2000 Solutions Euclid Contest
|
|
- Brianna Snow
- 5 years ago
- Views:
Transcription
1 Canadian Mathematics Competition n activit of The Centre for Education in Mathematics and Computing, Universit of Waterloo, Waterloo, Ontario 000 s Euclid Contest (Grade) for The CENTRE for EDUCTION in MTHEMTICS and COMPUTING wards 000 Waterloo Mathematics Foundation
2 000 Euclid s (a) If + 7 = 5, what is the value of? 5 = 5, 7 = Therefore, = 5 = (b) The line = a+ c is parallel to the line = and passes through the point (, 5) What is the value of c? Since the two given lines are parallel, the line = a+ c has slope and is of the form, = + c Since (, 5) is on the line, 5 = ()+ c c = (c) The parabola with equation = ( ) 6 has its verte at and intersects the -ais at B, as shown Determine the equation for the line passing through and B O B For = 0, 6 0 ( ) = [( ) ]( )+ 4 [ 4]= 0 Therefore = 6 or = Thus, the -intercepts of the parabola are and 6, and B has coordinates 6, 0 The verte of the parabola is at (, 6) Equation of line containing ( 6, 0) and (, 6) has slope 6 = 4 6 Thus the line has equation, 0 = 4 = ( ) (a) Si identical pieces are cut from a board, as shown in the diagram The angle of each cut is The pieces are assembled to form a heagonal picture frame as shown What is the value of?
3 000 Euclid s Each interior angle of a regular heagon is 0 Putting the frame together we would have the following = 0 (in degrees) = (b) If log = + log, what is the value of? 0 0 log 0 log0 = log0 = = 0 = 000 (c) If + = 6, determine all values of + Squaring both sides + = ; squaring = = = = 6 Creating a quadratic equation and solving = = 6 + 6= 0 0 ( )( )=
4 000 Euclid s 4 = or = For =, + = = = 6 97 = 6 For =, 9 = + 4 = (a) circle, with diameter B as shown, intersects the positive -ais at point D( 0, d) Determine d D 0, d, 0 O B 8, 0 The centre of the circle is, 0 Thus d + = 5 d = 5 d = 6 Therefore d = 4, since d > 0 ( ) and the circle has a radius of 5 Since B is a diameter of the circle, DB = 90 and OD = 90 DO ~ DBO Therefore, OD BO = O OD and d 8 = () d 6 = d = 4, since d > 0
5 000 Euclid s 5 DB = OD = BOD = 90 In OD, D = 4 + d In BOD, DB = 64 + d In DB, ( 4 + d )+( 64 + d )= 00 d = d = 4, d > 0 (b) square PQRS with side of length is subdivided into four triangular regions as shown so that area + area B = area C If PT = and RU = 5, determine the value of P T S C U B 5 Q R Since the side length of the square is, TS = and VS = 5 rea of triangle = ()() rea of triangle B= ( )( ) 5 rea of triangle C= ( 5)( ) From the given information, 5 ( )+ ( )= ( 5)( ) + 5 = = 0 ( )( ) = 5 0 Thus = 5 or = Therefore = 5 since = is inadmissible Labelled diagram P T S C 5 U B 5 Q R 4 (a) die, with the numbers,,, 4, 6, and 8 on its si faces, is rolled fter this roll, if an odd number appears on the top face, all odd numbers on the die are doubled If an even number appears on the top face, all the even numbers are halved If the given die changes in this wa, what is the probabilit that a will appear on the second roll of the die?
6 000 Euclid s 6 There are onl two possibilities on the first roll - it can either be even or odd Possibilit The first roll is odd The probabilit of an odd outcome on the first roll is fter doubling all the numbers, the possible outcomes on the second roll would now be,, 6, 4, 6, 8 with the probabilit of a being Thus the probabilit of a on the second roll would be = 9 Possibilit The first is even The probabilit of an even outcome on the first roll is fter halving all the numbers, the possible outcomes on the second roll would be,,,,, 8 The probabilit of a on the second die would now be 6 Thus the probabilit of a on the second roll is 6 = 9 The probabilit of a appear on the top face is = 9 b) The table below gives the final standings for seven of the teams in the English Cricket League in 998 t the end of the ear, each team had plaed 7 matches and had obtained the total number of points shown in the last column Each win W, each draw D, each bonus bowling point, and each bonus batting point B received w, d, a and b points respectivel, where w, d, a and b are positive integers No points are given for a loss Determine the values of w, d, a and b if total points awarded are given b the formula: Points = w W + d D+ a + b B Final Standings W Losses D B Points Susse Warks Som Derbs Kent Worcs Glam
7 000 Euclid s 7 There are a variet of was to find the unknowns The most efficient wa is to choose equations that have like coefficients Here is one wa to solve the problem using this method For Susse: 6w+ 4d+ 0a+ 6b= 0 For Som: 6w+ 4d+ 0a+ 54b= 9 Subtracting, 9b = 9 b = If b = For Derbs: 6w+ 4d+ 8a+ 55 = 9 6w+ 4d+ 8a= 6 () For Susse: 6w+ 4d+ 0a+ 6 = 0 6w+ 4d+ 0a= 8 () Subtracting, () () a = a = We can now calculate d and w b substituting a =, b = into a pair of equations n efficient wa of doing this is b substituting a =, b = into Som and Worcs For Som: 6w+ 4d+ 84 = 9 6w+ 4d = 08 () For Worcs: 6w+ d+ 85 = 00 6w+ d = 05 (4) Subtracting, () (4) d = Substituting d = in either () or (4), 6w + 4()= 08 (substituting in ()) 6w = 96 w = 6 Therefore w = 6, d =, a= b= 5 (a) In the diagram, D = DC, sin DBC = 0 6 and CB = 90 What is the value of tan BC? D B C Let DB = 0 Therefore, DC = D =6 B the theorem of Pthagoras, BC = 0 6 = 64 Therefore, BC = 8
