Canberra EM 2024 Prelims Paper 1 MS

Canberra Secondary School Mathematics 4052/01 2023 Preliminary Examination Secondary 4 Express / 5 Normal Academic 4E5N Prelim Exam 2024 Maths P1- Marking Scheme 1 19 252 1 162 32 x x x −= −= =− 2 Total sum of money $900 103 $3000 = = 3 32 2 1 1 2 32 2 1 2 1 5 21 aa aa a −−− =+ −− = − 4 3, 3, 4, 5, 10 5a 5b AB  6a a = 3, b = 7 6b Since the minimum point of y is at (-2, -25), y will not have a value smaller than - 25. 7a ( )( ) 22 22 3 33 32 x y y x xy x y xy x xy y −+ = + − − = + − 7b ( ) ( )( ) 2 2 3 12 3 1 4 3 1 2 1 2 a a aa − =− = − + 8a 3.5 A B

Canberra Secondary School Mathematics 4052/01 2023 Preliminary Examination Secondary 4 Express / 5 Normal Academic 8b 2 2 2 2 4 2 24 4 4 x x x x −  =  =  = 9a ( ) 2 2 2 2 2 10 27 10 5 5 27 52 xx xx x −+ = − + − + = − + p = -5, q = 2 9b (5, 2) 10 70 2 35 ACB= = (angle at centre = twice the angle at circumference) 180 70 110 ABC = − = (angles in opposite segments) 180 110 35 35 BAC = − − = (angles sum of triangle) Since angle ACB = angle BAC, triangle ABC is an isosceles triangle. 11a 18 360 20 x x = = 11b 3 6 9 180 the angles are interior angles. 6 5 11 180 x x x x x x += =  +=  The quadrilateral is a trapezium and not a parallelogram as it has 2 parallel sides and 2 non-parallel sides. 12a 3 3 2 300 1 2400 L L V Vm = =

Canberra Secondary School Mathematics 4052/01 2023 Preliminary Examination Secondary 4 Express / 5 Normal Academic 12b ( ) 2 2 2 2 3 3 1 3003 900 Volume of pyramid 12=333 2 2 900m 1800m xy xy xy xy = =  = = = 13a 78 13b The interquartile range = 87 – 73 = 14 14a 2215 12 81 9 9cos 15 3 5 AB ABD =− = =   = = 14b 20 12 15 BC BE DA DB BC = = BC = 16 cm 15a Total population = (33.57 + 5.64) million = 39.21 million 6 7 39.21 10 3.921 10 = = 15b Population of Singapore in 2013 6 5.64 100 million105 =5.37 million (or 5370000 or 5.37 10 ) =  16a 15 15 x− 16b P(green marbles) = P(yellow marbles)

Canberra Secondary School Mathematics 4052/01 2023 Preliminary Examination Secondary 4 Express / 5 Normal Academic 1 2 51 20 2 5 x x = + = = 17a(i) 232 3 5 17a(ii) 23 2 1500 2 3 5 23 23 1 and 7 mpn HCF mn =   =   =  = = 17b 323 24 LCM = = 18a 10 km 18b 5 cm: 2.5 km 1 cm: 0.5 km 1 cm: 50000 cm 1 : 50000 18c 22 22 1 : 0.25 30 : 7.5 cm km cm km 19a 12 3Gradient 15 0 3 5 3 equation of PQ: 3 5yx −= − =  = + 19b Gradient of R (15, 12)=− Perimeter of OPQR ( ) ( ) ( ) 22 223 15 0 12 3 12 2 15 12 63.70223 63.7 = + − + − +  + + = 20 The y-axis of the line chart does not start from zero and the reader may be misled to think that the sales performance has increased greatly from Jan to Jun. 21a 72 days 21b 12 men will take 54 days. 18 men will take 54 12 3618  = days. No. of days saved = (54 – 36) days = 18 days 22 Area of shaded region in triangle ABC

Canberra Secondary School Mathematics 4052/01 2023 Preliminary Examination Secondary 4 Express / 5 Normal Academic ( ) ( ) 222 2 115 sin 60 222 25 3 24 cm cm   =− =− tan1.2 2 tan1.2 4 tan1.2 EF OE EF DF = = = Area of shaded region in triangle ODF ( ) ( ) 2 2 2 11 2 4 tan1.2 2 2.422 4 tan1.2 4.8 cm cm =   − =− 2 2 2 area of shaded region of the logo 25 3= 2 4 tan1.2 4.84 10.030739 10.0 cm cm cm    − + − = 23a Let the speed of the car after 9s be v m/s. 9 20 15 12 v v = = 23b(i) ( ) ( ) 11 15 20 20 25 90022 1 20 25 7502 40 u u u    + +  = +  = = 23b(ii) 2 2/ms 24a 233 104   24b 233 M= 104   3.50 0.40 2.70 0.20 5.00 0.50   − −

Canberra Secondary School Mathematics 4052/01 2023 Preliminary Examination Secondary 4 Express / 5 Normal Academic 30.10 1.30 23.50 1.60 −= − 24c Azza should buy from Café B as she would save $1.30. 25 22 1 tan 35 75 75 tan 35 40 75 7225 tan 75 tan 35 7225 75 tan 35tan 7225 31.70902 31.7 FC FC AC FCFAC AC FAC − = = =+ = = =  =  = 26a Let the radius of the water surface be r cm. 2 4 58 5 2 15volume of water = 432 25 (Shown)3 r r   = =   = 26b Surface area 2 2 2 2 55 422 37.04716 37.0 rl cm cm   =    =+        = 26c Let the height of the water above the hemisphere be h cm.

Canberra Secondary School Mathematics 4052/01 2023 Preliminary Examination Secondary 4 Express / 5 Normal Academic 23 3 2 3 25 2 3 2 3 9 73 4 12 192 cm27 height of water in the test-tube 19 3= 2 cm27 2 114 cm (or 4.20 cm)54 h h h      +=       = =   + =