Dunearn Prelim 2009 Em P2 Solutions

Dunearn Secondary School Preliminary Examination 2009 Secondary Four Express Mathematics Paper 2

1（a）

QRP 

74  37 ( at centre) 2

B1

（b）

QSP  37 ( in same segment )

B1

（c）

OPQ 

180  74  53 (isos. ) 2 OPA  90 (tgt  rad )

M1

QPA  90  53 (complementary )  37

（d）

PQA  180  37  36 ( sum of )  107 PQR  180  107 (adj.  on str . line)  73 PSR  180  73 ( in opp. segment )  107

A1

M1

A1

OR

PQA  180  37  36 ( sum of )  107 PSR  107 (ext.  of cyclic quad ) （e）

2

M1 A1

RPQ  180  37  73 sum of RPQ   70

M1

RPO  70  53  17

A1

28 x 28 (ii) x 1

（a)(i)

B1

B1

2 2 (b) 10  28 5  28   55 x x 1 280 x  280  140 x  55 x( x  1)

M1

M1 M1

2

420 x  280  55 x  55 x 0  55 x 2  475 x  280 （c)

x x

95 

 952  41156  211

95  6561 22

x  8 or

x

A1 7 (reject); refer to condition in part(a)(ii 11

d) 8 kg of aluminium costs $28. 1 kg of aluminium costs $3.50

3(a)

M1

$43.08  $0.75  20  $0.90  20  $10.08 $10.08  $1.05  9.6

B1

M1

Volume of water used = 20 + 20 + 9.6 m3 = 49.6 m3 (b)

A1

(i) Total chargeable income = $(55 800 + 750 – 3635) = $52 915

B1

(ii) Income tax

 $40 000  2.25%  $12 915  8.5%  $1 997.78 (c)

5800 1.16  A$5 000

M1 A1

(i) S $ 5 800  A$

(ii) Amount withdrawn  5.5  2  = A$50001    A$ 5 550 100  

B1

M1

3 A$ 5 550 = S$ 5 550  1.09  S $6 049.50 Gain = S $ 6 049.50  5 800  = S$ 249.50

4(a)

M1 A1

(i) $370

B1

(ii) $280

B1

(iii) $450 − $280

M1

= $170

A1

(iv) % of workers earning $350 or less a week 320  100 760  42.1% (3 s. f .) 

OR 42 (b)

2 % 19

M1 Either answer A1

Prob. He earns more than $350 a week

760  220 760 27  38 

(c)

(i) The median of Supermarket B ($410) is

B1

B1

higher than the median of Supermarket A ($370). The interquartile range of Supermarket B

B1

($250) is wider than the interquartile range of Supermarket A ($170). (ii)

Supermarket B.

The median weekly wages is higher in Supermarket B than in Supermarket A.

B1

4 5(a)

1 432   162  h 3 h  8cm

(i)

B1

r 2  162 22 7 r  7.179516317cm

(ii) r  162 

Slant height =

(b)

8 2  7.179516317 2 = 10.74920716  10.7cm

M1

A1

Volume of solid figure

 vol of hemisphere  vol of cylinder - vol of cone

(c)

1 4        r 3   base area  height   432 2 3   1 4 22       7.179516317 3   (162  25)  432 2 3 7   775.3877622  4050  432  4393.387762

M1

 4390cm 3

A1

M1

Surface area of solid figure = surface area of hemisphere +surface area of cylinder + curved surface area of cone 1     4    r 2   2rh   rl 2  22 1   22  M1    4   7.179516317 2    2   7.179516317  25  7 7 2    M1  22     7.179516317  10.74920716  M1  7   324  1128.209707  242.5471972  1694.756904  1690cm 2

A1

5 6(a)

(i) 2(2a  5)(2a  5) (ii)

3 y ( x  3b)  5( x  3b)  ( x  3b)(3 y  5)

(b) (i) G 

M1 A1

3  4048  40  7

= 11.7 (3 s.f.)