8 000 Euclid s 8 Thus, tan BC = = 8 (b) On a cross-sectional diagram of the Earth, the and -aes are placed so that O( 00, ) is the centre of the Earth and C( 640, 000 ) is the location of Cape Canaveral space shuttle is forced to land on an island at ( 549, ), as shown Each unit represents 000 km Determine the distance from Cape Canaveral to the island, measured on the surface of the earth, to the nearest 0 km O 0, 0 54, 9 C 640, 000 Calculating OC Calculating arc length Distance tan OC = 9 54 OC = tan 9 = The arc length C = [ ( π )( 6 40 ) ]= 57 units 60 The distance is approimatel 570 km 6 (a) Let represent the greatest integer which is less than or equal to For eample, =, 6 = If is positive and = 7, what is the value of? We deduce that 4 < < 5 Otherwise, if 4, 6, and if 5, 5 Therefore = 4 Since = 7 4 = 7 = 45
9 000 Euclid s 9 (b) The parabola = + 4 has verte P and intersects the -ais at and B The parabola is translated from its original position so that its verte moves along the line = +4 to the point Q In this position, the parabola intersects the -ais at B and C Determine the coordinates of C P Q 4 O B C The parabola = + 4 has verte P( 04, ) and intersects the -ais at ( 0, ) and B( 0, ) The intercept B( 0, ) has its pre-image, B on the parabola = + 4 To find B, we find the point of intersection of the line passing through B( 0, ), with slope, and the parabola = + 4 The equation of the line is = Intersection points, = = 0 ( + )( )= 0 Therefore, = or = For =, = = 5 Thus B has coordinates (, 5) If (, 5) (, 0) then the required general translation mapping = + 4 onto the parabola with verte Q is (, ) ( + 5, + 5 ) Possibilit Using the general translation, we find the coordinates of Q to be, P( 04, ) Q( 0+ 54, + 5)= Q( 59, ) If C is the reflection of B in the ais of smmetr of the parabola, ie = 5, C has coordinates ( 80, ) Possibilit If B has coordinates (, 5) then C is the reflection of B in the -ais Thus C has coordinates (, 5) If we appl the general translation then C has coordinates ( + 5, 5 + 5) or ( 8, 0) Thus C has coordinates ( 8, 0) Possibilit Using the general translation, we find the coordinates of Q to be, P( 04, ) Q( 0+ 54, + 5)= Q( 59, ) The equation of the image parabola is = ( 5) + 9
10 000 Euclid s 0 ( ) + = To find its intercepts, ( ) = 5 9 5=± Therefore = 8 or = Thus C has coordinates ( 8, 0) The translation moving the parabola with equation = + 4 onto the parabola with verte Q is Ttt (, ) because the slope of the line = +4 is The pre-image of B is ( t, t) Since B is on the parabola with verte P, we have t = ( t) + 4 t = 4+ 4t t + 4 t 5t = 0 tt ( 5)= 0 Therefore, t = 0 or t = 5 Thus B is (, 5) Let C have coordinates ( c,0) The pre-image of C is ( c 5, 5) Therefore, 5= ( c 5) + 4 ( ) = Or, c 5 9 Therefore c 5= or c 5= c = 8 or c = Thus C has coordinates ( 8, 0) The translation moving the parabola with equation = + 4 onto the parabola with verte Q is T( p, p) because the slope of the line = +4 is Q will have coordinates p, p+ 4 ( ) Thus the equation of the image parabola is p p Since, 0 ( ) is on the parabola, 0 p p 4 p 5p= 0 pp ( 5)= 0 Therefore p = 0 or p = 5 The coordinates of Q are ( 5, 9) = ( ) + + = ( ) + + s in solution, we can use either reflection properties or the equation of the parabola to find that C has coordinates ( 8, 0) 4
11 000 Euclid s 7 (a) cube has edges of length n, where n is an integer Three faces, meeting at a corner, are painted red The cube is then cut into n smaller cubes of unit length If eactl 5 of these cubes have no faces painted red, determine the value of n If we remove the cubes which have red paint, we are left with a smaller cube with measurements, ( n ) ( n ) ( n ) Thus, ( n ) = 5 n = 6 (b) In the isosceles trapezoid BCD, B = CD = The area of the trapezoid is 80 and the circle with centre O and radius 4 is tangent to the four sides of the trapezoid Determine the value of B O C D Using the tangent properties of a circle, the lengths of line segments are as shown on the diagram rea of trapezoid BCD = ()( BC + D) 8 = 4 ( b+ b) = 8 Thus, 8 = 80 Therefore, = 0 Bb b C b b 4 b O 4 b b b D 8 In parallelogram BCD, B = a and BC = b, where a> b The points of intersection of the angle bisectors are the vertices of quadrilateral PQRS (a) Prove that PQRS is a rectangle (b) Prove that PR = a b D E P S M N Q R F C B (a) In a parallelogram opposite angles are equal Since DF and BE bisect the two angles, let DF = CDF = BE = CBE = (in degrees) lso CDF = FD = (alternate angles) Let DM = BM = DCN = BCN = (in degrees)
12 000 Euclid s For an parallelogram, an two consecutive angles add to 80, + = 80 or, + =90 D E P S M N Q R F C B Therefore in PF, PF = 90 F P Using similar reasoning and properties of parallel lines we get right angles at Q, R and S Thus PQRS is a rectangle (b) Since M is a bisector of DB, let DM = BM = lso, DM = (alternate angles) This implies that DM is isosceles Using the same reasoning in CBN, we see that it is also isosceles and so the diagram ma now be labelled as: D N F a b b Q P R S b E M C b a B N = a b Thus DM and CBN are identical isosceles triangles lso, M NC (corresponding angles) or, P NR B using properties of isosceles triangles (or congruenc), P = NR impling that PRN is a parallelogram Thus N = PR and since N = a b, PR = a b (as required)
