Mu Alpha Theta 2004 State Bowl Questions - Alpha

Round 1

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?

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?

Mu Alpha Theta 2004 State Bowl Questions – Alpha

Round 2

3. If f x  x2 and g x  2x, what is the value of f g 3  g f 3 ?          

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?

Round 3

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?

49 35 21 6. Arrange in order from the smallest value to the largest value 2 , 3 , 5 .

Mu Alpha Theta 2004 State Bowl Questions – Alpha

Round 4

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?

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

Round 5

 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

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.

Round 6 11. Express as an integer: . 93 36 3  61 28 3

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    

Round 7

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.

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

Round 8

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?

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

Round 9 2 17. Express in the simplest form, the real number  3 75  12  .  

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.

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

x x2 20. Solve for x: 5  5  120 5 Mu Alpha Theta National Convention 2004 Alpha State Bowl Answers

# 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

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?

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

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?

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

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?

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%

49 35 21 6. Arrange in order from the smallest value to the largest value 2 , 3 , 5 .

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

Round 4

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

 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

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    

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.

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            

Round 6 11. Express as an integer: 93 36 3  61 28 3

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

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    

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

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.

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

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

 4  Answer: ,6,4  3 

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

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?

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?

Answer: 210 Solution: Let x = the number actually produced. 0.8x = the number supposed to be produced. 5x = 6(0.8x) + 42 x = 210

Round 9 2 17. Express in the simplest form, the real number  3 75  12  .  

1 Answer: 3 2 2 2  3  3 1 Solution:  3   3  2 1 75  12  3 3   3   3        3

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.

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

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

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.

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

x = 3.5