Mathematics Grade 12 Paper 1 KwaZulu Natal Mock 2022 Memo

othe RE SABER. ¢ MATHEMATICS Bi PREPARATORY EXAMINATION Sy oO a EE a a OF a BL SI at BE a SEPTEMBER 2022 Seeseegeeeee MARKS: 149 TIME: 3 hours This question paper consists of % pages and 1 information sheet, Copyright reserved Please tum over

Downloaded from hlayiso.com Matbenates*Downloaded from Stanmdrephysics Cone sen 20% INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. This question paper consists of 13 questions. 2, Read the questions carefully. 3. Answer ALL the questions. 4. Number your answers exactly as the questions are numbered. 5, Clearly show ALL calculations, diagrams, graphs, etc. which you have used in determining your answers. 6. Answers only will NOT necessarily be awarded full marks. G 7, You may use an approved scientific calculator (non-programmable and wii, unless stated otherwise. 8. If necessary, round off answers correct to TWO decimal places sins stated otherwise. « eS 5 cS 9. Diagrams are NOT necessarily drawn to scale. 10. Write neatly and legibly. Copyright Reserved ~ Please turn over

Downloaded from hlayiso.com Mathematips\Pin loaded from Stanmor ephysics. CORN Preparatory Examination 2022 NSC - GRADE 12 QUESTION i 1.1 Solve for x: LLL @+5)@x-1)=0 (2) 112 -3°~7x=-8 (correct to TWO decimal places) (4) 1130 Vx+5+l=x 6) 1.14 (2x ~3)(x+5)<0 GB) 1.2 Solve for x and y simultaneously if: & & ©) x+3y=5 and xy+y?-3=0 . 2” 1.3 Simplify fully, without the use of a calculator: x aT Ey) eed Eee vea where @., : (4) & [24] QUESTION 2 Ke Given the quadratic number pattern: 5 ; 9 cf 29 3. 21 Write down the 5" and 6” &. of the pattern. (2) 2.2 Show that the n'* tempafeshe quadratic pattern is given by 7 =2n*—2n+5 (4) 2.3 Is 2023 aterm OE patent Motivate your answer with relevant calculations, (4) [10] \ RY QUESTION 3 ) Evaluate: 2 (30-4k) [4] Copyright Reserved Please turn over mc mr BUEN H

Downloaded from hlayiso.com Mathematic# OWNloaded from Stanmorephysics. eayFreparatory Examination 2022 NSC — GRADE 12 QUESTION 4 4.1 Given the geometric series a + ar + a + ar +.., where a is the first term and r is the common ratio. Prove that the sum to » terms of this series is given by se a(r" -1) na srel rol 4) 42 The first two terms of a geometric sequence with constant ratio r , and an arithmetic sequence with constant first difference , d, is the same. The first term is 12. 4.2.1 Write down the second and third terms of EACH sequence in terms of d and r. (2) 4.2.2 Ifitis further given that the sum of the first three terms of the geometric sequence is three more than the sum of the first three terms of the arithmetic sequence. Determine two possible values of the common ratio, r, of the geometric sequence. (5) [7] QUESTION 5 Sketched below is the graph of A(x) =1-5 . Ais y-intercept and B is the x-intercept of A. x- 5.1 Write down the equations of the asymptotes of h. (2) 5.2 Calculate the coordinates of A and B. Q) 5.3 Write down the equation of the line of symmetry of h with positive gradient. (2) 5.4 Write down the range of A. Q) 18] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathem@tsfinloaded from S fanmor ephysics - CORN Preparatory Examination 2022 QUESTION 6 SC 2; 0) and T(6 ; 0) are the x-intercepts of the graph of f(x) = ax’ +bx+c; a#0. R is the y-intercept of f and g. The straight line through R and T has the equation g(x)=-2x+d. 6.1 Calculate the value of d. 6.2 Show that I(x) =-x? +4x412. 6.3 Calculate the coordinates of the turing point of / 6:4 Determine for which values of x will: 6.4.1 F(x)- g(x) 20 6.4.2 xf (x) <0 6.5 Determine the coordinates of R' »the image of R on p if p(x)= —f(x-2), Copyright Reserved (2) (4) (3) (2) (3) (2) {16} Please turn over eT IT |