13 000 Euclid s 9 permutation of the integers,,, n is a listing of these integers in some order For eample, (,, ) and (,, ) are two different permutations of the integers,, permutation ( a, a,, a n ) of the integers,,, n is said to be fantastic if a + a + + a k is divisible b k, for each k from to n For eample, (,, ) is a fantastic permutation of,, because is divisible b, + is divisible b, and + + is divisible b However, (,, ) is not fantastic because + is not divisible b (a) Show that no fantastic permutation eists for n = 000 (b) Does a fantastic permutation eist for n = 00? Eplain (a) (b) In our consideration of whether there is a fantastic permutation for n = 000, we start b looking at the 000th position Using our definition of fantastic permutation, it is necessar that 000 ( L + 000) Since L = ( )( ) = ( 000)( 00), it is required that ( 00) This is not possible and so no fantastic permutation eists for n = 000 ( 00)( 00) The sum of the integers from to 00 is = ( 00 )( 00 ) which is divisible b 00 If t, t,, t 00 is a fantastic permutation, when we remove t 00 from the above sum, and what remains must be divisible b 000 We now consider t + t + L + t000 and determine what integer is not included in the permutation t + t + L + t000 = ( )( ) t t = ( )( ) 00 ( )( )=, t 00 must be a number of the form k00 where k is odd Since The onl integer less than or equal to 00 with this propert is 00 Therefore t 00 = 00 So the sum up to t 000 is = When we remove t 000 we must get a multiple of 999 The largest multiple of 999 less than is ( 999)( 00)= This would make t 000 = = 00 which is impossible since t000 t00 If we choose lesser multiples of 999 to subtract from we will get values of t 000 which are greater than 00, which is also not possible Thus, a fantastic permutation is not possible for n = 00
14 000 Euclid s 4 0 n equilateral triangle BC has side length square, PQRS, is such that P lies on B, Q lies on BC, and R and S lie on C as shown The points P, Q, R, and S move so that P, Q and R alwas remain on the sides of the triangle and S moves from C to B through the interior of the triangle If the points P, Q, R and S alwas form the vertices of a square, show that the path traced out b S is a straight line parallel to BC S P R B Q C In essence, this solution establishes that the perpendicular distance from S to BC is s( sin θ+ cos θ) and then showing that this is a constant b finding s sin θ+ cos θ the base which is itself a constant length Let RQC = θ and from S draw a line perpendicular to the base at P Then TQB = 80 ( 90 + θ)= 90 θ Let s be the length of the side of the square From R draw a line perpendicular to BC at D and then through S draw a line parallel to BC From R draw a line perpendicular to this line at E T S ( ) as part of E R 0 B F P Q D C 90 From RQD, RD = s sin θ Since QRD = 90 θ then SRE = θ From SER, ER = s cos θ The perpendicular distance from S to BC is RD + ER = ssin θ+ s cos θ which we must now show is a constant We can now take each of the lengths DC, DQ, PF, FB and epress them in terms of s From RDC which is a triangle, DC RD = Since RD = s sin θ (from above) DC = ( s sin θ)= ssin θ 60
15 000 Euclid s 5 From RDQ, QD RQ = cos θ, QD = s cos θ From TFQ, sin θ= FQ s and cos θ= TF s or, FQ From TFB, BF TF = s sin θ and TF = s sin θ =, BF = TF = s cos θ= s cos θ Since DC + QD + FQ + BF =, ssin θ+ scos θ+ ssin θ+ scos θ= ( scos θ+ ssin θ)+ ( scos θ+ ssin θ)= scos θ+ ssin θ= + Thus scos θ+ ssin θ is a constant and the path traced out b S is a straight line parallel to BC Note: number of enquiries have been made about this question Several individuals have made the comment that it is not possible to do this under the given conditions What is not mentioned, and what is not realized, is that the size of the square changes This makes it possible for the square to eist under the given conditions
2002 Solutions Euclid Contest(Grade 12)
Canadian Mathematics Competition n activity of The Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 00 Solutions Euclid Contest(Grade ) for The CENTRE for EDUCTION
More information2008 Euclid Contest. Solutions. Canadian Mathematics Competition. Tuesday, April 15, c 2008 Centre for Education in Mathematics and Computing
Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 008 Euclid Contest Tuesday, April 5, 008 Solutions c 008
More information2007 Hypatia Contest
Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 007 Hypatia Contest Wednesday, April 18, 007 Solutions c
More informationQ.2 A, B and C are points in the xy plane such that A(1, 2) ; B (5, 6) and AC = 3BC. Then. (C) 1 1 or
STRAIGHT LINE [STRAIGHT OBJECTIVE TYPE] Q. A variable rectangle PQRS has its sides parallel to fied directions. Q and S lie respectivel on the lines = a, = a and P lies on the ais. Then the locus of R
More information1997 Solutions Cayley Contest(Grade 10)
Canadian Mathematics Competition n activity of The Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 199 Solutions Cayley Contest(Grade 10) for the wards 199
More informationPREPARED BY: ER. VINEET LOOMBA (B.TECH. IIT ROORKEE) 60 Best JEE Main and Advanced Level Problems (IIT-JEE). Prepared by IITians.
www. Class XI TARGET : JEE Main/Adv PREPARED BY: ER. VINEET LOOMBA (B.TECH. IIT ROORKEE) ALP ADVANCED LEVEL PROBLEMS Straight Lines 60 Best JEE Main and Advanced Level Problems (IIT-JEE). Prepared b IITians.
