No. 32. R.E. Woodrow. As a contest this issue we give the Junior High School Mathematics

Transcription

1 334 THE SKOLIAD CORNER No. 32 R.E. Woodow As a contest this issue we give the Junio High School Mathematics Contest, Peliminay Round 1998 of the Bitish Columbia Colleges which was witten Mach 11, My thanks go to the contest oganize, Jim Totten, the Univesity College of the Caiboo, fo fowading the 1998 contest mateials to me. Students ae given 45 minutes to espond to the 15 questions. BRITISH COLUMBIA COLLEGES Junio High School Mathematics Contest Peliminay Round 1998 Time: 45 minutes 1. A numbe is pime if it is geate than one and divisible only by one and itself. The sum of the pime divisos of1998 is: (a) 5 (b) 14 (c) 42 (d) 66 (e) Successive discounts of 10% and 20% ae equivalent to a single discount of: (a) 15% (b) 25% (c) 28% (d) 30% (e) 32% 3. Suppose that A k = A 2 and A 2 B = A, 2B. Then the value of is: k7 2 k 3 (a) 1 (b) 16 (c) 961 (d) 43 (e) The expession that is not equal to the value of the fou othe expessions listed is: (a) 1 p9 +9,8 (b) (1+9)(, p 9+8) (c),1 9+ p 9+8 (d) (1, p 9) (9, 8) (e) 19, 9, 8 5. The sum of all of the digits of the numbe 10 75, 75 is: (a) 8 (b) 655 (c) 664 (d) 673 (e) 675

2 A cicle is divided into thee equal pats and one pat is shaded as in the accompanying diagam. The atio of the peimete of the shaded egion, including the two adii, to the cicumfeence of the cicle is: (a) 1 (b) 2 3 (c) 1 3 (d) 3+ 3 (e) 3 7. The value of 1 2, 1 2, 1 2, 1 2 is: (a) 3 4 (b) 4 5 (c) 5 6 (d) 6 7 (e) If each small squae in the accompanying gid is one squae centimete, then the aea in squae centimetes of the polygon ABCDE is: C D B A E (a) 38 (b) 39 (c) 42 (d) 44 (e) A point P is inside a squae ABCD whose side length is 16. P is equidistant fom two adjacent vetices, A and B, and the side CD opposite these vetices. The distance PAequals: (a) 8:5 (b) 6 p 3 (c) 12 (d) 8 (e) Agoup of 20 students has an aveage mass of 86 kg pe peson. It is known that 9 people fom this goup have anaveage mass of 75 kg

3 336 pe peson. The aveage mass in kilogams pe peson of the emaining 11 people is: (a) 94 (b) 95 (c) 96 (d) 97 (e) none of these 11. In the following display each lette epesents a digit: 3 B C D E 8 G H I The sum of any thee successive digits is18. The value of H is: (a) 3 (b) 4 (c) 5 (d) 7 (e) In the accompanying diagam \ADE = 140. The sides ae conguent as indicated. The measue of\ead is: D A E (a) 30 (b) 25 (c) 20 (d) 15 (e) The aea (in squae units) of the tiangle bounded by the x-axis and the lines with equations y =2x+4and y =, 2 3 x +4is: (a) 8 (b) 12 (c) 15 (d) 16 (e) Two diagonals of a egula octagon ae shown in the accompanying diagam. The total numbe of diagonals possible in a egula octagon is: (a) 8 (b) 12 (c) 16 (d) 20 (e) 28

4 A local baseball league is unning a contest to aise money to send a team to the povincial championship. To win the contest it is necessay to detemine the numbe of baseballs stacked in the fom of a ectangula pyamid. The fth and sixth levels fom the base of the stack of baseballs ae shown. If the stack contains a total of seven levels, the numbe of baseballs in the stack is: (a) 100 (b) 112 (c) 166 (d) 168 (e) 240 In the May numbe of the Cone we gave the poblems of the 15 th W.J. Blundon Contest witten by students in Newfoundland and Labado. Next we give the \ocial" solutions. My thanks go to Buce Shawye fo fowading the contest and solutions to me. 15 th W.J. BLUNDON CONTEST Febuay 18, (a) Find the exact value of 1 log log 3 36 : 1 log log 3 36 = log log 36 3 = log 36 6= 1. 2 (b) If log 15 5=a, nd log 15 9 in tems of a. 1 = log = log 15 (53) = log 15 5+log 15 3=a+log log 15 3=1,a=)log 15 9 = log = 2 log 15 3 = 2(1, a). 2. (a) If the adius of a ight cicula cylinde is inceased by 50% and the height is deceased by 20%, what is the change in the volume? V 2 = (1:5) 2 (:8h) =1:8( 2 h)=1:8v 1. So the volume is inceased by 80%. (b) How many digitsae thee in the numbe ? = = , which has = 1992 digits. 3. Solve: 3 2+x +3 2,x =82.

