng B

OTH

pie

ces

of in

fo, d

ecid

e w

hich

qua

dran

t you

nee

d to

w

ork

in

z

Mak

e a

sket

ch, d

raw

ing

the

trian

gle

in th

e co

rrect

qua

dran

t. z

Fill

in th

e tw

o kn

own

side

s fro

m th

e gi

ven

info

z

Use

Pyt

hago

ras

to fi

nd th

e th

ird s

ide

z

Sum

mar

ise

the

info

you

now

kno

w re

gard

ing

wha

t x, y

and

r a

re

all e

qual

to

(Be

care

ful o

f sig

ns h

ere!

) z

Use

this

info

rmat

ion

to c

ompl

ete

the

ques

tion

usin

g su

bstit

utio

n.

NB:

Nee

d to

kno

w tr

ig ra

tios

in te

rms

of x

, y a

nd r

Don

’t ev

en c

onsi

der t

he ‘q

uest

ion’

(find

…) u

ntil

the

grou

ndw

ork

is

done

.

Trig

gra

phs

Sine

gr

aph

A fu

nctio

n of

the

form

si

ny

x=

with

a p

erio

d of

360

°

90º

-11

180º

270º

360º

y =

sin

xy

x

Cos

ine

grap

hA

func

tion

of th

e fo

rm

cos

yx

= w

ith a

per

iod

of 3

60°

y =

cos

x

90º

-11

180º

270º

360º

y

x

Tang

ent

grap

hA

func

tion

of th

e fo

rm

tan

yx

= w

ith a

per

iod

of 3

60°

y =

tan

x

90º

-11

180º

270º

360º

y

x

Perio

d: T

he n

umbe

r of d

egre

es it

take

s fo

r the

gra

ph to

com

plet

e a

patte

rn b

efor

e it

gets

repe

ated

Ampl

itude

: The

max

imum

dev

iatio

n fro

m th

e x-

axis

. C

an b

e fo

und

by u

sing

: ½ (d

ista

nce

betw

een

max

imum

and

min

imum

va

lues

)

Verti

cal s

hifts

of t

he s

ine,

cos

ine

and

tang

ent g

raph

s

si

nco

sta

ny

xq

yx

qy

xq

=+

=+

=+

‘q’ r

epre

sent

s th

e un

its th

e ba

sic

grap

h sh

ifts

verti

cally

(up

or d

own)

It w

ill ch

ange

the

max

imum

and

min

imum

val

ue a

nd th

eref

ore

the

rang

e.




It w

ill N

OT

chan

ge th

e am

plitu

de o

r per

iod.

The

verti

cal d

ista

nce

(siz

e) re

mai

ns th

e sa

me.

Ampl

itude

shi

fts o

f the

sin

e an

d co

sine

gra

phs:

si

nco

sy

ax

ya

x=

=

The

grap

h is

‘stre

tche

d’ o

r ‘sq

uash

ed’ f

rom

its

orig

inal

pos

ition

. The

ve

rtica

l dis

tanc

e (s

ize)

cha

nges

– it

bec

omes

long

er o

r sho

rter.

The

valu

e of

‘a’:

z

give

s th

e ne

w a

mpl

itude

. If ‘

a’ is

neg

ativ

e, th

is a

ffect

s th

e di

rect

ion

of th

e gr

aph

z

chan

ges

the

max

imum

and

min

imum

val

ue a

nd th

eref

ore

the

rang

e.

z

does

NO

T ch

ange

the

perio

d. It

rem

ains

360

°.

aAm

plitu

deSt

retc

hes

(a >

1)

or s

quas

hes

(0 <

a <

1)

or fl

ips

over

if a

< 0

qVe

rtica

l shi

ftN

umbe

r of u

nits

shi

fted

up o

r do

wn

the

y-ax

is

EUCL

IDEA

N GE

OMET

RY A

ND M

EASU

REM

ENT

Fam

ily tr

ee o

f qua

drila

tera

ls s

how

ing

how

they

rela

te to

eac

h ot

her

NO

pai

rs o

f p

aral

lel s

ides

TWO

pai

rs o

f p

aral

lel s

ides

ON

E p

air

of

par

alle

l sid

es

Qua

dril

ater

al

Kite

Par

alle

logr

am

Rec

tang

leR

hom

bus

Sq

uare

Trap

ezoi

d

Isos

cele

sTr

apez

oid

Defi

nitio

ns o

f the

6 q

uadr

ilate

rals

Para

llelo

gram

A qu

adril

ater

al w

ith b

oth

pairs

of o

ppos

ite s

ides

pa

ralle

l

Rec

tang

leA

para

llelo

gram

with

4 ri

ght a

ngle

s

Rho

mbu

sA

para

llelo

gram

with

4 e

qual

sid

es




Squa

reA

para

llelo

gram

with

4 e

qual

sid

es a

nd 4

righ

t an

gles

Kite

A qu

adril

ater

al w

ith 2

pai

rs o

f adj

acen

t sid

es e

qual

an

d no

opp

osite

sid

es e

qual

.

Trap

eziu

mA

quad

rilat

eral

with

one

pai

r of o

ppos

ite s

ides

pa

ralle

l

Prop

ertie

s of

qua

drila

tera

ls

This

dia

gram

sho

ws

how

eac

h sh

ape

rela

tes

to a

noth

er. F

or e

xam

ple,

th

e rh

ombu

s is

insi

de th

e pa

ralle

logr

am w

hich

is in

side

the

quad

rilat

-er

al –

ther

efor

e, th

e rh

ombu

s is

a q

uadr

ilate

ral a

nd m

ore

spec

ifica

lly it

is

als

o a

para

llelo

gram

but

then

has

at l

east

one

ext

ra fe

atur

e (4

equ

al

side

s) th

at m

ake

it a

rhom

bus.

The

dia

gram

sho

ws

how

the

quad

ri-la

tera

l is

a ve

ry g

ener

al s

hape

whe

reas

the

squa

re is

a m

ore

spec

ific

shap

e. T

his

can

also

be

seen

in th

e su

mm

ary

of p

rope

rties

bel

ow.

Qua

drilateral

Kite

Parallelogram

Rec

tang

leRho

mbus

Squa

re

Trap

ezoid

Prop

erty

Para

llelo

- gr

amR

ecta

ngle

Rho

mbu

sSq

uare

Opp

osite

si

des

para

llel

Opp

osite

an

gles

equ

al

Opp

osite

si

des

equa

l

Dia

gona

ls

bise

ct e

ach

othe

r

Dia

gona

ls

are

equa

l

Dia

gona

ls

are

perp

endi

cula

r

Dia

gona

ls

bise

ct

oppo

site

an

gles

All s

ides

eq

ual

All a

ngle

s rig

ht a

ngle

s




How

to p

rove

a q

uadr

ilate

ral i

s a:

1. P

aral

lelo

gram

z

both

pai

rs o

f opp

osite

sid

es

para

llel

or z

both

pai

rs o

f opp

osite

sid

es

equa

l or

z

one

pair

of o

ppos

ite s

ides

eq

ual a

nd p

aral

lel

or z

diag

onal

s bi

sect

eac

h ot

her

or z

oppo

site

ang

les

equa

l

2. R

ecta

ngle

It m

ust b

e a

para

llelo

gram

with

: z

equa

l dia

gona

ls

or z

one

right

ang

le

3. R

hom

bus

It m

ust b

e a

para

llelo

gram

with

: z

4 eq

ual s

ides

or

z

diag

onal

s bi

sect

at r

ight

an

gles

4. S

quar

eIt

mus

t be

a z

rhom

bus

with

one

righ

t ang

le

or z

rect

angl

e w

ith 2

adj

acen

t si

des

equa

l

The

mid

poin

t the

orem

The

line

join

ing

the

mid

poin

ts o

f tw

o si

des

of a

tria

ngle

is p

aral

lel t

o th

e th

ird s

ide

and

equa

l to

half

the

leng

th o

f the

third

sid

e. (A

bbre

vi-

ated

reas

on –

mid

pt th

eore

m)

Giv

en:

Then

:A

DE

BC

A

DE

BC

of1 2

Con

vers

e of

mid

poin

t the

orem

The

line

draw

n fro

m th

e m

idpo

int o

f one

sid

e of

a tr

iang

le, p

aral

lel t

o an

othe

r sid

e, b

isec

ts th

e th

ird s

ide.