Downloaded from hlayiso.com Mathematied OWNIoaded from Stanmorephysics. saymreparatory Examination 2022 NSC — GRADE 12 QUESTION 7 Sketched below is the graph of p(x)=a";a>0; a% 1, The graph intersects the y-axisat A. The point TS ; 32) lies on p. a x 7A Write down the coordinates of the point A. (2) 7.2 Calculate the value of a. (2) 73 Write down the domain of p. (1) 7.4 Write down the equation of p', the inverse of p, in the formy=... (2) 75 Determine the values of x if p'(x) <5. (2) 9] Copyright Reserved Please turn over

Downloaded from hlayiso.com “Mateyeipffloaded from Stanmore HY SICS. COM KZN Preparatory Examination 2022 NS' — GRADE 12 QUESTION 8 8.1 8.2 Dipinda opened an account with an amount of R5000 on 1 June 2021. She then makes monthly deposits of R600 at the end of every month. Her first deposit is made on the 30 June 2021 and her last deposit on 30 April 2023. The account earns interest of 14.25% per annum compounded monthly. Calculate the amount that is in the account directly after her last deposit is made into the account, Molly wants to buy a house for her family for R800 000. She agreed to pay monthly instalments of R10 000 on the loan which incurred interest at a rate of 13,35 % per annum compounded monthly. The first payrnent was made at the end of the first month after the loan was granted. 8.2.1\ Show that the loan will be paid back in full after in 200 months, 8.2.2 Suppose Molly encountered unexpected expenses and was unable to pay any instalment at the end of the 120", 121", 122" and 123" months. At the end of the 124" month she increased her payment to still pay off the loan in 200 months by making 77 equal monthly payments. a) Calculate the balance on the loan after the 119** payment was made. b) Calculate the new monthly instalment Molly must pay from the 124% month to settle the loan in 200 months. QUESTION 9 9.1 9.2 Copyright Reserved Determine (x) from first principles if f(x) = = . x Determine: 92.1 g(x) if g(x)=(x+7) ay, 4 92.2 — if pave (6) (4) (3) (4) [17] 6) (5) (4) [14] Please turn over

Downloaded from hlayiso.com Mathematiesp1DOWNloaded From Stanmoerephysicszwerggnratory Examination 2022 NSC — GRADE 12 QUESTION 10 10.1 Given: f(x)=x° -12x-16 10.1.1 Calculate the x-intercepts of the graph of f- (5) 10.1.2 Determine the coordinates of the turning points of i (4) 10.1.3 Sketch the graph of f, indicating the intercepts with the axes and the coordinates of the turning points. (4) 10.1.4 Determine the values of x for which the graph of / is concave up. (2) 10.2 Given: p(x)=—-x° -8x Is it possible to draw a tangent with positive gradient to the graph of p? Show all calculations to justify your answer. GB) (18] QUESTION 11 The lead, L, in metres, of a runner in the comrades’ marathon in the last ¢ minutes of the race, where ft € [0:75] is given by equation: 2 L=1000+6¢-— 4 _ dL 11.1 Determine 7 2) 11.2 Calculate the time at which the runner has the greatest lead. (2) 11.3. At what sate is the runner’s lead decreasing when 1= 60 minutes? (2) 16] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathomppticy Pi lad) from Stanmorephysies, COMEZN Preparatory Examination 2022 QUESTION 12 Given the word “BRACKET”. The letters of this word are randomly arranged to form new arrangements of the letters, 12.1 How many unique arrangements of the letters can be made? (2) 12.2 Determine the number of unique arrangements of the'letters that aré randomly made if the letters R and A must be together. (3) {5} QUESTION 13 13.1 Two different events A and B are mutually exclusive. It is further given that: e 3P(B)=P(A) © P(A orB) = 0,63 Calculate P(B). (3) 13.2 Two cards from a regular pack of 52 playing cards (shown below) , are drawn at random one after the other, without replacement. a Ge ia Baa ar * HI : | * f Sgt Fi e & ey ; Hf oe ye aie 8} Calculate the probability that: 13.2.1 both cards are picture cards. (2) 13.2.2 at least one of the cards is a picture card. (3) 3] TOTAL: 150 Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematies/PIDOWNloaded from Stanméreph 1ysics. CONKZN Preparatory Examination 2022 NSC - GRA 1 INFORMATION SHEET: MATHEMATICS _-b tb? —4ac 2a A=P(+ni) A=P(I-ni) A=P(-i)’” A=P(I+iy’ T, =a+(n-N)d =" 2a+(n-Nd} my T, =ar"* S, mau ) rel S, = 3-l<r<l . r-l l-r pil) -1| pa2lt+0"] I L \ _ | f(xt+h)- f(x) = lim SE F®) 50 h d=y(x,-x) +(,-9,) M2427.) m= tan y=mxte Y-y =M(x-H) ee otal (xa) +(9-)'= b c In AABC; —2- =—- =— sinA sinB sinC a? =b* +c7 -2be.cosA area ABC=5absinC sin(a+§) =sina.cosB + cosa.sinB sin(o.—B) =sina.cosB —cosa.sinB cos(a +B) =cosa.cosB —sina.sinB cos(a—B) = cosa.cosB +sin a.sinB cos’ o—sin’ a _ 2 . . cos 20, =41-2sin* o sin 2a = 2sina.cosa 2cos’ a-1 1" 2 X(%,-* x= Le w= 2( ) nA n P(A) = 8) P(A or B)=P(A)+P(B) —P(A and B) - p= eH =P) ¥)y-F) yp=atbx “Sa =a ‘opyright Reserved

