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 Apr