(Abb

revi

ated

reas

on –

line

thro

ugh

mid

pt <

to 2

nd s

ide

Giv

en:

Then

:A

DE

BC

A

DE

BC

of1 2




MEA

SURE

MEN

TVo

lum

e

The

spac

e ta

ken

up b

y a

3D o

bjec

t. To

find

vol

ume,

the

area

of t

he

base

is m

ultip

lied

by th

e pe

rpen

dicu

lar h

eigh

t. Th

is o

nly

wor

ks fo

r rig

ht

pris

ms

VOLU

ME

OF:

AR

EA O

F B

ASE

◊ H

EIG

HT

Cub

e

ll

htl

ll

l3##

##

==

^h

Rec

tang

ular

pr

ism

lb

hlb

h##

=^

h

Tria

ngul

ar p

rism

()

bh

H21##

Not

e: T

he fi

rst ‘

h’ re

pres

ents

the

heig

ht o

f the

tri

angl

e in

ord

er to

find

the

area

of t

he b

ase.

Th

e 2n

d ‘H

’ rep

rese

nts

the

heig

ht o

f the

pris

m.

Cyl

inde

rr

htr

h2

2#

rr

=

Surf

ace

Are

a

The

area

take

n up

by

the

net o

f a 3

D s

olid

. The

sum

of t

he a

rea

of a

ll th

e fa

ces.

The

follo

win

g ba

sic

shap

e fo

rmul

ae a

re n

eede

d to

find

the

area

of t

he fa

ces

on a

ny 3

-dim

ensi

onal

sha

pe.

SHA

PEA

REA

FO

RM

ULA

Squa

re l

ll2

#=

Rec

tang

le l

b#

Tria

ngle

b

heig

ht21#

Circ

le

r2r

Con

es, p

yram

ids

and

sphe

res

(The

se fo

rmul

ae a

re g

iven

in a

n as

sess

men

t)

Shap

eSu

rface

Are

aVo

lum

e C

one

hs

r

rsr2

rr

+

(the

slan

t hei

ght i

s so

met

imes

nam

ed l)

rh

312

r

Sphe

re

r

r4

2r

r34

3r

Pyra

mid

h

bas

eArea

Sum

of t

he a

reas

of:

z

the

base

a

nd z

the

trian

gles

**th

e nu

mbe

r of

trian

gles

dep

ends

on

the

type

of b

ase

31 (a

rea

of b

ase)

h#

(rem

embe

r tha

t the

ba

se c

ould

be

any

poly

gon

but g

ener

ally

th

e sq

uare

, rec

tang

le

and

trian

gle

wou

ld b

e us

ed)




The

effec

t on

volu

me

whe

n m

ultip

lyin

g an

y di

men

sion

by

a co

nsta

nt fa

ctor

k:

z

If on

ly o

ne d

imen

sion

is c

hang

ed b

y a

valu

e of

k, t

he v

olum

e w

ill be

k ti

mes

big

ger

z

If on

ly tw

o di

men

sion

s ar

e ch

ange

d by

a v

alue

of k

, the

vol

ume

will

be k

2 tim

es b

igge

r

z

If al

l thr

ee d

imen

sion

s ar

e ch

ange

d by

a v

alue

of k

, the

vol

ume

will

be k

3 tim

es b

igge

r.




GRADE: 10 TERM: 4

REVISION

RESOURCE 5

REVISION WEEK 2: PAPER 2 2017




GRADE: 10 TERM: 4

REVISION

RESOURCE 6

REVISION WEEK 3: PAPER 1 EXEMPLAR

Copyright reserved Please turn over

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TIME: 2 hours

This question paper consists of 6 pages.

MATHEMATICS P1

EXEMPLAR 2012

NATIONAL SENIOR CERTIFICATE

GRADE 10



GRADE 10, TERM 4: RESOURCE PACKM

athe

mat

ics/

P12

DB

E/20

12N

SC –

Gra

de 1

0 Ex

empl

ar

Cop

yrig

ht re

serv

edPl

ease

turn

ove

r

INST

RU

CT

ION

S A

ND

INFO

RM

AT

ION

Rea

d th

e fo

llow

ing

inst

ruct

ions

car

eful

ly b

efor

e an

swer

ing

the

ques

tions

.

1. 2. 3. 4. 5. 6. 7. 8. 9.

This

que

stio

n pa

per c

onsi

sts o

f 7qu

estio

ns.

Ans

wer

ALL

the

ques

tions

.

Cle

arly

sho

w A

LL c

alcu

latio

ns,

diag

ram

s, gr

aphs

, et

cet

era

that

you

hav

e us

ed i

n de

term

inin

g yo

ur a

nsw

ers.

Ans

wer

s onl

y w

ill N

OT

nece

ssar

ily b

e aw

arde

d fu

ll m

arks

.

You

m

ay

use

an

appr

oved

sc

ient

ific

calc

ulat

or

(non

-pro

gram

mab

le

and

non-

grap

hica

l), u

nles

s sta

ted

othe

rwis

e.

If ne

cess

ary,

roun

d of

f ans

wer

s to

TWO

dec

imal

pla

ces,

unle

ss st

ated

oth

erw

ise.

Dia

gram

s are

NO

T ne

cess

arily

dra

wn

to sc

ale.

Num

ber

the

answ

ers

corre

ctly

acc

ordi

ng t

o th

e nu

mbe

ring

syst

em u

sed

in t

his

ques

tion

pape

r.

Writ

e ne

atly

and

legi

bly.

Mat

hem

atic

s/P1

3D

BE/

2012

NSC

–G

rade

10

Exem

plar

Cop

yrig

ht re

serv

edPl

ease

turn

ove

r

QU

EST

ION

1

1.1

Sim

plify

the

follo

win

g ex

pres

sion

s ful

ly:

1.1.

1(

) ()

22

62

nm

nm

nm

−−

−(3

)

1.1.

21

41

34

11

2

2

3

+−

−−

+−

+x

xx

xx

x(5

)

1.2

Fact

oris

e th

e fo

llow

ing

expr

essi

ons f

ully

:

1.2.

120

76

2−

−x

x(2

)

1.2.

2b

aba

a2

22

−−

+(3

)

1.3

Det

erm

ine,

with

out t

he u

se o

f a c

alcu

lato

r, b

etw

een

whi

ch tw

o co

nsec

utiv

e in

tege

rs

51lie

s.(2

)

1.4

Prov

e th

at

542,0

is ra

tiona

l.(4

)[1

9]

QU

EST

ION

2

2.1

Det

erm

ine,

with

out t

he u

se o

f a c

alcu

lato

r, t

he v

alue

of

xin

eac

h of

the

follo

win

g:

2.1.

121

42

=−

xx

(3)

2.1.

245

396

x=

(3)

2.1.

3Sx

R32

=(2

)

2.2

Solv

e fo

r p

and

qsi

mul

tane

ousl

y if:

52

37

6=

+=

+p

qp

q(5

)[1

3]




Mat

hem

atic

s/P1

4D

BE/

2012

NSC

–G

rade

10

Exem

plar

Cop

yrig

ht re

serv

edPl

ease

turn

ove

r

QU

EST

ION

3

3.1

3x+

1 ;

2x;

3x–

7…..

are

the

first

thre

e te

rms o

f alin

earn

umbe

r pat

tern

.

3.1.

1If

the

valu

e of

xis

thre

e, w

rite

dow

n th

e FI

RST

TH

REE

term

s.(3

)

3.1.

2D

eter

min

eth

e fo

rmul

a fo

r Tn,

the

gene

ral t

erm

of t

he se

quen

ce.

(2)

3.1.

3W

hich

term

in th

e se

quen

ce is

the

first

to b

e le

ss th

an –

31?

(3)

3.2

The

mul

tiple

s of t

hree

form

the

num

ber p

atte

rn:

3 ;

6

; 9

; 1

2 ;

...

