TMS 2023 1E Math P2 (ANS)

2023 Sec 1E FTE Paper 2 Mark Scheme 1 Largest = 728 649 999 m2 [B1] Smallest = 728 550 000 m2 [B1] 2a 30 3 18abc ab bc+− 3( 1 0 6)b ac a c= +− [B1] 2b 4 3(2 6)yz+− 4 6 18yz=+− [M1 – Expand, see either 6z or 18] 2(2 3 9)yz= +− [A1] 3 After discount 95 268.35100= × [M1 – discount] 254.9325= Total amount 254.9325 1.1 1.08= ×× [M1 – finding service charge OR GST] 302.86= [A1 – to 2 dp] 4a HCF = 227× [B1] 4b 11k = [B1] 4c Find LCM 4222571 1=××× [M1 – for attempt to find LCM] 2225q= × 100= [A1] 5 360 (5 15)DEA x∠ =−+ (angles at a pt.) [M1] 360 5 15x= −− 345 5 x= − Adding an additional parallel line at point E ( )26 180 3 29DEA x∠ = +−− (alt angles) (int. angle)[M1 – alt angle used] [M1 – int angle used] 26 180 3 29x=+ −+ 235 3 x= − DEA∠= 345 5 235 3xx−= − [M1 – form correct equation] 2 110x= 55x= [A1] Swapped answers – max 1m 118% on discounted price – max 2m B2 awarded if student can obtain 100 by observation. Poor presentation – deduct 1m if student get full marks for question

6a 5 km [B1] 6b 1.5 hours [B1, reject 1h 30min] 6c Gradient Rise Run= 20= [B1] 20 km/h represents the speed of Daniel travelling [B1] from home to supermarket. 7a 342abc×− 12 2ab c= − [B1] 7b 3( 2 )p qp−− 36pq p= −+ [M1 – expand, see either 3q or 6p] 73pq= − [A1] 7c 5 1 7( 3) 68 xx−++ 4(5 1) 21( 3) 24 24 xx−+= + [ B1 – change to same denominator] 20 4 21 63 24 24 xx−+= + 20 4 21 63 24 xx−+ += [M1 – attempt to expand numerator] 41 59 24 x+= [A1] 8a 2p=− [B1] 8b 2q= [B1] 8c Drawing the line [B1] 8d 0.5x=− [B1 – must show dotted line on grid] 8e Label R on graph [B1 – must see “R” on grid] R = (0.5, 0) [B1] Other accepted answers: “Rate of change of distance” “Distance travelled per hour”. 8d) If students calculated using the equation of graph without showing dotted line on grid  0m

9a Area of Triangle ABC = 1 8.66 102×× 43.3= cm2 [B1] 9b S.A of base 43.3 6= × [M1] 259.8= cm2 Volume of prism = Base area ×Height 259.8 63= × 16367.4= cm3 [A1] 9c Surface area of sides 63 10= × [M1] 630= cm2 Total S.A 259.8 2 630 6= ×+ × 4299.6= cm2 [A1] 9d 500 cm2 = $1.12 1 cm2 = 1.12 500÷ [M1] 0.00224= Total cost 0.00224 4299.6= × $9.63= [A1 – no FT] 9e Volume of cylinder 2rhπ= 2(8 )(4.5)π= [M1] 904.77868= cm3 Number of cylinders = 16367.4 904.77868÷ [M1*] 18.0899= 18≈ Will not be able as there are only 18 such cylinders [A1 – must state “18 cylinders” or he can create with the amount of plastic. “short of 1728.173cm3"] If students rounded exact value(s) for answer to 3s.f. for this question  deduct overall 1m fr