4016/01 October/November 2009

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ELEMENTRY MTHEMTIS Paper 1 Suggested Solutions 1. Topic: lgebra (a) 0xy 5y = 5y(4x 1) (b) 4x 4(x + ) = 4x 4x 1 = 1. Topic: rithmetic (Percentages) 100% $48 1% $48 100 5% $48 5 100 = $11.

1 ELEMENTRY MTHEMTIS Paper 1 Suggested Solutions 1. Topic: lgebra (a) 0xy 5y = 5y(4x 1) (b) 4x 4(x + ) = 4x 4x 1 = 1. Topic: rithmetic (Percentages) 100% $48 1% $ % $ = $11.80 Selling price = $ /01 October/November 009 nswer (a) 5y(4x 1) [1] (b) 1 [1] Selling price = ost price (100%) + Profit (15%) = 5% ost price nswer $ []. Topic: Statistics (Frequency Table & Pie hart) (a) Modal colour = lue Modal colour = Mode of frequency table = olour of car that appears most frequently (b) ngle representing the colour green = of sector in pie chart its given data 4. Topic: rithmetic (Percentages & Fractions) = 75 nswer (a) lue [1] (b) 75 [1] (a) % = 70 5 % 4 6 (b) Fraction of candidates who were not awarded an or grade = = Topic: lgebra (Indices) (a) p 5 = 40 p = 40 5 p = 8 p = p = nswer (a) (b) 1 % [1] [1] For tuition, exam papers & Last-Minute uddha Foot Hugging Syndrome treatment opyright 009 ϕ exampaper.com.sg. ll rights reserved. 1 / 11

2 (b) 1 x 4 = 1 1 x 4 = 1 x4 1 = x 4 1 am = a m nswer (a) p = [1] (b) x 4 [1] 7. Topic: Linear Inequalities < x 5 < < x < < x < 6 nswer (a) < x < 6 [] 6. Topic: ngle Properties of Polygon (a) x + E = 180 x = = 50 (b) E = 180 x = 10 (int. s) Sum of interior s = (5 ) 180 x + E + y = y = 540 y = 150 (int. s, E // ) Sum of int. s of an n-sided polygon = (n ) 180 nswer (a) x = 50 [1] (b) y = 150 [1] 8. Topic: rithmetic (ompound Interest) Total amount = = $ Interest = $ $5000 = $ ( d.p.) 9. Topics: Trigonometry and Mensuration (a) rea of = 1 ab sin c = 1 (7.4)(7.4) sin 8 = cm ( sig. fig.) Given in formula sheet (compound interest): Total amount = P 1 + r 100 n Total interest = Total amt. Principal amt. nswer $ [] For tuition, exam papers & Last-Minute uddha Foot Hugging Syndrome treatment opyright 009 ϕ exampaper.com.sg. ll rights reserved. / 11

3 (b) Volume of the prism = ase area height 10. Topic: Number Patterns (a) 1st term = 8 (b) T 1 = 8 nd term = 8 7 = 1 rd term = 1 7 = 4 T : 1= 8 7 T : 4= = rea of 0 = = 9.87 T n : 8 7(n 1) = 45 7n n th term = 45 7n 40 cm ( sig. fig.) nswer (a) 17.0 cm [1] (b) 40 cm [1] nswer (a) 1, 4 [1] (b) 45 7n [1] 11. Topic: rithmetic (pplication of Mathematics in Practical Situations) Small tin: 415 g $ g $ = $ /g Large tin: 815 g $1.98 $0.0051/g ( sig. fig.) 1 g $ = $ /g $0.004 /g ( sig. fig.) the large tin gives better value because it costs less per gram. 1. Topic: Kinematics (a) cceleration during the 1 st 40 seconds = 4 nswer The large tin gives better value [] = 0.6 m/s (b) Total distance travelled = Total area under the graph = 1 (60)(4) = 70 m 40 cceleration = Speed Time nswer (a) 0.6 m/s [1] (b) 70 m [1] For tuition, exam papers & Last-Minute uddha Foot Hugging Syndrome treatment opyright 009 ϕ exampaper.com.sg. ll rights reserved. / 11