Det

erm

ine

the

13th

num

ber i

n th

is p

atte

rnth

at is

eve

n.(3

)[1

1]

QU

EST

ION

4

4.1

Than

do h

as R

450

0 in

his

sav

ings

acc

ount

. The

ban

k pa

ys h

im a

com

poun

din

tere

st

rate

of 4

,25%

p.a

.Cal

cula

te th

e am

ount

Tha

ndo

will

rece

ive

if he

dec

ides

to w

ithdr

aw

the

mon

ey a

fter 3

0 m

onth

s.(3

)

4.2

The

follo

win

g ad

verti

sem

ent

appe

ared

with

reg

ard

to b

uyin

g a

bicy

cle

on a

hire

-pu

rcha

se a

gree

men

t loa

n:

Purc

hase

pri

ce

R5

999

Requ

ired

dep

osit

R600

Loan

term

Onl

y 18

mon

ths,

at 8

% p

.a.s

impl

e in

tere

st

4.2.

1C

alcu

late

the

mon

thly

am

ount

that

a p

erso

nha

s to

bud

get f

or in

ord

er to

pa

y fo

r the

bic

ycle

.(6

)

4.2.

2H

ow m

uch

inte

rest

doe

s one

hav

e to

pay

over

the

full

term

of t

he lo

an?

(1)

4.3

The

follo

win

g in

form

atio

nis

giv

en:

1ou

nce

= 28

,35

g$1

= R

8,79

Cal

cula

te th

e ra

nd v

alue

of a

1kg

gold

bar

, if 1

ounc

eof

gol

d is

wor

th $

978,

34.

(4)

[14]

Mat

hem

atic

s/P1

5D

BE/

2012

NSC

–G

rade

10

Exem

plar

Cop

yrig

ht re

serv

edPl

ease

turn

ove

r

QU

EST

ION

5

5.1

Wha

texp

ress

ion

BEST

repr

esen

ts th

e sh

aded

are

a of

the

follo

win

g V

enn

diag

ram

s?

5.1.

1(1

)

5.1.

2(1

)

5.2

Stat

e w

hich

of t

he fo

llow

ing

sets

of e

vent

s is m

utua

lly e

xclu

sive

:

A B C

Even

t 1:

The

lear

ners

in G

rade

10

in th

e sw

imm

ing

team

Ev

ent 2

: Th

e le

arne

rsin

Gra

de 1

0 in

the

deba

ting

team

Even

t 1:

The

lear

ners

in G

rade

8

Even

t 2:

The

lear

ners

in G

rade

12

Even

t 1:

The

lear

ners

who

take

Mat

hem

atic

s in

Gra

de 1

0Ev

ent 2

: Th

e le

arne

rs w

ho ta

ke P

hysi

cal S

cien

cesi

n G

rade

10

(1)

5.3

In a

cla

ss o

f 40

lear

ners

the

follo

win

g in

form

atio

n is

TR

UE:

•7

lear

ners

are

left-

hand

ed•

18le

arne

rs p

lay

socc

er•

4 le

arne

rs p

lay

socc

eran

d ar

e le

ft-ha

nded

•A

ll 40

lear

ners

are

eith

er ri

ght-h

ande

d or

left-

hand

ed

Let L

be

the

set o

f al

l lef

t-han

ded

peop

le a

ndS

be th

e se

t of

all l

earn

ers

who

pla

y so

ccer

.

5.3.

1H

ow m

any

lear

ners

in

the

clas

s ar

e rig

ht-h

ande

dan

d do

NO

Tpl

ay

socc

er?

(1)

5.3.

2D

raw

a V

enn

diag

ram

to re

pres

ent t

he a

bove

info

rmat

ion.

(4)

5.3.

3D

eter

min

e th

e pr

obab

ility

that

a le

arne

r is:

(a)

Left-

hand

ed o

r pla

ys so

ccer

(b)

Rig

ht-h

ande

d an

d pl

ays s

occe

r

(3)

(2)

[13]

AB

A




athe

mat

ics/

P16

DB

E/20

12N

SC –

Gra

de 1

0 Ex

empl

ar

Cop

yrig

ht re

serv

ed

QU

EST

ION

6

Giv

en:

1

3)

(+

=x

xf

and

4

2)

(−

−=

xx

g

6.1

Sket

ch th

e gr

aphs

of

fan

dg

on th

e sa

me

set o

f axe

s.(4

)

6.2

Writ

e do

wn

the

equa

tions

of t

he a

sym

ptot

es o

f f.

(2)

6.3

Writ

e do

wn

the

dom

ain

of f

.(2

)

6.4

Solv

e fo

r x

if

)(

)(

xg

xf

=.

(5)

6.5

Det

erm

ine

the

valu

esof

xfo

r whi

ch3

1<

≤−

)(xg

.(3

)

6.6

Det

erm

ine

the

y-in

terc

ept o

fk

if k

(x)=

2g(x

).(2

)

6.7

Writ

e do

wn

the

coor

dina

tes

of th

e x-

and

y-in

terc

epts

of

hif

his

the

grap

h of

gre

flect

ed a

bout

the

y-ax

is.

(2)

[20]

QU

EST

ION

7

The

grap

h of

q

axx

f+

=2

)(

is sk

etch

ed b

elow

.Po

ints

A(2

; 0) a

nd B

(–3

; 2,5

) lie

on

the

grap

h of

f.Po

ints

Aan

dC

are

x-in

terc

epts

of

f.

x

y

B(−3

; 2,5

)

A(2

; 0)

f

OC

7.1

Writ

e do

wn

the

coor

dina

tes o

f C

.(1

)

7.2

Det

erm

ine

the

equa

tion

of f

.(3

)

7.3

Writ

e do

wn

the

rang

e of

f.(1

)

7.4

Writ

e do

wn

the

rang

eof

h,w

here

(

)(

)2

hx

fx

=−

−.

(2)

7.5

Det

erm

ine

the

equa

tion

of a

nex

pone

ntia

l fun

ctio

n, g

(x) =

bx

+q,

with

rang

e y

>–

4an

d w

hich

pas

ses t

hrou

ghth

e po

int

A.

(3)

[10]

TO

TA

L:

100




GRADE: 10 TERM: 4

REVISION

RESOURCE 7

REVISION WEEK 3: PAPER 2 EXEMPLAR


MARKS: 100

TIME: 2 hours

This question paper consists of 10 pages and 1 diagram sheet.

MATHEMATICS P2

EXEMPLAR 2012

NATIONALSENIOR CERTIFICATE

GRADE 10




athe

mat

ics/

P22

DB

E/20

12N

SC–

Gra

de 1

0 Ex

empl

ar

Cop

yrig

ht re

serv

edPl

ease

turn

ove

r

INST

RU

CT

ION

S A

ND

INFO

RM

AT

ION

Rea

d th

e fo

llow

ing

inst

ruct

ions

car

eful

ly b

efor

e an

swer

ing

the

ques

tions

.

1. 2. 3. 4. 5.

This

que

stio

n pa

per c

onsi

sts o

f 9qu

estio

ns.

Ans

wer

ALL

the

ques

tions

.

Cle

arly

sho

w A

LL c

alcu

latio

ns, d

iagr

ams,

grap

hs, e

t cet

era

whi

ch y

ou h

ave

used

in

dete

rmin

ing

the

answ

ers.

Ans

wer

s onl

y w

ill N

OT

nece

ssar

ily b

e aw

arde

d fu

ll m

arks

.

You

m

ay

use

an

appr

oved

sc

ient

ific

calc

ulat

or

(non

-pro

gram

mab

le

and

non-

grap

hica

l), u

nles

s sta

ted

othe

rwis

e.

6. 7. 8.

If ne

cess

ary,

roun

d of

f ans

wer

s to

TWO

dec

imal

pla

ces,

unle

ss st

ated

oth

erw

ise.

Dia

gram

s are

NO

T ne

cess

arily

dra

wn

to sc

ale.

