Theta Individual Test - Mu Alpha Theta

Theta Individual TestMu Alpha Theta National Convention 2003The abbreviation NOTAdenotes “None Of These Answers.”6. Rectangle ABCD is drawn with AD=10 andBA=6. Line AE is drawn with slope 51 ,and AD has slope zero. Point E is on CD .1. For what value of k is x = 5 a solution toGive the length AE.the equation x2+ 3 = kx + 9 ?BCA. 2 261913A.B.E55B. 2 29C. 10141C.D. E. NOTA D. 401 AD55E. NOTA2. If w is increased by 400% of w and theresult is decreased by 80% (of the result)then 54 is the final value. What is 80%of w ?A. 13.5 B. 43.2C. 54.0 D. 67.5 E. NOTA1 1 13. If A + B = and A = 2B , then2 3 5what is the value of 100 ( A + B) ?7. For a = b +1, and ab ≠ 0 the expression⎛ 1 1 ⎞⎜ + ⎟− ⎜ a b ⎟ is equal to which of the following?⎜ 1 1 ⎟−⎝ a b ⎠A. 2 b +1 B. 2 b −1C. − 2 b + 1 D. − 2b −1E. NOTA8. AB bisects ∠CAD.If m∠CAB= 4x− 8and m ∠BAD= 2y+ x and m∠CAD = 100then find the value of y .A. 18 B. 45C. 60 D. 63 E. NOTAC4. The sum of two numbers is 2 and theA. 11.50Bdifference of the same two numbers is 4.B. 14.50What is the sum of the cubes of theC. 17.50same two numbers?D. 17.75E. NOTAADA. 26 B. 28C. 56 D. 72 E. NOTA 9. Lines l, m and n are three distinct lines.Line l is perpendicular to line m, and line5. A line with slope 4 and x-intercept − 8m is perpendicular to line n. Lines l and n maycontains the point P which is 40 units be ...above the x-axis. What is the x-coordinateof point P ?i) parallel ii) perpendicular iii) skewA. 2 B. 16 A. i only B. ii onlyC. 18 D. 32 E. NOTA C. i, iii only D. ii, iii only E. NOTAPage 1

Theta Individual TestMu Alpha Theta National Convention 200310. m∠BAD, inscribed in the circle as shown,15. If log 2 A = log8B and log 4 B = C , andlog log(2Kis 40 . The measure of arc A = ) then give the value of k.CA is 60 . Givethe measure of ∠ADC .33CA.B.A. 30B2C2B. 40C. 60D. 80E. NOTA11. A right triangle has sides of lengths6, 8 and 4 x + 2 . Which could be the valueof x ?AC40DC.23CD.2C3E. NOTA216. If = A + B C where B C is in2 − 3simplest radical form, then which is thevalue of B − A ?A.C.17 − B.232D.74A. − 1 B.−171C.D. 1 E. NOTA7 1 −72 217. Line L, perpendicular to the line withE. NOTA12. A rhombus has perimeter 40 and onediagonal of length 12. Give the area ofthe rhombus.A. 48 B. 96C. 120 D. 240 E. NOTA13. If x and y are positive integers andx + y < 150 , and x > 10 , then what is theleast possible value for x − y ?A. − 139 B. − 128C. −127D. − 125 E. NOTA14. An isosceles triangle has legs each oflength 4. The altitude to the base haslength 3. Give the length of the base.A. 5 B. 2 7C. 4 3 D. 10 E. NOTAequationy = 3x− 5 , contains the point(1, 4). What is the x-intercept of L?A. 12 B. 13C. 14 D. 15 E. NOTA18. If r is a solution to the equation( x −1)(x + 3) = 5 then 4 r + 1 could be ...A. 25 B. 16C. 7 D. 9 E. NOTA19. The number 401 five , written in base-five,is equal to the expression 100 A + A , abase ten expression. Give the value of A.A. 1 B. 2C. 3 D. 4 E. NOTAPage 2