More informationChapter 1 Coordinates, points and lines
Cambridge Universit Press 978--36-6000-7 Cambridge International AS and A Level Mathematics: Pure Mathematics Coursebook Hugh Neill, Douglas Quadling, Julian Gilbe Ecerpt Chapter Coordinates, points and
More information2001 Solutions Euclid Contest (Grade 12)
Canadian Mathematics Competition An activity of The Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 001 s Euclid Contest (Grade 1) for The CENTRE for EDUCATION
More information2007 Fermat Contest (Grade 11)
Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 007 Fermat Contest (Grade 11) Tuesday, February 0, 007 Solutions
More informationThe Coordinate Plane. Circles and Polygons on the Coordinate Plane. LESSON 13.1 Skills Practice. Problem Set
LESSON.1 Skills Practice Name Date The Coordinate Plane Circles and Polgons on the Coordinate Plane Problem Set Use the given information to show that each statement is true. Justif our answers b using
More informationInternational Examinations. Advanced Level Mathematics Pure Mathematics 1 Hugh Neill and Douglas Quadling
International Eaminations Advanced Level Mathematics Pure Mathematics Hugh Neill and Douglas Quadling PUBLISHED BY THE PRESS SYNDICATE OF THE UNIVERSITY OF CAMBRIDGE The Pitt Building, Trumpington Street,
More information1999 Solutions Fermat Contest (Grade 11)
Canadian Mathematics Competition n activity of The Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 999 s Fermat Contest (Grade ) for the wards 999 Waterloo
More informationChapter 3 Summary 3.1. Determining the Perimeter and Area of Rectangles and Squares on the Coordinate Plane. Example
Chapter Summar Ke Terms bases of a trapezoid (.) legs of a trapezoid (.) composite figure (.5).1 Determining the Perimeter and Area of Rectangles and Squares on the Coordinate Plane The perimeter or area
More information2009 Euclid Contest. Solutions
Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 009 Euclid Contest Tuesday, April 7, 009 Solutions c 009
More information2005 Euclid Contest. Solutions
Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 2005 Euclid Contest Tuesday, April 19, 2005 Solutions c
More informationExt CSA Trials
GPCC039 C:\M_Bank\Tests\Yr-U\3U 005-000 csa trials /09/0 Et 005 000 CS Trials )! Yearl\Yr-3U\cat-3u.05 Qn) 3U05-a sin Find the value of lim. 0 5 )! Yearl\Yr-3U\cat-3u.05 Qn) 3U05-b The polnomial P() is
More information0113ge. Geometry Regents Exam In the diagram below, under which transformation is A B C the image of ABC?
0113ge 1 If MNP VWX and PM is the shortest side of MNP, what is the shortest side of VWX? 1) XV ) WX 3) VW 4) NP 4 In the diagram below, under which transformation is A B C the image of ABC? In circle
More informationName: GEOMETRY: EXAM (A) A B C D E F G H D E. 1. How many non collinear points determine a plane?
GMTRY: XM () Name: 1. How many non collinear points determine a plane? ) none ) one ) two ) three 2. How many edges does a heagonal prism have? ) 6 ) 12 ) 18 ) 2. Name the intersection of planes Q and
More information2010 Euclid Contest. Solutions
Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 00 Euclid Contest Wednesday, April 7, 00 Solutions 00 Centre
More informationModule 3, Section 4 Analytic Geometry II
Principles of Mathematics 11 Section, Introduction 01 Introduction, Section Analtic Geometr II As the lesson titles show, this section etends what ou have learned about Analtic Geometr to several related
More informationMATH TOURNAMENT 2012 PROBLEMS SOLUTIONS
MATH TOURNAMENT 0 PROBLEMS SOLUTIONS. Consider the eperiment of throwing two 6 sided fair dice, where, the faces are numbered from to 6. What is the probability of the event that the sum of the values
More informationUNC Charlotte 2005 Comprehensive March 7, 2005
March 7, 2005 1. The numbers x and y satisfy 2 x = 15 and 15 y = 32. What is the value xy? (A) 3 (B) 4 (C) 5 (D) 6 (E) none of A, B, C or D Solution: C. Note that (2 x ) y = 15 y = 32 so 2 xy = 2 5 and
More informationMath 030 Review for Final Exam Revised Fall 2010 RH/ DM 1
Math 00 Review for Final Eam Revised Fall 010 RH/ DM 1 1. Solve the equations: (-1) (7) (-) (-1) () 1 1 1 1 f. 1 g. h. 1 11 i. 9. Solve the following equations for the given variable: 1 Solve for. D ab
More information0811ge. Geometry Regents Exam BC, AT = 5, TB = 7, and AV = 10.
0811ge 1 The statement "x is a multiple of 3, and x is an even integer" is true when x is equal to 1) 9 2) 8 3) 3 4) 6 2 In the diagram below, ABC XYZ. 4 Pentagon PQRST has PQ parallel to TS. After a translation
More informationANALYTICAL GEOMETRY Revision of Grade 10 Analytical Geometry
ANALYTICAL GEOMETRY Revision of Grade 10 Analtical Geometr Let s quickl have a look at the analtical geometr ou learnt in Grade 10. 8 LESSON Midpoint formula (_ + 1 ;_ + 1 The midpoint formula is used
More information1 k. cos tan? Higher Maths Non Calculator Practice Practice Paper A. 1. A sequence is defined by the recurrence relation u 2u 1, u 3.