ON

Edi

agra

m sh

eet f

orQ

UES

TIO

N 6

.1.1

and

QU

ESTI

ON

9is

atta

ched

at t

he e

nd o

f th

is q

uest

ion

pape

r. W

rite

your

cen

tre n

umbe

r and

exa

min

atio

n nu

mbe

r on

this

she

et

in th

e sp

aces

pro

vide

d an

d in

sert

the

shee

t ins

ide

the

back

cov

er o

f yo

ur A

NSW

ER

BOO

K.

9. 10.

Num

ber

the

answ

ers

corre

ctly

acc

ordi

ng t

o th

e nu

mbe

ring

syst

em u

sed

in t

his

ques

tion

pape

r.

Writ

e ne

atly

and

legi

bly.

Mat

hem

atic

s/P2

3D

BE/

2012

NSC

–G

rade

10

Exem

plar

Cop

yrig

ht re

serv

edPl

ease

turn

ove

r

QU

EST

ION

1

A b

aker

kee

ps a

reco

rd o

f the

num

ber o

f sco

nes

that

he

sells

eac

h da

y. T

he d

ata

for 1

9 da

ysis

show

nbe

low

.

1.1

Det

erm

ine

the

mea

n of

the

give

n da

ta.

(2)

1.2

Rea

rrang

e th

e da

ta in

asc

endi

ng o

rder

and

then

det

erm

ine

the

med

ian.

(2)

1.3

Det

erm

ine

the

low

er a

nd u

pper

qua

rtile

s for

the

data

.(2

)

1.4

Dra

w a

box

and

whi

sker

dia

gram

to re

pres

ent t

he d

ata.

(2)

[8]

3136

6274

6563

6034

4656

3746

4052

4839

4331

66




Mat

hem

atic

s/P2

4D

BE/

2012

NSC

–G

rade

10

Exem

plar

Cop

yrig

ht re

serv

edPl

ease

turn

ove

r

QU

EST

ION

2

Traf

fic a

utho

ritie

s ar

e co

ncer

ned

that

hea

vy v

ehic

les

(truc

ks) a

re o

ften

over

load

ed. I

n or

der t

o de

al w

ith th

is p

robl

em,a

num

ber

of w

eigh

brid

ges

have

bee

n se

t up

alon

g th

e m

ajor

rou

tes

in

Sout

h A

frica

. The

gro

ss (t

otal

) veh

icle

mas

s is

mea

sure

d at

thes

e w

eigh

brid

ges.

The

hist

ogra

mbe

low

show

s the

dat

a co

llect

ed a

t a w

eigh

brid

ge o

ver a

mon

th.

2.1

Writ

e do

wn

the

mod

al c

lass

of t

he d

ata.

(1)

2.2

Estim

ate

the

mea

n gr

oss v

ehic

le m

ass f

or th

e m

onth

.(5

)

2.3

Whi

ch o

f th

e m

easu

res

of c

entra

l ten

denc

y,th

e m

odal

cla

ss o

r th

e es

timat

ed m

ean,

w

ill b

em

ost a

ppro

pria

te to

desc

ribe

the

data

set?

Exp

lain

you

r cho

ice.

(1)

[7]

103

19

70

77

85

99

Mas

s of V

ehic

les (

in k

g)

Mat

hem

atic

s/P2

5D

BE/

2012

NSC

–G

rade

10

Exem

plar

Cop

yrig

ht re

serv

edPl

ease

turn

ove

r

QU

EST

ION

3

3.1

In t

he d

iagr

am b

elow

, D

(–3

; 3)

, E(

3 ;

–5)

and

F(–1

; k

) ar

e th

ree

poin

ts i

n th

e C

arte

sian

pla

ne.

3.1.

1C

alcu

late

the

leng

th o

f DE.

(2

)

3.1.

2C

alcu

late

the

grad

ient

of D

E.(2

)

3.1.

3D

eter

min

e th

e va

lue

of k

if °

=90

FED

ˆ.

(4)

3.1.

4If

k=

–8,d

eter

min

e th

e co

ordi

nate

s of M

,the

mid

poin

t ofD

F.(2

)

3.1.

5D

eter

min

e th

e co

ordi

nate

s of

a p

oint

G s

uch

that

the

quad

rilat

eral

DEF

G

is a

rect

angl

e.(4

)

3.2

C is

the

poin

t (1

; –2)

. The

poi

nt D

lies

in th

e se

cond

qua

dran

t and

has

coo

rdin

ates

(x; 5

). If

the

leng

th o

fC

D i

s 53

units

, cal

cula

te th

e va

lue

ofx.

(4)

[18]

x

y

D(–

3 ; 3

)

E(3

; –5)

F(–1

; k)

y

x




athe

mat

ics/

P26

DB

E/20

12N

SC–

Gra

de 1

0 Ex

empl

ar

Cop

yrig

ht re

serv

edPl

ease

turn

ove

r

QU

EST

ION

4

4.1

In th

e di

agra

m b

elow

, ∆A

BC is

righ

t-ang

led

atB

.

Com

plet

e th

e fo

llow

ing

stat

emen

ts:

4.1.

1A

BC

sin

=(1

)

4.1.

2B

CA

BA=

(1

)

4.2

With

out u

sing

a c

alcu

lato

r, d

eter

min

e th

e va

lue

of:

°°

° 4530

60 sec

tan

.sin

(4)

4.3

Inth

e di

agra

m, P

(–5

; 12)

is a

poi

nt in

the

Car

tesi

an p

lane

and

θ

POR

=.

Det

erm

ine

the

valu

e of

:

4.3.

1θ

cos

(3)

4.3.

21

cose

c2+

θ(3

)[1

2]

A BC

θ

P(–5

; 12

)

y

xR•

•

O

Mat

hem

atic

s/P2

7D

BE/

2012

NSC

–G

rade

10

Exem

plar

Cop

yrig

ht re

serv

edPl

ease

turn

ove

r

QU

EST

ION

5

5.1

Solv

e fo

r x, c

orre

ct to

ON

E de

cim

al p

lace

, in

each

of t

he fo

llow

ing

equa

tions

whe

re

0o ≤

x≤

90o .

5.1.

13

5=

xco

s(2

)

5.1.

2191

2,

tan

=x

(3)

5.1.

35

3-se

c4

=x

(4)

5.2

An

aero

plan

e at

J i

s fly

ing

dire

ctly

ove

r a

poin

t D

on t

he g

roun

dat

a h

eigh

t of

5

kilo

met

res.

It is

hea

ding

to la

nd a

t poi

nt K

. The

ang

le o

f dep

ress

ion

from

Jto

K is

8°

. S is

a p

oint

alo

ng th

e ro

ute

from

D to

K.

5.2.

1W

rite

dow

n th

e si

ze o

f D

KJˆ.

(1)

5.2.

2C

alcu

late

the

dist

ance

DK

, cor

rect

to th

e ne

ares

t met

re.

(3)

5.2.

3If

the

dist

ance

SK

is8

kilo

met

res,

calc

ulat

e th

e di

stan

ce D

S.(1

)

5.2.

4C

alcu

late

the

angl

e of

ele

vatio

n fro

m p

oint

Sto

J, co

rrect

to O

NE

deci

mal

plac

e.(2

)[1

6]

J DS

K

8°

5km




Mat

hem

atic

s/P2

8D

BE/

2012

NSC

–G

rade

10

Exem

plar

Cop

yrig

ht re

serv

edPl

ease

turn

ove

r

QU

EST

ION

6

6.1

Con

side

r the

func

tion

xy

tan

2=

.

6.1.

1M

ake

a ne

at s

ketc

h of

x

yta

n2

=fo

r °

≤≤

°36

00

xon

the

axes

pro

vide

d on

DIA

GR

AM

SH

EET

1.C

lear

ly in

dica

te o

n yo

ur s

ketc

h th

e in

terc

epts

w

ithth

e ax

es a

nd th

e as

ympt

otes

.(4

)

6.1.

2If

the

grap

h of

x

yta

n2

=is

ref

lect

ed a

bout

the

x-ax

is, w

rite

dow

n th

e eq

uatio

n of

the

new

gra

ph o

btai

ned

by th

is re

flect

ion.

(1)

6.2

The

diag

ram

belo

w sh

ows t

he g

raph

of

xa

xg

sin

)(

=fo

r °

≤≤

°36

00

x.