4 1. Topic: Volumes of Similar Figures (a) Let w be the width of the prism. Let v 1 = volume of water when d = 1 and v = volume of water when d = 4 cm v 1 v = (ase area when d=1) w (ase area when d=4) w v 1 = 1 v 4 w w v 1 = 1 4 v since v takes 8 seconds, v 1 takes seconds t = when d = 1 nswer (a) t = [1] (b) 14. Topic: Mensuration (Surface area) Surface area of hemisphere = 1 surface area of sphere (from formula sheet) = 1 (4πr ) = πr Surface area of the toy = πr + πrl = π(.8) + π(.8)(7.) cm ( sig. fig.) 15. Topic: reas & Volumes of Similar Figures v (a) (i) S = R S R L v L = R S R L urved surface area of a cone = πrl (from formula sheet) nswer 11 cm [1] = 640 R L R S 150 [1] (ii) = 4 5 Ratio of the smaller radius to the larger radius = 4 : 5 S L = R s R L = 4 5 = 16 5 Ratio of the surface area to the larger surface area = 16 : 5 For tuition, exam papers & Last-Minute uddha Foot Hugging Syndrome treatment opyright 009 ϕ exampaper.com.sg. ll rights reserved. 4 / 11

5 (b) M S M L = R S R L M S 5 = M S = 5 15 = 1.8 kg Mass of the smaller sphere = 1.8 kg nswer (a)(i) 4 : 5 [1] (ii) 16 : 5 [1] (b) 1.8 kg [1] 16. Topic: Geometry (a) (i) = = 60 is an equilateral Length of = 8 cm (ii) Since = 60, = = 70 = 180 (70 ) = 40 (sum of s in ) nswer(a)(i) = 8 cm [1] (ii) = 40 [1] (b) (i) POT = PQO ( at centre = s at circumference) = ( ) = 64 (ii) OPT = 90 (tan radius) OTP = = 6 (sum of s in ) 17. Topic: lgebra (a) (i) x + kx 15 = 0 (1) Sub x =, () + k 15 = k 15 = 0 k = k = 1 nswer(b)(i) POT = 64 [1] (ii) OTP = 6 [1] For tuition, exam papers & Last-Minute uddha Foot Hugging Syndrome treatment opyright 009 ϕ exampaper.com.sg. ll rights reserved. 5 / 11

6 (ii) Sub k = 1 into (1), x x 15 = 0 (x + 5)(x ) = 0 x + 5 = 0 or x = 0 x = 5 =.5 x = (given) (b) 6p pq 10ap + 5a = p(p q) 5a(p q) = (p q)(p 5a) 18. Topic: Factors and Multiples (a) 150 = 75 = 5 = 5 (b) 150 = 5 48 = 4 HF = = 6 nswer (a)(i) k = 1 [1] (ii) x =.5 [1] (b) (p q)(p 5a) [] (c) LM of 48 and 150 = 4 5 = 100 Least number of chocolate bars he could have bought = 100 = Topics: pproximation & Estimation, Trigonometry (a) (i) = = (ii) (1 d.p.) (b) tan 0 = 80 = 80 tan 0 = m ( sig. fig.) (b) 6 [1] (c) 8 [] nswer (a)(i) [1] (ii) 0.1 [1] 0 80 m nswer (b) 16 m [] nswer (a)150 = 5 [1] For tuition, exam papers & Last-Minute uddha Foot Hugging Syndrome treatment opyright 009 ϕ exampaper.com.sg. ll rights reserved. 6 / 11

7 0. Topic: Statistics (umulative Frequency) Total no. of students who took the exam Total no. of students scoring < 78 (a) = 10 6 [ ] = 10 6 [10 8] = = (b) verage number of per square kilometer living in frica = = 1. people per sq. km 1 million = billion = (a) Median mark = 50 th percentile of 800 students = 54 (b) Interquartile range = Upper quartile Lower quartile = = 1 (c) Number of students who are awarded a grade = = Topic: Standard Form Lower quartile Median mark Upper quartile nswer (a) 54 [1] (b) 1 [] (c) 80 [1] (c) Number of people living in Singapore Number of people living in hina = = = = 415 Ratio of no. of people living in Singapore : no. of people living in hina = 14 : 415 = 1 : nswer (a) [] (b) 1. [1] (c) 1 : [1] For tuition, exam papers & Last-Minute uddha Foot Hugging Syndrome treatment opyright 009 ϕ exampaper.com.sg. ll rights reserved. 7 / 11