Higher Maths Non Calculator Practice Practice Paper A. A sequence is defined b the recurrence relation u u, u. n n What is the value of u?. The line with equation k 9 is parallel to the line with gradient
More informationTopic 2 [312 marks] The rectangle ABCD is inscribed in a circle. Sides [AD] and [AB] have lengths
Topic 2 [312 marks] 1 The rectangle ABCD is inscribed in a circle Sides [AD] and [AB] have lengths [12 marks] 3 cm and (\9\) cm respectively E is a point on side [AB] such that AE is 3 cm Side [DE] is
More informationUNCORRECTED. To recognise the rules of a number of common algebraic relations: y = x 1 y 2 = x
5A galler of graphs Objectives To recognise the rules of a number of common algebraic relations: = = = (rectangular hperbola) + = (circle). To be able to sketch the graphs of these relations. To be able
More informationIdentifying second degree equations
Chapter 7 Identifing second degree equations 71 The eigenvalue method In this section we appl eigenvalue methods to determine the geometrical nature of the second degree equation a 2 + 2h + b 2 + 2g +
More information0811ge. Geometry Regents Exam
0811ge 1 The statement "x is a multiple of 3, and x is an even integer" is true when x is equal to 1) 9 ) 8 3) 3 4) 6 In the diagram below, ABC XYZ. 4 Pentagon PQRST has PQ parallel to TS. After a translation
More informationSample Problems For Grade 9 Mathematics. Grade. 1. If x 3
Sample roblems For 9 Mathematics DIRECTIONS: This section provides sample mathematics problems for the 9 test forms. These problems are based on material included in the New York Cit curriculum for 8.
More information2015 Canadian Team Mathematics Contest
The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 205 Canadian Team Mathematics Contest April 205 Solutions 205 University of Waterloo 205 CTMC Solutions Page 2 Individual Problems.
More informationSample Question Paper Mathematics First Term (SA - I) Class IX. Time: 3 to 3 ½ hours
Sample Question Paper Mathematics First Term (SA - I) Class IX Time: 3 to 3 ½ hours M.M.:90 General Instructions (i) All questions are compulsory. (ii) The question paper consists of 34 questions divided
More informationSpecial Mathematics Notes
Special Mathematics Notes Tetbook: Classroom Mathematics Stds 9 & 10 CHAPTER 6 Trigonometr Trigonometr is a stud of measurements of sides of triangles as related to the angles, and the application of this
More information2013 Grade 9 Mathematics Set A
2013 Grade 9 Mathematics Set A Copyright National Institute for Educational Policy Reserch All Right Reserved URL: https://www.nier.go.jp/english/index.html The English translation is prepared by the Project
More information2010 Fermat Contest (Grade 11)
Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 010 Fermat Contest (Grade 11) Thursday, February 5, 010
More informationnx + 1 = (n + 1)x 13(n + 1) and nx = (n + 1)x + 27(n + 1).
1. (Answer: 630) 001 AIME SOLUTIONS Let a represent the tens digit and b the units digit of an integer with the required property. Then 10a + b must be divisible by both a and b. It follows that b must
More informationMT - w A.P. SET CODE MT - w - MATHEMATICS (71) GEOMETRY- SET - A (E) Time : 2 Hours Preliminary Model Answer Paper Max.
.P. SET CODE.. Solve NY FIVE of the following : (i) ( BE) ( BD) ( BE) ( BD) BE D 6 9 MT - w 07 00 - MT - w - MTHEMTICS (7) GEOMETRY- (E) Time : Hours Preliminary Model nswer Paper Max. Marks : 40 [Triangles
More informationApplications. 12 The Shapes of Algebra. 1. a. Write an equation that relates the coordinates x and y for points on the circle.
Applications 1. a. Write an equation that relates the coordinates and for points on the circle. 1 8 (, ) 1 8 O 8 1 8 1 (13, 0) b. Find the missing coordinates for each of these points on the circle. If
More informationnumber. However, unlike , three of the digits of N are 3, 4 and 5, and N is a multiple of 6.
C1. The positive integer N has six digits in increasing order. For example, 124 689 is such a number. However, unlike 124 689, three of the digits of N are 3, 4 and 5, and N is a multiple of 6. How many
More informationchapter 1 vector geometry solutions V Consider the parallelogram shown alongside. Which of the following statements are true?
chapter vector geometry solutions V. Exercise A. For the shape shown, find a single vector which is equal to a)!!! " AB + BC AC b)! AD!!! " + DB AB c)! AC + CD AD d)! BC + CD!!! " + DA BA e) CD!!! " "
More informationVectors Higher Mathematics Supplementary Resources. Section A
Education Resources Vectors Higher Mathematics Supplementar Resources Section This section is designed to provide eamples which develop routine skills necessar for completion of this section. R I have
More informationZETA MATHS. Higher Mathematics Revision Checklist
ZETA MATHS Higher Mathematics Revision Checklist Contents: Epressions & Functions Page Logarithmic & Eponential Functions Addition Formulae. 3 Wave Function.... 4 Graphs of Functions. 5 Sets of Functions
More information2009 Math Olympics Level I
Saginaw Valle State Universit 009 Math Olmpics Level I. A man and his wife take a trip that usuall takes three hours if the drive at an average speed of 60 mi/h. After an hour and a half of driving at
More information1 What is the solution of the system of equations graphed below? y = 2x + 1
1 What is the solution of the system of equations graphed below? y = 2x + 1 3 As shown in the diagram below, when hexagon ABCDEF is reflected over line m, the image is hexagon A'B'C'D'E'F'. y = x 2 + 2x
More informationMathematics. Mathematics 2. hsn.uk.net. Higher HSN22000
Higher Mathematics UNIT Mathematics HSN000 This document was produced speciall for the HSN.uk.net website, and we require that an copies or derivative works attribute the work to Higher Still Notes. For
More informationThe University of the State of New York REGENTS HIGH SCHOOL EXAMINATION COURSE I. Tuesday, June 20, :15 to 4:15 p.m., only
The Universit of the State of New York REGENTS HIGH SCHOOL EXAMINATION THREE-YEAR SEQUENCE FOR HIGH SCHOOL MATHEMATICS COURSE I Tuesda, June 0, 000 :5 to :5 p.m., onl Notice... Scientific calculators must
More informationSMT 2011 General Test and Solutions February 19, F (x) + F = 1 + x. 2 = 3. Now take x = 2 2 F ( 1) = F ( 1) = 3 2 F (2)
SMT 0 General Test and Solutions February 9, 0 Let F () be a real-valued function defined for all real 0, such that ( ) F () + F = + Find F () Answer: Setting =, we find that F () + F ( ) = Now take =,
More informationAnswer Explanations. The SAT Subject Tests. Mathematics Level 1 & 2 TO PRACTICE QUESTIONS FROM THE SAT SUBJECT TESTS STUDENT GUIDE
The SAT Subject Tests Answer Eplanations TO PRACTICE QUESTIONS FROM THE SAT SUBJECT TESTS STUDENT GUIDE Mathematics Level & Visit sat.org/stpractice to get more practice and stud tips for the Subject Test
More informationJUST IN TIME MATERIAL GRADE 11 KZN DEPARTMENT OF EDUCATION CURRICULUM GRADES DIRECTORATE TERM
JUST IN TIME MATERIAL GRADE 11 KZN DEPARTMENT OF EDUCATION CURRICULUM GRADES 10 1 DIRECTORATE TERM 1 017 This document has been compiled by the FET Mathematics Subject Advisors together with Lead Teachers.
