( )( ) La Salle College Form Six Mock Examination 2013 Mathematics Compulsory Part Paper 2 Solution

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1 L Slle ollege Form Si Mock Emintion 0 Mthemtics ompulsor Prt Pper Solution 6 D 6 D 6 6 D D 7 D D 9 9 D 9 D 9 D D 5. = + + = + = = = + = =. D The selling price = $ ( ) ( + 0% ) = $ 58. r+ s = r+ s 5 ( r+ s) = ( r+ s) 5 0r+ 5s = 6r+ 8s r = s r = s r: s = :. D ( )( ) ( ) ( ) ( ) ( )( ) + + = = = 0 + = 0 = or = 0 0-LS-F6 MOK-MTH-P -Sol

2 5. Note tht α + β = = nd 5 αβ= = 5. α + β = α + β α β = 5= 6 6. D ( )( + ) = + = + ( + ) + ( + ) + 6+ compring like terms, + = 6 = = = = 9 7. The mount of work tht ngus cn complete on ech d = The mount of work tht Rmund cn complete on ech d = The mount of work to be completed when ngus nd Rmund work together = + Thus, the required time = = = Let T n be the number of dots in the nth pttern. T = 8 T = 8+ ( + ) = T = + ( + ) = 0 T = 0 + ( + ) = 9 Thus, the number of dots in the 8th pttern is ( ) ( 5 ) ( 6 ) ( 7 ) 85 T = + + = T = + + = T = + + = T = + + = 9. 0-LS-F6 MOK-MTH-P -Sol

3 0. EFD = (bse s, isos. Δ ) In Δ DEF, DEF + + = 80 ( sum of Δ ) DEF = 6 F = 6 (propert of //grm) = 50 (lt. s, // G) G = 6 50 = 66 G 50 D E F. = sin 6 sin sin 6 = sin D O = 80 (dj. s on st. line) O = 90 In Δ O, = O + O (Pth. Thm.) = km = 580 km =. km (cor. to sig. fig.) N 6 O 8. onstruct nd lbel the digrm s shown. = DE = 0 m In Δ D, 0 m tn 5 = D D = 0 m E = 55 0 m = 5 m In Δ E, 5 tn E = 0 E m E D 5 0 m Thus, the required ngle = 9.8 (cor. to sig. fig.) 0-LS-F6 MOK-MTH-P -Sol

4 . tn 5 sin 0 + = + = + cos θ + sin 90 + θ cos θ + cos θ cos θ + cos θ = ( + cos θ) + ( cos θ) ( cos θ)( + cos θ) = = cos θ sin θ 5. The possible outcomes re listed in the tble. nd dice st dice Sum The required probbilit = = 6 6. D 7. onstruct nd lbel the digrm s shown. O = + = (Pth. Thm.) tn O = O = 60 (, ) = (, ) = (, 00 ) O (, ) = = = LS-F6 MOK-MTH-P -Sol

5 b c= b c nd 6 b= b H..F. = b 5 5 L..M. = b c= b c. + = =. 8b b b b b b b b = = + + = > 5 > 5 > 9 < < > 6 >. D Let, s nd be the cost, number of stff nd number of customers respectivel. ks Thus, =, where k is non-zero constnt. ( + ) ( 0% ) ks 0% 0 ks 0 ' = = = ' Percentge chnge = 00% = 8 00% = 5.8% (cor. to sig. fig.) i 0 + i + i 0 + 0i+ i+ i = = i i + i i 0 + i i 8 = = = + i 9 i 9 Thus, the imginr prt is. 0-LS-F6 MOK-MTH-P -Sol 5 5

6 6. Let r cm be the rdius. Slnt height = r + 8 cm= r + 6 cm (Pth. Thm.) π r r π r = 96π r r r = 96 8 cm r r + 6 = 96 r ( r r + 6 ) = ( 96 r ) r r + 6 = 96 9r + r r cm r + 6r = 96 9r + r 56 r = 96 r = 6 r = 6 Volume = 6 8 cm 96 cm π = π 7. ( Q + 5 ) + 00 = 80 (opp. s, cclic qud.) Q = 6 Q = 6 (rcs prop. to s t ce ) In Δ Q, Q = 80 ( sum of Δ ) Q = = 80 (opp. s, cclic qud.) = Q 8. onstruct nd lbel the digrm s shown. Let r cm be the rdius. In Δ PQT, sin QPT sin 0 = 5 7 QPT 8.07 OPT = 90 (tngent rdius) OPQ = = In Δ OPQ, r = r + 7 r 7 cos r cos = 9 r = 5.66 (cor. to sig. fig.) Thus, the rdius is 5.66 cm. P T 0 O Q 0-LS-F6 MOK-MTH-P -Sol 6 6