8 . Topics: Mensuration and Trigonometry θ must be in radians (a) Length of the major arc = rθ = 10(π.) = 9.8 (b) sin x = cm ( sig. fig.) sin θ = sin(π θ). Topics: ngle Properties of Polygons and Simultaneous Equations (a) 80 y x sin π = 1 and sin π π = sin π = x = π 5π or 6 6 = 0.55 or or.6 ( sig. fig.) Sin +ve π Tan +ve π ll +ve os +ve 0 y x = 180 x = 100 y (1) y x π nswer (a) 9.8 cm [] (b) x = 0.54 or.6 [] y + x = 60 (int. s of quadrilaterals) y + x = 11. () (b) Sub (1) into (): y y = 11 y = 1 Sub y = 1 into (1), x = 100 (1) = 8 x = 8 and y = 1 nswer (a) x = 100 y y + x = 11 [] (b) x = 8 y = 1 [] For tuition, exam papers & Last-Minute uddha Foot Hugging Syndrome treatment opyright 009 ϕ exampaper.com.sg. ll rights reserved. 8 / 11

9 4. Topic: ongruency and Similarity (a) (b) SSS: = since X X in (a) X X nswer (a) In triangles X and X, X = X (given) X = X (given) X = X (vertically opp. s) y SS property, X and X are congruent. [] (b) Triangles and or Triangles and [1] (c) Triangles X and X [1] 5. Topic: oordinate Geometry (a) Gradient of = = 6 Gradient of straight line passing through (x 1, y 1 ) and (x, y ) = y y 1 x x 1 = 1 (c) (b) Equation of : y = 1 x + c Sub (0, 1), y = 1 x + 1 Equation of straight line with gradient m & y-intercept c: y = mx + c X X For tuition, exam papers & Last-Minute uddha Foot Hugging Syndrome treatment opyright 009 ϕ exampaper.com.sg. ll rights reserved. 9 / 11

10 (c) rea of = = 9 units (d) Two possible points of are ( 6, 4), (6, 0) or (6, 6) (ny two) y x 4 Expressing a vector in terms 6 of its position vectors LL = OO OO nswer (a) 1 cm [] (b) y = 1 x + 1 [1] (c) 9 unit [1] (d) ( 6, 4 ) ( 6, 0 ) [] ( 6, 6 ) 6. Topic: Vectors in Two imensions (a) = O + O = 6b + 6a (b) (c) (i) L = 1 = 1 [ 6b + 6a] = b + a = a b (ii) OL = O + L = 6b + a b = a + 4b (iii) LL = LO + OL = a 4b + a = a 4b LL = LL OL OL = (4b a) OL = 1b a + a = 1b OL = 1b = (6b) = O = k, are collinear (straight line) O, and P are collinear and OP is twice of O. O O O a 6a L 6b 6b a 4b M a b M For tuition, exam papers & Last-Minute uddha Foot Hugging Syndrome treatment opyright 009 ϕ exampaper.com.sg. ll rights reserved. 10 / 11

11 (d) N L M Since LMN is a parallelogram, N = LL O ON = 4b a ON = O 4b + a = 6b 4b + a = b + a = a + b NN and LL are equal vectors same direction and magnitude nswer (a)(i) L = a b [] (ii) OL = a + 4b [1] (iii) LL = a 4b [1] (b) OL = 1b [1] (c) (d) O, and P are collinear, OP is twice of O. [] ON = a + b [1] For tuition, exam papers & Last-Minute uddha Foot Hugging Syndrome treatment opyright 009 ϕ exampaper.com.sg. ll rights reserved. 11 / 11

4016/01 October/November 2011

ELEMENTARY MATHEMATICS Paper 1 Suggested Solutions 1. Topic: Arithmetic (Approximation & Estimation) 4.51 4016/01 October/November 2011 19.6.91 2 1.05 ( sig. fig.) Answer 1.05 [2] 2. Topic: Integers 2