Downloaded from hlayiso.com Downloaded from Stanmorephynes.-com education Department: Education PROVINCE OF KWAZULU-NATAL NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICS P1 PREPARATORY EXAMINATION SEPTEMBER 2022 MARKING GUIDELINE MARKS: 150 TIME: 3 hours NOTE: If a candidate answered a QUESTION TWICE, mark only the FIRST attempt. If a candidate crossed out an answer and did not redo it, mark the crossed-out answer. Consistent accuracy applies to ALL aspects of the marking guidelines. Assuming values/answers in order to solve a problem is unacceptable. This marking guideline consists of 12 pages. Copyright Reserved Please Turn Over

Mathemaownloaded from Stanmorephysics.com Downloaded from hlayiso.com NSC Marking Guideline September 2022 Preparatory Examination QUESTION 1 _ _i vo- vy i Lit |x=-5 or x=5 A 5 A ; 0 1.1.2 | -3x?-7x+8=0 AY © standard form —b +~vb? — 4ac (Penalize 1 mark x= a if rounding off is | | CA¥ correct substitution into —(-7) + JC)? — 4€3)(8) incorrect — once quadratic formula x= 2(—3) here for entire x= 3,17 or 0,84 paper) CAY CAY answers (4) 13 |Veee41=x Vvx+5=x-1 . . (vexs)" = («—1) AVY _ isolating surd x+5=x7-2x4+1 x? —3x-4=0 CAY standard form (x + 1)(x-4) =0 CAY factors x=-lorx=4 CAY x#-1 nla CAY x=4 () 114 |5<z7<3 AY critical value — 5 i. AY critical value ; CA¥Y interval notation (3) Copyright Reserved Please Turn Over