7 9. log = mn = log ( mn ) log log = log m+ log n log = log n + log m When = 0, log =, log m = m = When = 9, log = 0, 9 O ( n) log 9 + log = 0 9log n = log log n = log 9 n = 9 n = 0.,0 7,0 Note tht the center lies on the -is. Thus, is dimeter of the circle. onstruct nd lbel the digrm s shown. oordintes of the center = = = Rdius = = 5 = 0 0-LS-F6 MOK-MTH-P -Sol 7 7 = 90 ( in semi-circle) Let =. In Δ, = 0 = 00 (Pth. Thm.) = = 0 ( ) 00 = 0 ( ) 00 = = = = 0 ( )( ) =.6 = 0 or = 90 =.6 or = 9.9 (rejected) O 7,0 + + = 0

8 . Sub = 0 into : = 0, 8+ = 0 6 = 0 = or = 6 ( )( ) oordintes of = (,0) O L divides the circle into two equl hlves is dimeter of the circle nd the center is the mid-point of. 8 6 oordintes of the center =, = (,) Let the coordintes of be (, ). + = + = 8 = = = 6 Thus, the coordintes of re ( 6,6 ). 6 0 Slope of L = = 6 The tngent to the circle t L (tngent rdius) Let m be the tngent to the circle t. m = m = The required eqution is 6 = = = 0 ( ) = ( ) : = 0 L 0-LS-F6 MOK-MTH-P -Sol 8 8

9 . Slope of PG = = 0 0 Slope of PH = = 5 5 PG PH = 5 ( ) = ( 5) 5 = = 0. T5 = S5 S = 5 = 9 lterntive Solution Let nd d be the first term nd the common difference of the sequence respectivel. T5 = + ( 5 ) d = + d n ( ) + n d = n n + d = n Sub n = 9 in (*) : n + d = n 9 + d = 9 + d = 9 T = 9 5 (*). D b ommon rtio = b First term = = b Sum to infinit = b = b = b b b b 0-LS-F6 MOK-MTH-P -Sol 9 9

10 5. 6. D The coordintes of, nd re ( 0, ), (, ) nd Let P= t O ( 0,0), P = = t ( 0,), P = = t (,), P = = t ( 7,0), P = = 5 Thus, the mimum vlue is. 7,0 respectivel. O 7. Probbilit = + = + = Probbilit = = Probbilit = 5 = Probbilit = + + = Probbilit = + = 8 9 D. Probbilit = 5 = 0-LS-F6 MOK-MTH-P -Sol 0 0

11 0. D Let be the re of Δ DEH. ΔHF ~ ΔDHE () E D re of Δ HF = = re of Δ DHE re of Δ HF = E : ED = : (intercept thm.) re of Δ HE E = = re of Δ DHE DE re of Δ HE = ΔHE ~ ΔGD () F H G re of Δ HE E = = = re of Δ GD D re of Δ GD = = 9 re of Δ HGD = = re of ΔFH : re of Δ DGH = : = 8 : 0-LS-F6 MOK-MTH-P -Sol

12 . = nd E psses through the centroid G E is the ngle bisector of E = 90 = 5 Let = =. In Δ E, E. sin 5 = E E = sin 5 = D G θ E cos 5 = E E = cos 5 = G is the centroid. E EG = = In Δ EG, 0 G = + = E sin GE = G ( θ ) sin 80 = 0 0 sin θ = 0 0-LS-F6 MOK-MTH-P -Sol

13 . m F Lbel the digrm s shown. sin θ = b m = sinθ m θ φ b d m cos θ = D b = cosθ c sin φ = b c= bsin φ = cos θ sin φ d cos φ = b d = bcos φ = cos θ cos φ Volume = cd m = cos θ sin φ cos θ cos φ sin θ m = sinθ cos θ sinφ cosφ m c m E. Scores lss oundries lss Mrk Frequenc Rnge = = Men = = = Stndrd devition = 50 = 6 (cor. to the nerest integer). D Let d be the stndrd devition of the originl set of numbers. Then, v= d fter is subtrcted from ech number, men = m, inter-qurtile rnge = q, stndrd devition = d nd vrince = v. fter the resulting number is doubled, men ( m ) devition = d nd vrince = d = d = v. =, inter-qurtile rnge = q, stndrd 5. 0-LS-F6 MOK-MTH-P -Sol