-4-3-2-11234

x

y

6.2.

1D

eter

min

e th

e va

lue

of a

.(1

)

6.2.

2If

the

grap

h of

gis

tran

slat

ed 2

units

upw

ards

to o

btai

n a

new

gra

ph h

,w

rite

dow

n th

e ra

nge

ofh.

(2)

[8]

g

090°

180°

270°

360°

Mat

hem

atic

s/P2

9D

BE/

2012

NSC

–G

rade

10

Exem

plar

Cop

yrig

ht re

serv

edPl

ease

turn

ove

r

QU

EST

ION

7

7.1

The

roof

of

a ca

nvas

tent

is in

the

shap

e of

a r

ight

pyr

amid

hav

ing

a pe

rpen

dicu

lar

heig

ht o

f0,

8 m

etre

s on

a sq

uare

bas

e. T

he le

ngth

of o

ne si

de o

f the

bas

e is

3m

etre

s.

7.1.

1C

alcu

late

the

leng

th o

fAH

.(2

)

7.1.

2C

alcu

late

the

surf

ace

area

of t

he ro

of.

(2)

7.1.

3If

the

heig

ht o

f th

e w

alls

of

the

tent

is

2,1

met

res,

calc

ulat

e th

e to

tal

amou

nt o

f can

vas r

equi

red

to m

ake

the

tent

if th

e flo

or is

exc

lude

d.(2

)

7.2

A m

etal

bal

l has

a ra

dius

of8

mill

imet

res.

7.2.

1C

alcu

late

the

vol

ume

of m

etal

used

to

mak

e th

is b

all,

corr

ect

to T

WO

de

cim

al p

lace

s.(2

)

7.2.

2If

the

radi

us o

f the

bal

l is d

oubl

ed, w

rite

dow

n th

e ra

tio o

f th

e ne

w v

olum

e : t

he o

rigin

al v

olum

e.(2

)

7.2.

3Y

ou w

ould

like

this

bal

l to

be s

ilver

pla

ted

to a

thic

knes

s of

1 m

illim

etre

. W

hat i

s th

e vo

lum

e of

silv

er re

quire

d?G

ive

your

ans

wer

cor

rect

to T

WO

de

cim

al p

lace

s.(2

)[1

2]

0,8

m

3m

A

B

C

D

EFG

H




athe

mat

ics/

P210

DB

E/20

12N

SC–

Gra

de 1

0 Ex

empl

ar

Cop

yrig

ht re

serv

ed

Giv

ere

ason

s for

you

r st

atem

ents

in th

e an

swer

s to

QU

EST

ION

S8

and

9.

QU

EST

ION

8

PQR

Sis

a k

ite su

ch th

at th

e di

agon

als i

nter

sect

inO

. O

S =

2 cm

and

°=

20SP

Oˆ

.

8.1

Writ

e do

wn

the

leng

th o

f OQ

. (2

)

8.2

Writ

e do

wn

the

size

of

QOPˆ

.(2

)

8.3

Writ

e do

wn

the

size

of

SPQ

.(2

)[6

]

QU

EST

ION

9

In th

e di

agra

m, B

CD

Ean

d A

OD

E ar

e pa

ralle

logr

ams.

9.1

Prov

e th

at O

F || A

B.(4

)

9.2

Prov

e th

atA

BOE

is a

par

alle

logr

am.

(4)

9.3

Prov

e th

at∆A

BO

≡ ∆

EOD

.(5

)[1

3]

TO

TA

L:

100

S

P

Q

RO

2cm

20°

A

C

B

D

E

O

F

Mat

hem

atic

s/P2

DB

E/20

12N

SC–

Gra

de 1

0 Ex

empl

ar

Cop

yrig

ht re

serv

ed

CE

NT

RE

NU

MB

ER

:

EX

AM

INA

TIO

N N

UM

BE

R:

DIA

GR

AM

SH

EE

T 1

QU

EST

ION

6.1

.1

9018

027

036

0

-6-5-4-3-2-1123456

x

y

QU

EST

ION

9

A

C

B

D

E

O

F




GRADE: 10 TERM: 4

REVISION

RESOURCE 8

REVISION WEEK 3: MEMORANDUM PAPER 1 EXEMPLAR


MARKS: 100

This memorandum consists of 7 pages.

MATHEMATICS P1

EXEMPLAR 2012

MEMORANDUM

NATIONALSENIOR CERTIFICATE

GRADE 10




athe

mat

ics/

P1

2

DB

E/20

12

N

SC –

Gra

de 1

0 –

Mem

oran

dum

Cop

yrig

ht re

serv

ed

P

leas

e tu

rn o

ver

QU

EST

ION

1

1.1.

1(

) ()

22

62

nm

nm

nm

−−

−

32

23

32

22

23

211

8

212

26

nm

nn

mm

nm

nn

mm

nn

mm

++

−=

++

−−

−=

ex

pans

ion

3

m;

32n

+

2

211

8m

nn

m+

−(3

)1.

1.2

14

13

41

12

2

3

+−

−−

+−

+x

xx

xx

x () (

)(

)()

2

211

41

14

11

1(

1)

2xx

xx

xx

xx

xx

+−

++

−=

−+

−+

=+

−−

=

(

) () 1

12

+−

+x

xx

(

)() 1

14

−+

xx

)1

(1

−−

+x

x

an

swer

(5)

1.2.

1

()(

) 52

43

207

62

−+

=−

−x

xx

x(

) 43

+x(

) 52

−x(2

)1.

2.2

)2

)(1(

)1(

2)1

(2

22

ba

aa

ba

ab

aba

a

−+

=+

−+

=−

−+

gr

oupi

ng

(1 +

a)

(

) ba

2−

(3)

1.3

Sinc

e 49

72=

and

6482

=an

d 64

5149

<<

,8

517

<<

i.e.

51lie

s bet

wee

n 7

and

8

64

5149

<<

an

swer

(2)

1.4

Let

542,0

=x

Then

542,24

510

00

=x

i.e.

245

999

=x

i.e.

999

245

=x

Ther

efor

e x

is a

ratio

nal n

umbe

r.

in

trodu

ce v

aria

ble

542,

245

1000

=

x

245

999

=x

99

924

5=

x

(4)

[19]

Mat

hem

atic

s/P1

3

D

BE/

2012

NSC

–G

rade

10

–M

emor

andu

m

Cop

yrig

ht re

serv

ed

P

leas

e tu

rn o

ver

QU

EST

ION

2

2.1.

1

()(

)0

73

021

4

214

2

2

=−

+=

−−

=− x

xx

x

xx 3

or0

3−

==+

xx

707==

−x

x

st

anda

rd fo

rm

fact

ors

an

swer

s

(3)

2.1.

2

()

()

162232

32

396

454545

5454

======

x

xx

45

32x

=

(

)5432

=x

an

swer

(3)

2.1.

3

49

2332

22S

Rx

xRS

SxR ===

M

ultip

ly b

y 3S

and

di

vide

by

2

Squa

ring

both

side

s (2)

2.2

2Eq

uatio

n ..

......

....

......

....

52

1Eq

uatio

n ..

......

....

......

....

37

6=

+=

+p

qp

q 67

3....

......

......

......

Equa

tion

114

735

......

......

......

..mul

tiply

Equ

ati o

n 2

with

7 ..

......

Equa

tion

3q

pq

p+

=+

=

Equa

tion

3 –

Equa

tion

1:

4328

==qq ()

354

2−

==+

pp

35

714

=+

pq

32

8=

q

4=

q

su

bstit

utio

n

3−=

p(5

)[1

3]




Mat

hem

atic

s/P1

4

D

BE/

2012

NSC

–G

rade

10

–M

emor

andu

m

Cop

yrig

ht re

serv

ed

P

leas

e tu

rn o

ver

QU

EST

ION

3

3.1.

110

; 6

; 2

10

6

2(3

)3.

1.2

d=

–4

T n=

–4n

+ 14

–

4n

14

(2)

3.1.