Mathemapownloaded from Stanmorephysics.com NSC Marking Guideline Downloaded from hlayiso.com September 2022 Preparatory Examination 120 [x+3y=5 (1) xy+y?-3=0 _+(2) From (1): x =5=3y > (3) Av making x the subject Substituting (3) into Q): yS-3y)+y?-3=0 CAV’ substitution —2y?+5y-3=0 2y? —5y+3=0 CAY standard form (2y —3)(y—1) =0 3 CAY factors yaa ory=1 CAY y-—values x5 Or x= 2 CAY x-—values (6) OR x+3y=5 > (1) OR xy+y?-3=0 -(2) From (1): y= = > (3) V : . Substituting (3) into (2): Av making y the subject _ 2 x ( 5 *) + ( 3 *) —3=0 CAV’ . substitution — — 2 _ = 3x(5—x)+25—-10x+x*-27=0 CAV’ standard form —2x7+5x-2=0 2 = 2x°—Sx+2=0 CAV factors (2x —1)(x -2) =0 x=5 orx=2 CAY x-—values 3 1 CAY ] =s ory= - y=5 y y— values 6) 13 | afton + ane2 52n + 4,57 _* 27,5" + 2", 22 ~ | 524 4,57 _" 2(5" + 4) AY factorising numerator 57(5" + 4) AY factorising denominator n/2n = lon CAY’ simplifying 2 == CAY answer : (4) [24] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathemaparv/nlioaded from Stanmorephysics. com September 2022 Preparatory Examination NSC Marking Guideline QUESTION 2 2.1 45 ; 65 AAY Y answers (2) 2.2 5 9 17 29 1D \ / 8 12 2D 4 4 AY 2a=4 2a=4 .a=2 AY az 3atb=4 «b=-2 AY 3a+b=4 atb+c=5 “~c=5 AY a+b+c=5 Ty = 2n? —2n+5 (4) 2.3 Tn = 2n? — 2n +5 = 2023 AY — equating n" term to 2023 2n* — 2n — 2018 = 0 n? —n—1009 =0 CAV standard form 1+¥v1+ 4036 n= —_,= 32.27 or — 31.27 CAY n-—values Since n is not a Natural Number, 2023 is not a term of the sequence. | CA Y conclusion (4) [10] QUESTION 3 26;22;18;... AV n~—value _? _ AY — a~—value Sn = ze + (a Dd] AY d-value Sso = [2(26) + (50 ~ 1)(-4)] CAW answer my Ss0 = —3600 OR OR AY n—value Sn = n [a+ Tn] AY a—value 2 AY Tso — value 50 v Sso = > [26 + (-170)] CAV answer (4) Sgo = 3600 [4] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathemapgwnloaded from Stanmorephysics.com NSC Marking Guideline September 2022 Preparatory Examination QUESTION 4 41 Sy =atartar?+are+--+ar™ + (1) AY — equation (1) rS, =ar+ar? +ar?+--+ar™1+ar" = (2) AY — equation (2) (2)— (@): TS, — Sy =ar™-—a AY © subtracting LHS and Sn(r — 1) = a(r™ — 1) RHS terms a(r™ — 1) Sn=— aq ir el AY factorising 7 (4) 4.2.1 | AS: 12+d;12+ 2d)... AY AS set up of terms GS: 12r;12r?;... Av — GS set up of terms (2) 4.2.2 |12+d=12r > (1) AY equation (1) and (2) 36+ 3d4+3=124+ 1274 12r2 = (2) From (1): d=12r—12 ~ (3) AY — making d the subject Substituting (3) into (2), we have 36 + 3(12r — 12) + 3 = 12 42 + 12r2 12+ (12r -—12)+1=4+4r+4r? 4r2 —8r+3=0 CAV’ standard quadratic form (2r —1)(2r -—3) =0 CAV factors 1 3 r= orr=5 CAY answers (5) [11] QUESTION 5 5.1 yell AY x=2 =2andy=1 x andy AY y=1 Q) 1 v i 5.2 y — intercept : ( 0:1 ) A y-intercept x — intercept: x = 3 AY x=3 (3; 0) CAY coordinate form (3) 5.3 }y=x-1 AY Gradient value (2) AY y-— intercept 5.4) yER y#1 AY answer qd) OR OR y € (—0; 1) U (1; 00) AY answer () [8] Copyright Reserved Please Turn Over

