WYSE 2012 Math Sectional Solution Set. = Where. parametric relationship. That means that its graph would have dual lines.

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1 x 1. Ans E: Think of f ( x) = uv, where u = x and v = e. The derivative of uv, y the x x x x produt rule, isu' v + v 'u = e + x e = e + 0xe.. Ans A: Using Cramer s Rule we have x 1 1 = Where a a 1 1 a = 1, a = 4, =, =, = 1, =. The system is ax+ y= ax+ y= x + y = 1 4x + y =. We see that. Ans E: As x = sin t and y sin = t, y = x. Thus, y = xand y = xoth satisfy this parametri relationship. That means that its graph would have dual lines. 4. Ans B: Interhange variales and solve for y. We have log x 7 y y = log x + 7. ( + ) = +. ( ) x = y + 7 x + 7 = y + 5. Ans A: We an split this figure left to right into an isoseles right triangle, a retangle, and a seond isoseles right triangle. The two right triangles will eah have area of 0.5. The retangle will have area of. Add these up to get 1+, whih rounds to.4. Irok Irok 6. Ans E: For the rok onert, = log. So1 = log. Thus, 1 1 and Irok = 1 W / m. For the whisper, whisper 1 I = 1 rok 1 I Iwhisper 0 = log. So = log. Thus, I whisper 9 = and I 1 whisper = W / m. So the ratio etween the two is given as follows: Irok 1 9 = = = 1,000,000, I whisper 7. Ans C: Taking any point on the hyperola, say (4,0), and the foi (-5,0) and (5,0), we find the differene of the distane from (4,0) to (-5,0) and the distane from (4,0) to (5,0). We have = Ans C: The old standard deviation was, so the old variane was 0, and the sum of the squared distanes from the mean was 0*4 = 400. The new mean is still 0. All of the old distanes and squared distanes will remain the same, and the new numer adds no distane. The new sum of the squared distanes is 400, ut the new variane will e 80. This means the standard deviation will e 80, roughly 8.94, rounded to Ans D: For a sinusoidal funtion of the form f ( t) = a sin(kt), let p e the period and f e the frequeny. 1 1 π k 1 f = = = = = 55. t is in seonds, so frequeny is in Hz. p ππ π k 1

2 . Ans D: = Ans B: The area of a setor of a irle is given y A = θ r, where r is the radius of the irle and θ is the radian measure of the entral angle. We an use the relationship θd θr etween degrees and radians ( 180 = 8π π ) to see that θ = =. So, if we fill in the π known quantities in our area equation, we get π = π r. If we multiply oth 5 5 sides y π, we get 144 = r. Sor = 1". 1. Ans E: Let t represent taxi time, p the plane time and n the onnetion time. The 55 t p n = t p n = 0 system of equations is Plaing this into matrix 1 t + 0 p 1 1 n = 0 1 t + 1 p n = 11 form and solving we find the following: At this point, working akwards, we find that p =, t +1 p = 11 so t = 4. Using the top equation we an find the value of. Finally, sine =, the value of n an e found using the last equation, so n =. 1. Ans B: The equilateral triangle an e split into six idential right triangles, eah a triangle. The shorter leg has a length equal to the smaller irle s radius, and the hypotenuse has a length equal to the larger irle s radius. Sine the hypotenuse is twie as long as the shorter leg, the radius of the larger irle is twie the radius of the smaller, making its area four times that of the smaller irle. 14. Ans D: There are three types of meals listed here, main ourse-dessert-salad-appetizer, main ourse-dessert-salad-soup and main ourse-dessert-soup-appetizer. Sine all three types are disjoint, we may find the numer of eah type and then add those together to get the total numer of meal options. For eah meal type, multiply together the numer of options for eah dish type (i.e. the numer of main ourse-dessert-saladappetizer meals is found y taking = 8064, the numer of main oursedessert-salad-soup meals is = 60 and the numer of main ourse-dessertsoup-appetizer meals is = 880 ). Adding up those produts, we reeive a slate of 14,04 potential meals. 15. Ans A: Making a hart of values we find θ 0 π π π π 5π π 7π π r

