First Round Solutions Grdes 4, 5, nd 1) There re four bsic rectngles not mde up of smller ones There re three more rectngles mde up of two smller ones ech, two rectngles mde up of three smller ones ech, nd the outer rectngle (which is in fct squre) mde up of the four smller ones Thus there re 4 + 3 + 2 + 1 = 10 rectngles in ll 2) The first, third, fifth, etc, terms re 1, 2, 3, etc The second, fourth, sixth, etc, terms re 4, 8, 12, etc Following these progressions seprtely, the next two terms re 20 nd 3) Reding from right to left, the there re two choices for the lst digit, two for the middle digit, nd one for the first digit (which must be 5) Then there re 1 2 2=4 such numbers in totl, which re 500, 505, 550, nd 555 4) If hlf pound is removed from ech box then totl of 10 lbs is removed from the 20 boxes The remining ppys will weigh 1590 lbs 5) In the digrm below, the squre intersects the circle in 8 points This is the mximum becuse no side of squre cn cut circle in more thn two points Since squres hve four sides, no squre cn intersect circle in more thn eight points ) If there re ten trees then there re nine spces between them Since ech spce is 15 m, the distnce between the first nd lst trees is 135 m 1
7) Suppose Mri gives gift to An Then An cnnot give one to Mri or else Ros must give gift to herself So An must give gift to Ros nd Ros must give one to Mri If Mri doesn t give gift to An then she must give one to Ros In this cse, Ros will give one to An nd An will give one to Mri Thus there re two wys of exchnging gifts t the prty 8) If there re n dogs nd n chickens then there re 4n legs on the dogs nd 2n legs on the chickens The totl number of legs is n The only possible number of legs is 24 becuse tht is the only -multiple mong the choices offered 9) After n yers, Mrio ws 27 + n nd Pedro ws 3 + n Mrio would hve been three times s old s Pedro when 27 + n = 3(3 + n) In this cse, 27 + n =9+3n; 18 = 2n; 9 = n Thus fter 9 yers Mrio ws three times s old s Pedro, nd this hppened when Mrio ws 3 nd Pedro ws 12 10) Note tht AH cn be no lrger thn 99 nd A cn be no lrger thn 9 Then HEE must be number between 100 nd 108 In prticulr, H = 1 Also, A = 9 becuse if it were less thn 9 then the sum of AH nd A would be less thn or equl to 89 Then A =9,AH = 91, nd HEE = 100 Thus H + E =1 11) Suppose one side, sy the top, is red nd n djcent side, sy the front, is blue A side djcent to both of them must be nother colour such s green However, with these three colours the rest of the cube cn be coloured s shown (The bck of the cube would be blue) The lest number of colours needed is 3 Ṛ R B G B G R 2
12) The re of rectngle BCGE is hlf the re of ABCD nd the re of tringle EFG is hlf the re of BCGE Then the re of the tringle EFG is 9 squre units B E F C G A D 13) If you remove ll the smll cubes which cn be seen from the outside, there would remin 3 3 3 cube which could not hve been seen from the outside Thus there re 27 smller cubes which cnnot be seen from the outside 14) Luis wrote down 1 digit for ech number from 1 to 9, 2 digits for ech number from 10 to 99, nd three digits for the number 100 Thus he wrote down 9(1) + 90(2) + 1(3) = 192 digits in totl 15) Let b nd h be the originl bse nd height, respectively, nd B nd H the new ones Then B = b + 01b = 11b nd H = h 01h = 09h The new re is 1 2 BH = 1 099 2 (11b)(09h) = 2 bh Since the originl re is 1 2bh, the new re is 099 or 99% of the originl one 1) The first bulb will flsh fter 2, 4,, 8, 10, 12, 14,, minutes The first time on this list tht is multiple of 3 1 minutes is 14 minutes It follows tht the bulbs flsh 2 together every 14 minutes This mens tht they flsh together t 12:14 m, 12:28 m, 12:42 m, 12:5 m, nd 1:10 m The first time fter 1 m tht they flsh together is 1:10 m 3
