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

AMC WARM-UP PAPER SENIOR PAPER 7 SOLUTIONS c 2009 Australian Mathematics Trust 1.3 x+1 =81→3 x+1 =3 4→x=3, hence (C). 2. m m−n+n n−m=m m−n−n m−n =m−n m−n =1, hence (D). 3.If each of the 12 houses gets 4 letters then there are 9 letters left. Giving one more letter to 8 houses (including George’s) means that George must also get the last letter, so he must get at least 6 letters, hence (D). 4.From the data, f(x)=1 1+x f(f(x)) =1 1+ 11+x =x+1 x+1+1 =x+1 x+2, hence (B).

Senior 7 Solutions Page 2 5.Alternative 1 Now 70 = 2×5×7, so we only have to look for two factors differing by 2 (as the difference in the capacity of the cases is 2 kg), and in this instance it must be 5 and 7, so the capacity of the standard case is 5 kg, hence (B). Note:in this particular example we do not need to know that 4 less cases are used. Alternative 2 Let the capacity of the smaller cases bex, then the capacity of the larger isx+2 and 70 x−70 x+2=4 70(x+2)−70x=4x 2+8x x 2+2x−35 = 0 (x+7)(x−5) = 0 x= 5 (neglecting the negative root), hence (B). 6.Landing within 1 m from the hole is landing within a circle radius 1 m, i.e. with areaπ×1 2=π. The area of the green isπ×12 2= 144π. Thus the probability of the ball landing within 1 m from the hole is π 144π=1 144, hence (E). 7.The average of thennumbers iskso their sum iskn.Whenxis added, there are n+ 1 numbers and their average becomesk+1. So kn+x n+1=k+1 kn+x=(n+1)(k+1) =kn+k+n+1 thusx=k+n+1, hence (A).

Senior 7 Solutions Page 3 8.JoinRT. This passes through the centre of the circleOas it subtends a right angle at the circumference. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . .. . . . . .. . . .. . . .. . .. . .. .. . .. .. . .. .. . . .. ... .. ... .. ... .... ... ..... ...... ................................................................... ................................................................................................. .... ... .... .... .... ... ... .. ... .. .. .. .. .. .. .. .. . .. .. . .. . .. . . .. . . .. . . . . .. . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .......................................................... ..................................................................................................... ..................................................................................................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..................................................................................................... ..................................................................................................... ................................................................................................... P S RO Q T M 1 30 ◦ Thus TRQ=30 ◦. So, from the right triangleTRQ TQ=2sin30 ◦=2× 12=1. Also,RQ=2cos30 ◦=√ 3. Thus the area of the rectangleRS T Q=1×√ 3=√ 3, hence (E). 9.Alternative 1 The total number of ways of choosing 3 numbers from 10 is 10 3 = 120. There are 8 triples containing the pair 1,2; 8 containing the pair 2,3; . . . 8 containing the pair 9,10, i.e 72 such triples. However we have counted twice the triples 123, 234,···, 8910, so only 64 triples are excluded. This leaves 120−64 = 56 triples available, hence (C). Alternative 2 Suppose 1≤a

Senior 7 Solutions Page 4 10.Label the sides of the octagon as shown. ha b cd ge f Label the sides of the octagon as shown. Ifg=7, thenh=8 and{a, c, e}={1,2,4}. Sinceb+f=d+h,thend=3. To maximise the area of the hexagon, we must havea=1,c=4,e=2,b=6and f= 5, giving an area of 30. Ifg=8, thenh=7 and{a, c, e}={1,2,5}or{1,3,4}. Sinceb+f=d+h, d=4 and{h, f}={4,6}. To maximise the area of the hexagon, we must havea=1,c=5,e=2,b=6and f= 4, giving an area of 36, hence (E).