<p>Mu Alpha Theta 2004 State Bowl Questions - Alpha</p><p>Round 1</p><p>1. The maximum possible number of identical rectangular blocks are placed inside a rectangular carton. The rectangular blocks each have a length of 7, a width of 5, and a height of 4 and the rectangular carton has inside dimensions that are length of 16, a width of 15, and a height of 14. What is the total surface area of the rectangular blocks that are inside the rectangular carton?</p><p>2. When a right triangle of area 3 is rotated 360o about its shorter leg, the solid that results has a volume of 30. What is the volume of the solid that results when the same right triangle is rotated about its longer leg?</p><p>Mu Alpha Theta 2004 State Bowl Questions – Alpha</p><p>Round 2</p><p>3. If f x  x2 and g x  2x, what is the value of f g 3  g f 3 ?          </p><p>4. A committee of 3 women and 2 men is to be formed from a pool of 11 women and 7 men. How many different committees can be formed?</p><p>Mu Alpha Theta 2004 State Bowl Questions - Alpha</p><p>Round 3</p><p>5. During the eighteenth century , the population of country X increased by 1,000 percent. During the nineteenth century, the population increased by 500 percent. By what percent did the population increase over the two century period? </p><p>49 35 21 6. Arrange in order from the smallest value to the largest value 2 , 3 , 5 .</p><p>Mu Alpha Theta 2004 State Bowl Questions – Alpha</p><p>Round 4</p><p>7. Rachel drives one-third of the distance from A to B at x kilometers per hour, and she drives the other two-thirds of the distance at y kilometers per hour. What is Rachel’s average rate of speed, in kilometers per hour and in terms of x and y, for the entire trip?</p><p>8. Find all integral solutions x, y of the equation: x2  xy  5 ordered pair s required .       Mu Alpha Theta 2004 State Bowl Questions – Alpha</p><p>Round 5</p><p> 3  3  3  3  3  3   3  9. Find the numerical value of the product: 1 1 1 1 1 ... 1 1  4  5  6  7  8  26  27</p><p>10. If a wooden right circular cylinder with radius 2 meters and height 6 meters has a cylindrical hole of diameter 2 meters drilled through the center, what is the entire surface area (including the top and bottom faces), in square meters, of the resulting figure.</p><p>Mu Alpha Theta 2004 State Bowl Questions - Alpha</p><p>Round 6 11. Express as an integer: . 93 36 3  61 28 3</p><p>3 12. The roots of the equation x  19x  30 are a, b, and c. Find the numerical value of the product: a  1 b  1 c  1    </p><p>Mu Alpha Theta 2004 State Bowl Questions - Alpha</p><p>Round 7</p><p>13. In a sequence of positive integers, each term after the first term is one more than the previous terms. If the eleventh term is 2560, find the first term.</p><p>14. Find the real solution x, y,z to the following system of equations:   2 3 1 1 12 4   2   1   3 x y x z y z</p><p>Mu Alpha Theta 2004 State Bowl Questions - Alpha</p><p>Round 8</p><p>15. A train 100 meters long travels at a speed of 50 kilometers per hour. The train passed through a tunnel. The end of the train emerged from the tunnel exactly 12 minutes after the front of the train entered it. How long, in kilometers, is the tunnel?</p><p>16. A bicycle factory was supposed to fill a certain order for bicycles in 6 days. The factory produced 25% more bicycles each day than it was supposed to, and as a result, it filled the order in 5 days, producing 42 extra bicycles besides. If the same number of bicycles was produced each day, what was that number? Mu Alpha Theta 2004 State Bowl Questions – Alpha</p><p>Round 9 2 17. Express in the simplest form, the real number  3 75  12  .  </p><p>18. The second, sixth, twenty-second and last term of an increasing arithmetic progression, taken in this order, form a geometric progression. Find the number of terms in the arithmetic progression.</p><p>Mu Alpha Theta 2004 State Bowl Questions - Alpha</p><p>Round 10</p><p> 19. If A is not a multiple of , find the simplified numerical value of 2 1 1 1 1 2  2  2  2 . 