3

1225

11454

3114

4

=>−

<−

−<

+−

nnnn

,

31

144

−<

+−

n

n>1

1,25

an

swer

(3)

3.2

7813

66

13===

)(

Tn

T nO

R78

2633

26===

)(

Tn

T n

6n

subs

titut

ion

of 1

3

answ

er(3

)O

R

3n

subs

titut

ion

of 2

6

answ

er(3

)[1

1]

QU

EST

ION

4

4.1

47.49

9310

025.41

4500

)1(

5.2

R

iP

An

=

+

=

+=

n

= 2.

5

subs

titut

ion

an

swer

(3)

4.2.

1Lo

an a

mou

nt =

R5

999

–R

600

= R

5 39

9

Tota

l am

ount

ow

ed =

5 3

99[1

+(0,

08)(1

,5)]

= R

6 04

6,88

Mon

thly

inst

alm

ent

=

1888.

6046

= R

335,

94

y

= 0

5

399

n

= 1,

5

Subs

titut

ion

R

604

6,88

÷

18

R33

5,94

(6)

4.2.

2R

6 04

6,88

-R

5 39

9=

R64

7,88

an

swer

(1)

4.3

1kg

= 1

000

g

35,2810

00=

35,2

7336

861…

ounc

es

35,2

7336

861…

x 97

8,34

x 8

,79

= R

303

337,

16

co

nver

sion

di

visi

on

mul

tiplic

atio

n

answ

er(4

)[1

4]

Mat

hem

atic

s/P1

5

D

BE/

2012

NSC

–G

rade

10

–M

emor

andu

m

Cop

yrig

ht re

serv

ed

P

leas

e tu

rn o

ver

QU

EST

ION

5

5.1.

1A∩

BO

RA

and

B

answ

er(1

)5.

1.2

A/

OR

not A

an

swer

(1)

5.2

B

answ

er(1

)5.

3.1

19le

arne

rs a

re ri

ght-h

ande

dan

d do

not

pla

y so

ccer

.

answ

er(1

)5.

3.2

15

4

2

19

(4)

5.3.

3 (a

)P(

L O

R S

)14

43

40++

=

21 40=

15

+ 4

+ 2

40

an

swer

(3)

5.3.

3 (b

)P(

R A

ND

S)

14 40=

7 20=

4015

an

swer

(2)

[13]

143

4

SL

19




athe

mat

ics/

P1

6

DB

E/20

12

N

SC –

Gra

de 1

0 –

Mem

oran

dum

Cop

yrig

ht re

serv

ed

P

leas

e tu

rn o

ver

QU

EST

ION

6

6.1

-10

-9-8

-7-6

-5-4

-3-2

-11

23

45

67

89

10

-7-6-5-4-3-2-112345678

x

y

fg

f

sh

ape

of f

x-

int o

f f

x-

inte

rcep

t of g

y-

inte

rcep

t of g (4

)

6.2

x=

0 an

d y

= 1

an

swer

an

swer

(2)

6.3

);

0()0;

(∞

∪−∞

va

lues

no

tatio

n(2

)6.

4

0)1

)(32(

03

52

52

3

52

3

42

13

2

2

=+

+=

++

−−

=

−−

=

−−

=+

xx

xx

xxx

x

xx

23−

=x

or

1−=

x

4

21

3−

−=

+x

x

st

anda

rd fo

rm

fact

ors

answ

ers

(5)

6.5

5,15,3

5,35,1

72

33

42

1

−≤

<−

−>

≥−

<−

≤<

−−

≤−

xxxx

OR

]5,1

;5,3

(−

−∈x

3

42

1<

−−

≤−

x

72

3<

−≤

x

an

swer

(3)

6.6

84

)42

(2)

(−

−=

−−

=x

xx

k y-in

terc

ept:

(0 ;

–8)

eq

uatio

n of

k(x

)

answ

er(2

)

6.7

x-in

terc

ept:

(2

; 0)

y-in

terc

ept:

(0

; –

4)

x-in

terc

ept

y-

inte

rcep

t(2

)[2

0]

Mat

hem

atic

s/P1

7

D

BE/

2012

NSC

–G

rade

10

–M

emor

andu

m

Cop

yrig

ht re

serv

ed

QU

EST

ION

7

7.1

C(–

2 ; 0

)

answ

er(1

)7.

2

()

() 4

21)

(

2155,2

)4)3

((5,2

4)

(

)(

22

22

−=

=

=−

−=

−=

+=

xx

fa

aa

xa

xf

qax

xf

(

)16

2−

=x

ax

f)

(

su

bstit

utio

n of

(–5

; 2,2

5)

an

swer

(3)

7.3

Ran

ge o

f f:

);

2[

∞−

an

swer

(1)

7.4

Ran

ge o

f h:

( ;

0]−∞

no

tatio

n

criti

cal v

alue

s(2

)7.

52 2

()

40

44

2(

)2

4

x x

gx

bb b

b gx

=−

=−

= ==

−

(

)4

xg

xb

=−

su

bstit

utio

n

an

swer

(3)

[10]

TO

TA

L: 1

00




GRADE: 10 TERM: 4

REVISION

RESOURCE 9

REVISION WEEK 3: MEMORANDUM PAPER 2 EXEMPLAR


MARKS: 100

This memorandum consists of 10 pages.

MATHEMATICS P2

EXEMPLAR 2012

MEMORANDUM

NATIONAL SENIOR CERTIFICATE

GRADE 10




athe

mat

ics/

P22

DB

E/20

12N

SC –

Gra

de 1

0 Ex

empl

ar–

Mem

oran

dum

Cop

yrig

ht re

serv

edPl

ease

turn

ove

r

NO

TE

:•

If a

cand

idat

e an

swer

s a q

uest

ion

TWIC

E, o

nly

mar

k th

e FI

RST

atte

mpt

.•

If a

cand

idat

e ha

s cro

ssed

out

an

atte

mpt

of a

que

stio

n an

d no

t red

one

the

ques

tion,

mar

k th

e cr

osse

d ou

t ver

sion

.•

Con

sist

ent a

ccur

acy

appl

ies i

n A

LLas

pect

s of t

he m

arki

ng m

emor

andu

m.

•A

ssum

ing

answ

ers/

valu

es in

ord

er to

solv

e a

prob

lem

is N

OT

acce

ptab

le.

QU

EST

ION

1

1.1

Mea

n =

nxn i

i∑ =1

=89

481992

9,

=

19929

an

swer

(2)

1.2

31 ;

31 ;

34 ;

36 ;

37 ;

39 ;

40 ;

43 ;

46 ;

46 ;

48 ;

52 ;

56 ;

60 ;

62 ;

63 ;

65 ;

66 ;

74.

Med

ian

= 46

ar

rang

ing

in

as

cend

ing

orde

r

m

edia

n(2

)1.

3Lo

wer

qua

rtile

= 3

7U

pper

qua

rtile

= 6

2

low

erqu

artil

e

uppe

r qua

rtile (2

)1.

4

bo

xw

ithm

edia

n

whi

sker

(2

)[8

]

7525

3545

5565

3174

3746

62

Mat

hem

atic

s/P2

3D

BE/

2012

NSC

–G

rade

10

Exem

plar

–M

emor

andu

m

Cop

yrig

ht re

serv

edPl

ease

turn

ove

r

QU

EST

ION

2

2.1

The

mod

al c

lass

is

4500

2500

<≤

x

4500

2500

<≤

x(1

)2.

2G

ross

Veh

icle

M

ass (

GV

M)

(in k

g)Fr

eque

ncy

Mid

poin

tFr

eque

ncy

×m

idpo

int

4500

2500

<≤

x10

335

0036

050

0

6500

4500

<≤

x19

5500

104

500

8500

6500

<≤

x70

7500

525

000

1050

085

00<

≤x

7795

0073

150

0

1250

010

500

<≤

x85

1150

097

750

0

1450

012

500

<≤

x99

1350

01

336

500

Sum

453

403

550

0

Estim

ated

mea

n )

(X=

39,89

0845

340

3550

0=

kg.

m

idpo

ints

frequ

enci

es ×

m

idpo

int

4

035

500

an

swer

(5)

2.3

The

estim

ated

mea

n.It

is m

ore

at th

e ce

ntre

of t

he d

ata

set.

