2024 TPSS EMATH P2 PRELIM MS

Tampines Secondary School Sec 4 Prelim Math Paper 2 2024 Marking Scheme Total Marks: 90 √ = follow through No. Answers Marks 1a 3 2 2 6( 1) 28 7 9( 1) 2( 1) 4 13 8( 1) 3 ab ba a a + × + += × += B1 (for 2( 1)a+ seen) B1 (for 8 3 seen) 1bi 2 23( 4)2 11.65 bvxa k x = + −= += B1 1bii 2 2 2 2 () () bvxa k bvxa k k x a bv kx a vb = + −= −= − = ()kx av b −=± M1 (elimination of fraction) M1 (square root) A1 (± seen) 1c 4 7 23 ------- (1) 6 2 3 ---------- (2) xy xy += − −= (1) 3 :12 21 69 ----- (3) (2) 2 :12 4 6 ---------- (4) xy xy × += − × −= (3) (4) : 25 75 3 0.5 y y x −= − =− = − M1 (correct method to eliminate one variable) A1 A1 1d 2 2 22 2 ( )( ) ( )( 3 ) () ( )( 3 ) ( )( 3 ) x x yx y x yx y x xy x yx y y x yx y −− + +− −−= +− = +− M1 A1 (accept expanded denominator) 11 marks

No. Answers Marks 2a 925000 121500 100 8 R R ×× = = Rate is 8% M1 (or equivalent) A1 2b $1 = 26.77 THB $500 = 500 x 26.77 = 13385 THB Money left = 13385 – 10600 = 2785 THB 26.88 THB = $1 2785 THB = 2785 26.88 =$103.61 M1 (get remaining amount of money left in THB) A1 (nearest cent) 2c Deposit = 15 2700 $405100×= Total instalments = 68 x 36 = $2448 Total paid = 405 + 2448 = $2853 Interest = 2853 – 2700 = $153 % required = 153 2100 5.67% (5 %)2700 3×= M1 (calculate both deposit & instalments) M1 (their interest √ ÷ cash price x 100) A1 2d 100%  $180 110%  110 180 $198100×= 75%  $198 100%  100 198 $26475 ×= M1 (find discounted price) M1 ( √ from above discounted price) A1 10 marks

No. Answers Marks 3a length of box = 3 1 3 3 (3 5) width of box = 13 3 3 ( 7) height of box = 3 x x cm x x cm cm +−−= − + −−= + 2 2 2 3(3 5)( 7) 930 = 3(3 21 5 35) 310 = 3 16 35 3 16 345 0 volume x x x xx xx xx =−+ + −− +− +−= B1 (Length & width) M1 √ M1 (expansion √) A1 (establish the equation) 3b 216 16 4(3)( 345) 2(3) 8.38 or 13.72 x xx −± − −= = =− B1 B1 B1 3c Length of box = 3(8.38) – 5 = 20.14cm Width = 8.38 + 7 = 15.38 cm Length of SQ = 2220.14 15.38 25.3+= cm M1 (√ find either length or width with Q3b value) M1 & A1 √ 10 marks

No. Answers Marks 4ai 22 3 13 78 10 15 10 10 ( 15) 18.015 AB AO OB= + −  = +  −−   =−  = +− =−       M1 or use length formula A1 4aii 2 10 2( )15 53 = + 7.5 7 2 14.5 BA AP AO OP OP OP = − = + −−      −   −=             P (-2, 14.5) M1 A1 4b(i) 3 4 3 (4 2)4 3 =3 2 CP CB ab ab = =−+ −+       B1 for CB    B1 4b(ii) 3 23 2 3 = 2 AP AC CP aa b ab = + =−+ −+       B1 4b(iii) 23 32( ) 2 2 AD a b AD a b AP = −+ = −+ =       is a scalar multiple of , and are para llel with A as the common point. AD A P AD AP 