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