PLMGS 4E 5NA Math P2 Prelim 2017 Worked Solutions

Paya Lebar Methodist Girls’ School (Secondary) Department of Mathematics 2017 Preliminary Examination Mathematics Paper 2 (4048/2) Worked Solutions Qns No. Working 1(ai) 222 )54(254016  xyxyyx 1(aii) kxyLHS  2)54( kxy  )54( x ky 4 5 1(b) 0.5or 2 1 36 24276 46 32)152(2 46 32 3 152 6 3243 152       x x x xx xx xx 1(c) x x x x xx xx xx xx xx x xx x xx x              6 )2(2or 6 )2(2 )2)(6( )2)(2(2or )2)(6( )2)(2(2 )124 )4(2or )124( )4(2 412 28 2 2 2 2 2 2

2 Qns No. Working 1(d) 5 3 5 3 2)3( 3)3(2 2)( 3)(2 2 232 into (1) Sub (1) ------- 3 311            xy xy xyxy xyxy xyxy xyxy yxyx yxy-x xyxy yx 2(ai) 15 48 720 483601080 '163 360360 3 36016360 :(2) into (1) Sub (2) ------ 3 36016Ext -(1)----- 360Ext        n n n nn nn nx nx    2(aii) line)str aon s (adj 1728180Int 8)15(3 360    x 2(b) In triangle QRS and triangle PST, QR = PS (opp sides of rhombus) ) ofmidpt theis ( ) s, ding(correspon RTSSTRS SP//RQTEPSRQ   Hence, triangle QRS is congruent to triangle PST. (SAS) 2(c) QP = ST and QP // ST (RS & QP are opp sides of rhombus and RST is a str. line) QS = PT and QS // PT (Corresponding sides of congruent triangles, proven in (i)) Hence, PQST is a parallelogram. (2 pairs of equal and // sides)

3 Qns No. Working 2(e) In triangle POQ is similar to triangle RQT, equal) are rhombus of angles (opp ) s,(alt ) s,(alt QRTOPQ QP//RTQTROQP SP//RQTQRQOP    Hence, triangle POQ and triangle RQT are similar. (2 pairs of corresponding angles are equal) 3(a) a = 90 b = 36 c = 55 3(b) Numbers in the R column is made up of the sum of consecutive odd-number factors, i.e. 1 + 3 + 5 + 7 +... And 1 + 3 + 5 + 7 +...+ 17 = 100, hence 99 cannot appear in column R. 3(c) P = T – R + 1 3(d) nnT nnT n 32 )32( 2   3(ei)             34 32372 3232 3721312 22 1 22 22 1       p ppppTT ppppT ppppT pp p p 3(eii) 341  pTT pp , a common factor 4 cannot be derived from 4p + 3. 3(f) P = 275 – 121 + 1 = 155 4(ai) AOC 2 26 2 rad3     4(aii) Area of circle 2r Area of AOB 2 2 2 1 sin23 13 22 3 4 r r r     Or 0.433r2  Area of hexagon ABCDE 2 2 36 4 33 2 r r   or 2.598 r2

4 Qns No. Working Total shaded area 2 2 2 33 2 33( ) units sq.2 rr r     or 0.544 r2 4(b) Connect point C to point F, then  180CFAABC (  s in opp seg are suppl)  180EFCCDE (  s in opp seg are suppl) Adding up:  360EFCCFACDEABC  360EFACDEABC (Shown) 5(a) tan 32 TB DB tan32TB DB tan 24 6 TB DB  = tan 32 6 DB DB   tan32 6tan24 tan24DB DB    6tan 24 14.87064tan32 tan 24DB    m Height of the flagpole = 14.87064tan32 or 20.87064tan 24o = 9.292209 = 9.29 m (3 s.f) 5(b) AB = 6 + 14.87064 = 20.87064 ≈ 20.9 m By Cosine Rule,    2 2 2 50 20.87064 2 50 20.87064 cos38BC     BC = 35.92986 = 35.9 m (3 s.f) 5(c) s.f) (3 m 4.16 4463.16 38cos87064.20 87064.2038cos     AE AE AE AE  

