CHIJ SJC 4048 02 AS

CHIJ St Joseph Convent Elementary Mathematics Preliminary Examinations Paper 2 Solutions Qn Solution 1(a) ( )( ) 22 2 3 2 3 6 5 6 3 2 2 3 1 = 32 x y x y x xy y x y x y xy −− =− − + − + 1(b) ( )( ) ( )( ) ( ) ( ) ( )( ) ( )( ) ( )( ) 22 13 24 13 2 2 2 13 2 2 2 1 2 3 22 23 22 33 22 x y x x y x y x x y x y x x y x y x y x y x x y x y x y x x y x y xy x y x y +−−− +=− − − + −+=− − − + − + − += −+ − − − −= −+ − − −= −+ 1(c) ( ) 22 2 2 2 1 2 (1 )3 1 4 19 1 449 36 36 36 36 xy x y x y x y x y x x y x xy x =− =− =− =− −= 2a) b) Amount of water used in January = 100 x Amount of water used in December = 100 0.05x+

c) d) e) 2 2 2 100 10020.05 100 2( 0.05) 100 0.05 100 2 0.1 100 5 2 0.1 5 0 20 50 0 (shown) xx x xx x x x x xx xx +=+ ++ =+ + + = + + − = + − = 21 1 4(20)( 50) 2(20) 1 4001 40 1.5563 or -1.6063 1.56 or -1.61 x −  − −= −= =  Therefore price of water in January = $1.56/m3 Amount of water used in December = 3 100 1.5563 0.05 62.254 62.3 m = + = 3a) b) c) d) a = 25, b = 36, c = 60 No of dots = 2(25 1) 676= + = 2( 1)yn=+ 2(2 1) / 2 2 / 2 ( 1)z n n n n n n n= + + + +

e) 2 2 2 2 2 5100 2550 0 50 or -51 (reject) therefore D = (50+1) = 2601 nn nn n += + − = = 4(ai) 1.85, 1.85, 1.6a b c= = = 4(aii) 1.85 0 0 0.70 0 1.85 0 0.80 0 0 1.60 1.40 1.295 1.48 2.24 1.30 1.48 2.24 S S S      =        =  = 4(b) ( ) ( ) ( ) 25 14 0 1.295 1 1 1 18 20 19 1.48 13 20 7 2.24 1.295 56 54 26 1.48 2.24 210.68 T T T      =         =  = 4(c) Total number of workers in the company = 56 + 54 +26 + x = 136 + x

( ) ( ) ( )( ) ( )( ) 2 40 18 12 136 135 329 40 18 329 12 136 135 236880 = 12 136 135 271 1380 = 0 5 276 = 0 5, 276 (re j) xx xx xx xx xx x    =  ++     =  +  + ++ +− −+ =− Total number of workers in the company = 136 + 5 =141 ( ) ( ) ( )( ) 2 2 2 40 18 12 136 135 239 172080 = 12 136 135 14340 = 18360 271 271 4020 = 0 271 271 4 1 4020 = 2(1) xx xx xx xx x    =  ++    +  + ++ ++ −  − 15.749, 255.25 15.7, 255 x x =− − =− − 5(ai) 4a 3b AB AO OB=+ =− + 5(aii) ( ) 1 2 1 4a 3b2 3 = 2a b 2 AM AB= = − + −+ 5(aiii) ( ) 3 4a 2 b 2 3 2 b 2 1 4 3b2 OM OA AM a a or a =+ = − + =+ +

5(aiv) 7 6 PQ PO OQ ab =+ =− + 5(av) ( ) 3 2 6 2 9 2 2 1 4 9 2 MQ MO OQ a b b ab Or a b =+ =− − + =− + −+ 5(avi) ( ) 3 2 7 2 3 5 2 1 10 32 MP MO OP a b a ab Or a b =+ =− − + =− − 5(bi) 1 ()area of 2 1area of ()2 4 3 OA hOAM AMP AP h  = = 5(bii) 1 1 area of area of 1 () 12 1 1()2 7 4 OMP OMP OAM OMB OAM OMB OP h OA h  = = =

5(c) ( ) 7 15 7 7 615 49 14 15 5 PR PQ ab ab = = − + =− + ( ) 49 14 7 15 5 56 14 15 5 14 4 315 OR OP PR a a b ab ab =+ = − + =+ =+ 1 2 14 15 15 28 OM hOR OM hOR OMh OR h h = = = = = 6ai) ii) iii) median = 3.6 to 3.7 Lower quartile = 2.6 Upper quartile = 4.5 to 4.7 Interquartile range = 1.9 to 2.0 10% at least x hours 90% of 600 = 540 ➔ 7 to 7.2 Therefore x = 7 to 7.2

bi) ii) ci) ii) 7ai) ii) median = 4.2 Lower quartile = 2.8 Upper quartile = 7.4 Interquartile range = 7.4 – 2.8 = 4.6 Disagree, because the median of 4.2 hours for Group B is higher than the median of 3.6 (3.6 -3.7) hours for Group A, implying that people in Group B spent more time on their smart phones than Group A. Group A is more consistent on the use of the smart phones than Group B because Group A had a smaller interquartile range of 2 compared with 3.6 for Group B Area of segment = 240 2012 = Area of sector – area of triangle AOB = 20 22 2 2 11 (1.5) (sin1.5) 2022 1 (1.5 sin1.5) 202 40 1.5 sin1.5 = 79.601 r = 8.9219 8.92 rr r r −= −= = −  2 2 2 2 AB 8.9219 8.9219 2(8.9219)(8.9219) cos1.5 = 147.93 AB = 12.162 surface area of water = 12.162 12 = 145.94 146 cm = + −  

b) Lateral height of cone: ( ) 222 2 2 10 8.9219 13.401 cm area of container that is in contact with water 1 1 1= (2 (8.9219) 2 (8.9219)12 (8.9219)(13.401)2 2 2 250.07 336.34 187.80 774.21 774 cm l l    =+ = +  +  = + + =  8a) b) Amount received for 10 years =10(12 x 800) = $96 000 Total amount received after 15 years = 96 000 + 80 000 = $176 000 Total interest earned = 176 000 – 100 000 = $76 000

c) Amount received after 5 years = 20 100000100 =$20 000  Interpretation 1 Interpretation 2 10 1480000 1 100 $296577.71 Tot Amt = 20000 296577 71 316577 71 Interest = 316577 71 100000 216577 71 . $. $. $. =+  = + = − = 20 780000 1 100 $309574.76 Tot Amt = 20000 309574 76 329574 76 Interest = 329574 76 100000 229574 76 . $. $. $. =+  = + = − = 15 15 5180000 = P 1 100 5P= 180000 1 100 = $86 583.08 + + 9a) b) ci) a = 37 Refer to graph Draw graph of y = x + 20 x = 1.05 or 2.8

cii) ciii) civ) d) k = 17 – 17.5 Draw line y = 5x or any line with gradient = 5 x = 2.2 to 2.3 Draw the line of y = 2x + 20 1 < x < 3.1 2 32 32 202 5 1 2 20 5 2 5 20 0 xx x x x x x x x + = + + = + − − + = A= -5, B= -1, C=20 10(ai) Angle N1CA = 180 – 52 (Interior angles, parallel lines) = 128 Angle N1CB = 360 - 128 – 112 (angles at a point) = 120 Angle N1BC = 180 - 120 (Interior angles) = 60 Therefore, bearing of C from B North C B A 42 m 26 m 112 52 N1 N2