The

mod

al c

lass

is fo

und

at

the

extre

me

left-

hand

side

of t

he d

ata

set.

es

timat

ed m

ean

with

reas

on

(1)

[7]




Mat

hem

atic

s/P2

4D

BE/

2012

NSC

–G

rade

10

Exem

plar

–M

emor

andu

m

Cop

yrig

ht re

serv

edPl

ease

turn

ove

r

QU

EST

ION

3

3.1.

1

1010

0

53

3)3

(D

E2

2

==

−−

+−

−=

))(

(

subs

titut

ion

into

di

stan

ce fo

rmul

a

an

swer

(2)

3.1.

2

343

33

5

−=

−−−

−=

)(

DE

m

subs

titut

ion

into

gr

adie

nt fo

rmul

a

an

swer

(2)

3.1.

343

=EF

mEF

⊥ D

E

8324

124

2043

4543

13

5

−==

−=

−−

=−

−

=−

−−

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43

=EF

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13

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−)

(k

si

mpl

ifica

tion

8−

=k

(4)

3.1.

4

−−

=

−+

−+

−

252

28)

32

1)3)

(M

;

(;

(

subs

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x

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D(–

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)

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x

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3.1.

5If

DEF

G is

a re

ctan

gle

then

M is

als

o th

e m

idpo

int o

f EG

.Le

t the

coo

rdin

ates

of G

be

(x;y

)

−−

=

−

++

252

25

23

;)

(;

yx

74

3

22

3 −=

−=

+

−=

+

xxx

and

05

525

25

=−

=−

−=

−

yyy

∴G( –

7; 0

)

OR

The

trans

latio

n th

at se

nds E

(3 ;

–5) t

o F(

–1; –

8) a

lso

send

s D

(–3

; 3) t

o G

.(–

1 ; –

8) =

(3 –

4 ; –

5–

3)∴

G =

(–3

–4

; 3 –

3) =

(–7

; 0) OR

The

trans

latio

n th

at se

nds E

(3 ;

–5) t

o D

(–3

; 3) a

lso

send

s

F(–1

; –8)

to

G.

(–3

; 3) =

(3 –

6 ; –

5 +

8)∴

G =

(–1

–6

; –

8 +

8) =

(–7

; 0)

2

23

−=

+x

7−

=x

25

25

−=

−y

0

=y

(4)

m

etho

d

x–

4

y–

3

answ

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)

m

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x–

6

y+

8

answ

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)

3.2

()

()

22

2

2 2

15

(2)

53(

1)49

532

149

530

23

0(

1)(

3)0

1or

3

but D

is in

the

seco

nd q

uadr

ant

onl

y 1

is v

alid

xx x

xx

xx

xx

x

x

−+

−−

=−

+=

−+

+−

=−

−=

+−

==−

=

∴=−

eq

uatio

n us

ing

dist

ance

form

ula

st

anda

rd fo

rm

fa

ctor

isat

ion

answ

er m

ust

excl

ude

3(4

)[1

8]




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oran

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QU

EST

ION

4

4.1.

1AC

CA

Bsi

n=

A

C(1

)

4.1.

2B

CA

B=

Aco

t

Aco

t(1

)

4.2

sin6

0.ta

n30

sec4

5

31

23

21 2 2

11

22

12

22

2

2 4

°°

°

= = =×

=×

=

subs

titut

ion

si

mpl

ifica

tion

an

swer

(4)

4.3.

1(

)(

)

13169

125

2

22

2 ==

+−

=

rrr

135co

s−

=θ

(

)(

)22

212

5+

−=

r

13

=r

an

swer

(3)

4.3.

2

144

313

144

144

144

169

11213

1co

sec

2

2

=

+=

+

=

+θ

1213

=

si

mpl

ifica

tion

an

swer

(3)

[12]

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QU

EST

ION

5

5.1.

1

°=

===

− 1,53

53co

s

53co

s

3co

s5

1

xxxx

53

=x

cos

an

swer

(2)

5.1.

2

°=

°===

−

25...

..95

845

,49

2)

19,1(ta

n2

19,12

tan

1

xxxx

°=

....

,958

492x

an

swer

(3

)5.

1.3

°=

======−

−

6021

cos

21co

s

21se

c12

sec

8se

c4

53

sec

4

1

xxxxxxx

2

sec

=x

in

verti

ng b

oth

side

s

co

sx

an

swer

(4)

5.2.

1°

=8

DKJˆ

alte

rnat

e an

gles

answ

er(1

)5.

2.2

m57

735

DK

km...

35,5

7684

..D

K8

tan 5

DK

DK5

tan8

==°

=

=°

D

K58

=°

tan

°

=8

tan 5

DK

an

swer

(3)

5.2.

3D

S=

35,5

8–

8 =

27,5

8 km

an

swer

(1)

5.2.

4

°=

=

=

−

10,3

JSD

27,5

85

tan

JSD

27,5

85

JSD

tan

1

27

,58

5JS

D=

ˆta

n

an

swer

(2)

[16]




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2012

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Exem

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QU

EST

ION

6

6.1.

1

-6-5-4-3-2-1123456

x

y

corre

ct x

-inte

rcep

ts

corre

ct y

-inte

rcep

t

asym

ptot

es

sh

ape

(mus

t pas

s th

roug

h (4

5° ;

2))

(4)

6.1.

2x

yta

n2−

=

answ

er(1

)

6.2.

1

4)1(

490

sin

4si

n)

(

==°

==

aaa

xa

xg

4

=a

(1)

6.2.

2R

ange

is

62

≤≤

−y

.

–2

6

(2)

[8]

090°

180°

270°

360°

Mat

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QU

EST

ION

7

7.1.

1A

H2

= 0,

82+

1,52

AH

2=

2,89

AH

= 1,

7

A

H2

= 0,

82+

1,52

A

H=

1,7

(2)

7.1.

2Su

rface

are

a of

roof

=)

,(

713

214

××

= 10

,2 m

2

)

,(

713

214

××

an

swer

(2)

7.1.

3Su

rface

are

a of

wal

ls =

4×

3×

2,1

= 25

,2 m

2

Tota

l sur

face

are

a =

10,2

m2 +

25,2

m2

= 35

,4 m

2

25

,2m

2

an

swer

(2)

7.2.

1V

olum

e =

()3

834 π

= 21

44,6

6 m

m3

()

38

34 π

an

swer

(2)

7.2.

2N

ew v

olum

e : o

rigin

al v

olum

e =

23: 1

= 8

: 1

23

an

swer

(2)

7.2.

3V

olum

e in

clud

ing

silv

er =

()3

934 π

= 3

053,

63m

m3 .

Vol

ume

of si

lver

= 3

053,

63 –

2144

,66

= 90

8,97

mm

3

()3

934 π

an

swer

(2)

[12]

QU

EST

ION

8

8.1

OQ

= 2

cm

2cm

co

rrect

reas

on(2

)

8.2

°=

90Q

OP

°90

co

rrect

reas

on(2

)

8.3

°=

°+°

=∴

°=

4020

20SP

Q

20OP

Q

°=

20OP

Qw

ithco

rrect

reas

on

°

=40

SPQ

(2)

[6]

…. (

the

long

dia

gona

l of a

kite

bis

ects

th

e sh

orte

r dia

gona

l)

…. (

the

diag

onal

s of a

kite

inte

rsec

t at

right

ang

les)

…. (

the

long

er d

iago

nal b

isec

ts th

e an

gles

of a

kite

)




athe

mat

ics/

P210

DB

E/20

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0 Ex

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ar –

Mem

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Cop

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ht re

serv

ed

QU

EST

ION

9

9.1

O is

the

mid

poin

t of B

D.

F is

the

mid

poin

t of O

E.

∴O

F || A

B

O

is th

e m

idpo

int

of B

D

reas

on –

diag

onal

s of p

arm

F

is th

e m

idpo

int

of O

E

re

ason

–m

idpo

int t

heor

em (4

)9.

