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Class: MARKING SCHEME Candidate Index Number:

SHUQUN SECONDARY SCHOOL2014 Mid-Year ExaminationSecondar !"ree Ex#re$$

and !"ree Norma% Exce#tiona%

MA!HEMA!&CS 401'(01 Ma 2014

)a#er 1 1 "o*r +0 min*te$

Candidates answer on the Question Pa er

READ !HESE &NS!RUC!&ONS ,&RS!

!rite "our #lass$ index number and name on all the wor% "ou hand in&!rite in dar% blue or bla#% en&'ou ma" use a en#il (or an" dia)rams or )ra hs&*o not use sta les$ a er #li s$ hi)hli)hters$ )lue or #orre#tion (luid&

Answer a%% +uestions&

I( wor%in) is needed (or an" +uestion it must be shown with the answer&,mission o( essential wor%in) will result in loss o( mar%s&Cal#ulators should be used when a ro riate&I( the de)ree o( a##ura#" is not s e#i(ied in the +uestion$ and i( the answer is not exa#t$ )i-ethe answer to three si)ni(i#ant (i)ures& Gi-e answers in de)rees to one de#imal la#e&.or π $ use either "our #al#ulator -alue or /&012$ unless the +uestion re+uires the answer interms o( π &

3he number o( mar%s is )i-en in bra#%ets 4 5 at the end o( ea#h +uestion or art +uestion&3he total o( the mar%s (or this a er is 67&

3his +uestion a er #onsists o( 14 rinted a)es&

!*rn o.er Mathematical Formulae

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Compound Interest

3otal amount

nr

P

+=100

1

Mensuration

Cur-ed sur(a#e area o( a #one rl π =

Sur(a#e area o( a s here24 r π =

8olume o( a #onehr 2

31 π =

8olume o( a s here3

34

r π =

Area o( trian)le ABC ab

21=

C sin

Ar# len)thθ r =

$ whereθ

is in radians

Se#tor areaθ 2

21

r =$ where θ is in radians

Trigonometry

C

c

B

b

A

a

sinsinsin==

Abccba cos2222 −+=

Statistics

Mean f fxΣΣ=

Standard de-iation

22

ΣΣ−ΣΣ= f fx

f fx

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Answer a%% the +uestions&

0 a; E-aluate ( 1

5 π − 4.21

2)(√ 3.812 − 2.1325 ) $ )i-in) "our answer to 1

si)ni(i#ant (i)ures&

b; .ind the (ra#tion that is exa#tl" between5

11 and6

11 &

Answer: a; − 22.89 <<<<<< A0

b;1

2 <<<<<<<<<< A0

<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

2 a; 02&=> o( a number is ?7& .ind the number&

b; Ex ress the sum o( =02 i#ose#onds and 6&@ nanose#onds$ inse#onds$ )i-in) "our answer in standard (orm&

Answer: a; 617 A0b; 7.41 × 10 − 9 A0

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<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

/ .a#torise 2 r (s− 3 t )+3 t − s &

2 r (s− 3 t )− 1 (s− 3 t )(s− 3 t )(2 r − 1 )

M0

Answer: (s− 3 t )(2 r − 1 ) 0

<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

1 3he tem erature at the to o( the mountain is x℃ & 3he tem erature atthe bottom o( the mountain is − 9 ℃ & !rite down an ex ression$ in terms

o( x $ (or

a; the di((eren#e between the tem eratures$

b; the mean o( the two tem eratures&

Sin#e the tem erature at the to is smaller than the tem erature atthe bottom o( the mountain

Answer: a; − 9 − x∨−( x+9 )℃ A0

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b; x− 9

2℃

A0

<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

= .a#torise #om letel"&

a; x y2 − 49 x $

b; p− q− 1 + pq &

a;

b;

x( y2 − 49 )

p− 1 − q+ pq p− 1 +q (− 1 + p)

Answer: a; x( y+7 )( y− 7 ) A0

b; ( p− 1 )(q+1 ) A0

<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

6 Sim li("(− xy)5

3 x− 3 ×3√ 27 y

6

2 &

− x5 y

5

3 x− 3 × 3 y

2

2

M0

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Answer: − x8 y

7

2

0

<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

B I( the hei)ht o( a trian)le is de#reased b" 27> while the area remainedun#han)ed$ (ind the er#enta)e in#rease in the len)th o( the base&

Area 1

2×b×h =

1

2× 0.8 h×

1

0.8×b M0

Answer: 2= > A0

<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

? Gi-en that x+2 y

x = 7

3 $ (ind the -alue o( y x

&

3 x+6 y= 7 x,r 6 y= 4 x M0

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Answer:2

3 A0

<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

@ Gi-en = #m on a ma re resents 07 %m on the )round& Cal#ulate&

a; the s#ale o( the ma in the (orm 1: n.

b; 3he a#tual len)th$ in %m$ o( the road that is re resented b" 02 #m onthe ma &

Answer: a; 1: 200 000 A0

b; 21 %m A0

<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

07 Gi-en x4

= √3 y+4

1 − 2 y$ ex ress y in terms o( x

&

x2

16= 3 y+4

1 − 2 y

x2

− 2 x2

y= 48 y+642 x

2 y+48 y= x

2 − 64

M0

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y= x2 − 64

2 x2 +48

Answer : y= x2

−64

2 x2 +48 0

<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

00 Sol-e the simultaneous e+uations3 y= 15 x−

1

2

3 x− 2 y= 5

6 y= 30 x− 1

3 x− 2 y= 5

30 x− 6 y= 1

9 x− 6 y= 15

21 x=− 14

x= − 2

3

y= − 7

2∨−3.5

M0

Answer: x= ¿ − 2

3 y= ¿

− 7

2∨− 3.5 A2

<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

02 a; Cal#ulate the total time$ in hours$ re+uired (or = Dourne"s$ ea#h o(

0 hour and /1 minutes lon)&b; A marathon ra#e started at 00 == and one o( the runners rea#hed the

