[PDF] Australian Mathematics Competition AMC Warm-Up Paper 7 Intermediate Solutions.pdf | Plain Text

AMC WARM-UP PAPER INTERMEDIATE PAPER 7 SOLUTIONS c 2009 Australian Mathematics Trust 1.From adjacent angles on a straight line and the angle sum of a triangle we have x+ 65 + 55 = 180 thusx= 60, hence (C).     115 ◦ x◦ 125 ◦ 65◦ 55◦ 2. 5n+7>100 5n>93 n>183 5 thusn=19, hence (B). 3.The lightest sweets will weigh at 30 per kilogram. Thus the minimum weight for 240 is 240 30=8 kg, hence (C). 4.Barbara’s share (in dollars) is 2 3+2×250 =2 5×250 = 100, hence (B).

Intermediate 7 Solutions Page 2 5.Since all numerators are 4, we need to choose the fraction with the smallest de- nominator and this is clearly (0.4) 2=0.16, hence (C). 6.SincePSRis equilateral, each angle is 60 ◦. ..................................................................................................... ..................................................................................................... ..................................................................................................... .................. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .. .. . . . . . . . . . . ....... ...... . . . . . . .. . . . . QSP R 60◦ 60◦ 60◦ 30◦ 30◦ Since the area ofPRSis half that ofPQRit has the same area asSRQ.As these triangles have the same area and the same altitude fromRtoPQ, the bases PSandSQare equal and henceRS=SQ. Now QS R= 120 ◦and SRQ= SQR=30 ◦.So PRQ=60 ◦+30 ◦=90 ◦, hence (C). 7.Let the radius of the quadrant ber. Then from Pythagoras’ Theorem inYOX,YX= √ r2+r 2=r√ 2. T+Sis a quadrant of a circle radiusr,so T+S=1 4πr 2 S+Cis a semi-circle with a diameter ofr√ 2, so S+C=1 8π(r√ 2) 2=πr 2 4 Now (T + S)−(C + S)=T−C=πr 2 4−πr 2 4=0, i.e.T=C, and thus the ratio areaT:areaC=1, hence (B). Note:The same result is obtained by calculating the area of each ofTandCto be 12r2.

Intermediate 7 Solutions Page 3 8.Originally, 98% of the watermelon is water, so the solids are 2% or 150 of its weight, i.e. 150 ×20 = 25kg. When the water becomes 95% of its weight, these solids become 5% or 120 of its weight. Thus its weight then is 20× 25=8kg, hence (E). 9.Alternative 1 The third entry in the first row must be 3, then the third column completes to 3, 2, 4, 5, 1 and the last row to 3, 5, 1, 2, 4. The first column can now be completed to 1, 4, 5, 2, 3. In the last column, 2 can only go in the third position, then that columncompletesto2,3,1,4,5,sox=1, hence (A). Alternative 2 The bottom left hand corner must be 3. Suppose the number in the indicated position is not 1, then it must be 3 as it has 2 above it, 5 underneath and 4 to the right. This then forces the number to the left ofxto be 5, which then results in two 4s in the second row, which is impossible, thusxmust be 1, hence (A). Note:This solution does not use the number 5 given in the middle column. If it were not given, there would be two ways to complete the square, though the number in the required position must still be 1. The reader is invited to verify this. 10.Since there are 8 teams, there are 7 rounds of four matches and thus a total of 7×8 = 56 points available. Consider a team with 10 points. It is possible to have 5 teams on 10 points and 3 teams on 2 points when each of the top 5 draws with each other, each of the bottom 3 draws with each other and each of the top 5 wins against each of the bottom 3. So 10 points does not guarantee a place in the top 4. Consider a team with 11 points. If this team was fifth then the number of points gained by the top 5 teams is≥55. This is impossible as the number of points shared by the bottom 3 teams is then 1, as these 3 teams must have at least 3×2 = 6 points between them for the games played between themselves. Hence 11 points is sufficient to ensure a place in the top 4. Thus 11 points are required, hence (D).

Intermediate 7 Solutions Page 4 Generalisation Suppose we want to be guaranteed to be in the topkofnteams, for 1≤k≤n−1. Withnteams there are  n n−1  games and hencen(n−1) =n 2−npoints available. (a) Consider a team with 2n−k−2points. Put the teams in two divisionsAandBwithk+ 1 teams in DivisionA andn−k−1teamsinDivisionB. Let all games within each division be ties (draws). All games across divisions are won by the teams in DivisionA. Then each team in DivisionAhas 2n−k−2 points and none is guaranteed to be in the topk.So2n−k−2 points is not enough. (b) Consider a team with 2n−k−1points. There aren 2−npoints altogether. Suppose at leastk+1 teams have at least 2n−k−1 points each. Then together they have at least (k+ 1)(2n−k−1) = 2nk+2n−k 2−2k−1 points. This leavesn 2−2nk−3n+k 2+2k+ 1 points for the remainingn−k−1 teams. These teams must have between them at least (n−k−1)(n−k−2) =n 2−2nk−3n+k 2+3k+2 points (for the n−k−1 n−k−2  games they have between themselves). This is a con- tradiction so at least one of the topk+ 1 teams must have less than 2n−k−1 points and hence 2n−k−1 points guarantees a position in the topkteams. Q10 is a particular case wheren=8andk= 4. The required points are then 16−4−1 = 11, hence (D).