2A

E || O

D∴

AE

|| OB

OF

|| AB

∴O

E || A

B

∴A

BOE

is a

par

alle

logr

am

A

E || O

B

reas

on

O

E || A

B

re

ason

–op

psi

des p

aral

lel

(4)

9.3

In ∆

AB

O a

nd∆E

OD

1.A

B =

EO…

(Opp

side

s of p

arm

ABO

E ar

e eq

ual)

2.A

O =

ED

…(O

pp si

des o

f par

m A

OD

E ar

e eq

ual)

3.BO

= D

O…

(Dia

gona

ls o

f par

m B

CD

E bi

sect

eac

h ot

her)

∴∆A

BO

≡∆E

OD

(S, S

, S)

A

B =

EO

AO

= E

D

reas

on –

opp

side

s are

equ

al

BO =

DO

re

ason

–di

agon

also

f par

m (5)

[13]

TO

TA

L:

100

…. (

Dia

gona

ls o

f par

m A

OD

E b

isec

t ea

ch o

ther

)

…. (

Opp

side

s of p

arm

AO

DE

are

pa

ralle

l)

…. (

prov

en a

bove

)

…. (

both

pai

rs o

f opp

osite

side

s of

quad

are

par

alle

l)

…. (

Dia

gona

ls o

f par

m B

CD

E b

isec

t ea

ch o

ther

)

…. (

The

line

join

ing

the

mid

poin

ts o

f tw

o si

des i

n a

∆ is

|| to

third

side

)

A

C

B

D

E

O

F




Gr 10

MATHEMATICS TESTQUESTION DESCRIPTION MAXIMUM MARK ACTUAL MARK

1 Algebra 32

2 Probability 18

TOTAL 50

QUESTION 1 32 MARKS

1.1 Factorise the following:

1.1.1 x x32 - (1)

1.1.2 x px mx mp2 + - - (2)

1.1.3 x x2 5 32 - - 2)

1.1.4 x27 83 - (3)

1.2 Simplify the following:

1.2.1 3 2- (1)

1.2.2 )x

x42

0

1 2- -^

^

h< F (2)

1.2.3 x x x2 5 2 32+ - +^ ^h h (3)

1.3 Solve for x :

1.3.1 x4 2 5 28- =^ h (2)

1.3.2 x x2 3 6+ - =^ ^h h (4)

1.3.3 ax b c- = (2)

1.3.4 .2 3 54x = (2)

1.3.5 x x3 4 5$- + (2)

1.4 Three chocolates and five packets of chips cost R54. One packet of chips and 10 chocolates cost R86.

1.4.1 If x y3 5 54+ = represents part of the above information, write down an equation to represent the second part of the information. (1)

1.4.2 Use the equations to solve for x and y and state the cost of one packet of chips and one chocolate. (5)




QUESTION 2 18 MARKS

2.1 A survey was done with 150 learners to determine how many had been to a live soccer match and a music festival. The following results were obtained:

• 85 had been to a music festival

• 125 had been to a live soccer match

• 70 had been to both a music festival and a live soccer match

2.1.1 How many learners had been to a music festival or a live soccer match? (3)

2.1.2 Represent the information provided in a Venn diagram (4)

2.1.3 Determine the probability that a learner chosen at random (simplifying is not essential):

(i) Had attended a music festival

(ii) Had attended a live soccer match but not a music festival

(iii) Had not attended a live soccer match or a music festival (3)

2.2 Complete the following statements:

2.2.1 If A and B are complementary events, then P A …=^ h (2)

2.2.2 If P and Q are mutually exclusive events, then ( ) ...andP P Q = (2)

2.3 Consider the word SEPTEMBER. A letter is chosen at random. What is the probability that the letter is an:

2.3.1 R (2)

2.3.2 E (2)




Gr 10 Mathematics Test

MEMOQUESTION DESCRIPTION MAXIMUM MARK ACTUAL MARK

1 Algebra 32

2 Probability 18

TOTAL 50

QUESTION 1 32 MARKS

1.1 Factorise the following:

1.1.1 ( )

x x

x x

3

3

2

{

-

= - (1)

1.1.2

( )( )

x px mx mp

x x p m x p

x p x m

2

{

{

+ - -

= + - +

= + -

^ ^h h

(2)

1.1.3 ( )( )

x x

x x

2 5 3

2 1 3

2

{{

- -

= + - (2)

1.1.4 ( )( )

x

x x x

27 8

3 2 9 6 4

3

2

-

= - + +

(3)

1.2 Simplify the following:

1.2.1 3

91

2

{=

-

(1)

1.2.2

)x

x

x

x

x

42

2

2

4

0

1 2

1 1 2

2 2

2

{

{

=

=

=

- -

- - -

^

^

h<

6

F

@

(2)

1.2.3

x x x

x x x x x

x x x

2 5 2 3

2 4 6 5 10 15

2 4 15

2

3 2 2

3 2

{

{ {

+ - +

= - + + - +

= + - +

^ ^h h

(3)




1.3 Solve for x :

1.3.1

x

x x

x

x

4 2 5 28

8 20 28

12 28

37

{

{

- =

- =

- =

=-

^ h

(2)

1.3.2

or

x x

x x

x x

x x

x x

2 3 6

6 6 0

0

1 0

0 1

2

2

{

{

{ {

+ - =

- - - =

- =

- =

= -

^ ^

^

h h

h

(4)

1.3.3

ax b c

ac c b

x ac b

{

{

- =

= +

= + (2)

1.3.4

.2 3 54

3 27

3 3

x

x

x 3

{

=

=

= x 3 {= (2)

1.3.5 x x

x x

x

x

3 4 5

4 5 3

5 2

52{

{

$

$

$

#

- +

- - -

-

-

(2)

1.4 Three chocolates and five packets of chips cost R54. One packet of chips and 10 chocolates cost R86.

1.4.1 If x y3 5 54+ = represents part of the above information, write down an equation to represent the second part of the information.

x y10 86 {+ = (1)

1.4.2 Use the equations to solve for x and y and state the cost of one packet of chips and one chocolate.

x y

y x

10 86

86 10 {

+ =

= -

x y

x x

x x

x

x

3 5 54

3 5 86 10 54

3 430 50 54

47 376

8

{

{

+ =

+ - =

+ - =

- =-

=

^ h y 86 10 8` = - ^ h

y 6 {=

` one chocolate costs R8 and one packet of chips costs R6 (5)




QUESTION 2 18 MARKS

2.1 A survey was done with 150 learners to determine how many had been to a live soccer match and a music festival. The following results were obtained:

• 85 had been to a music festival

• 125 had been to a live soccer match

• 70 had been to both a music festival and a live soccer match

2.1.1 How many learners had been to a music festival or a live soccer match?

n M or S n M n S n M and S

85 125 70

140

{{

{

= + -

= + -

=

^ ^ ^ ^h h h h

(4)

2.1.2 Represent the information provided in a Venn diagram

( )n s 150=

S M

55 70 15

10

(4)

2.1.3 Determine the probability that a learner chosen at random (simplifying is not essential):

(i) Had attended a music festival 15085 {

(ii) Had attended a live soccer match but not a music festival 15055 {

(iii) Had not attended a live soccer match or a music festival 15010 { (3)

2.2 Complete the following statements:

2.2.1 If A and B are complementary events, then ( )P A P B1 {{= -^ h (2)

2.2.2 If P and Q are mutually exclusive events, then ( )andP P Q 0 {{= (2)

2.3 Consider the word SEPTEMBER. A letter is chosen at random. What is the probability that the letter is an:

2.3.1 R

P R 91 {{=^ h (2)

2.3.2 E

P E 93

31 {{= =^ h (2)




Question Knowledge Routine Complex Problem Solve

1.1.1 1

1.1.2 2

1.1.3 2

1.1.4 3

1.2.1 1

1.2.2 2

1.2.3 3

1.3.1 2

1.3.2 4

1.3.3 2

1.3.4 2

1.3.5 2

1.4.1 1

1.4.2 5

2.1.1 3

2.1.2 4

2.1.3(i)-(iii) 3

2.2.1 2

2.2.2 2

2.3.1 2

2.3.2 2

Totals 10 18 16 6 50


resource pack gr10-term4

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