HIHS Sec 2 Math 2024 EOY P1 (Ans key)

]2024 2E EOY P1 Marking Scheme Every 3 Presentation or Accuracy errors deduct one mark. Question Solution Marks awarded /Remarks 1a ( )2 4 5 2 20 4 5 22 24 4.45 −  − −  −   cc cc c c or c M1 A1 1b 4 B1 2a 3 20 B1 2b 82 20 5= B1 3a 54 3 35 4 33 3.754 = = = x x x or B1 3b ( ) ( ) 13 832 2 1 3 3 8 6 2 2 3 9 48 5 41 18 8.25 −++= − + + =  − + + = = = xx xx xx x x or M1 A1 4 2 ky x= New value of y ( ) 2 2 4 16 1 16 k x k x y = = = Percentage change M1

1 16 100% 15 100%16 93.75% − = =−  =− yy y M1 A1 5 By Elimination Method ( ) (1) 2 : 6 10 6 ____(3) 2 3: 9 6 51____(4) (3) (4) : (6 10 ) (9 6 ) 6 51 19 57 3  + =  − = + + + − = + = = xy yx x y y x y y Substitute y = 3 into 3 5 3xy+= 3 5(3) 3 3 12 4 x x x += =− =− By Substitution Method 3 5 3 3 3 5 35 3 xy xy yx += =− −= Substituting 35 3 yx −= into 3 2 17yx−= 353 2( ) 17 3 9 2(3 5 ) 51 19 57 3 yy yy y y −−= − − = = = Substituting 3y= into 3 5 3xy+= 3 5(3) 3 3 12 4 x x x += =− =− M1 for showing the elimination A1 A1 M1: Substitution A1 A1

6a 22 2 2 2 2 35 1225 (28) (21) 1225 AB BC CA = = + = + = Since 2 2 2 1225AB BC CA= + = , by the converse of Pythagoras Theorem, triangle ABC is a right-angled triangle. } } } M1 } A1 for conclusion including underlined part. 6b Let the shortest distance be x cm. 11 35 28 2122 17.5 294 16.8   =   = = x x x Shortest distance is 16.8 cm. M1 A1 7a 22 2a ab b−+ B1 7b 2 2 2( ) 2a b a ab b− = − + ( ) 2 22 22 22 3 2(40) 80 9 89 = − + + = + += ab ab ab M1 A1 8a 53.1 B1 8b 66 24 24 30 3 24 = += = CB AC CD EC BC BC = 72 cm M1 A1 8c By using similar triangles 66 24 18 30 AB AC ED EC AB = += AB = 54 cm Alternatively, by using Pythagoras theorem, 2 2 2 2290 72 AB BC AC AB += =− M1

AB = 54 cm Therefore, perimeter of quadrilateral ABDE 66 18 (72 30) 54= + + − + = 180 cm M1 A1 9a 90 B1 9b Equilateral Triangle B1 9c 180 30 44 106 BAD BAD  = − −   =  ADE BAD = (Alternate angles) Therefore 106ADE =  M1 A1 10a Speed 72 1000 60 60 20 m/s =  = B1 10b Total distance 1.5 72 40 148km =  + = Total time taken 1.5 0.75 2.25 h =+ = Average speed 148 2.25 65.8 km/h (to 3 sig fig) = = M1 M1 A1 11a Common difference 19 14 3 11 −−= =− p = 25 M1 A1 11b 36 11nTn=− B1 11c 36 11 101 11 137 137 11 51211 − =− − =− −= − = n n n Since n is not an integer, -101 is not a term in this sequence. M1 ecf Note: Marks will not be awarded if working is not clear to the marker (eg angles not labelled).

A1 for conclusion and only if n is 512 11 12a 736 hours B1 12b 723 724 2 723.5 hours + = B1 12c 30% of batteries 30 30100 9 = = t = 716 hours M1 A1 12d 7 30 B1 13a Exterior angle 180 160 20 = −  = Number of sides 360 20 18 =  = M1 A1 13b 160 90 70 x=− = B1 13c 180 70 2 55 CBQ − = = 160 55 105 =− = y M1 A1 14a Total surface area 2 112 18 12.5 2 7 15 18 722 456 cm    =   +   +        = M2 for all 3 correct area computed. Deduct 1 mark for any 1 area wrongly computed. A1 14b Height of pyramid 2215 9 12 cm =− = M1

V olume ( ) 3 1 18 7 123 504 cm =    = M1 A1 15a $4 $10( 0.3) 2[4( 0.3)] 4 10 3 8 2.4 22 5.4( ) x x x x x x x shown + − + − = + − + − =− B1 for (x – 0.3) B1 for 4(x – 0.3) or (4x – 1.2) 15b 22 5.4 22.1 22 27.5 1.25 x x x −= = = B1 15c Percentage of price dropped 13.95 11.25 10013.95 −= % = 19.4% < 20% Disagree with durian seller. M1 A1 [Must mention percentage decrease is less than 20% ] or “< 20%” seen.