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[PDF] Australian Mathematics Competition AMC Warm-Up Paper 7 Junior Solutions.pdf | Plain Text
AMC WARM-UP PAPER JUNIOR PAPER 7 SOLUTIONS c 2009 Australian Mathematics Trust 1.1 2of1 3=1 2×1 3=1 6, hence (D). 2.x+ 45 + 105 + 125 = 360, sox= 360−275 = 85, hence (C). 3.The four payments total 4×$65 = $260. Hence the amount saved is $260−$249 = $11, hence (A). 4.The total areaAis given by A= areaofall5circles−area of the overlap =5− 4×1 8 =5−1 2=41 2, hence (B). 5.If one side is twice the other, the perimeter is equivalent to 6 times the shorter side, so the shorter side is 246 = 4 cm and the longer side is then 8 cm. Thus the area, in square centimetres, is 8×4 = 32, hence (E).Junior 7 Solutions Page 2 6.Volume of the carton is given byV=s 2h,wheresis the length of the square base andhis the height. 1L = 1000 mL = 1000 cubic centimetres. Thus 7×7×h= 1000 h=1000 49 ≈1000 50=20, hence (B). 7.The minimum scorexcan be gained by one student when each of the other nine students achieve the maximum score, i.e have a combined score of 900. Then, for this scorex, 900 +x 10=92, andx= 20, hence (A). 8.The area of the square is 5×5 = 25 square units. Counting the shaded right-angled triangles (each half a square), we get a total of 20, which have an area of 10 square units. Thus, as a fraction of the square, the portion shaded is10 25=0.4, hence (B). 9.Let the number of large bags bex. Then the number of small bags is 46−x,and 20x+8(46−x) = 560 20x+ 368−8x= 560 12x= 192 x=16, hence (B). 10.The total number of edges on all the faces is (20×3) + (30×4) + (12×5) = 240, but this counts each edge twice, as it occurs on exactly two adjacent faces. Hence the number of edges is 120, hence (B).