1 sin A 1 cos A 1 sec A 1 csc A</p><p> x x2 20. Solve for x: 5  5  120 5 Mu Alpha Theta National Convention 2004 Alpha State Bowl Answers</p><p># Answer # Answer 1-1 3,984 7-13 4 4 2  4  1-2 7-14  ,6,4 5  3  2-3 18 8-15 9.9 2-4 3,465 8-16 210 1 3-5 6,500% 9-17 3 3-6 521,249,335 9-18 86 3xy 4-7 10-19 2 2x  y (1,4), (5,-4), 4-8 10-20 3.5 (-1,-4), (-5,4) 5-9 203 5-10 42 6-11 4 6-12 -48 Round 1</p><p>1 The maximum possible number of identical rectangular blocks are placed inside a rectangular carton. The rectangular blocks each have a length of 7, a width of 5, and a height of 4 and the rectangular carton has inside dimensions that are length of 16, a width of 15, and a height of 14. What is the total surface area of the rectangular blocks that are inside the rectangular carton?</p><p>Answer: 3,984 Solution: The placement of the blocks can be based on the factors of 16, 15, and 14. 14 16 15  2  4  3 There are 2 4 3  24 blocks Each inside block 7 4 5     has surface area of 274 275 245 56  70  40  166  Total surface area = 24 166 =3984   2. When a right triangle of area 3 is rotated 360o about its shorter leg, the solid that results has a volume of 30. What is the volume of the solid that results when the same right triangle is rotated about its longer leg? 4 Answer:  2 5 1 6 1 Solution: Area of the triangle = 3  rh  h  Volume of cone  V  r 2h 2 r 3 1 6 15 6 2  r 2  2r  30  r=  h   When the triangle is revolved 3 r  15 5  2 1  2  15 4 2 about the longer side, the radius and height are switched. V    3  5   5</p><p>Round 2 3. If f x  x2 and g x  2x, what is the value of f g 3  g f 3 ?           Answer: 18 2 g3 23 6 f 3 3  9  f g-3 f 6 36 Solution:  g f 3  g 9  18  f g 3  g f 3  36  18  18            4. A committee of 3 women and 2 men is to be formed from a pool of 11 women and 7 men. How many different committees can be formed?</p><p>Answer: 3465 11! 11109 Combinations of 11 women taken 5 at a time is   165 3!11-3! 321 Solution: 7! 76 Combinations of 7 men taken 2 at a time is   21 2!7-2! 21 The number of combined combinations is 21 165 =3,465    Round 3</p><p>5. During the eighteenth century , the population of country X increased by 1,000 percent. During the nineteenth century, the population increased by 500 percent. By what percent did the population increase over the two century period? </p><p>Answer: 6,500% Solution: If the original population is P, then 1000% is 10P. After an increase of 1000% we have 11P at the beginning of the nineteenth century. 500% is 5 times and 500% of 11P is 55P. 11P + 55P = 66P 66P = P + 65P  an increase of 6500% </p><p>49 35 21 6. Arrange in order from the smallest value to the largest value 2 , 3 , 5 .</p><p>21 49 35 Answer: 5 , 2 , 3 7 7 7 7 7 7 Solution: 249  27  128  335  35  243  521  53  125             35 49 21 Therefore 3  2  5</p><p>Round 4</p><p>7. Rachel drives one-third of the distance from A to B at x kilometers per hour, and she drives the other two-thirds of the distance at y kilometers per hour. What is Rachel’s average rate of speed, in kilometers per hour and in terms of x and y, for the entire trip? 3xy Answer: 2x  y 1 2 Solution: Since the distance is divided into thirds, let d be the of the distance and 2d be of the 3 3 total distance 3d 3d 3dxy 3xy Avg rate      distance. total time d 2d dy  2dx dy  2dx y  2x  x y xy 8. Find all integral solutions x, y of the equation: x2  xy  5 ordered pair s required .       Answer: 1, 4 , 5,4 , 1,4 , -5, 4         Solution: x2  xy  x x  y  5  factors of 5 are -1, -5, 1, 5    giving solutions 1, 4 , -1, -4 , 4, 1 , -5, 4 , 5,-4           Round 5</p><p> 3  3  3  3  3  3   3  9. Find the numerical value of the product: 1 1 1 1 1 ... 1 1  4  5  6  7  8  26  27</p><p>Answer: 203  7  8  9  10  11  27  28  29  30 282930 Solution: Rewriting ...   