3 1 Therefore the graph is a limaon with radius. 16. Ans E: Start y multiplying oth sides y AB to get A B = AB. Rearrange the equation to get A AB B = 0. Use quadrati formula to get ( B) ± ( B) 4( B ) B ± 5B B ± B 5 A =. This an simplify down to A = = 17. Ans B: In reating the list of PINs, our first step might e to hoose the two digits that need to e zeroes. There are C(5, ) = ways to do that. The next steps would e to fill in eah of the remaining three spaes. There are only 9 options at eah. By the multipliation priniple, we an multiply y the ue of 9 in order to get 7,90 possile 5-digit PINs with exatly two zeroes. You may take solae in the fat that she is highly unlikely to guess orretly efore the mahine, as a seurity measure, eats your ard. dz 18. Ans B: x + 40 = z. When z = 50, y Pythagorean theorem, x = 0. 6 dt =. dx dz dx Plugging these values into x = z we find dt dt dt =. 40 z x Ans C: The area of a regular polygon is given y the equation A = ap, where a is the length of the apothem (a segment emanating from the enter of the polygon whih happens to serve as oth an altitude and a isetor to one of the sides) and p is the perimeter. Sine an 11-gon has 11 sides, the perimeter of this partiular figure would e 176 inhes (11 times 16). The easiest way to find the area of a non-triangular, nonsquare, non-hexagonal regular polygon is likely found y forming an isoseles triangle etween the enter thereof and two adjaent verties, the vertex angle of whih measures Thus, the other two angles of the triangle are. 11 =. So, if one draws an altitude from the vertex angle to the other side, they have formed the apothem a as the long leg of a right triangle, the short leg of whih is

4 (half of the side length of the original figure). By the Law of Sines, we an reate the 8 8 sin a 8 following relationship: =. So a =. Thus, y our sin sin sin 8 8 sin 1 equation for the area of a regular polygon, A = 176,98. Whew. 180 sin (There is an alternative solution in whih the same triangle is used, ut you note the size of the halved angle, and use a tangent approah in order to find the apothem.) 0. Ans C: Using Law of Cosines we have ( )( ) = os 45. Therefore = 6.7. The desired distane is approximately 6 ft. 9 in. 1. Ans D: This an e modeled using the funtion P = 00 * 0.5. Set P = and solve for t to get = 00 * 0.5 t 0.01= 0.5 t = t t ln0.01 ln0.5 ln0.01= ln0.5 ln0.01 t = We an also model this using a P = ln0.5 would e the same. Ae kt t funtion, ut the results. Ans C: If we multiply.85 (15% off),.65,.9,.85,.8 and 1.06 (6% sales tax) together, we get, to the nearest hundredth,.7, whih would orrespond to a 6% disount.. Ans D: The prie range within two standard deviations for eah store is given elow: Kohl s - $18 to $50, Sears - $ to $4, May s - $19.75 to $4.75, Dress Barn - $6 to $4 and JC Penney - $ to $9. Therefore, Kohl s will have the lowest prie. 4. Ans A: If all sides have length x, the surfae area must e x + x + x + x + x. Solve ( ) x = 0 for x to get so insert the value of x to get a volume of approximately ui inhes. x The volume ends up eing ( ) x x, 4 5. Ans A: The geometri mean of the numers is x x = 8. So 11x = Thus x =. So the two possiilities for x are x = ±. The sum of those two potential x-values is Ans E: The side of the square is 144 = 1. So the perimeter of one-fourth irle is π ( 1) given y = 6π. Sine eah side of the square is 1 feet long, the perimeter of 4 the fish pond will e (6π + (1)) ft.

