Rocket City Math League
Senior Division
2014-2015
Inter-School Solutions
Answers must be written inside the corresponding box on the answer sheet. All answers must be written in exact, reduced, simplified, and
rationalized form. Also, figures are not necessarily drawn to scale. No calculators, books, or other aides may be used.
1. To find the number of total distinct arrangements, divide the “number of letters” factorial by the “amount of same letters
repeated” factorial for each letter:
2.
9!
 60480
3!
f ( x)  f (4  1)  (4) 3  2(4) 2  4  28
3. Solve by expressing the dot product:
(5)(3)+(2x+1)(2)+(4)(9)=61
15+4x+2+36=61
4x=8
x =2
4. Use the distance formula of
( x 2  x1 ) 2  ( y 2  y1 ) 2 ; (6  5) 2  (3i  2i ) 2  121  25  146
5. The question is asking for the possible squares in a 6 6 grid. There are 6  6  36 1 1 squares, 5  5  25 2  2 squares,
4  4  16 3 3 squares, 3  3  9 4  4 squares, 2  2  4 5 5 squares, and 1 1  1 6 6 squares. Add up the number
of squares to find there are 91 total squares.
4 5 x 4  810 x 8
2 2(5 x 4)  2 3(10 x 8)
6. 10 x  8  30 x  24
 16  20 x
4
x
5
7. The number of yoga mats wanted is 231 in base 10. The problem wants this to be converted to base 6.
38
6 236 R3
6
6 38 R2
1
6 6 R0
0
6 1 R1
The number of mats in base 6 is 1023 mats.
8.
982alienbytes
578alienbytes
 t minutes +
 t minutes =8245 alienbytes
min
min
982t+578t=8245
T=
60 sec
8245
1649
min 
=
sec
1 min
26
312
9. Set the equations equal to each other to find where the graphs intersect:
6 x  8  x 3  7 x 2  16 x  8
x 3  7 x 2  10 x  0
x( x 2  7 x  10)  0
x( x  2)( x  5)  0
x  0,2,5
Plug x into either of the original equations to find the values of y. The points then are (0,8) , ( 2,20) , and (5,38) .
10. In the trapezoid, AE  AH , EB  BF , FC  GC, and DG
 DH . Since BF  6, EB  6. Since
AB  3 x  5, AE  3 x  5  6  3x  11. Since DH  13, DG  13.
Since CD  4 x  5, GC  4 x  5  13  4 x  8.
The perimeter is thus:
13  (3x  11)  (3x  11)  6  6  (4 x  8)  (4 x  8)  13  84
14x  84
x6
When x is plugged into the unknowns, AH  7, DC  29, GC  16, and FC  16. Simplify:
7  29
203  256 459
 16 

16
16
16
11. The x-values are equivalent to 21 cos t and the y-values are equivalent to 18 sin t. To solve, isolate each trigonometric function
x2
y2
x
y
 cos 2 t and
 sin 2 t . Since
 cos t and
 sin t . Square both sides of each equation to have
21
18
441
324
2
2
x
y
cos 2 t  sin 2 t  1 , then

 1. This is an equation for an ellipse. To find the area of the ellipse, multiply:
441 324
( 441)( 324 )  378
to one side:
12. For f to be continuous at x=-2, both functions have to be equivalent to each other at x=-2
6(-2)4+17(-2)3-23(-2)2-l(-2)+8=2(-2)k2+7(-2)2=81
96-136-92+2k+8=-4k2+109
4k2+2k-233=0
 2  4  4(4)(233)
8
=
 2  3732  2  2 933  1  933
=
=
8
8
4
13. Use the equation y  Ce where C=first amount of microorganisms, k= the growth constant, t=time in hours; and y= total
amount of microorganisms after t hours. Plug in the information given:
kt
7e 26  6e13e 7 k => 7e 26  6e137 k
Use exponential and logarithmic rules to solve for k:
7 e137 k
7
ln 7  ln 6  13
 26  ln( )  13  7k  26  ln 7  ln 6  7k  13  ln 7  ln 6  13  7k 
k
6
6
7
e
14. Finding the probability of at least one is equivalent to:
C8  6 C 0

17 C8
11 10  9  8  7  6  5  4
439
1

17  16  15  14  13  12  11 10 442
1  (none)  1 
11
15. To solve the problem, find how many total angles are used for the expression. The first four angles are
 2 7
6
,
3
,
6
, and
5
.
3
Since there are 12 total angles in all, there are a total of three revolutions around the unit circle. After finding the sum of the first
four angles substituted into the function, multiply the answer by three to find the sum of all values. To make the function more
simple, decompose it into partial fractions:
3
5
3 cos  10 sin   6 3 cos  10 sin   6 3 cos  10 sin   6




2 sin  cos   2
sin 2  4 sin 
2 sin  cos  4 sin 
2 sin  (cos   2)
3
 0 because the values are all in opposite quadrants of each other. The next step is to plug in
All values when plugged into
2 sin 
the values for the second half of the decomposition, add them, and multiply the sum by 3.

 
 
 


 
 


 5( 3  2)  5( 3  2)
5

5

5

5

5
16
2
2





 




  1


1
cos  2  3
3



3
3
3
(
 2)( 
 2)
2  2 
2  2

2
2
  2

 2
  2
 2
16  10  10
16
80
448
 
  
3 3 4
3
13
39
4
Remember that this is only the sum of the first four values. Multiply the answer by 3 to find the total sum.

448
448
3  
39
13
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