B1

3 AD  A 7

M1

(ii) G 2 

7G 2  3 AD  3 A 2 D OR D  (c)

B2

7G 2  3 A 2 3A

7G 2 A 3A

15 units: 180ْ 2 units: 24ْ

M1 Either answer A1

M1 A1 A1

15sides 7(a)

(i)

sin PRS sin 78  7. 2 9.5

M1

sin PRS  0.741332918 PRS  47.84508384  47.8 (1 dec pl )

A1

(ii) PRQ  180  PRS  132.1549162

Area of PQR 

1  9.5  6.4  sin PRQ 2



1  9.5  6.4  sin 132.1549162 2  22.53652069

M1

 22.5 cm 2 (3 s. f .)

A1

6

(iii)

PQ 2  9.5 2  6.4 2  29.56.4  cos132.154...  212.8203165 PQ  212.8203165 PQ  14.58836237 PQ  14.6 cm (3 s. f .)

M1

M1 A1

(iv)Let U be the perpendicular from R to PQ. RU is the shortest distance from R to PQ. Since area of PQR  22.53652069cm 2 , 1  RU  14.588...  22.53652069 2 RU  3.089657375

M1

 3.09 cm (3 s. f .)

A1 7(b)

Greatest angle of elevation of T occurs at U. 8. 4 3.08965... RUT  69.8 (1 dec pl )

tan RUT 

8(a)

(c)

A1

Arc length PQR = 26(1.5) =39 cm

(b)

M1

r 26  r 26 sin 0.75  r sin 0.75  r r 1  sin 0.75  26 sin 0.75 26 sin 0.75 r 1  sin 0.75 r  10.5388911 r  10.5 (3 s. f .)

sin 0.75 

  Reflex AOC  2  2  0.75  2 

B1 M1

M1

A1

7

 2    1.5  4.641592654 rad .

M1

Area of major sector AOCQ 

1 10.53889112 4.641592654  2

 257.7667301 cm 2

M1 A1

 258 cm 2

(= 255.8677951 cm2 if using r = 10.5) (= 256 cm2)

8(d)

Area of sector BPQR =

1  26 2  1.5 2

 507 cm 2

10.5388911 BA 10.5388... BA  tan 0.75  11.31272128 cm ( 11.27097456 if u sin g r  10.5)

M1

tan 0.75 

M1

Area of shaded region 1 =507 − 257.7667..− 2   10.5388  11.312... 2

M1

=130.0097323 cm2 (= 132.786972 cm2 if r is

A1

10.5) =130 cm2 (3s.f.) 9(a)

(i)

(= 133 cm2 )

M C

(ii) Teenagers who play computer games but

B1 B2

do not play games on the playstation nor on the M-Box.

(iii) Teenagers who play either computer games or games on the playstation but do not play games on the M-Box.

B1

8 9(b) (i)

(ii)

 80     60  B  50    35     80     220 410 535 335  60    AB    260 410 495 355  50   35     80 675     82 575  

(iii)

B1

The elements of AB represents the total

B1 B1

amount collected from ticket sales for Saturday and Sunday respectively.  1    220 410 535 335 1   (iv) D   260 410 495 355  1  1   1500     1520  (v) The elements of D represents the total

B1 B1

number of tickets sold for Saturday and Sunday respectively. 1500   (vi) F  1 1 1520  = 3 020 F shows the total number of tickets sold for both nights of performances.

B1

9

10(b) (c)

360 books (20) (i) gradient 

26  9.5 100  800

= − 0.02 (up to  0.03)

B1 M1 A1

26.5  9.5 25.5  9.5 to ) 100  750 100  820 (0.0263 to  0.0222)

Range (

(ii) Cost is reduced at the rate of 2 cents per copy.

Either one B1

OR Represents the rate of reduction of the cost of production per copy of book (d)

(ii) 240 (20)  x  770 (20)

B2