2014 MYE 3EX P1 ms
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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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