Hougang Sec 2022 Prelim Paper 1 - Marking Scheme

Marking Scheme for 4NA Prelim Paper 1 – 2022 Qn Working Marks Remarks 1 5.07 76.1 2.32 5 80 2 10   = M1 A1 2 3 20 20 3 2 6 3 x x x −  − − Smallest integer is 6− M1 A1 3(a) Substitute ( 2,6) into 3 4 12 3( 2) 4(6) 12 Hence, ( 2,6) does not lie on the line 3 4 12 xy yx − + = − +  − + = B1 Show that the equation is not balanced. 3(b) 3 4 12 4 12 3 33 4 Hence, coordinates of -intercept is (0,3 ) xy yx yx y += =− =− B1 4

2 8 5 2 15 15 180 18055 15 =60 60 u u b + + = = =   =  M1 A1 Find 1 unit 5 B2 B1 – Correct horizontal & vertical length B1 – Correct slanting length 6 Gradient 11 3= 68 1 + −− =− M1 A1 7(a) P(number 3) 3 8= B1 7(b) P(prime number) 5 8= B1 7(c) P(number 6) = 0 B1

3 8(a) sin 90  120% 20 8 1 3.1415… 1.2 2.5 Smallest : sin 90 , 120% , 20 8 ,  B2 Minus 1 mark for any 2 incorrect order. 8(b) 1 tan 0.5 tan 0.5 = 26.6 (1 dp) A A − = =  B1 9(a) 249 1 (7 1)(7 1) y yy − = − + B1 9(b) 8 4 2 4 (2 ) (2 ) (2 )(4 ) ax ay bx by a x y b x y x y a b + − − = + − + = + − M1 A1 10(a) 0 Hence, accept any values of that is les s than zero a a  B1 10(b) Line of symmetry is 28 2 3 x x −+= = B1 11

4 (radii) Hence, and are both isosceles 180 20 2 80 80 2 160 Exterior =180 160 =20 360No of sides = 20 = OA OB AOB BOC ABO ABC =    − = =  =  =  −     18 Alternatively, Number of sides = 360 20   (angles at a point) = 18 M1 A1 M1 A1 12(a) 7 11 10 2 3 13 x y x y xy − − + − =− B1 12(b) 2 2 3 5 20 86 56 8 20 3 16 x xy y x y xy y  = = M1 A1

5 13 Perimeter of major sector 240= 2 (6) 2(6)360 8 12   + =+ M1 A1 Award mark for finding length of major arc 14 121 30 = 91 ABC = −   No because 91 not 90 (angle at semicircle is 90 ). ABC =    M1 A1 Workings for = 91ABC Explain angle at semicircle is 90o 15 Total Distance from A to C = 120 + 180 = 300 km Total Time from A to C = 45 180 201 h + h + h60 75 60 Average Speed M1 M1 Total Distance Total Time

6 300 45 180 20160 75 60 66.9144... 66.9 km/h (3 sf) = ++ =  A1 16 M1 M1 A1 Perpendicular bisector Angle bisector Shaded region 17(a) Number of hospitalized patients in April 100 1350360 375 =  = M1 A1 17(b) Mr Tay might be wrong because the number of patients in the month of May might be different

7 from the month of April (Eg. Number of patients in the month of May is 3600). Hence, the number of hospitalized patients in the month of May might be more than the hospitalized patients in the month of April although the angle represented in the pie chart is lesser (500 > 340 patients). Eg. 50 3600 500360  = patients B1 Accept any other logical reasoning. 18(a) PQRS is similar to TXYZ Hence, 110 and 75ZTX XYZ =   =  360 110 62 75 = 113 TXY = − − −   M1 A1 18(b) 33 14 22 33 1422 = 21 XY XY = = M1 A1 19 For answer (p) 110o 75o

8 95% $10080 560 95% $10640 106405% 5 95 $560 →+ →  =  = For answer (q) 560 100%10080 560 560 5% = ++ = For answer (r) 100% $1120 1120800% 800100 = $8960 →  →  For answer (s) 1120 100%8960 1120 1120 10% = ++ = B1 B1 B1 B1 20(a)

9 Modal shoe size is 36 B1 20(b) Middle position is 8th and 9th Hence, median shoe size is 37 38 2 37.5 += = B1 20(c) He should stock up the shoe size 36 because it is the modal shoe size which is most commonly sold in the shop. B1 Explain using statistical evidence of mode. 21(a) 4 100 18 26 4 56 xy xy + = − − += (shown) B1 Evidence of showing subtracting values from the total figure of 100. 21(b) 3(18) 4(4 ) 5(26) 6( ) 4.56100 184 16 6 456 16 6 272 8 3 136 xy xy xy xy + + + = + + = += += (Shown) M1 A1 Form equation involving mean

10 21(c) 4 56 ----------[1] 8 3 136 --------- [2] [1] 2 8 2 112 ---------- [3] 8 3 136 --------- [2] [3] [2] 24 24 Substitute 24 into [1] 4 56 4 32 8 xy xy xy xy y y y xy x x += +=  += += − − =− = = += = = M1 A1 A1 Manipulation to eliminate variables 22(a) 2210 15 18.02775 18.0 (3 sf) PQ PQ PQ =+ = = M1 A1 22(b) 18.02775