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   tan3