Boyle’s and Charles’ Laws

SoluUons 1. Answer (D): The average of the five values is 6 + 0.5 + 1 + 2.5 + 10 5 = 20 5 = 4. 2. Answer (В): Let s Ье the side length of the smaller square. Then the length of the rectangle is 4s, and the width is 4s - s = 3s. Hence the rectangle length is :: = f times as large as its width. 3. Answer (D): Let х Ье the number of marЫes that Тyrone gave to Eric. Then 97 - х = 2(11 + х). Solving this equation yields х = 25. 4. Answer (В): Весанsе 412-;- 56 is between 7 and 8, the reading will need 8 discs. Therefore each disc will contain 412-;- 8 = 51.5 minutes of reading. 5. Answer (Е): Because the circumfe1·ence is 27Гr Therefore the area is 1Тr 2 = 1447Г, and k = 144. 6. Answer (С): Note that •(2, 2) = 2 - ½ = !- = 247Г, the i-adiнs r is 12. Tl1erefore 7. Answer (С): When Crystal ti-avels one mile northeast she travels :{}- miles north and 4, miles east. Simila1:ly, when she travels southeaзt for one mile she travels :{}- miles soнth and miles east. Just before the last portion of her run she has tr-aveled а net ot· 1 + 4,- - {f- = 1 miles north, and :{}- + :{}- = v'2 miles east. Ву the Pythagorean Theorem, the last portion of her run is 1, J12 + (V2) 2 = JI+2 = Jз miles. 8. Answer (D): Топу w01·ked for 2 • 50 = 100 hours. Нis aveiage eaп.1i11gs per hour during this period is $16cf0° = $6.30 . Hence his a,rerage age during this period was = 12.6, and so at the end of the six month period 11 was 13. i�:�g Solutions 9. Answer (Е): Let х + 32 Ье written in the fшm CDDC. Because х has three digits, 1000 < х + 32 < 1032, and so С = 1 and D = О. Hence х = 1001 - 32 = 969, and the suш of the digits of х is 9 + 6 + 9 = 24. 10. Answer (Е): А пon-leap year has 365 days, and 365 = 52 • 7 + 1, so there are 52 weeks апd 1 day in а non-leap year. Because Мау 27 was after leap day in 2008, Marvin's Ьirthday fell on Wednesday in 2009, and will fall on Thursday in 2010 and Friday in 2011. I -lis Ьirthday will Ье on Sunday in the leap year 2012, Monclay in 2013, Thesday in 2014, Wednesday in 2015, Friday in 2016, and Saturday in 2017. 11. Answer (D): The solution of the inequality is Ь-3 а-3 --<х<--. 2 - - 2 If ь23 - а23 = 10, then Ь - а = 20. 12. Answer (С): The volнme scale fOl' Logan's model is 0.1 : 100,000 = 1 : 1,000,000. Therefore the linear scale is 1 : {1/1,000,000, which is 1 : 100. Logan's water tower should stand 1�00 = 0.4 meters tall. 13. Answer (А): Angelina drove 80t km before she stopped. After her stop, she drove (3 - ½ - t) hours at an average rate of 100 kph, so she covered 100( t) km in that time. Therefore 80t + 100( t) = 250. Note that t = ¾-s 1- 1- 14. Answer (С): Let а = L.BAE = L.ACD = L.ACF. equilateral, it follows tl1at L.CFА= 120° анd thcн Because 6CFE is L.FАС= 180 ° - 120° - L.ACF = 60 ° - а. Therefore L.BAC = L.BAE + L.FAC =а+ (60° - а)= 60 ° . Because АВ = 2-АС, it f'ollows that !::.ВАС is а 30-60-90° triangle, and thus L.ACB = 90° . Solпtions А 15. Answer (D): LeRoy and Chris cannot both Ье frogs, because their statements would Ье tше and frogs lie. Also LeR,oy and Chris сашюt both Ье toads, because then their statemeпts would Ье false, апd toads tell the truth. Невсе betweeв LeRoy and Chris, exactly one must Ье а toad. If Brian is а toad, then Mike must Ьс а fr-og, but tl1is is а contradictioв as Mike's statemeвt would then Ье true. Hence Brian is а frog, so Brian's statement must Ье false, and Mike mнst Ье а frog. Altogether there are 3 frogs: Brian, Mike, авd either LeRoy or Chris. 16. Answer (В): Ву tl1e Angle Bisector Тlieorem, 8 · БА = 3 · БС. Thus БА must Ье а multiple of 3. If БА= 3, the triangle is degenerate. If БА= 6, then БС = 16, and tl1e perimeter is 6 + 16 + 11 = 33. 17. Answer (А): The volнme of the solid cube is 27 in3. The fi1·st hole to Ье cut removes 2 х 2 х 3 = 12 iв3 from the volume. The other holes remove 2 х 2 х 0.5 = 2 iп 3 fr-om each of the foнr remaining f"aces. The volнme of the remaining solid is 27 - 12 - 4(2) = 7 in3 . 