341 THE SKOLIAD CORNER No. 48 R.E. Woodow This issue we give the peliminay ound of the Senio High School Mathematics Contest of the Bitish Columbia Colleges witten Mach 8, 2000. My thanks go to Jim Totten, The Univesity College of the Caiboo, fo sending them fo use in the Cone. BRITISH COLUMBIA COLLEGES Senio High School Mathematics Contest Peliminay Round Mach 8, 2000 1. Antonino sets out on a bike ide of 40 km. Afte he has coveed half the distance he nds that he has aveaged 15 km/h. He decides to speed up. The ate at which he must tavel the est of the tip in ode to aveage 20 km/h fo the whole jouney is: (a) 25 km/h (b) 30 km/h (c) 35 km/h (d) 36 km/h (e) 40 km/h 2. O (3 2) B Acicle with cente at(3 2) intesects the {ais at the oigin, O, and at the point B. The tangents to the cicle at O and B intesect at the point P. The y{coodinate ofp is: P (a) ;3 1 2 (b) ;4 (c) ;4 1 2 (d) ;5 (e) none of these 3. Fom ve students whose ages ae 6, 7, 8, 9, and 10, twoaeandomly chosen. The pobability that the dieence in thei ages will be at least 2 yeas is: (a) 1 2 (b) 2 5 (c) 3 5 (d) 7 10 (e) 3 4
342 4. The centes of thee cicles of adius 2 units ae located at the points (0 0), (12 0) and (0 5). If the cicles epesent pulleys, what is the length of the belt which goes aound all 3 pulleys as shown in the diagam? (0 5) (0 0) (12 0) (a) 30 + (b) 30 + 4 (c) 36 + (d) 60 ; 4 (e) none of these 5. If Mak gets 71 on his net quiz, his aveage will be 83. Ifhegets 99, his aveage will be 87. How many quizzes has Mak aleady taken? (a) 4 (b) 5 (c) 6 (d) 7 (e) 8 6. u u u u u u u u u u While 10 pin bowling (see diagam) Sam left 3 pins standing which fomed the vetices of an equilateal tiangle. How many such equilateal tiangles ae possible? (a) 15 (b) 14 (c) 12 (d) 10 (e) none of these 7. If I place a6 cm 6 cm squae on a tiangle, I can cove up to 60% of the tiangle. If I place the tiangle on the squae, I can cove up to 2 3 of the squae. What is the aea, in cm 2, of the tiangle? (a) 22 4 5 (b) 24 (c) 36 (d) 40 (e) 60 8. Two cicles, each with adius 10 cm, ae placed so they aetangent to each othe and a staight line. A smalle cicle is nestled between them so that it is tangent to the lage cicles and the line. What is the adius, in centimetes, of the smalle cicle? (a) p 10 (b) 2:5 (c) p 2 (d) 1 (e) none of these
343 9. Aange the following in ascending ode: 2 5555 3 3333 6 2222 (a) 2 5555 3 3333 6 2222 (b) 2 5555 6 2222 3 3333 (c) 6 2222 3 3333 2 5555 (d) 3 3333 6 2222 2 5555 (e) 3 3333 2 5555 6 2222 [Edito's note: Astute eades will notice that this is the same question as Question 10 fom the Junio Contest given last issue. The solution appeas in this issue.] 10. Given that 0 < < y < 20, the numbe of intege solutions ( y) to the equation 2 +3y =50is: (a) 25 (b) 16 (c) 8 (d) 5 (e) 3 11. Suppose A, B, and C ae positive integes such that 24 5 = A + 1. B + C +1 1 The value of A +2B +3C equals: (a) 9 (b) 12 (c) 15 (d) 16 (e) 20 12. A bo contains m white balls and n black balls. Two balls ae emoved andomly without eplacement. The pobability one ball of each colou is chosen is: (a) mn mn 2mn (b) (c) (m+n)(m+n;1) (m+n) 2 (m+n;1)(m+n;1) (d) 2mn (m+n)(m+n;1) (e) m(m;1) (m+n)(m+n;1) 13. If it takes buildes y days to build z houses, how many days would it take q buildes to build houses? Assume these buildes wok at the same ate as the othes. (a) qy z (b) yz q (c) qz y (d) y qz (e) z qy 14. If 2 + y + =14and y 2 + y + y =28, then a possible value fo the sum of + y is: (a) ;7 (b) ;6 (c) 0 (d) 1 (e) 3