20. You have 42 inches of string. You want to cutit to form an equilateral triangle and a square,using all of the string for both. If the sidesof the equilateral triangle and the squareare congruent and distinct, what is the areaof the square, in square inches?Theta Individual TestMu Alpha Theta National Convention 200324. Let A be the total combined area of theshaded region, which consists of the area ofa trapezoid and a semicircle. The parallelbases of the trapezoid are vertical. Ifthen2 A = B + Cπ B + C =A. 24 B. 36C. 49 D. 70.56 E. NOTA(0,4)(5,6)21. A team played its first 20 games and won15 of them. No ties are allowed. They thenplayed five more consecutive games and lostall five. What is the minimum number ofconsecutive games that this team must nowplay and win to bring its winning percentageback to at least what it was before the fivegame losing streak?A. 2 B. 12C. 15 D. 17 E. NOTA(11,0)A. 29.5 B. 43C 59 D. 109 E. NOTA25. In the diagram shown, AC=12 and BE=5,m∠ACB = 30and m∠ECD = 45.B, E and C are collinear. Give the value of CD.A22. Lines BD and AC are parallel, and AC=10.If m∠DAC= 30, m∠BAC= 60and thearea of ∆ABC is 75, which is the area of∆DAC ?B DA. 37. 5B. 37 .5 2C. 37 .5 3D. 75E. NOTAACA. 2B. 6 6 − 5 2BC. 3 6 − 2.5 2D. 6 6 − 5E. NOTAED3045C23. The graph of 9 x2+ 25y2= 225 has fociat points P and Q. The graph ofx2+ y2− 8x+ 4y− 2 = 0 has center R. Ofthe distance PR, and QR, give the distancethat is the larger.A. 2 B. 6C. 2 17 D. 2 34 E. NOTA26. The graph of a parabola opens downward,with y-intercept 10 and x-intercepts − 1and 5. If the point P(8, k) lies on the graphof the parabola, what is the value of k?A. − 60 B. − 54C. − 27 D. − 8 E. NOTAPage 3

27. A sphere of radius R inches is inscribedin the cylinder shown so that the sides andbases of the cylinder touch the sphere. Thevolume of the region outside of the sphereand inside the cylinder is 18 π cubic inches.Give the area of the great circle of thesphere.Theta Individual TestMu Alpha Theta National Convention 200329. Arc BC has a degree measure that is3 that of arc EF, shown. The area of the4sector shaded in circle A, bounded by arcBC , is 81 that of the sector (shaded) incircle D, bounded by arc EF. If AB haslength x then what is the length of DE interms of x ? (diagram not drawn to scale)BA. 36π B. 16πC. 9π D. 5.76πE. NOTAA. 48xB. 4 3xC. 6 x4 2D. x3E. NOTADAECF28. X liters of a mixture with 20% acid aremixed with four liters of pure acid, toresult a new mixture that is 25% acid.How many total liters (X) were in theoriginal solution before the pure acid wasadded?A. 15 L B. 24 LC. 60 L D. 80 L E . NOTA30. A triangle has angles of measures( 9 x + 9) degrees, ( 4 x + 4)degrees,and kx degrees, for some constant k.Each angle is between 0 and 180 degrees.If the triangle is a right triangle, then whatis the largest possible value for k ?A.7713B. 10C.117077D. There is no upper bound for k.Page 4E. NOTASolutions:

Theta Individual TestMu Alpha Theta National Convention 200313. x may be between 10 and 149. To minimize1. 25 + 3 = 5k + 9. k = 19/5. Choice A.x-y, and maximize y we use 11-138 for the2. ( w + 4 w)0.20= 54. w = 54. 0.80w= 43.2difference. This is -127, choice C.Choice B.14. Using the Pythagorean Th. 4 43. Multiply the first equation by 30. 15A+10B=6.gives half the base is 73Substitute to get 30B+10B=6, B=3/20 and so the base is 2 7 , choice B.A=3/10. 100 times the sum is 45. Choice B.log A log B15. Using the change of base rule, =4. x+y=2 and x-y=4. Adding gives 2x=6. x=3log2 3log2and y= -1. Sum of cubes is 27+ -1 = 26. Choice A.which gives 3 log A = log B . In the second5. Using slope 4, and point ( -8,0) giveslog B 3log Ay-0=4(x+8). Substituting (k,40), 40=4(x+8)equation, = C and thus = C2log22log 2and x=2. Choice A.2C6.ED 1= , =ED and ED=2. Using the2log25 AD 5 10so log A = ⋅C= log 2 3 which gives3Pythagorean Theorem we get AE= 100 + 4a power of 2C/3, choice D.which is 2 26 or choice A. 16. Multiplying numerator and denominator by7. Multiply numerator and denominator by ab2 + 3 2a + bthe conjugate 2 + 3 gives sogives − and substituting gives− 7b − aB= -3/7 and A= -2/7, and B-A= -1/7. Choice Bb + 1+b= 2b+ 1 which is choice A.b + 1−b17. The slope of the perpendicular line is -1/3.8. 4x-8=50, x=14.5, and 2y+14.5=50 so y=17.75. The equation is then x+3y=13, and if y=0, x=13.Choice D. Choice B.9. Since the lines are not necessarily coplanarchoices i, ii and iii can be true. The lines may18. x2+ 2x− 3 = 5, x2+ 2x− 8 = 0, (x+4)(x-2)=0be perpendicular if all three lines intersect atone point. Choice E.gives x = r = -4 or 2. So 4r+1= -15 or 9.10. Since arc AC is 60, angles ABC and ADC areChoice D.both 30 degrees. Choice A.11. If 4x+2 is the hypotenuse then 4x+2=10 and19. 4(25)+0(5)+1=101. 101A=101 gives A=1.x=2, not a choice. If 8 is the hypotenuse, thenChoice A.4 x + 2 = 64 − 36 = 2 7 so20. Seven sides for 42 inches gives 6 inches for2 7 − 2 7 −1x= = which is choice D. each. Area of the square is then 36. Choice B.4 212. The sides are each 10 and diagonals are15 + A 3perpendicular so we have four 6-8-10 right21. = (which is 75%) gives A=15.25 + A 4triangles in the interior. So each has area Choice C.1(6)(8) which is 24. And 24X4=96. Choice B. 22. Same base and same height gives the same2area. The angles are irrelevant. Choice D.Page 5

Theta Individual TestMu Alpha Theta National Convention 200323. 2 2+ yANSWERS:= 1 gives a=5 and b=3 and25 91. Aa2− b2= c2and c is the distance from2. Bcenter to focus. So foci are (4,0) and (-4,0).3. BCompleting the square for the second conic4. Agives ( x − 4)2+ ( y + 2)2= 22 so center R is5. Aat (4, -2). The distances PR and PQ are then6. C7. A2 and 82+ 22= 2 17 , the larger. Choice C.8. D124. Trapezoid = (4 + 6)5 = 25 . Semicircle= 9. E210. A1π (9) = 4. 5π . 2A=50 + 9π so B+C=59.11. D212. BChoice C.13. C25. EC= 6 3 − 5 so CD = ( 6 3 − 5) 2 which 14. Bequals 6 6 − 5 2 , choice B.15. D26. The equation of the parabola is16. By = −a( x + 1)( x − 5) . If the y-intercept is 1017. B(let x=0) we get a=2. Set x=8, we get y= -54.18. Dwhich is choice B.19. A420. B27. Volume = πr2h − πr3. Setting h=2r, and 21. C32 22. Dsimplifying gives V = πr3 . Setting this 23. C3equal to 18 π gives r=3, and the area of the24. C25. Bgreat circle is π r2 = 9π . Choice C.26. B28. 0 .20x + 4 = 0.25( x + 4) so x=60. Choice C.27. C3m28. C129. 4 2 m( π x ) = ( πR2)( ) . This solves to29. C360 360 830. C6x 2= R2and so R = 6x. Choice C.30. One of the angles must be 90 degrees.If 9x+9=90 and then x=9. If 4x+4=90 thenx=21.5. If kx=90 then the sum of the othertwo angles is 90 and then 13x+13=90 givesx=77/13. Either way, the sum of all angles is180 and 13x + 13 + kx = 180. For x=9, k=5.55and for x=21.5, k is negative. We discard thisanswer. If x=77/13 then k= 1170/77 whichis the larger answer for k, which is choice C.Page 6