Mathemapownloaded from Stanmorephysics.com Downloaded from hlayiso.com NSC Marking Guideline September 2022 Preparatory Examination QUESTION 6 6.1 0=-2(6)+d AY © substitution of point T(6 ; 0) d-12 AY — answer (2) Answer only full marks 6.2 c=12 y = a(x + 2)(x — 6) AY - subst. x — intercepts 12 = a(0+ 2)(0- 6) AV © substitution R(0 ; 12) 12 =—-12a a=-1 AY a-—value y= 1x + 2)(x— 6) AY — a—value and factors f(x) = —-x? +4x +412 (4) OR OR x = 2...Axis of symmetry AY axis of symmetry value y= a(x —2)?+q.....(A) Substituting (0;12) into (A) 12 =a(0—2)?+q >12=4a+q...(1) | AY © substitution R(O ; 12) Substituting (6;0) into (A) 0=a(6-2)?+q 70=16a+q.....(2) (1) -(@): 12 =-12a y -t=a A a—value 16=q y = —1(x — 2)? + 16 AY — a—value and TP (4) f(x) = —-x? + 4x4 12 = 77 - <— 6.3 2x 4+4=0 or x= or x= 2+6 A Using calculus/formula/midpoint AD 2 x=2 . ae CAY x=2 (provided it is +ve) y = —(2)? + 4(2) +12 = 16 CAY y-value (2 ; 16) (3) 6.4.1 |O<x<6 AAY “answer (2) (penalty 1 mark for incorrect notation) 6.4.2 —2<x<0 or x>6 AAYY-2<x<0 AY x>6 (3) 6.5 |R/(2;-12 ) Avx— value (2) AY y—value [17] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathemap@wnloaded from Stanmorephysics.com September 2022 Preparatory Examination NSC Marking Guideline QUESTION 7 7.1 A(0;1) AAY V answer (2) 7.2 y=a* 32=a° AY _ substitution of point T(5;32) 5 a=2 AVY answer (2) 7.3 xeER AY answer (1) OR OR 1 x € (—00; 00) AVY answer i) 7.4 y = log, x CACAV V answer (2) 75 log,x=5 CAY end points AY _ interval x= 25 =32 O0<x< 32 Can be solved by log inequalities. (2) Answer Only — Full marks [9] QUESTION 8 8.1 1 Jun 31 Jul 31Aug | oe 30 Apr — 2021 2021 2023 A— formula 600 | 600 | 600 | sree 600 Av value of n Dipinda’s final amount in the account: AY value of i x[1+i"-1 =P(1+i)"+ IC D ] FV — formula 03 AY — value ofn 14.25% 23 600 1+ = 1| = 5000 (1 + ae + ( 2 ) CAY correct substitution into A 12 = CA correct substitution into Fv v = R22 321,54 CAY answer (6) 821 |) x-a+o 7) i 10000 [1 — (1+ 235%)" AY value of P, x and 800 000 = u value of i 13.35% x = AY ‘substitution into 13.35%\" 11 formula 1+ =) =—=0,11 ( 12 100 —n= log, ,2225%) 0.11 AY use of logs n = 199,5083362 AY decimal value Therefore the loan will be paid off in 200 months. (4) N.B. Candidates can also substitute the value of 200 into the Pv formula to show that the loan will be paid in 200 months. Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathemaparv/nlioaded from Stanmorephysics. com September 2022 Preparatory Examination NSC Marking Guideline 8.2.2a pe x{1-(4+i)] ~ i 13.35% 12 AY — value ofn a AY — value of i (3) = R530 009,55 CAY answer 10 000 [.- (1+ ee] If n = 81 is used and P = R532 010,58 Give a maximum of 2/3 marks N.B. Candidates can also use the method of A -Fv 8.2.2b [A = P(1 +i)” 13.35 %)\* ) AY — value of n A = R530 009,55 (1 += A = R553 991,4839 CAY answer 1-a4+i7 p20 +04 13.35 %\~77 x[1-(1+ 12 ) | AY — value ofn 553 991,4839 = 13.35% 12 CAY answer (4) x = R10 748,55 [17] Copyright Reserved Please Turn Over

