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 from Qn 9 9d) Alternative Answer: 4299.6 8.5992500 ≈ [M1] 9≈ (whole number) Cost 1.12 9= × = $10.08 [A1] *9e) Alternative for 2nd M1: Volume required for 20 cylinders 904.77868 20= × = 18095.573 Difference 18095.573 16367.4 1728.173 = − = [M1]

10a March and April [B1] 10b Temperature difference 31.0 30.1= − 0.9 C=  [B1] 10c 14 [B1] 10d Percentage = 22 18 19 12 11 14 100%181 +++++ × 53.0%= (to 3 s.f) [B1] 10e Disagree [B1] The y axis of the graph starts from 10 instead of 0. [B1] The height of the bars should not be used to compare the number of times. 10f Agree. From the graph, in January 2023, the number of rainy days is the highest and the temperature is the lowest in 2023. [B1 – Agree + comparison done for only 1 month] While in May 2023, the number of rainy days is the lowest, resulting in the highest maximum temperature. This proves that the trend of number of rainy days is linked to the daily maximum air temperature. [B1 – Comparing at least 2 different months] If students circled disagree without attempt to explain  0m