More information2017 Canadian Team Mathematics Contest
The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 017 Canadian Team Mathematics Contest April 017 Solutions 017 University of Waterloo 017 CTMC Solutions Page Individual Problems
More informationRegd. Office : Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi Ph.: Fax :
Regd. Office : akash Tower, Plot No.-, Sec-, MLU, Dwarka, New Delhi-007 Ph.: 0-766 Fax : 0-767 dmission-cum-scholarship Test (Sample Paper) First Step Course for JEE (Main & dvanced) 0-07 (Syllabus of
More informationSTUDY KNOWHOW PROGRAM STUDY AND LEARNING CENTRE. Functions & Graphs
STUDY KNOWHOW PROGRAM STUDY AND LEARNING CENTRE Functions & Graphs Contents Functions and Relations... 1 Interval Notation... 3 Graphs: Linear Functions... 5 Lines and Gradients... 7 Graphs: Quadratic
More informationMATHEMATICS HSC Course Assessment Task 3 (Trial Examination) June 21, QUESTION Total MARKS
MATHEMATICS 0 HSC Course Assessment Task (Trial Eamination) June, 0 General instructions Working time hours. (plus 5 minutes reading time) Write using blue or black pen. Where diagrams are to be sketched,
More informationBRITISH COLUMBIA COLLEGES High School Mathematics Contest 2004 Solutions
BRITISH COLUMBI COLLEGES High School Mathematics Contest 004 Solutions Junior Preliminary 1. Let U, H, and F denote, respectively, the set of 004 students, the subset of those wearing Hip jeans, and the
More information5 Find an equation of the circle in which AB is a diameter in each case. a A (1, 2) B (3, 2) b A ( 7, 2) B (1, 8) c A (1, 1) B (4, 0)
C2 CRDINATE GEMETRY Worksheet A 1 Write down an equation of the circle with the given centre and radius in each case. a centre (0, 0) radius 5 b centre (1, 3) radius 2 c centre (4, 6) radius 1 1 d centre
More informationPractice Test Geometry 1. Which of the following points is the greatest distance from the y-axis? A. (1,10) B. (2,7) C. (3,5) D. (4,3) E.
April 9, 01 Standards: MM1Ga, MM1G1b Practice Test Geometry 1. Which of the following points is the greatest distance from the y-axis? (1,10) B. (,7) C. (,) (,) (,1). Points P, Q, R, and S lie on a line
More information0809ge. Geometry Regents Exam Based on the diagram below, which statement is true?
0809ge 1 Based on the diagram below, which statement is true? 3 In the diagram of ABC below, AB AC. The measure of B is 40. 1) a b ) a c 3) b c 4) d e What is the measure of A? 1) 40 ) 50 3) 70 4) 100
More informationHIGHER SCHOOL CERTIFICATE EXAMINATION MATHEMATICS 4 UNIT (ADDITIONAL) Time allowed Three hours (Plus 5 minutes reading time)
N E W S O U T H W A L E S HIGHER SCHOOL CERTIFICATE EXAMINATION 996 MATHEMATICS 4 UNIT (ADDITIONAL) Time allowed Three hours (Plus 5 minutes reading time) DIRECTIONS TO CANDIDATES Attempt ALL questions.
More informationSOLUTIONS SECTION A [1] = 27(27 15)(27 25)(27 14) = 27(12)(2)(13) = cm. = s(s a)(s b)(s c)
1. (A) 1 1 1 11 1 + 6 6 5 30 5 5 5 5 6 = 6 6 SOLUTIONS SECTION A. (B) Let the angles be x and 3x respectively x+3x = 180 o (sum of angles on same side of transversal is 180 o ) x=36 0 So, larger angle=3x
More information0609ge. Geometry Regents Exam AB DE, A D, and B E.
0609ge 1 Juliann plans on drawing ABC, where the measure of A can range from 50 to 60 and the measure of B can range from 90 to 100. Given these conditions, what is the correct range of measures possible
More informationNumber Plane Graphs and Coordinate Geometry
Numer Plane Graphs and Coordinate Geometr Now this is m kind of paraola! Chapter Contents :0 The paraola PS, PS, PS Investigation: The graphs of paraolas :0 Paraolas of the form = a + + c PS Fun Spot:
More informationLinear Equation Theory - 2
Algebra Module A46 Linear Equation Theor - Copright This publication The Northern Alberta Institute of Technolog 00. All Rights Reserved. LAST REVISED June., 009 Linear Equation Theor - Statement of Prerequisite
More information0610ge. Geometry Regents Exam The diagram below shows a right pentagonal prism.
