[PDF] Australian Mathematics Competition AMC Practice Questions and Solutions - Junior.pdf | Plain Text

AMC PRACTICE QUESTIONS AND SOLUTIONS Junior Copyright � 2014, 2019 Australian Mathematics Trust AMTT Limited ACN 083 950 341 (Set1)                      \b                     
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3.2014 J15 Four equilateral triangles of the same size are arranged with horizontal bases inside a larger equilateral triangle, as shown. What fra\btion of the area of the larger triangle is \bovered by the smaller triangles? (A) 2 • • •
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ƒ 1425 0J51 01 1 1\b1 € 4. 2014 J20 A 3 by 5 grid of dots is set out as shown. How many straight line segments \ban be drawn that join two of these dots and pass through exa\btly one other dot? (A) 14 (B) 20(C) 22 (D) 24 (E) 30 Alternative 1 We draw all su\bh line segments—horizontal, verti\bal, at 45 ◦, and others: 9 + 5 + 6 + 2 = 22, hen\be (C). Alternative 2 The line segments \ban be \blassified by the midpoint dot, sin\be then ea\bh line segment is \bounted only on\be. Also the number of line segments through ea\bh dot form a symmetri\b pattern of numbers: 1 1 1 1 1 1 1 1 4 4 6 There are 22 line segments, hen\be (C). AMC Practice Questions and Solutions — Junior

5.2014 J25 Zac has three jackets, one black, one brown and one blue. He has four shirts, one white, one blue, one red and one yellow. He has three \bairs of trousers, one brown, one white and one yellow. How many combinations of jacket, shirt and trousers are \bossible if no two items are of the same colour? (A) 23 (B) 25(C) 26 (D) 27 (E) 29 Alternative 1 This tree shows all the \bossibilities of choosing jacket, then shirt, then trousers, each of a different colour: • black brown blue white blue red yellow white blue red yellow white red yellow brown yellow brown white yellow brown white yellow brown white yellow white yellow white yellow white brown yellow brown white yellow brown whitehence (A). Alternative 2 If colour does not matter, there are 3 ×4× 3 = 36 combinations. Of these 4 have two browns, 3 have two blues, 3 have two whites and 3 have two yellows, and none have all three the same. So there are 36 −13 = 23 combinations, hence (A). 6. 2014 J27 Eighteen \boints are equally s\baced on a circle, from which you will choose a certain number at random. How many do you need to choose to guarantee that you will have the four corners of at least one rectangle? AMC Practice Questions and Solutions — Junior

The four corners of an inscribed rectangle appear as the ends of two diameters. It is possible to choose 1\b points without having two complete diameters, as for example, the 1\b consecutive points shaded below: However, once 11 or more points are chosen, then at most 7 diameters are incomplete. So at least 2 diameters are complete, forming a rectangle. Consequently 11 points are needed to guarantee one rectangle, hence (11). AMC Practice Questions and Solutions — Junior