PLMGS 4E 5NA Math P1 Prelim 2017 Worked Solutions

Paya Lebar Methodist Girls’ School (Secondary) Department of Mathematics 2017 Preliminary Examination Mathematics Paper 1 (4048/1) Worked Solutions Qns No. Solution 1(a) )5(2)3(3  xyx = 10239  xyx = 1037  yx 1(b) 36)12( 2 x = )612)(612(  xx = )72)(52(  xx OR 36)12( 2 x = 36144 2  xx = 3544 2  xx = )72)(52(  xx 2 aybybxax 10142115  = )57(2)75(3 abybax  = )23)(75( yxba  3(a) 3(b) P = {1, 2, 3} Q = {2, 3, 4, 5, 6} Elements in P  Q are 2 and 3. 4 )516()34( 22  nn = 51692416 22  nnn = 424 n = )16(4 n Since the expression is a multiple of 4, it is divisible by 4. 5 Let the width be x cm. 288)9)(2)(( xx 28818 2 x 162 x 4x (4 is rejected) Length of base = 8 cm.  A B

Qns No. Solution 6 8.13 52sin 4.17 sin XZY       8.13 52sin4.17sin XZY = 83.5 or 96.5 7 3 1 45 45 3       h 3 3 1 45 45  h 453 145 3       h 80.13h (2 dp) OR Let the height of small pyramid be x cm. 3 1 45 3      x       3 3 145x 2013.31x 2013.3145h = 13.80 (2 dp) 8  Title is biased – Does not allow reader to make their own judgement  The width of cylindrical bars are not equal – exaggerates the difference between the years 9 xxx 21 2 472 3 2  = xxx 21 2 )4)(12( 3  = 12 2 )4)(12( 3  xxx = )4)(12( )4(23   xx x = )4)(12( 112   xx x 10(a)  50DAC ( at centre = 2 at circumference) 10(b)  70180ADO (s in the opp segs) = 110 10(c)  11050180ADO ( sum of ) = 20 2 100180 ADO (base  of isos )

Qns No. Solution = 40  2040ACO = 20 11(a) 21215 xx = 15122  xx = 2 2 2 1215)6(     x = 21)6( 2x 11(b) Minimum value is 21 11(c) Equation is 6x . 12(a) dP 100941001.1 5  d100941001.11022.2 55  10094 1001.11022.2 55 d = 12.0 m (3 sf) Depth of diver = 12.0 m 12(b) 1 5 1 100941001.1 dP  2 5 2 100941001.1 dP  )100941001.1()100941001.1(105.3 2 5 1 55 dd  )(10094105.3 21 5 dd  10094 105.3 5 21  dd = 34.7 m Difference in depths is 34.7 m 13 In Singapore, 10 grams of gold cost $567.40 1 gram costs $56.74 USD 1275.10 = $1.382  1275.10 = $1762.19 In Los Angeles, 31.10 grams of gold cost $1762.19 1 gram of gold costs $56.67 Gold is cheaper in Los Angeles. OR 1.382 Singapore dollars = 1 US dollar S$567.40 = USD 382.1 40.567 = USD 410.56 In Singapore, 10 grams of gold cost USD 410.56

Qns No. Solution 1 gram costs USD 41.06 In Los Angeles, 31.10 grams of gold cost USD 1275.10 1 gram of gold costs USD 41 Gold is cheaper in Los Angeles. 14 C(46, 24) 15(a) 5 6tan BCP  194.50BCP PCQ = 2  50.194 = 100.4 (1 dp) 15(b)  4.1009090360PBQ ( sum of quad) = 79.6 Perimeter of APDQA = )6(2360 6.79360)5(2360 4.100360     = 52.0 cm (3 sf) 15(c) Area of APDQA =       )6)(5(2 12)6(360 6.79360)5(360 4.100360 22  = 175 cm2 16(a)  115DCB (alt s, AB //CD) 16(b)  115125360DCF (s at a point) = 120 Since DCF = EFG = 120, therefore EF // DC. Since DC // BA, therefore EF // BA. 17(a) 533322540  = 532 32  17(b) 540 is not a perfect cube as power of 2 and 5 are not multiples of 3. 17(c) m = 3, n = 5 18(a) Subst )3,1( A into bxaxy  2 )1()1(3 2  ba ba 3 ------------ (1) Subst )33,3(B into bxaxy  2 )3()3(33 2 ba  ba 3933  ------------(2) (1)  3: ba 339  ------------(3)

Qns No. Solution (2) + (3): a1224 2a Subst a = 2 into (1) b 23 5b 18(b) Gradient of AB = )1(3 )3(33   = 9 Equation of AB is xy 9 19 For carpark A, Total charge = 3$20.1$420.2$  = $10 For carpark B, Total charge = )45603(04.0$  = $9 She should park in carpark B 20(a) 1 : 20 000 1 cm : 0.2 km Distance on map = 2.0 66 = 330 cm 20(b) Area of the lake in Town Y in map = 34.0100 150 = 0.51 cm2 1 cm : 0. 2 km 1 cm2 : 0. 04 km2 Actual area of lake in Town Y = 0.51  0.04 = 0.0204 km2 21(a) Bearing of C from B =  1201 21(b) No. It is not the point of intersection of the 2 bisectors. Refer to attachment at the last page 22(a) Let x be the probabilty of Bernice winning. 122  xxx 15 x 5 1x P(Ann winning) = 5 2

Qns No. Solution P(Bernice winning) = 5 1 P(Carol winning) = 5 2 22(b) P(Bernice or Carol wins) = 5 2 5 1 = 5 3 22(c) P(a player wins both games) =                 5 2 5 2 5 1 5 1 5 2 5 2 = 25 9 23(a)            4 3 0 kPQ =         4 3k 23(b) OQOP  222 4)3( k 252 k k = 5 or 5 23(c) POOR 3        4 33OR =       12 9 R(9, 12) 24(a) Circumference of circle = )6(2 x = x12 cm 24(b) Perimeter of shaded region =   xxx 4)2(22 1)4(2   = xxx 428   = xx 410  = )25(2 x cm 24(c) Area of shaded region =  22 )2(2 1)4( xx   = 22 216 xx   = 214 x cm2 Area of circle = 2)6( x

Qns No. Solution = 236 x cm2 Fraction not shaded = 2 22 36 1436 x xx    = 2 2 36 22 x x   = 18 11 25(a)       401 123N 25(b) T = NM =                   2.07.0 25.08.2 1.06.1 401 123 =        4.040.4 6.01.11 25(c) The elements represent the amount Ben spent in bookstore A and the difference in amount spent in the two bookstores respectively. 25(d) Charles has to pay = $4.40 + $0.40 = $4.80 25(e) Increased priced payment = 10.11$100 110 = $12.21 Amount Ben has to pay after discount = 21.12$100 70  = $8.55