HIHS Sec 2 Math 2021 EOY P1 (Ans key)

Holy Innocents’ High School End Of Year Examination 2021 Secondary 2 Express Mathematics Paper 1 2021 EOY 2E Math Paper 1 Marking Scheme Qn Solution Marks 1a 22180 2 3 5 2, 1xy =   == 1b 2236 2 3 HCF 36 = = 2 Sleep : Study : Play = 14 : 6 : 3 11.5Time spent studying 6 14 6 3 3 hours = ++ = 3a 0.961529697 0.96 (2sf)= 3b Cara is not correct, because the degree of accuracy is different / 2.50 is more accurate with one more decimal place / 2.50 is 3sf while 2.5 is 2sf, 2.50 more accurate. 4a 2 2 2 22 22 ( ) 2 16 2( 2) 20 (shown) a b a b ab ab ab + = + + = + + − += 4b 2 2 2( ) 2 20 2( 2) 24 a b a b ab− = + − = − − = 5a 0.64 m3 = 0.64 x 1003 cm3 = 640 000 cm3 5b 60 1000 km60m/s 1 60 60 h 216 km/h =  = 6a AC2 = 292 = 841 AB2 + BC2 = 202 + 212 = 841 = AC2 By converse of Pythagoras’ Theorem, this is a right-angled triangle. 7a ( )( ) 24 9 2 3 2 3x x x− = − + 7b ( ) ( ) ( )( ) 6 3 8 4 3 2 4 2 3 4 2 px qx py qy x p q y p q x y p q + − − = + − + = − + B1 o.e. (grouping) A1

Holy Innocents’ High School End Of Year Examination 2021 Secondary 2 Express Mathematics Paper 1 8a 15Gradient 5 3 = = 8b y-intercept = 15 Equation of line AB is y = 3x + 15 8c y = 3(–2) + 15 = 9 9a 22 (19) 2(9)(1800) 32761 181 v v =+ = = 9b 22 22 2 2 2 2 v u as u v as u v as =+ =− = − 10a 5, 5, 1, 2, 4, 3 10b Mean will be 2 cm less 11a (i) ( ) ( )2 2 5 3 3 4 10 3 9 19 xx xx x −  + −  +  11a (ii) 19 11b 2x− 12a 2 3 1 -----(1) 1 4 -----(2)5 xy xy += −= Elimination Method 10 (2): 2 10 40 -----(3)xy − = Sub (3) to (1): 2(20 5 ) 3 1 40 13 1 13 39 3 yy y y y + + = += =− =− From (1): 2 3( 3) 1 2 10 5 x x x + − = = = Substitution Method 0 –2

Holy Innocents’ High School End Of Year Examination 2021 Secondary 2 Express Mathematics Paper 1 From (2): 20 5 -----(3)xy=+ (3) (2): (2 10 ) (2 3 ) 40 1 13 39 3 x y x y y y − − − + = − −= =− Sub to (3): 20 5( 3) 5 x= + − = 12b Suggests alternative method, elimination or substitution, to solve (refer to alternative method above) OR Substitute x and y values back into both equations, ensure both equations are satisfied. 13a ( ) ( )( ) 2 2 External surface area 2 5 2 5 18 230 or 722.566 723 cm =  +  = = 13b Method 1 (Volume) ( ) ( ) ( ) ( ) 22 1Total cups 5 18 2.5 9 3 24 =   = Method 2 (Proportion) Since volume of cone is one third that of a cylinder, and the conical cup has half the radius and height. 2 1 1 1propotion of cup 3 2 2 1 of cylinder24 =   = Therefore, Ben is correct as it can fill 24 cups. 14a Method 1 (Interior angle) 160 ( 2) 180 20 360 18 nn n n  = −  = = Method 2 (Exterior angle) Each exterior angle 180 160 20 Number of sides 360 20 18 =− = = =

Holy Innocents’ High School End Of Year Examination 2021 Secondary 2 Express Mathematics Paper 1 14b ( ) ( )( ) ( )( ) Number of sides left 6 Sum of exterior angle 360 6 30 6 360 6 180 180 6 n nx nx x n =− = + − = −= = − 15a (i) 4 150 600 600 kV P k k V P = = = = 15a (ii) 3 60010 60 N/m P P = = 15b (i) Since distance = speed x time, n = 1 15b (ii) Since vol 32 3 r= n = 3 16a ( ) ( ) ( ) ( ) 1Area of trapezium 3 3 6 2 7.52 187.5 3.75 10 20 50 10 20 10 70 7 xx x x x x = − + − =− =− = = 16b ( ) ( ) ( ) ( ) ( ) Perimeter 3 3 6 2 2 2 5 3 15 5 2 9 68 cm x x x= − + − + − = + + = 16c oo o (alternate angles) =180 125 (interior angles) 55 CBE BEF = − = 17a 22 By Pythagoras' Theorem, 42 12 or 3.4641 =3.46 cm BC =− =

Holy Innocents’ High School End Of Year Examination 2021 Secondary 2 Express Mathematics Paper 1 17b 9.6Scale factor 12 2.7712 2.77 = = = 17c 4 2.771 4 4 =7.08 cm DE CE=− =  − 18a 90 120360 30  = 18b o o 15Angle (EL) 360120 45 = = 18c 7.5 units represents (120 – 30 – 15) students Number of students whose fav. subject is Science 75= 6.57.5 65  = OR Angle represented by Science o 360 90 45 6.57.5 195 −−= = Number of students whose fav. subject is Science 195= 120360 65  = 18d Can show the relative size of a part in relation to the whole / Can show proportion compared to the whole