18. Answer (В): The probabllity that Bernardo picks а 9 is � = ½- In this case, his three-digit number will begin with а 9 and will Ье larger thaп Silvia's three-digit number. If Bernardo does not pick а 9, theп Bernardo апd Silvia will f"orm the same number with probabllity 1 1 56" Ш- If they do rюt, f"orm tl1e same nншЬеr tlien Beпшrdo's пurnber will Ье larger ½ of the time. Solutioпs Hence the probaЬility is 1 2 1 3+3• 1 ( 2 l 56) = 111 168 = 37 56. 19. Answer (Е): Тriangles АБС, CDE апd EFA are congruent, so дАСЕ is eqнilateral. Let Х Ье the int,ersectiori of t,he liпes АВ and EF апd defirie У ancl Z similarly as shown in the figure. Because ABCDEF is equiangular, it follows tliat L.X AF = L.AFХ = 60° . Tlшs дХAF is equilatcr·al. I�ct Н Ье thc midpoint of XF. Ву the Pythagorean Theorem, Thus, the area of дАСЕ is The area of hexagon ABCDEF is equal to [XYZ] - [XAF] - [УСЕ] - [ZED] = � ((2r + 1) 2 - Зr 2 ) = � (r2 + 4r + 1) Because [АСЕ]= /0 [ABCDEF], it follows that r2 from which r· 2 r is 6. - 6т + 1 +r+1 = 7 2 r 10 ( + 4т + 1) = О and т = 3 ± 2)2. The sum of all possiЫe values of z I I I I ,, \ 1� / / о I ,', ' xj- - - - :: ,...,:.------4 ,l Solutjoпs 20. Answer (D): Each of the 8 lines segments оп the fly's path is an edge, а face diagoпal, 01· an iпterior diagoнal of tlic cube. TJ:1ese thr·ee typc [ lirte segm1шLs have lengths 1, J2, and у'З, respectively. Because each vertex o'f the спЬе is visited онlу онсе, thc two liнc scgmcпts that шееt at а ve1·tex l1av• а eombluc t � 2 • 67 = 134, so tl1e шiнinшш valнe of а is 78. 22. Answer (А): Three chшds Cl'eate а triaнgle if and онlу if they iнtersect pairwise inside the ci1·cle. Two chords iнtersect iнside the circle if анd only if their endpoints alterнate in order al'ouнd the circle. Therefore, if points А, В, С, D, Е, and F are in order around the circle, then only the chords AD, ЕЕ, CF all iнtersect pairwise iнsidc Н1с circlc. Tlшs every set of 6 points determines а unique triangle, and there al'e (:) = 28 such triangles. 23. Answer (А): If Isabella reaches the k th Ьох, she will draw а white marЬle from it with probabllity k�l. For п � 2, the probability that she will draw white marЫes from each of tl1e first п - 1 boxes is 1 2 3 2 3 4 п-1 1 п so the probability that she will draw her first red marЫe from the п Lli Ьох is 2 Р(п) = n(n:+i). The conditio11 Р(п) < 1/2010 is equivalent to п +п-2010 > О, Solutioпs from which п > ½ (-1 + v®мl) and (2n + 1)2 > 8041. The smallest positive odd iнtegel' wl10se squa1·c cxceeds 8041 is 91, анd the coпespoнdiнg value of п is 45. 24. Answer (А): There are 1 8 factors of 90! that are multiples of 5, 3 factors that are multiples of 25, анd no factors that are multiples of higher powers of 5. Also, there are more than 45 factors of 2 in 90!. Thus 90! = 1021 N where N is an integer not divisiЫe Ьу 10, and if N п (mod 100) with О < п � 99, then п is а multiple of 4. = Let 90! = АВ where А coнsists of the factors tliat are relatively prime to 5 and В consists of the factors that are divisiЬle Ьу 5. Note that П�=l (5k + j) 5k(l + 2 + 3 + 4) + 1 • 2 • 3 • 4 24 (mod25), thus = = А (1 • 2 • 3 • 4) • (6 · 7 • 8 • 9) • • • • • (86 • 87 • 88 • 89) 2418 (-1)18 1 (mod25). = = Similarly, В = (5 • 1 0 • 1 5 • 20) • (30 • 35 • 40 • 45) • (55 • 60 • 65 • 70) • (80 • 85 • 90) • (25 • 50 • 75), thus в (1 • 2 • 3 • 4) • (6 • 7 • 8 · 9) • (11 • 12 • 13 • 14) • (16 • 17 • 18) • (1 • 2 · 3) 521 = (-1) • 1 = -1 (mod25). = 2 • (1024) = 2 • (-1) = 2 (mod25), so 13 • 2 = 243 • (-9). (-8) • (-7) • 6 Fiнally, 221 = 2 • (21 0)2 1 3 • 2 1 (mod25). Therefore = N 90! 521 12 (mod25). (13 · 221 )N -13 = = 13 • 3 2 2 = 1 3 ·А· в21 5 21 = 13 • 1 · (-1) (mod25) Thus п is equal to 12, 37, 62, or 87, and because п is а rnultiple of 4, it follows that п = 12. 25. Answer (В): Let the sequence Ье (а1, а2, ... , а8). For j > 1, aj-l = aj + т2 for some т sнch that aj < (т + 1)2 - rn2 = 2m + 1. 1Ъ miнimize the value of а1, coнstruct the sequence in reverse order and choose the smallest possiЬle value of m for each j, 2 � j � 8. The terms in reveгse order are а8 = О, а7 = 1, а5 = 1 + 12 = 2, а5 = 2 + 12 = 3, а4 = 3 + 22 = 7, а3 = 7 + 42 = 23, а2 = 23 + 122 = 167, and N = а 1 = 167 + 842 = 7223, wl1ich has the unit digit 3.