(inishin) line at 0= /B& Cal#ulate the time ta%en$ in hours and minutesb" this runner&

#; Another runner too% 2= minutes (or the (irst ? %m& I( he #ontinues torun at the same s eed$ #al#ulate the time ta%en$ in minutes (or thenext 02 %m&

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Answer: a; B&?/ hours A0

b; / hours 12 minutes A0

#; /B&= minuteS A0

<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

0/Sol-e the ine+ualit"

3 x+4

− 3 ≤ 20 x− 13

4<¿ ; & Re resent the solution on the

number line shown in the answer s a#e&

− 3 ≤ 20 x

−13

4 <(3 x+4 )− 12 ≤ 20 x− 13 <12 x+16

1 ≤ 20 x<12 x+29

1 ≤ 20 x and 20 x<12 x+29

1

20≤ x and 8 x<29

1

20≤ x <

29

8

M0

Answer:1

20≤ x <3

5

8 A0

+/1

20

7

0

<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

01 3a A ta%es = minutes to (ill u a tan%& It ta%es 1 minutes (or 3a A and3a to (ill u the same tan% to)ether& Cal#ulate the amount o( time itwould ta%e (or 3a to (ill u the tan% b" itsel(&

1

4

=1

5

+1

x1

x= 1

20

M0

M0

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3a #an (ill u in 27 minutes b" itsel(

Answer: 27 minutes A0

<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

0= It is )i-en that 300 = 22× 3 × 5

2 $ when ex ressed as a rodu#t o( its rime(a#tors&

a; Ex ress 0?7 as a rodu#t o( its rime (a#tors&

b; .ind the lar)est inte)er that is a (a#tor o( both 0?7 and /77&

#; .ind the smallest inte)er -alue o( x $ su#h that the lowest #ommonmulti le o( 0?7$ /77 and x is 0?77&

Answer: a; 180 = 22× 3

2× 5 A0

b; 67<<<<<<<< A0

#; ? <<< A0

<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

06 In 0@==$ the o ulation o( a state in India was = B01 @77& In 277@$ theo ulation was @ 7?B 777&

a; i; !rite = B01 @77 #orre#t to / si)ni(i#ant (i)ures&

ii; !rite @ 7?B 777 in standard (orm&

b; !or% out the er#enta)e in#rease in the o ulation (rom 0@== to277@&

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Answer: a; i; = B07 777 A0

ii; 9.087 × 106 A0

b; =@&7> A0 <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

0B PQ,QR and RS are adDa#ent sides o( a re)ular ol")on& Gi-en that RPQ = 20 ° $

S

R

Q

P27°

Cal#ulate$

a; the exterior an)le o( the ol")on$

b; the number o( sides o( the ol")on$

#; ∠ PRS &

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Answer: a; 40 ° A0

b; @ A0

#; 120 ° A0

<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

0? Sol-e the (ollowin)

a; (

x−

2

) ( x

+2

)=12

$b; ( x+3 )2 − 49 = 0 &

a;

b;

x2 − 4 = 12

( x+3 )= ± 7❑

M0

M0

Answer: a; x= 4∨ x=− 4 A0

b; x= 4∨ x=− 10 A0

<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

0@ a; It is )i-en that y is in-ersel" ro ortional to x2 & .ind the

er#enta)e in#rease in y when x is de#reased b" =7>&

b; Mr& i de osited F 0777 in a ban% that a"s sim le interest at r er annum& A(ter 6 months i( he re#ei-ed F070= (rom the ban%$ (ind

the -alue o( r &

a;

y= k x

2

M0

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b;

New y= k 1

4 x

2

15 = 1000 × r100 ×

1

2

M0

Answer: a; /77> A0

b; r = 3 A0 <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

27 A uni-ersal set and its subsets A and B are )i-en b"

¿{ x: x is aninteger ∧

10< x<

20} $ A={ x : x isa prime n mber } $

B={ x : x is an integerthat is a per!e"t sq are } &

a; *raw a 8enn dia)ram showin) , A and B and la#e ea#h o( themembers in the a ro riate art o( the dia)ram&

A

0@

0?

0B06

0= 01

0/

02

00

ε

A2

b; .ind

i; B A ′∩ $

ii; )( ′∪ B An

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Answer: b; i; 00$ 0/$ 0B$ 0@ A0

ii; 1 A0

<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

20 3hree oints A , B and # are shown below&

A C B=& 77°

C*

A

.or drawin) C*

.or drawin) C C0C0

a; Constru#t the arallelo)ram AB$# &

b; Measure the siJe o( B$# &

#; i; Constru#t the er endi#ular bise#tor o( AB & C0

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ii; Constru#t the an)le bise#tor o( ∠*A & C0

iii; 3he two bise#tors meet at % & Com lete the senten#e below&

3he oint % is e+uidistant (rom the lines A# < and AB <<<<< and e+uidistant (rom the oints A and B

A0 A0

Answer: b; 75 °± 1 ° A0

<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

End o )a#er

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