5 x +3 2,x = x x = 82 9(3 x ) 2, 82(3 x )+9 = 0 (9 3 x, 1)(3 x, 9) = 0 3 x = 1 9 ; 3x =9 x=,2 x=2 4. Find all odeed pais of integes such that x 6 = y x 6 = y x 6,y 2 = 53 (x 3, y)(x 3 + y) = 53 x 3, y =53 x 3,y=1 x 3,y=,53 x 3, y =,1 x 3 + y =1 x 3 +y=53 x 3 +y=,1 x 3 +y=,53 x =3 x=3 x=,3 x=,3 y=,26 y =26 y=26 y=,26 The pais ae(3;,26); (3; 26); (,3; 26); (,3;,26). 5. When one-fth of the adults left a neighbouhood picnic, the atio of adults to childen was 2 : 3. Late, when 44 childen left, the atio of childen to adults was 2:5. How many people emained at the picnic? Let A be the numbe of adults and C be the numbe of childen initially at the picnic. Afte one-fth of the adults left, fou-fths emain. So 4 5 A C = 2 =) 6A =5C: 3 Afte 44 childen left C, A = 2 =) 8A =25C,1100: 5 Solving the two equations gives A =50, C =60. The numbe emaining is then 4 (50) + (60, 44) = = 56: 5

6 Find the aea of a hombus fo which one side has length 10 and the diagonals die by b+2 b 10 b b 10 (b+2) 2 +b 2 = 100 2b 2 +4b,96 = 0 b 2 +2b,48 = 0 (b, 6)(b +8) = 0 b=6; b6=,8 Since the aea of a hombus is one half the poduct of the diagonals we get A = 1 2 (2b)(2b +4)= 1 (12)(16) = 96: 2 7. In how many ways can 10 dollas be changed into dimes and quates, with at least one of each coin being used? Let q be the numbe of quates and d be the numbe of dimes. Then 25q +10d= 1000 d = 100, 5 2 q: Since d must be an intege, q must be even. Also d must be positive. So 100, 2 5 q>0 q<40: So q must be an even positive intege less than 40, of which thee ae19.

7 Solve: p x p x+10=12. Let y = 4p x +10. Then y 2 = p x +10, and the equation becomes y 2 + y =12 Then: p 4 x +10=3 y 2 +y,12=0 x+10=81 (y+ 4)(y, 3)=0 x=71 y6=,4; y =3 9. Find the emainde when the polynomial x 135 +x 125,x 115 +x 5 +1 is divided by the polynomial x 3, x. x x 125, x x 5 +1 = (x 3,x)Q(x)+ax 2 + bx + c = x(x, 1)(x +1)Q(x)+ax 2 + bx + c This must be valid fo all values of x. Substituting in x = 0, x = 1, and x =,1 gives: Solving the system x = 0 : 1=0+c =) c=1 x= 1 : 3=0+a+b+c =) a+b=2 x=,1:,1=0+a,b+c =) a,b=,2 a + b = 2 a, b =,2 gives a =0,b=2. So the emainde is 2x Quadilateal ABCD below has the following popeties: (1) The midpoint O of side AB is the cente of a semicicle; (2) sides AD, DC and CB ae tangent to this semicicle. Pove that AB 2 =4AD BC. Fist join the obvious lines in the given gue: E D F C G A O B By the popeties of tangents, DE = DF and CF = CG. Theefoe \EDO = \FDO = and \FCO = \GCO =. Since OA = OB, we have \EAO = \GBO =.

8 341 Summing the angles of quadilateal ABCD, we get = 360. Hence + + = 180 ; that is, they ae the angles of a tiangle. Consideing tiangles AOD, DOC and COB, we get \AOD =, \DOC = and \COB =. Thus the thee tiangles ae simila. Consideing the tiangles ADO and BOC, we have AD AD BC = AO OB. Since AO = OB = 1 AB, we get the esult. 2 AO = OB BC,o Last issue we gave the poblems of the U.K. Intemediate Mathematical Challenge. Hee ae the solutions. 1. C 2. B 3. A 4. A 5. E 6. C 7. C 8. A 9. C 10. B 11. C 12. E 13. D 14. D 15. D 16. D 17. D 18. C 19. A 20. E 21. E 22. D 23. A 24. A 25. B That completes the Skoliad Cone fo this numbe. Send me you comments, suggestions, and most impotantly, suitable contest mateials fo use in futue issues. Advance Announcement The 1999 Summe Meeting of the Canadian Mathematical Society will take place at Memoial Univesity in St. John's, Newfoundland, fom Satuday, 29 May 1999 to Tuesday, 1 June The Special Session on Mathematics Education will featue the topic What Mathematics Competitions do fo Mathematics. The invited speakes ae Ed Babeau (Univesity of Toonto), Ron Dunkley (Univesity of Wateloo), Tony Gadine (Univesity of Bimingham, UK), and Rita Janes (Newfoundland and Labado Senio Mathematics League). Requests fo futhe infomation, o to speak in this session, as well as suggestions fo futhe speakes, should be sent to the session oganizes: Buce Shawye and Ed Williams CMS Summe 1999 Meeting, Education Session Depatment of Mathematics and Statistics, Memoial Univesity St. John's, Newfoundland, Canada A1C 5S7