KSS 2021 P1 (MS)

Marking Scheme to 4 NA Prelims Paper 1 No. Workings and Steps 1 2 >10 2x > 7 x > 3.5 x = 5 3a When x = 0, y = 4 point A = (0. 4) b Gradient = -1.5 4 = 1 unit reps $30 12 units rep 12 x $30 = $360 5 6 arc length 4.26 8.53 97.1 49 100 0.6 ´ ´= = 23x+ 345 33 ++ - 4 32 32 4 32 3 35 5 12 6 35 6 12 5 7 2 pr r pp q pr p q pr pq r ÷ =´ = 360 110 21 5360 125 6 p p -=´ ´ =

perimeter = 7a tan angle MAD = b tan angle CMN = CN = cm 8 -3, 3.03, 9a October and November b percentage increase = = 1.80% 10a 125 306 p+ 2310 CN 2 3 10 CN= 210 6.673´= 13,3- 133 1.7 1.67 100%1.67 - ´

bi scale factor = ii y = 4 x 1.8 = 7.2 11a b 12a b OR 13ai 64 aii 2.4 b 14 total distance = km 181.810= 4 10 3 6 4 yx y x yx -- + =- 25 3 25 2(2 5) 6 10 4 25 10 xx xx x -- --= -= 190 156´= (45 15) 3 2 20 -÷ = (45 15) 3 2 20 -÷ ´ = 3920 7.5 20.560´+ =

total time taken = h average speed = km/h 15a angle DCA b AC = 44.8 cm BC = 44.8 – 40 = 4.8 cm 16a 324 = 22 x 34 bi LCM = 23 x 34 x 52 x 7 bii) HCF = 2 x 32 = 18 17a value of handbag = = S$179.21 b watch in A$ = $255 x 1.054 = A$268.77 total cost = A$268.77 x 1.0205 = A$274.28 It is cheaper to buy in Singapore. 18 39 1 23 60 2 20+= 2320.5 17.820÷= 286440 AC CD BF BA AC = =´ 189 1.054657 2 2 82 3 ( 2)( 3) 8 68 14 0 1 1 4(1)( 14) 2 1 57 2 e e ee ee ee e e -= + -+ = +- = +- = -± - -= -±=

e = 3.27 or - 4.27 19a angle ACE = 180º - 80º = 100º Angle ACE = 100º because angle BAC and angle ACE are interior angles, AB // CE OR Angle ACE = 100º because BCE = 47º (alt. s, AB // CE) and ACE is the sum of ACE and BCE b 20 P(late) = 1 – 0.3 – 0.52 = 0.18 21 p = 80% 22 slant height perpendicular height = 21.81742 Volume ÐÐÐÐÐ100 53 47 123 47 76 BCD CBD Ð= ° - ° = ° Ð= ° - ° = ° 22.40100%28p=´ 2226 10 24 =- = 2224 10 476 =- =

23 1 int angle of pentagon = 108º 1 int angle of heptagon = º x = 360º - 108º - x = 117º 24a b c 5x + 3y = 136 ………………….(1) 8x + 9y = 268 ………………….(2) (1) × 3 15x + 9y = 408 …………………(3) (3) – (2) 7x = 140 x = 20 subst x = 20 into (1) 5(20) + 3y = 136 y = 12 x = 20 y = 12 120 20 4763 2910 =´ ´ ´ = (5 2) 180 5 -´ °= (8 2) 180 8 -´ °= 135135°25 15 680 531 3 6 xy xy += += 16 18 536 8 9 268 xy xy += +=

Alternative Method(Substitution) 5x + 3y = 136 ………………….(1) 8x + 9y = 268 ………………….(2) From (1) 3y = 136 – 5x ………..(3) Subst (3) into (2) 8x + 3(136 – 5x) = 268 8x + 408 – 15x = 268 -7x = -140 x = 20 Subst x = 20 into (3) y = 12 x = 20 y = 12 d diff = $20 – $12 = $8

25a b Refer to picture above c Refer to picture above d PA = 5.3 cm

26 Volume of water discharged by the 2 pipes per second = 47123.9 cm3 Since 47123.9 < 60 000 The two pipes are not able to discharge the water as quickly as it enters as the volume of water discharged per second by the two pipes is less than the volume of water that enters the tank. 253 0 0 2p=´ ´ ´