0610ge 1 In the diagram below of circle O, chord AB chord CD, and chord CD chord EF. 3 The diagram below shows a right pentagonal prism. Which statement must be true? 1) CE DF 2) AC DF 3) AC CE 4) EF CD
More informationThe CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca Cayley Contest. (Grade 10) Tuesday, February 27, 2018
The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 018 Cayley Contest (Grade 10) Tuesday, February 7, 018 (in North America and South America) Wednesday, February 8, 018 (outside of
More informationADDITIONAL MATHEMATICS
00-CE MTH HONG KONG EXMINTIONS UTHORITY HONG KONG CERTIFICTE OF EDUCTION EXMINTION 00 DDITIONL MTHEMTICS 8.0 am.00 am (½ hours) This paper must be answered in English. nswer LL questions in Section and
More informationThe region enclosed by the curve of f and the x-axis is rotated 360 about the x-axis. Find the volume of the solid formed.
Section A ln. Let g() =, for > 0. ln Use the quotient rule to show that g ( ). 3 (b) The graph of g has a maimum point at A. Find the -coordinate of A. (Total 7 marks) 6. Let h() =. Find h (0). cos 3.
More informationCircle. Paper 1 Section A. Each correct answer in this section is worth two marks. 5. A circle has equation. 4. The point P( 2, 4) lies on the circle
PSf Circle Paper 1 Section A Each correct answer in this section is worth two marks. 1. A circle has equation ( 3) 2 + ( + 4) 2 = 20. Find the gradient of the tangent to the circle at the point (1, 0).
More informationThe CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca Euclid Contest. Thursday, April 6, 2017
The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 017 Euclid Contest Thursday, April 6, 017 (in North America and South America) Friday, April 7, 017 (outside of North America and
More informationExercises for Unit V (Introduction to non Euclidean geometry)
Exercises for Unit V (Introduction to non Euclidean geometry) V.1 : Facts from spherical geometry Ryan : pp. 84 123 [ Note : Hints for the first two exercises are given in math133f07update08.pdf. ] 1.
More informationMathematics. Mathematics 2. hsn.uk.net. Higher HSN22000
hsn.uk.net Higher Mathematics UNIT Mathematics HSN000 This document was produced speciall for the HSN.uk.net website, and we require that an copies or derivative works attribute the work to Higher Still
More informationChapter (Circle) * Circle - circle is locus of such points which are at equidistant from a fixed point in
Chapter - 10 (Circle) Key Concept * Circle - circle is locus of such points which are at equidistant from a fixed point in a plane. * Concentric circle - Circle having same centre called concentric circle.
More information1 is equal to. 1 (B) a. (C) a (B) (D) 4. (C) P lies inside both C & E (D) P lies inside C but outside E. (B) 1 (D) 1
Single Correct Q. Two mutuall perpendicular tangents of the parabola = a meet the ais in P and P. If S is the focus of the parabola then l a (SP ) is equal to (SP ) l (B) a (C) a Q. ABCD and EFGC are squares
More informationMODEL QUESTION PAPERS WITH ANSWERS SET 1
MTHEMTICS MODEL QUESTION PPERS WITH NSWERS SET 1 Finish Line & Beyond CLSS X Time llowed: 3 Hrs Max. Marks : 80 General Instructions: (1) ll questions are compulsory. (2) The question paper consists of
More informationPOINT. Preface. The concept of Point is very important for the study of coordinate
POINT Preface The concept of Point is ver important for the stud of coordinate geometr. This chapter deals with various forms of representing a Point and several associated properties. The concept of coordinates
More information5.2 Solving Quadratic Equations by Factoring
Name. Solving Quadratic Equations b Factoring MATHPOWER TM, Ontario Edition, pp. 78 8 To solve a quadratic equation b factoring, a) write the equation in the form a + b + c = b) factor a + b + c c) use
More informationThe CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca Euclid Contest. Wednesday, April 15, 2015
The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 015 Euclid Contest Wednesday, April 15, 015 (in North America and South America) Thursday, April 16, 015 (outside of North America
More informationAnswers Investigation 3
Answers Investigation Applications. a., b. s = n c. The numbers seem to be increasing b a greater amount each time. The square number increases b consecutive odd integers:,, 7,, c X X=. a.,,, b., X 7 X=
More information[Class-X] MATHEMATICS SESSION:
[Class-X] MTHEMTICS SESSION:017-18 Time allowed: 3 hrs. Maximum Marks : 80 General Instructions : (i) ll questions are compulsory. (ii) This question paper consists of 30 questions divided into four sections,
More information21. Prove that If one side of the cyclic quadrilateral is produced then the exterior angle is equal to the interior opposite angle.
21. Prove that If one side of the cyclic quadrilateral is produced then the exterior angle is equal to the interior opposite angle. 22. Prove that If two sides of a cyclic quadrilateral are parallel, then
More information0612ge. Geometry Regents Exam
0612ge 1 Triangle ABC is graphed on the set of axes below. 3 As shown in the diagram below, EF intersects planes P, Q, and R. Which transformation produces an image that is similar to, but not congruent
More informationMathematics. Mathematics 1. hsn.uk.net. Higher HSN21000
Higher Mathematics UNIT Mathematics HSN000 This document was produced speciall for the HSN.uk.net website, and we require that an copies or derivative works attribute the work to Higher Still Notes. For
More informatione x for x 0. Find the coordinates of the point of inflexion and justify that it is a point of inflexion. (Total 7 marks)
Chapter 0 Application of differential calculus 014 GDC required 1. Consider the curve with equation f () = e for 0. Find the coordinates of the point of infleion and justify that it is a point of infleion.