203  4  5  6  7   8   24  25  26  27 4 5 6    </p><p>10. If a wooden right circular cylinder with radius 2 meters and height 6 meters has a cylindrical hole of diameter 2 meters drilled through the center, what is the entire surface area (including the top and bottom faces), in square meters, of the resulting figure.</p><p>Answer: 42 Solution: The entire surface area consists of the lateral area of the outside cylinder, the lateral area of the inside cylinder, and the top and bottom of a circle minus the inner circle. 2Rh+2rh+2 R 2  r 2  2 2 6  2 1 6  2  4   1  24  12  6  42            </p><p>Round 6 11. Express as an integer: 93 36 3  61 28 3</p><p>Answer: 4 2 2 Solution: Using the principle square For 93+36 3  a  b  93 2ab  36 3  2 ab  18 3  a  9, b  2 3 93 36 3  9  2 3 For 61 28 3  a2  b2  61 2 2ab  28 3  ab  14 3 a  7 b  2 3 61 28 3  7  2 3</p><p>2 2 93 36 3  61 28 3  9  2 3  7  2 3  9  2 3  7  2 3  16  4     3 12. The roots of the equation x  19x  30 are a, b, and c. Find the numerical value of the product: a  1 b  1 c  1    </p><p>Answer: -48 Solution : a  1 b  1 c  1  abc  ab  bc  ac  a  b  c  1 For x3  19x  30  0         abc  30 ab  bc  ac  19 a  b  c  0 a  1 b  1 c  1  30  19  0  1  48       Round 7</p><p>13. In a sequence of positive integers, each term after the first term is one more than the previous terms. If the eleventh term is 2560, find the first term.</p><p>Answer: 4 Solution: Let x = the first term. The successive terms are x + 1, 2x + 2, 4x + 4, . . . , 512x + 512 where 512x + 512 is the eleventh term. 512x + 512 = 2560 x = 4</p><p>14. Find the real solution x, y,z to the following system of equations:   2 3 1 1 12 4   2   1   3 x y x z y z</p><p> 4  Answer: ,6,4  3 </p><p>1 1 1 a  , b  , c  , to obtain 2a  3b  2, a  c  1, 12b  4c  3. Substituting x y z 3 1 1 Solution: a  c  1, we obtain by linear combinations a  , b  , c   . Therefore 4 6 4 4  4  x  , y  6, z  4  ,6,4 3  3    Round 8</p><p>15. A train 100 meters long travels at a speed of 50 kilometers per hour. The train passed through a tunnel. The end of the train emerged from the tunnel exactly 12 minutes after the front of the train entered it. How long, in kilometers, is the tunnel?</p><p>Answer: 9.9 kilometers 1 Let d  the length of the tunnel in kilometers. The train travels d  kilometers 10 1 d  1 distance 10 1 Solution: in 12 minutes = hour. Since rate =  50=  10  d  5 time 1 10 5 9 d  9 or 9.9 kilometers 10 16. A bicycle factory was supposed to fill a certain order for bicycles in 6 days. The factory produced 25% more bicycles each day than it was supposed to, and as a result, it filled the order in 5 days, producing 42 extra bicycles besides. If the same number of bicycles was produced each day, what was that number?</p><p>Answer: 210 Solution: Let x = the number actually produced. 0.8x = the number supposed to be produced. 5x = 6(0.8x) + 42 x = 210 </p><p>Round 9 2 17. Express in the simplest form, the real number  3 75  12  .  </p><p>1 Answer: 3 2 2 2  3  3 1 Solution:  3   3  2 1 75  12  3 3   3   3        3</p><p>18. The second, sixth, twenty-second and last term of an increasing arithmetic progression, taken in this order, form a geometric progression. Find the number of terms in the arithmetic progression.</p><p>Answer: 86 Represent the terms as a  d, a  5d, a  21d, and a  n  1d. Since they a  5d a  21d form a G.P.,   4d 2  12ad  d  0, and d  3a, Solution: a  d a  5d d  0 is meaningless. The terms are now 4a, 16a, 64a, and a  n  13a   2a  3an  256a  3an  258a  n  86</p><p>Round 10  19. If A is not a multiple of , find the simplified numerical value of 2 1 1 1 1 2  2  2  2 . 1 sin A 1 cos A 1 sec A 1 csc A</p><p>Answer: 2 Solution: You could use a trigonometric or an algebraic approach. Using an algebraic one we get 1 2  x2  1 1 x2 2    1 . sinA and cscA are reciprocals and cosA and secA are 1 x 1 2 1 1 2 2  x  2 x x also. The sum = 2.</p><p> x x2 20. Solve for x: 5  5  120 5 Answer: 3.5  24 Solution: 5x  5x2  120 5 5x (1 52 )  120 5 5x  120 5 5x  125 5 5x  53.5  25</p><p> x = 3.5</p>