17) Referring to the digrm below, +2 + +2 = 18 Thus = 18 nd hence =3 It follows tht the perimeter of the originl squre is 24 2 2 18) Let A, B, C, nd D be the number of ptrons t the four shows tht dy Then C =2(A + B) nd D =2C =4(A + B) The money the thetre received ws 350A +350B +7C +7D =350(A + B) + 14(A + B)+28(A + B) =4550(A + B) Thus 455(A + B) = 1183 nd hence A + B = 2 There were 4(2) = 104 ptrons t the fourth show 19) Ech side of the originl tringle is 25 cm long After removing the smller tringles ech longer side hs length 15 cm Since ech shorter side hs length 5 cm, the perimeter of the resulting region is 0 cm 5 15 15 5 5 15 20) There re two possibilities to consider: three offices contin one plnt ech or one office contins two plnts while nother hs one In the first cse, one plnt my be plced in ny of 5 offices, nother plced in ny of the 4 remining offices, nd the third in ny of the lst three offices Thus there re 5 4 3 = 0 wys of plcing the plnts this wy In the second cse, there re three wys of piring two plnts to be plced in the sme office: the cctus nd the zle, the cctus nd the orchid, nd the zle nd the orchid Then the pir my be plced in ny of 5 offices nd the single plnt plced in ny of the remining 4 offices Then there re 3 5 4 = 0 wys of plcing the plnts in this cse Combining the two cses, there re 0 + 0 = 120 wys ll together of plcing the plnts 4
Second Round Solutions Grdes 4, 5, nd 1) Rosit hs 1 lrge box, 4 medium-sized boxes, 8 smll boxes, nd 24 tiny boxes The totl number of boxes is 1 + 4 + 8 + 24 = 37 2) After the first term, we my dd 2, 4, nd 8, respectively, to generte the next three terms If we then dd 1, 32, nd 4 we would obtin 31, 3, nd 127 Since these lst two numbers re correct, the missing number is evidently 31 3) There re five bsic tringles which re not mde up of smller ones There re four more tringles mde up of two smller ones, nd one more tringle mde up of three smller ones There re totl of 10 tringles in the digrm 4) We know tht Betrice is in Chir 3 Also, since Betrice nd Crlos re not beside ech other, Crlos is in Chir 1 Since Antonio is not between Betrice nd Crlos, he is in Chir 4 This mens tht Dine in Chir 2 5) If T denotes tiger nd L denotes lion, the nimls must be lined up in the order T, L, T, L, T There re 3 choices for the first position, 2 for the second, 2 for the third, 1 for the fourth, nd 1 for the fifth The number of possible rrngements is 3 2 2 1 1 = 12 ) If n is the number then 3n = 3 4 120 = 90 Then n = 30 7) Ech white squre is ; suppose tht ech shded rectngle is s shown Then ech squre hs perimeter 24 nd ech shded rectngle hs perimeter 2 + 12 Since the rectngles hve double the perimeter of the squres, 2 + 12 = 2(24) = 48 Solving for gives = 18 By pplying this informtion to the originl digrm, we see tht perimeter of the originl figure is 84 30 12 5
8) Suppose there re n dys in week Then there re n n dys in month nd n n n dys in yer It follows tht n 3 = 1000 nd hence n = 10 There re 10 dys in week 9) After ech cut, the mount of cke remining is 2 3 of wht it ws After 3 cuts, the mount remining is 2 3 2 3 2 3 = 8 of the originl cke 27 10) Suppose p is the originl price After pplying the 10% discount, the price is p 010p = 090p After pplying the tx, the price becomes 090p +(010)(090)p =090p +009p =099p The finl price is 099 or 99% of the originl price 11) If we multiply 1 3 5 7 9 we obtin, in order, 1, 3, 15, 105, 945, In fct, since 5 is fctor in the overll product its lst digit must be 0 or 5 Since the nswer cnnot be even, the lst digit is 5 12) The lst set in Row 1 is number 24, the lst one in Row 2 is number 48, the lst one in Row 3 is number 72, nd so on It follows tht the lst set in Row 15 is number 30, nd so the 15th set in Row 1 is number 375 The nswer is Row 1 13) Since it is not possible for Jun to hve 0 for the hundreds digit, the smllest possible hundreds digit is 1 Strting with this, the smllest possible number Jun cn obtin is 108 14) The powers of 2 re 2, 4, 8, 1, 32, 4, 128, 25, 512, etc The lst digits of these powers re 2, 4, 8,, 2, 4, 8,, 2, etc The 2008th power of 2 will end in After subtrcting 2, the lst digit of 2 2008 2 will be 4
15) Since is opposite n ngle of 40, it is 40 Since s is the supplement for 100 ngle, it is 80 Since + s + y = 180, y =0 Since x nd y re supplements, x = 120 40 x y s 100 7