Mathemaownloaded from Stanmorephysics.com Downloaded from hlayiso.com NSC Marking Guideline September 2022 Preparatory Examination QUESTION 9 (penalize 1 mark once for incorrect notation in this question) h) — v 9.1 f/Ge) = imo +h) — f(x) A formula h>0 h 202 f/@) = lim AY © substitution 2x-2(xth) 1 CAY LCD / = lim —_* x = MOO =lim—Seah “<h CAV simplification of Ie) =i 2x —2x—2h x 1 numerator P@)= jim—se@th) A —2 CAY answer N@= x5 ™ 3x (5) OR OR AVY — value of f(x + h) +h)= forth) 3(x +h) CAY value of 2 h) — f(x) h)— =—* + fart FOr +h) — FO) 3(x+h) 3x fO+h-fG)_— -2 h 3x(x +h) CAY value of Fonte) (x) = lim ——— £@) h-0 3x(x + h) AY — formula —2 "y) - (5) fo) 3x2 CAY _ answer 9.2.1 | g(x) =(x+7)3 g(x) = x? + 21x? + 147x + 343 AAV V (two terms correct 1 g! (x) = 3x? + 42x +147 mark, all terms correct 2 marks) CACACAYY V each term (5) 922 | pe_ 4 ye 9x2 s 4 0, AAVY y=xz-=—x an . 9 writing in exponential form dy 93 8s (4) dx 2 9 CACAY Y each term [14] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathemaparv/nlioaded from Stanmorephysics. com September 2022 Preparatory Examination NSC Marking Guideline QUESTION 10 10.1.1 | x? —12x -16 =0 (x + 2)(x? — 2x -8) =0 AY _ binomial factor (x + 2)(x + 2)(x -4) =0 AAV ¥ factors x=-2 orx=4 CA CA ¥ ¥ answers (5) 10.1.2 | f(x) =x? — 12x —16 f/ (x) = 3x? -12=0 AY derivative and x?-4=0 equating to 0 (x +2)(x-2)=0 x=-2 or x=2 CAV factors y=0 or y=~-32 CAY x—values CAY_y—values (4) 10.1.3 (-2:0) , a of o CAY Maximum and ji Minimum points \ CAY x-—intercepts \ f AY — y—intercept us AY — shape \ \ of (32) (4) 10.1.4 | f//(x) = 6x > 0 AVY —_.2™ derivative x>0 Answer only: full marks AY answer @) 10.2 | p/(x) = —3x4*7+8 Av derivative 2 —3x° <0 forall xeR AY reasoning —3x?-8<0 The gradient of all tangents to the graph of p is always AV reasoning negative. (3) [18] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathemapownloaded from Stanmorephysics.com NSC Marking Guideline September 2022 Preparatory Examination QUESTION 11 11.1 t? L = 1000 + 6t — — 4 aL 1 —=6-- AAY ¥ for each term 2 dt 2 (2) 11.2 | For greatest lead: “=0 cAv 4=9 at dt 1 or 6—-=t=0 equating derivative 2 to0 t = 12 minutes (2) CAY answer 11.3 | dL 1 CAV’ substitution of t = 60 — =6--5(60) ; a dt t=60 2 into derivative and dL _ os value of — 24 dt r=60 The runner’s lead is decreasing at 24 metres per minute CAY conclusion (provided the derivative is — ve) Q) [6] QUESTION 12 12.1 7! or 5040 AY AY 7! or 5040 (2) 12.2 6! x 2! AAY ¥6! x 2! = 1440 Av 1440 (3) [5] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathemaporwnloaded From Stanmorephysics.com _ September 2022 Preparatory Examination NSC Marking Guideline QUESTION 13 13.1 | P(A or B) = P(A) + P(B) AY — condition for mutually 0,63 = 3P(B)+P(B) exclusive events 4P(B) = 0,63 AY — correct substitution P(B) = 0,16 Av P(B) value (3) 2. 12 #11 12 v1 13.2.1 P(Both Picture cards) = — x — Ay 52° 54 52 51 1 i = = 557 = 000498 = 4,98 % AY 337 = 0,0498 = Q) 4,98 % 13.2.2 fo 280 | Outcomes Probabilities 12.11 5 parade ie 11/51 P PP 52 x 51 P - ae Pie | : ay 40517) NP | PNP 28 §2 51 40 12 ? 12/51 P NPP —x— war 52° 51 MS — os 39/51 NP NPNP ga x 39 52°51 P (at least 1 picture card) = 1-P (mo picture card) AY Method 40 39 =1- (5 x =) AY Correct Substitution AY —=0,4118= 7 17 => 0,4118 = 41,18 % 41,18 % OR OR (3) P (at least 1 picture) 12 11 12 40 40 12 AAY Y probabilities = (6*a) + * a) * (a) , 7 AY oid 0,4118 = = 0,4118 = 41,18 % 41,18 % (3) [8] TOTAL: 150 Copyright Reserved