344 15. Two conguent ectangles each measuing 3 cm 7 cm ae placed as in the gue. The aea of ovelap (shaded), in cm 2, is: (a) 87 7 (b) 29 7 (c) 20 7 (d) 21 2 (e) none of these The poblems given last issue wee those of the peliminay ound of the Junio High School Contest of the Bitish Columbia Colleges. My thanks fo these \ocial solutions" to Jim Totten, The Univesity College of the Caiboo. BRITISH COLUMBIA COLLEGES Junio High School Mathematics Contest Peliminay Round Mach 8, 2000 1. Afte 15 lites of gasoline was added to a patially lled fuel tank, the tank was 75% full. If the tank's capacity is28 lites, then the numbe of lites in the tank befoe adding the gas was: Answe. (d). Let be the numbe of lites of gasoline in the tank pio to lling. Then +15= 3 28, o =6. 4 2. The following gues ae madefom matchsticks. If you had 500 matchsticks, the numbe of squaes in the lagest such gue you could build would be: Answe. (b). The st gue iscomposed of one squae of side 1 (consisting of 4 matchsticks) plus 2 squaes of side 1 each missing 1 matchstick, fo a total of 4+2 3=10matchsticks. Each subsequent gue consists of the pevious gue plus 2 squaes of side 1 each missing 1 matchstick. Thus, the n th gue in the sequence contains 4+2 3 n =6n +4matchsticks. The lagest value n fo which 500 matchsticks is sucient is thus 82 (which uses up 6 82 + 4 = 496 matchsticks). Now the numbe of squaes in the
345 st gue is 3 and each subsequent gue contains 2 moe squaes than the pevious one. Theefoe the numbe of squaes in the n th gue is2n +1. Fo n =82this means that 165 squaes would be in the lagest gue made with 500 matchsticks. 3. The peimete of a ectangle is 56 metes. The atio of its length to width is 4:3. The length, in metes, of a diagonal of the ectangle is: Answe. (b). Let ` and w be the length and width (in metes) of the ectangle in question. Since the peimete is 56 metes, we have 2` +2w = 56,o` + w = 28. We ae also told that ` : w = 4 : 3, o ` = 4 w. Using this in the st equation we get 3 4 3 w + w = 28 7 3 w = 28 w = 12 which implies that ` =16. By the Theoem of Pythagoas the length of the diagonal is p 12 2 +16 2 = p 400 = 20. 4. If Apil 23 falls on Tuesday, then Mach 23 of the same yea was a: Answe. (a). Since thee ae 31 days in Mach, thee ae 31 days between Mach 23 and Apil 23. That is, the peiod in question is 4 weeks and 3 days. Since Apil 23 is a Tuesday, we must have Mach 23 a Satuday, namely 3 days ealie in the week. 5. Conside the dat boad shown in the diagam. If a dat may hit any point on the boad with equal pobability, the pobability it will land in the shaded aea is:
346 Answe. (d). The total aea of the boad is25 2 squae units. The aea of the shaded egion is 4 + 3 =7 2 squae units. Theefoe, the pobability of hitting the shaded aea is 7 2 25 2 = 7 25 = 0:28. 6. The pope divisos of a numbe ae those numbes that ae factos of the numbe othe than the numbe itself. Fo eample, the pope divisos of 12 ae 1, 2, 3, 4 and 6. An abundant numbe is dened as a numbe fo which the sum of its pope divisos is geate than the numbe itself. Fo eample, 12 is an abundant numbe since 1+2+3+4+6> 12. Anothe eample of an abundant numbe is: Answe. (c). Let us compute the sum of the pope divisos of each of the 5 possible answes in the list: 13 : 1 < 13 16 : 1 + 2 + 4 + 8 = 15 < 16 30 : 1 + 2 + 3 + 5 + 6 + 10 + 15 = 42 > 30 44 : 1+2+4+11+22 = 40 < 44 50 : 1+2+5+10+25 = 43 < 50 The only one of these which qualies as an abundant numbe is 30. 7. The gue below is a ight tapezoid with side lengths 4 cm, 4 cm, and 6 cm as labelled. The cicle has adius 2 cm. The aea, in cm 2, of the shaded egion is: 4 cm 2 cm 4 cm 2 cm 6 cm Answe. (d). The aea in question is the aea of a tapezoid less the aea of a semicicle. The aea of the semicicle is obviously 1 2 22 =2 cm 2. The aea of the tapezoid is 1 2 4(4 + 6) = 20 cm2. Thus, the shaded aea is 20 ; 2 cm 2.