5 Qns No. Working 6(ai) Since EF : EO = 4: 7, hence FC : OB = 4: 7. Therefore FC = 4b 6(aii) CB = CF + FO + OB = −4b − 4a + 7b = 3b − 4a 6(aiii) EC = EF + FC or 3 4 CB = − 3 16 a + 4b = 3 4 (−4a + 3b) = 3 4 (−4a + 3b) 6(aiv) Since ED // AB and ED : AB = 4: 3, Therefore ED = 3 8 b 6(b) E, C and B are collinear. EC = CB3 4 6(ci) Since triangle EFC is similar to area of triangle EOB, Therefore 49 16 7 4 2        EOB EFC A A Area of triangle EFC : area of triangle EOB = 16 : 49 6(cii) Since triangle EDC is similar to area of triangle ABC, Therefore 9 16 3 4 2        ABC EDC A A Area of triangle EDC : area of triangle ABC = 16 : 9 6(ciii) 3 2 4 3/8 ))((5.0 ))((5.0    hFC hED A A EFC EDC 48 32 3 2    EFC EDC A A 147 48 49 16    EOB EFC A A Therefore, 99 80 48147 3248  -A A FCBO EDCF area triangle EDCF : area of triangle OBCF = 80 : 99 7(a) 22 3538 y y = 41 7(bi) Time taken = x 37 h

6 Qns No. Working 7(bii) Time taken = 4 37 x h 7(biii) (Shown) 05924 4)148(4 4 1 )4( 37)4(37 60 15 4 3737 2 2      xx xx xx xx xx 7(biv) Using quadratic formula or completing the square 2 23844 )1(2 )592)(1(4)4(4 05924 2 2    x x xx 4.22x or 4.26x (3 s.f) 7(bv) Time taken =  65.14.22 3737 x 1 hour 39 minutes (Correct to nearest minute) 8(a) p = 5.71 8(b) 3 8(ci) Min value = −0.2 8(cii) When y = 4, x = 0.5 or x = 3 8(d) Draw tangent at @ (2, 1) m = 5.223 15.3   (Accept 0.473 to 2.52) 8(e) Drawing line 52  xy a = 5 and b = 2 9(aia) 20 + 16 + x + 10 + x = 50 2x = 4 x = 2 9(aib) Percentage = 10050 30 = 60% 9(aiia) Mean = 50 5420 = 108.4 cm 9(aiib) Standard deviation = 24.10850 603400 = 17.8 cm 9(aiii) Second group of students have a higher mean than first group of students. Hence, 2nd group of students jumped further/longer distance. Also, 2nd group of students has a lower standard deviation. This implies that the distance achieved by all the students is more consistent than students in the 1st group.

7 Qns No. Working 9(bi) 9(biia) P(Gold from bag B) 56 15 8 2 7 2 8 2 7 4 8 3 7 1               9(biib) P(Bronze from bag B) = 28 1 8 1 7 2      10(a) U-Taxi offers the cheapest fare. Percentage difference between U-taxi & C-taxi = 10080.9 02.680.9  = %7 438 or 38.6% 10(bi) Base Fare = $3.00 Distance Fare = $0.45 (8 km) = $3.60 Time Fare = $0.20 (12 min) = $2.40 Total Fare = $3.00 + $3.60 + $2.40 = $9.00 10(bii) The driver can get reach the destination in time because of the relatively clear roads during midnight. 10(c) C-Taxi  Base Fare = $3.20 Metered Fare =           45 60522.0400 850022.0 = 0.22 (21 or 21.25 or 22) + 0.22(7) = $4.675 + $1.54 = $6.215 or $6.16 or $6.38 S G B S G S G S G B Bag A Bag B 7 1 7 4 7 2 4 1 8 2  8 5 8 3 4 3 8 6  4 1 8 2  8 5 8 1

8 Qns No. Working Peak Hour Surcharge = ($3.20 + $6.215)(25%) = $2.35 or $2.34 or $2.95 Total Fare = $3.20 + $6.215 + $2.35 = $11.77 or $11.70 or $11.96 G-Taxi  Base Fare = $3.00 Distance Fare = $0.80 (9.5 km) = $7.60 Total Fare = 7.60 + 3 = $10.60  $11 U-Taxi  Base Fare = $3.00 Distance Fare = $0.45 (9.5 km) = $4.275 Time Fare: $0.20 (15 min) = $3.00 Total Fare = $3.00 + $4.275 + $3 = $10.28 Ms Seet should use G-Taxi. Although U-taxi is slightly cheaper than G-taxi, the 15 minutes journey time was based on no traffic jam during the morning peak hour when she commutes to work.