More information9. AD = 7; By the Parallelogram Opposite Sides Theorem (Thm. 7.3), AD = BC. 10. AE = 7; By the Parallelogram Diagonals Theorem (Thm. 7.6), AE = EC.
3. Sample answer: Solve 5x = 3x + 1; opposite sides of a parallelogram are congruent; es; You could start b setting the two parts of either diagonal equal to each other b the Parallelogram Diagonals Theorem
More informationTime : 2 Hours Preliminary Model Answer Paper Max. Marks : 40. [Given] [Taking square roots]
.P. SET CODE MT - w 05 00 - MT - w - MTHEMTICS (7) GEOMETRY - (E) Time : Hours Preliminary Model nswer Paper Max. Marks : 40.. ttempt NY FIVE of the following : (i) BC ~ PQ [Given] ( BC) ( PQ) BC PQ [reas
More informationDiagnostic Assessment Number and Quantitative Reasoning
Number and Quantitative Reasoning Select the best answer.. Which list contains the first four multiples of 3? A 3, 30, 300, 3000 B 3, 6, 9, 22 C 3, 4, 5, 6 D 3, 26, 39, 52 2. Which pair of numbers has
More informationHanoi Open Mathematical Competition 2017
Hanoi Open Mathematical Competition 2017 Junior Section Saturday, 4 March 2017 08h30-11h30 Important: Answer to all 15 questions. Write your answers on the answer sheets provided. For the multiple choice
More information46th ANNUAL MASSACHUSETTS MATHEMATICS OLYMPIAD. A High School Competition Conducted by. And Sponsored by FIRST-LEVEL EXAMINATION SOLUTIONS
46th ANNUAL MASSACHUSETTS MATHEMATICS OLYMPIAD 009 00 A High School Competition Conducted by THE MASSACHUSETTS ASSOCIATION OF MATHEMATICS LEAGUES (MAML) And Sponsored by THE ACTUARIES CLUB OF BOSTON FIRST-LEVEL
More information(b) the equation of the perpendicular bisector of AB. [3]
HORIZON EDUCATION SINGAPORE Additional Mathematics Practice Questions: Coordinate Geometr 1 Set 1 1 In the figure, ABCD is a rhombus with coordinates A(2, 9) and C(8, 1). The diagonals AC and BD cut at
More informationreview math0410 (1-174) and math 0320 ( ) aafinm mg
Eam Name review math04 (1-174) and math 0320 (17-243) 03201700aafinm0424300 mg MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Simplif. 1) 7 2-3 A)
More information4-4. Exact Values of Sines, Cosines, and Tangents
Lesson - Eact Values of Sines Cosines and Tangents BIG IDE Eact trigonometric values for multiples of 0º 5º and 0º can be found without a calculator from properties of special right triangles. For most
More informationSUMMATIVE ASSESSMENT-1 SAMPLE PAPER (SET-2) MATHEMATICS CLASS IX
SUMMATIVE ASSESSMENT-1 SAMPLE PAPER (SET-) MATHEMATICS CLASS IX Time: 3 to 3 1 hours Maximum Marks: 80 GENERAL INSTRUCTIONS: 1. All questions are compulsory.. The question paper is divided into four sections
More informationHIGHER SCHOOL CERTIFICATE EXAMINATION MATHEMATICS 2/3 UNIT (COMMON) Time allowed Three hours (Plus 5 minutes reading time)
HIGHER SCHOOL CERTIFICATE EXAMINATION 998 MATHEMATICS / UNIT (COMMON) Time allowed Three hours (Plus 5 minutes reading time) DIRECTIONS TO CANDIDATES Attempt ALL questions. ALL questions are of equal value.
More information0114ge. Geometry Regents Exam 0114
0114ge 1 The midpoint of AB is M(4, 2). If the coordinates of A are (6, 4), what are the coordinates of B? 1) (1, 3) 2) (2, 8) 3) (5, 1) 4) (14, 0) 2 Which diagram shows the construction of a 45 angle?
More information3.5. Did you ever think about street names? How does a city or town decide what to. composite figures
.5 Composite Figures on the Coordinate Plane Area and Perimeter of Composite Figures on the Coordinate Plane LEARNING GOALS In this lesson, ou will: Determine the perimeters and the areas of composite
More informationIntroductory Chapter(s)
Course 1 6 11-12 years 1 Exponents 2 Sequences Number Theory 3 Divisibility 4 Prime Factorization 5 Greatest Common Factor Fractions and Decimals 6 Comparing and Ordering Fractions 7 Comparing and Ordering
More informationQUESTION BANK ON STRAIGHT LINE AND CIRCLE
QUESTION BANK ON STRAIGHT LINE AND CIRCLE Select the correct alternative : (Only one is correct) Q. If the lines x + y + = 0 ; 4x + y + 4 = 0 and x + αy + β = 0, where α + β =, are concurrent then α =,
More informationThe equation 8(9x + 7) 7(6x 5) = 1 has the solution x = k, where k is a positive integer. Pass on the value of k.
A1 The equation 8(9x + 7) 7(6x 5) = 1 has the solution x = k, where k is a positive integer. Pass on the value of k. A3 Y is proportional to the reciprocal of the square of X. Y = 20 when X = 6. Pass on
More informationMATHEMATICS EXTENSION 2
Sydney Grammar School Mathematics Department Trial Eaminations 008 FORM VI MATHEMATICS EXTENSION Eamination date Tuesday 5th August 008 Time allowed hours (plus 5 minutes reading time) Instructions All
More informationUKMT UKMT UKMT. IMOK Olympiad. Thursday 16th March Organised by the United Kingdom Mathematics Trust. Solutions
UKMT UKMT UKMT IMOK Olympiad Thursday 16th March 2017 Organised by the United Kingdom Mathematics Trust s These are polished solutions and do not illustrate the process of failed ideas and rough work by
More information