Mu Alpha Theta National Convention: Hawaii, 2005
Solutions – School Bowl – Theta Division
5
1. Using the binomial theorem, we have  (3 x 2 ) 2 (1)3  90 x 4 ; the coefficient is -90.
2
 
1
i
1 i2
i
2
 2
 2.
2. i
=
2
i
i 2 1
i
i
3. We can call the numbers x and y. Then we can form the following equations:
x  y  12
x
4

y
5
. We see that  4 y  5 x ; multiply both sides by y, and we have
4 2
y  xy .
5
We can solve for y: since x = –12 –y then  4 y  5(12  y) or y = -60. The product of x
& y is
4
(60)2 = –2880.
5
4. Difference of squares: 3022-2992 = (302-299)(302+299) = 3(601) = 1803.
5. Divisibility rules tell us 1,2, and 4 evenly divide the number.
6. We solve (r  .25r )  2(r  .25r )  22.5 and r = 10.
7. Solving the system of equations tells us that x =1, y2 = 4 , or y = 2, so the expression
evaluates to -1.
8. The distance formula yields 36  16  36  88  2 22 .
9. Let x = rate they work together. Then
x x
99
  1 so x 
.
11 9
98
6 4
10. Since the LHS is an infinite sum, we can rewrite the equation as
subsequently, y = 3.
11. Possible choices for a sum of 8:
On the standard six-sided die: 1
2
3
4
5
On the fair nine-sided die:
7
6
5
4
3
Then the probability is
2 y  2  2 and
6
2
6 1
 .
54 9
12. From 100-299: 133, 233; from 300-399: 303,313,323,330-339, 343, 353, 363, 373,
383, 393; from 400-999: 433,933. There are 27.
( y  3) 2 ( x  1) 2

 1 so |ka|+|hb| = 12 + 2 = 14.
42
22
1
1
14. f (-1)= -1 ; g (-1)=
; f ( )= 0.
2
2
13.
15. Let r = number of regular tickets sold and d = #deluxe tickets sold. Then we have the
following system of equations:
r  d  60
20r  30d  1400
Solving this gives us r = 40 & d= 20.
16.
6!
 180 .
2!2!
17. We look at 7 p2  28  0 to see when h is undefined. This happens at p=+/-2.
Mu Alpha Theta National Convention: Hawaii, 2005
Solutions – School Bowl – Theta Division
y4
)
2
2  3x  y  4
x  log 3 (
18.
f 1 ( x )  2  3x  4
f 1 (2)  2  32  4  14
19. The set has mode -1 and median 0. -1 + 0 = -1.
20. Distance2 = 25-9 = 16, so the distance is 4.
21. We know P ([ A  B]C )  1  P ( A  B)  0.1 and P ( A  B)  P ( A)  P ( B)  P ( A  B) .
Then 0.9  .7  .5  P ( A  B) or P ( A  B ) = 0.3.
22. Factor by grouping:
x 5  x 3  x 2  1 x 2 ( x 3  1)  ( x 3  1) ( x 2  1)( x 3  1)
 2

 x2  x  1.
3
2
2
x  x  x 1
x ( x  1)  ( x  1)
( x  1)( x  1)
23. We set up a system of equations:
l  2w  3
( l  2)( w  1)  28
9
2
Here we can substitute and get w   ,3 . Substituting w = 3 back into the first equation,
the length is 9. Thus our answer of (9,3).
24. We want the number of glasses g such that 3g+40 = 5g. This happens when g =20.
25. The vertex is at (-4,0) on the x-axis; this information gives us h and k.. To find a we
pick a point and plug into our parabola equation: 2  a (2  4)2 and we find out that
1
1
. Then our equation is y  ( x  4)2 (0) .
2
2
26. 102211 3  31910  477 8 .
a
 2  1
1 0 
1 1 1 
T
1
T
1 2 , M   1 1  so M M  0 1 .
2 1 





2
12  54
18  6
)  log 6 (
)  log 6 (6)  1 .
28. Using log rules we have log 6 (
108
18  6
27. M 1 
 a 5b 4
29.  6
 c




3

c 18
.
a 15 b12
30. This is really saying 6 x  4  40 and 0.5 x  2 y  40. From the first equation we have
x= 6, and plugging this in for x in the second equation gives us y = 18.5.
 210 
31. A = 10(21) = 210 and B = -4. Then 
   52.5  52 .
4
32.
2 x ( x  4)  1( x  2)( x  4)  3( x  2)
x  5x  2  0
2
, so the sum of solutions is
5
= -5.
1
33. After we square both sides we get 2 x  5  1  8 x  4 2 x . We isolate the square root
on one side and then square both sides again:
3
x  1  2x
2
9 2
x  3x  1  2 x
4
Mu Alpha Theta National Convention: Hawaii, 2005
Solutions – School Bowl – Theta Division
2
9
We simplify into 9 x 2  20 x  4  0 . Then x  ,2 are solution candidates; checking both
2
is the only solution.
9
1
34. The slope of our line is  . Then the equation of the line is
2
1
1
7
y  3   ( x  1)  y   x  .
2
2
2
we see that x =
35. The orange hits the ground when h(t)=0. This happens for t = 2.5 seconds.
36. Rearranging the equations gives
y  2x  3
, where we can see the second equation
3y  6x  9
is clearly a multiple of the first. We thus have infinitely many solutions that satisfy the
system.
37. The information from the problem gives us this Venn diagram:
We can see that football is the most popular sport with 85 participants.
38. There are 18 non-prime (composite) numbers between 1 and 30.
39. Volume of the cylinder = 160 , the volume of the sphere is = 288 so the volume
outside of the cylinder but inside the sphere is 288  160  128 .
40. We can easily confirm that (x+1) is a factor of the cubic. Then we can recover the
other factors and write the cubic as ( x  1)( x  3)( x  8) . Then a – b = 8 –(-3) = 11.