5 7. Ans C: Every omplex numer an e thought of as an ordered pair (r,θ), where r is the unique distane etween that point in the omplex Cartesian plane and the origin and θ is an angle etween the line segment etween that numer s orresponding point in the plane and the origin and the positive x-axis. De Moivre s theorem says that n an e thought of as the ordered pair ( r,nθ ). An angle that the segment from the origin 1 1 to+ imakes with the positive x-axis an e given y θ = tan. So, y de Moivre s theorem, an angle that the segment from the origin to ( + i) 50 makes with the positive x- axis an e given y φ = 50θ = 50 tan , the terminal edge of whih would e in qiii. 8. Ans A: π r = 7.5. So r = Sine A = 1 1 π r then A = Ans B: Car B would ath ar A when they have traveled the same distane. If x is the numer of hours after noon, then solving 60x = 70(x 0.5) would give us the orret time. Solving gives us x =.5, whih means at :0 PM. The distane for oth would e miles, so ar B does ath ar A. 0. Ans C: First, it must e noted that 5 yards is 5 feet. Then, if we form a right triangle whose legs are from my vantage point to the point on the uilding diretly aross from me and from that point to the top of the uilding, we an note that we have a right triangle with a 15 degree angle, whose adjaent side is 5 feet and whose opposite a side is unknown. So tan15 = and the distane from the point diretly aross from 5 me to the top of the uilding, a, is5 tan15. If we susequently form a right triangle whose legs are from my vantage point to the point on the uilding diretly aross from me and from that point to the ottom of the uilding, we an note that we have a right triangle with a 5 degree angle, whose adjaent side is 5 feet and whose opposite side is unknown. So tan5 = and the distane from the point diretly aross from 5 me to the ottom of the uilding,, is5 tan5. Thus the height of the uilding is a + = 5 tan tan5 '. 1. Ans B: From the given, we know that m CAB = 0 sine m EAC + m CAB = 180. This makes m DCB = 0. ABC is the alternate interior angle of DCB. Therefore the m ABC = 0. The final vertex angle in the triangle, ACB, must e for the vertex angle sum of the triangle to add to 180. Therefore, the triangle must e an otuse isoseles triangle.. Ans E: Turn eah rate value into tank per minute. Fauet 1 s rate is 1/50 = 0.0 tank per minute, fauet s rate is 1/40 = 0.05 tank per minute, and the drain s rate is 1/0 = 0.05 tank per minute. At 1:, the tank is 0.0* = 0., or 0% full. At 1:0, the tank is 0. + *( ) = 0.65, or 65% full. Next, we notie that = In other words, the tank will lose half a perent eah minute. Sine the tank is 65% full, it will take another 10 minutes to empty, or until :0 PM. n

6 . Ans D: From the denominator we know that x annot e 0 or as oth values make the denominator 0. Multiplying the inequality y the LCD 4x( x + ) and setting the right side to 0 we have 6x 9x + 6 < 0. Dividing y produes the inequality x + x > 0. The ritial values found here are 1 and. These values must e exluded sine the inequality does not ontain values equal to 0. Plugging in values around the exluded values, we find that only numers hosen in,,0 1, will work. ( ) ( ) ( ) 4. Ans D: Let u = 4 + 9os( 8 + 9t). The osine of a linear funtion an take on all values etween -1 and 1. This means that u an take on all values etween 4 and 5. Sine this overs a potential range of 18 units (muh larger than the period of a sine funtion, f t an thus vary etween 116 whih is only π ), sin u an vary etween -1 and 1, and ( ) and 18. So the differene etween the peak and trough is twie the amplitude of the sinusoidal funtion desried therein, $1M. 5. Ans A: There are 4 quarts in a gallon, and thus 4 peks in a ushel. So a ushel and a pek would e equivalent to five peks. He must then pik a total of 45 peks of pikled peppers, 44 more than he already has. Peter Piper is swithing to the metri system at the next opportunity. 6. Ans D: ( ) Ans B: On eah roll, there is a 5/6 hane that the person does not roll a doule. To get this result three times, we take 5/6 * 5/6 * 5/6 = Ans A: Let x e the proaility that Clyde makes his shot (as a deimal). Then Bonnie has a 1.5x proaility of sinking her shot. The proaility that Clyde misses is then 1 x and the proaility that Bonnie misses is 1 1.5x. We know that the proaility that they oth miss, expressed as a deimal, is (as it s the omplement of the event where 1 x 1 1.5x = and, if we at least one makes the shot). So we have that ( )( ) expand the left-hand side, we get1.5x.5x + 1 = Bringing everything over to one side for quadrati formula purposes, the equation eomes1.5x.5x = 0. If we go through the quadrati formula and simplifiation, we get:.5 ± (.5) 4(1.5)(.515).5 ± ± ± 1.75 x = = = = (1.5) If we simplify at the end, we get either x =.5 or x = Sine the latter one of those solutions is not a possile proaility, Clyde has a.5 = 5% hane of making his shot. 9. Ans E: ( ) x ( x 5 (7 ( x x ) x)) 1( x) x x x + x 6x 4 + 1x x x (7 x x x) 4 1x x.

7 40. Ans C: Based on the given eams and total, we have the following four equations: A + B = C + D, D + A = B, A + B + C + D = 0, and D= C+. This an e solved slowly with sustitution, or rather quikly with matrix row redution. Either way, we end up with A = 1, B = 11, C =, and D = 5.

2012 Academic Challenge

2012 Academic Challenge 01 Academic Challenge MATHEMATICS TEST - SECTIONAL This Test Consists of 40 Questions Mathematics Test Production Team Kevin Boyer, Illinois State University Author/Team Leader Linda Wiggins, Illinois