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PEOPLE COMMITTEE OF NGO QUYEN DISTRICT NGOQUYEN EDUCATION AND EDUCATION DEPARTMENT NGO QUYEN ENGLISH MATHEMATICS AND SCIENCE COMPETITION FOR GRADE STUDENTS SCHOOL YEAR: 2021 – 2022 Time allowance: 120 minutes PART 1: MULTIPLE – CHOICE (100 mark) p p 3 Question 1: If p 2q then q equals A B C D E None of the above x y 2 ; y x 2 ; Question 2: If A 10 B 15 C 20 2 and x y What is the value x y ? D 25 E None of the above Question 3: How many zerof are there in the last digits of the following number P 1112 13 88 89 A 16 B 17 C 18 D 19 2 E 20 Question : What is the smallest possible value of M x y – x – y – xy A 1; B 1; C 2; D 2; E None of the above Question 5: Find the unit digit of 2021 A B C D E · Question 6: In the diagram, PS 5, PQ PQS is right-angled at Q QSR 30 and QR RS The length of RS is: ( A) ; ( B ) 3; (C ) 2; ( D) ; (E) None of the above A B.28 C 36 D 45 E 90 2 2 2 Question 7: Calculate 1999 1998 1997 1996 A 1999 B.2 C 1999000 D -2 Question 8: Two of the three side of a triangle are 25 and 15 Which of the folloowing numbers is not a possible perimeter of the triangle? A 52 B 57 C 62 D 67 E 47 x Question 9: The value of A 45; B 60; C 90; D 105 Question 10 The diagram beside shows the isosceles trapesiod ABCD, which is with BD cm, ·ABD 45 Determine the area of the one 2 B 22.5cm C 49cm A 24,5cm PART II: COMPOSE (200 mark) D 28.5cm E None of the above 3 2 Problem Given a 3ab 9; b 3a b 46 Find the value of P a b Problem Let n be a prime numbe A 2021n which is divisible by Problem Find all pairs n Prove that the value of expression x, y of integers such that x xy – y · Problem Let ABCD be a rectangular, BDC 30 Draw the straight line through point C , perpendicular to BD, intersects BD at E and intersects the bisector of angle ADB at M a) Prove that AMBD is an isosceles trapezoid b) Let N and K be the projections of point M on DA and AB respectively Prove that N , K and E are collinear THE END PART 1: MULTIPLE – CHOICE (100 mark) ANSWERS AND MARKS Questio n Answer B B C Marks 10 10 10 PART II: COMPOSE (200 mark) problem 10 B 10 E 10 D 10 C 10 E 10 D 10 A 10 Answer Marks We have a 3ab 9; b 3a b 46 1(50 mark) Therefore ( a3 – 3ab2)2 + (b3 – 3a2b)2 = 2197 20 Hence we have (a2 +b2)3 = 2197 20 So a2 + b2 =13 10 We have 2(50 mark) 2021n 2016n n 1 n 1 2016n Since 20 8 M 10 n 2 , n 1 n 1 M8 And n be the prime number so 2021n Therefore 3 M Where n P; n We have x + xy – y = So we have (x -1)(y+1) = 3(50 mark) x-1 y+1 x y So (x; y) 20 20 -7 -1 -6 -2 (8;0);(2;6);( 6; 2);(0;8) -1 -7 -8 20 10 ( 50 mark) ả a, Easily proved A1 B2 30 so AM is parallel to BD 10 10 o · · easily proved we have ADB MDB 60 Hence AMBD is an isosceles trapezoid b, Since M lies on the bisector of angle ADB so we have MN = ME So MNE isosceles triangle at M, also have · · NME 120o so MNE 30o 10 o · µ Have quadrilateral MNAK is a rectangle, so MNK A1 30 From (1) and (2) we have · · MNK MNE So three points N, K, E are collinear 10 ... PART 1: MULTIPLE – CHOICE (100 mark) ANSWERS AND MARKS Questio n Answer B B C Marks 10 10 10 PART II: COMPOSE (200 mark) problem 10 B 10 E 10 D 10 C 10 E 10 D 10 A 10 Answer Marks We have a ... – 3a2b)2 = 2197 20 Hence we have (a2 +b2)3 = 2197 20 So a2 + b2 =13 10 We have 2(50 mark) 2021n 2016n n 1 n 1 2016n Since 20 8? ?? M 10 n 2 , n 1 n 1 M8 And n be... y+1 x y So (x; y) 20 20 -7 -1 -6 -2 (8; 0);(2;6);( 6; 2);(0 ;8) -1 -7 -8 20 10 ( 50 mark) µ ¶ a, Easily proved A1 B2 30 so AM is parallel to BD 10 10 o · · easily proved we have ADB MDB
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