Mathematics Secondary IV May May Practice Exam. Competency Two Uses Mathematical Reasoning. Science Option ANSWER KEY

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1 Mathematics Secondary IV May 015 May Practice Exam Competency Two Uses Mathematical Reasoning Science Option ANSWER KEY

2 ANSWER KEY FOR THE EXAMINATION PARTS A AND B Part A Questions 1 to 6 4 marks or 0 marks 1. B 4 0. D D A D C 4 0 Part B Questions 7 to 10 4 marks or 0 marks 7. The distance between the Frisbee and the ground decreased for seconds. 8. The length of line segment EF is 31 cm The diameter of the cylinder is 8 cm The solutions of the inequality are the values belonging to the interval ]0, 9[. 4 0 Note: Accept equivalent answers such as the values in between 0 and 9. Give marks if the student wrote the two critical values (i.e. 0 and 9). Page

3 Mathematics 46 SN Competency Administration and Marking Guide PART C 11. ANGLE EFD EXAMPLE OF AN APPROPRIATE SOLUTION LENGTH OF LINE SEGMENTS CE AND DE AND BD cos (m EFB) = mbf mbe { Cosine ratio in right triangle BFE cos 40 o = 18cm mbe 18cm m BE = o cos 40 = m 3.5 cm m CE = m BE m BC = 3.5 cm 14 cm = 9.5 cm m DE = m CE = 9.5 cm m BD = m BE + mde = 3.5 cm cm = 33 cm LENGTH OF LINE SEGMENT DF (m DF ) = (m BD ) + (m BF ) (m BD )(m BF ) cos (m FBD) { Cosine law applied in triangle BFD (m DF ) = (33 cm) + (18 cm) (33 cm)(18 cm) cos 40 o (m DF ) = 1089 cm + 34 cm cm (m DF ) = m m DF = m MEASURE OF ANGLE DFE m FBE + m BFE + m BEF = 180 o {sum of angles in a triangle 40 o + 90 o + m BEF = 180 o m BEF = 50 o m FED + m BEF = 180 o {the angles are supplementary m FED + 50 o = 180 o m FED = 130 o med = sin(m DFE) mdf sin(m DEF) { Sine law applied in triangle EFD 9.5cm = sin(m DFE).4cm o sin130 sin (m DFE) = m DFE = o CONCLUSION To the nearest degree, the measure of angle DFE is 19 o. Page 3

4 Mathematics 46 SN Competency Administration and Marking Guide 1. TWO PARABOLAS EXAMPLE OF AN APPROPRIATE SOLUTION RULE OF FUNCTION f f ( x) = a( x h) + k f ( x) = a( x 30) Using the vertex (30, 1.5) 0 = a (40 30) Using f(40) = = a (100) 0.15 = a Rule of function f : f ( x) = 0.15( x 30) Y-INTERCEPT OF FUNCTIONS f AND g + f (0) = 0.15(0 30) 1.5 f (0) = 0.15(900) +1.5 f (0) = 100 g (0) = f (0) = 100 RULE OF FUNCTION g Function g passes through (0, 100), (10, 0) and (40, 0). g(x) = a (x 10) (x 40) Using the two zeros 10 and = a (0 10) (0 40) Using g (0) = = a (400) 0.5 = a Rule of function g : g(x) = 0.5 (x 10) (x 40) VERTEX OF FUNCTION g The x-coordinate the vertex of function g: = 5 g (5) = 0.5 (5 10) (5 40) g (5) = 0.5 (15) ( 15) = 56.5 The coordinates of the vertex of function g are (5, 56.5). DISTANCE BETWEEN THE TWO VERTICES d (V f, V g) = ( 30 15) + ( ) = = units CONCLUSION Page 4

5 Mathematics 46 SN Competency Administration and Marking Guide To the nearest tenth of a unit, the distance between the vertices of the parabolas is 46.3 units. Page 5

6 Mathematics 46 SN Competency Administration and Marking Guide 13. A PENTAGON EXAMPLE OF AN APPROPRIATE SOLUTION VALUE OF X m GC mgd Area of triangle CDG = ( 5x + 3 )( 8x 1 ) 40x + 19x 3 = = Area of rectangle AFDE = Area of triangle GCF {Since equivalent plane figures are 40x + 19x 3 6x + 41x + 30 = equal in area. 1x + 8x + 60 = 40x + 19x 3 0 = 8x 63x 63 7(4x 9x 9) = 0 7(x 3) (4x 3) = 0 3 x = 3 or x = {this value is impossible because the length of line segment GC 4 would be less than zero. Thus, x is equal to 3. NUMERICAL VALUES OF THE LENGTH OF LINE SEGMENTS GC AND GF Length of line segment GC = 8(3) 1 = 3 units Area of rectangle AFDE = m AF m FD 6x + 41x + 30 = (x + 6) (6x + 5) m FD = 6x + 5 Length of line segment GF = m FD m GD Length of line segment GF = (6x + 5) (5x + 3) = x + Length of line segment GF = 3 + = 5 units AREA OF RECTANGLE FBCG Area of rectangle FBCG = m GC m GF Area of rectangle FBCG = 3 u 5 u Area of rectangle = 115 u CONCLUSION The numerical value of the area of rectangle FBCG is115 u. Page 6

