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

AMC PRACTICE QUESTIONS AND SOLUTIONS Intermediate Copyright � 2014, 2019 Australian Mathematics Trust AMTT Limited ACN 083 950 341 (Set1)                 \b       
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Alternative 2 Q R S T U V W X Y Z 108 ◦ The pentagonQS U W Yis regular, so that ∠QY W= 108 ◦. The quadrilateral QT W Yis c\bclic, so opposite angles add to 180 ◦. Thus ∠QT W = 180 ◦−108 ◦= 72 ◦, hence (D). 4. 2014 I21 Standard six-sided dice have their dots arranged so that the opposite faces add up to 7. If 27 standard dice are arranged in a 3 ×3× 3 cube on a solid table what is the maximum number of dots that can be seen from one position? (A) 90 (B) 94(C) 153 (D) 154 (E) 189 There are at most three 3 ×3 faces of the cube visible, and the maximum will occur with exactl\b three cube faces visible. A total of 153 is possible. dice face count pips 1 4 7 35 19 114 153 To see that no greater total is possible, of the 19 visible dice, one has 3 faces visible, sixhave 2 faces visible and 12 have one face visible. So the sum of all visible faces cannot exceed 1 ×15+6× 11 + 12×6 = 153, hence (C). 5. 2014 I25 Thanom has a roll of paper consisting of a ver\b long sheet of thin paper tightl\b rolled around a c\blindrical tube, forming the shape indicated in the diagram. Initiall\b, the diameter of the roll is 12 cm and the diameter of the tube is 4 cm. After Thanom uses half of the paper, the diameter of the remaining roll is closest to (A) 6 cm (B) 8 cm(C) 8.5 cm (D) 9 cm (E) 9.5 cm AMC Practice Questions and Solutions — Intermediate

4π 16π 16π r 2 6 Working in centimetres, let the half-roll’s radi\bs be r. From end on, the f\bll roll had area π(6 2− 2 2) = 32π , so half the roll has area 16π . Incl\bding the t\bbe, the end of the half-roll has area 20 π= πr 2. Then r 2= 20, b\bt 4 .5 2= 20. 25 and 4. 4 2= 19.36, so that 4. 4