347 8. Thee vetices of paallelogam PQRS wee P (;3 ;2), Q(1 ;5), and R(9 1) with P and R diagonally opposite. The sum of the coodinates of vete S is: Answe. (e). Letthecoodinates of the point S be ( y). Since PSkQR, they must have the same slope: y +2 +3 = ;5 ; 1 1 ; 9 o 4y ; 3 = 1. = 3 4 Since RSkQP, we also have (by the same agument): y ; 1 ;5 +2 = ; 9 1+3 o 4y +3 = 31. = ; 3 4 Fom these two equations in 2 unknowns we easily solve fo = 5 and y =4. Thus, + y =9. 9. Which shape cannot be lled, without any ovelapping, using copies of the tile shown on the ight? Answe. (b). The diagam below shows how gues (a), (c), (d), and (e) can be lled with copies of the \T" tile. No matte how one ties gue (b) cannot be lled with copies of it. (a) (c) (d) (e) 10. Aange the following in ascending ode: 2 5555 3 3333 6 2222 Answe. (e). Notest that 2 5555 = (2 5 ) 1111 = 32 1111 3 3333 = (3 3 ) 1111 = 27 1111 6 2222 = (6 2 ) 1111 = 36 1111 Since 27 < 32 < 36, wehave 27 1111 < 32 1111 < 36 1111, which means 3 3333 < 2 5555 < 6 2222.
348 11. 2000 days, 2000 hous, 2000 minutes, and 2000 seconds would be equivalent to N million seconds. Of the choices oeed, the closest appoimation of N is: Answe. (d). Let us st compute the numbe of seconds in 1 day, 1 hou, 1 minute, and 1 second, and then multiply by 2000. Now 1 day plus 1 hou is clealy 25 hous. Then 1 day, 1 hou, plus 1 minuteis2560 + 1 = 1501 minutes. Epessed in seconds this is 1501 60 = 90060 seconds. Thus, 1 day, 1 hou, 1 minute, and 1 second is 90 061 seconds. The answe to the poblem is this gue multiplied by 2000 that is, 180 122 000, which to the neaest million is 180 000 000. 12. A thee-digit decimal numbe abc may be epessed as 100a + 10b + c whee each of the digits is multiplied by its espective place value and subsequently summed. If a = b = c and a > 0, which of the following numbes must be a facto of the thee-digit numbe abc? Answe. (e). Ifa = b = c, then 100a +10b + c =100a +10a + a = 111a. Since a can be any digit, in ode fo a numbe to be a facto of the thee-digit numbe, it must be a facto of 111. The factos of111 ae 1, 3, 37, and111. The only one of these appeaing in the list is 37. 13. If( + y) 2 ; ( ; y) 2 > 0, then Answe. (a). ( + y) 2 ; ( ; y) 2 > 0 (=) 2 +2y + y 2 ; 2 +2y ; y 2 > 0 (=) 4y > 0 (=) y > 0. The last condition clealy holds if and only if and y have the same sign that is, both ae positive o both ae negative. 14. Conside all non-conguent tiangles with all sides having whole numbe lengths and a peimete of 12 units. The following statements coespond to these tiangles. (i) Thee ae only thee such tiangles. (ii) The numbe of equilateal tiangles equals the numbe of scalene tiangles. (iii) None of these tiangles ae ight angled. (iv) None of these tiangles have a side of length 1 unit. Of the fou statements made, the numbe of tue statements is: Answe. (d). Let a, b, c be the thee sides of the tiangle. Let us assume that a b c. Since the peimete is 12, we have a + b + c =12. Let us now list all possible sets of integes (a b c) satisfying the above conditions: (1 1 10), (1 2 9), (1 3 8), (1 4 7), (1 5 6), (2 2 8), (2 3 7), (2 4 6), (2 5 5), (3 3 6), (3 4 5), (4 4 4).
349 Howeve, it is clea that some of these \tiangles" do not actually eist, since in any tiangle the sum of the lengths of the two shote sides must be geate than the length of the longest side. With this additional condition we have only the following tiangles (a b c): (2 5 5), (3 4 5), (4 4 4). We can now eamine the fou statements and conclude that (i), (ii) and (iv) ae clealy tue. As fo (iii), we see that tiangle (3 4 5) above is ight-angled hence, (iii) is false. 15. An altitude, h, of a tiangle is inceased by a length m. How much must be taken fom the coesponding base, b, so that the aea of the new tiangle is one-half that of the oiginal? Answe. (e). The aea of the oiginal tiangle is 1 bh. The new tiangle 2 has altitude h+m and base b;. We need to nd such that the aea of the new tiangle is 1 bh. Clealy the aea of the new tiangle is 1 (h + m)(b ; ). 4 2 Thus, 1 bh 4 = 1 (h + m)(b ; ) 2 bh 2(h + m) = b ; = bh b ; 2(h + m) = 2bh +2bm ; bh 2(h + m) = b(2m + h) 2(h + m) That completes the Skoliad Cone fo this issue. Send me you contest mateials and any communications about the Cone.