7 Mathematics 46 SN Competency Administration and Marking Guide 14. THE JEWELLERY PACKAGES EXAMPLE OF AN APPROPRIATE SOLUTION SYSTEM OF EQUATIONS REPRESENTING THE INFORMATION ON THE MASS OF THE JEWELLERY PLACED IN PACKAGES A AND B x: mass of a bracelet, in grams y: mass of a necklace, in grams 0x + 5y = x + 48y = 9600 MASS OF A BRACELET AND MASS OF A NECKLACE 0x + 5y = 4500 X x + 100y = x + 48y = 9600 X 5 (1600x + 100y = ) 640x = 4000 x = (37.5) + 5y = y = y = 3750 y = 150 Mass of a bracelet : 37.5 g Mass of a necklace : 150 g MASS OF PACKAGE C Total mass of package C: g g = g Mass of all three packages together : 4500 g g g = g COST OF SHIPPING Cost of shipping package A: Cost of shipping package B: f (4500) = f (9600) = f (4500) = [ 9] + 0 f (9600) = [ 19.] + 0 f (4500) = f (9600) = 3 ( 0) + 0 f (4500) = $33.50 f (9600) = $50 Cost of shipping package C: Cost of shipping 3 packages together: f (15 600) = f (9 700) = f (15 600) = [ 31.] + 0 f (9 700) = [ 59.4] + 0 f (15 600) = 3 ( 3) f (9 700) = ( 60) + 0 f (15 600) = $68 f (9 700) = $110 Cost of A + B + C separately : $ $50 + $68 = $ > $110 CONCLUSION If the jewellery maker combined all three packages into one package, the shipping cost would be less than the sum of the shipping cost of the three packages separately. Page 7

8 Mathematics 46 SN Competency Administration and Marking Guide 15. THE GARDEN EXAMPLE OF AN APPROPRIATE SOLUTION REGION WHERE VEGETABLES WILL BE PLANTED Equation of the line defining the half-plane: x y Or y x + 30 Test point (0, 0) So, 0 (0) ? False. Let P be the point of intersection of the line defining the half-plane and line segment FG. Let Q be the point of intersection of the line defining the half-plane and line segment GH. Pentagon EFPQH represents the region where the vegetables will be planted. COORDINATES OF POINTS P AND Q Equation of line FG: a = = y = x +10 or 4x + 3y = Coordinates of point P by solving the system of equations: 4x + 3(x + 30) = x + 90 = x = 70 x = 7 Coordinates of point P: (7, 84) y =(7) + 30 = 84 Coordinates of point Q: ( 15, 0) AREA OF THE PORTION OF THE GARDEN WHERE TINA WILL PLANT VEGETABLES Area of pentagon EFPQH = Area of trapezoid Area of triangle PGQ Area = area of trapezoid area of triangle PGQ ( ) 10 m GQ (dis tance _ between _ P _ and _ GQ) Area = ( ) dm 84dm Area = Area = dm 4410 dm = 6990 dm CONCLUSION The area of the portion of the garden where Tina will plant vegetables this year is 6990 dm. Page 8

9 Mathematics 46 SN Competency Administration and Marking Guide 16. LINE SEGMENT DF EXAMPLE OF AN APPROPRIATE SOLUTION EQUATION OF LINE SEGMENT BE Slope of BE slope of AB = 1 3 Slope of BE = Slope of BE = 3 Since line segments BE and AB are perpendicular So, y = 3 4 x + b Using B (360, 384), 384 = 3 4 (360) + b 864 = b The equation of line segment BE is y = 3 4 x 864 LENGTHS OF LINE SEGMENTS BE AND AB Coordinates of point E : 0 = 3 4 x = 3 4 x x = 648 Coordinates of point E : (648, 0) m BE = ( ) + (0 ( 384)) = 480 units mab tan ( m AEB) = mbe {Tangent ratio applied in right triangle ABE mab tan.6 o = 480 m AB = 00 units SIMILARITY OF TRIANGLES ABD AND CDF AB // DF ABC CDF ACB DCF because lines with equal slopes are parallel. because two alternate interior angles formed by two parallel lines (AB and DF) cut by a transversal (BD ) are congruent. because vertically opposite angles are congruent If two angles of one triangle and the two corresponding angles of another triangle are congruent, then the triangles are similar. Therefore, Δ FGH ~ Δ JKG. LENGTH OF LINE SEGMENT DF In similar figures, the lengths of corresponding segments are proportional. m DF mcd m DF 75 = = mdf = 440 units mab mbc CONCLUSION mdf = 440 units Page 9

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14 Appendix C