BSS 2023 Maths Sec 4E Prelim MS - P1

Broadrick Secondary School Prelims P1 2023 Elementary Mathematics - Secondary Four Express Mark Scheme Qn Mark Scheme 1(a) 31.764 1(b) 32 2(a) 681p 2(b) 7 3(a) A 3(b) D 4 The vertical axis of the bar chart does not start from zero. From the bar chart, it seems like the exchange rate for MYR is more than double for every 1 Singapore dollar in 2022 as compared to 2018. However, by calculation, the increase is only about 7%. 5 1st equation: 8 y ax=− + 23 56 xy+ = 2 3 30xy+= 3 2 30yx=− + 2nd equation: 2 103yx=− + By comparison, 2 3a= . 6 7(a) 223 = 12 7(b) 2 2 22 3 5m=   = 900 8(a) 31 x−  + and 51 2 xx −+ 31 x− −  2 2 5xx+  − 4x− 33x 1x Solution: 41 x−   8(b) Depth (cm) Time (s) – 4 – 3 – 2 – 1 0 1

9 (3 1) 2(1 )p q q− = + 3 2 2pq p q− = + 3 2 2pq q p− = + (3 2) 2q p p− = + 2 32 pq p += − 10(a) 10(b) x = – 1 11 28 23 (3 )(3 )x x x+=+ − + 2(3 ) 8 2( 3)(3 ) x xx −+ =+− 26 2 8 2(3 9 3 )x x x x− + = − + − 214 2 2 18xx− =− + 22 2 4 0xx− − = 2 20xx− − = ( 1)( 2) 0xx+ − = 1x=− or 2x= 12 Volume of remaining cube = 21110 10 10 32 xx  −    31987.652 1000 6 x=− 31 1000 987.6526 x =− 3 6(12.348)x = 3 74.088x= 4.2x= 13(a) 225( 2 )( 2 )y x y x= + − 13(b) 3 ( 2 ) ( 2 )a x y b x y= − − − ( 2 )(3 )x y a b= − − 14 A : B : C 1.5 : 1 1 : 0.9 15 : 10 : 9 Box A : Box C 15 : 9 O y x 1 (-1, 2)

15 50YXZ =  (vert. opp.  s) 90XZY =  (adj.  s on a st. line) 180 90 50XYZ = − − (  sum of  ) = 40º 180 40XYC = − (adj.  s on a st. line) = 140º EDC = (5 2) 180 5 − = 108º 180 108 2ECD CED − = = (base  s of isos.  ) 36= º 180 36 140CEY = − − (  sum of  ) = 4º 16(a) 6 4 0 4 8 6 y x −− =−− (Grad. of AB and DC) 1 32yx=− + --- (1) 4 0 6 8 6 4 y x −− =−− (Grad. of AD and BC) 6 2 8yx− = − 22yx=− --- (2) Sub (2) into (1), 12 2 3 2xx− =− + 4 4 6xx− =− + 5 10x= 2x= When x = 2, from (2), y = 2 Coordinates of D = (2, 2) (Also accept use of vectors or mid-points) 16(b) Length of AB = 22(6 4) (4 8)− + − = 20 cm OR Length of CD = 22(2 0) (2 6)− + − = 20 cm Length of BC = 22(4 0) (8 6)− + − = 20 cm OR Length of DA = 22(6 2) (4 2)− + − = 20 cm Since ABCD is a parallelogram and their adjacent lengths are equal, ABCD is also a rhombus. 17(a) Retardation is gradient of the speed-time graph. Grad = 0 2 1 12 6 3 − =−− Hence retardation is 1 3 km/min2. 17(b) Distance travelled is area under the graph.

( )11 12 2 12 222 v  =  +  6 14v= 7 3v= 18 1T k L= 1Tk L = 1 9new TTL L = 13newTT = % increase = 31 100%1 −  = 200 % 19(a)  = {1, 2, 3, 4, 6, 9, 12, 18, 36} A = {3, 6, 9, 12, 18, 36} B = {6, 12, 18, 36} 19(b) ( )'AB = {1, 2, 3, 4, 9} 19(c) 20(a) $148 20(b) C = $40 x = $36 20(c) Method 1: Draw the straight-line graph to show Lionel’s plumbing cost. Lionel’s claim is valid for plumbing repairs that is less than 2.8 hours but not valid if it is more than 2.8 hours. Method 2: Show working for cost of Lionel’s plumbing cost for 2 hours and 3 hours, with the assumption that rate is charged on an hourly basis. Lionel’s claim is valid for plumbing repairs that is 2 hours or less but not valid if it is 3 hours or more. 21(a) Original value = 289000 10085  = AUD$340 000 21(b) Interest = 32 4 / 2289000 1 289000100  +− = 36460.94 = AUD$36461 (nearest AUD$)

21(c) Price of printer in AUD$ = 4500 1.52 = AUD$6840 Price of printer in SGD$ = 6840 1.13 = SGD$6053.10 (to nearest cents) 22(a) Row Number Triangle Sum of row Last element of row 6 31 33 35 37 39 41 216 41 22(b) 3n 22(c) 2 1 159nn+ − = 2 160 0nn+ − = It is not possible for the last element of the row to be 159 as the quadratic equation cannot be solved to get integer value for n. 23(a) C = 9 10 22    23(b) V = 95 15 2 103 6 1 22     = 45 150 44 27 60 22 ++  ++ = 239 109   23(c) The elements in V represents the total cost, in dollars, to make 1 table and 1 chair respectively. 23(d) W = (3 )12 24(a) Grad = 10 3 30 − − = 7 3 Eqn of PSR : 7 33yx=+ 24(b) R = ( - 1.5 , - 0.5 ) 24(c) Area of triangle PQS = 1 872 = 28 units²

25(a) 6OCD = rad. sin 6 r OC  = 1 2 r OC= 2OC r= 2R r r=+ 3Rr= (Shown) 25(b) Area of 2 triangles = 1 2 sin 223 rr      = 21.732r Area of unshaded major sector = 21 (2 2 )23r  − = 2 2 3 rr  − = 22 3 r Area of sector ABC = 21 (3 )23 r   = 21 (3 )23 r   = 29 6 r = 23 2 r Area of shaded region = 23 2 r – ( 21.732r + 22 3 r ) = 2 32( (1.732 ))23r −+ = 20.886r (to 3SF) 26(a) Mean mass of female members = 58 59 62 63 65 65 68 70 8 + + + + + + + = 63.75 kg 26(b) Modal mass = 65 kg 26(c) P(1 male>80 and 1 female <60) = 22 78 = 1 14 26(d) 75 88x *If the male member is 75 kg, he can be the 4th, 5th or 6th person in ascending order of mass, with median mass remaining at 75kg. If the male is more than 75kg and less than 88kg, he is not the heaviest and he can be the 6th person in ascending order of mass, with median mass remaining at 75kg.