Download Senior Division Solutions Inter-School Test

Rocket City Math League
2009-2010
Inter-School Test
Senior Division Solutions
1. To find the sum of the roots of an equation of the form: ax 2  bx  c  0 the formula is
b
=sum of roots.
a
2
This comes from the quadratics formula  b  b  4ac . So you add the roots and you get
2a
 b  b  4ac  b  b  4ac  2b  b . In this case, the equation is



2a
2a
2a
a
2
2010
2
x 2  2010 x  2009  0 , so the sum
of the roots is –(-2010)/1=2010. You could also factor the polynomial on the left side of the equation and get
x  1x  2009  0 and add the 2 roots, 1+2009=2010.

1

( 21)
2
3  6  5  33
2. (3  3  2)  5  3
=
 2 9  5  27  23  22  50
1/ 2
0.5
3. The heights that the super ball drops form an infinite series starting with 60 with a common ration of 5/8. After
every drop height except the initial 60 foot drop, the ball travels back up the same distance. The sum of an infinite
series =
first _ term
260 
 60  260
. Therefore, the total distance traveled 
5
1  ratio
1
8
260
4. If 2 people take all 4 AP science classes, then 10 people take exactly 3 AP science classes, and 9 people take
exactly 2 AP science classes. The sum of the people taking AP Chemistry, people taking AP Environmental
Science, people taking AP Biology, and people AP Physics (157) is equal to the number of people on the math
team with the number of people taking exactly 2 AP science classes included twice, exactly 3 classes included 3
times, and all 4 AP science classes included 4 times. To find the number of people on the Grissom Math Team,
you must subtract the people taking exactly 2 AP science classes once, exactly 3 classes twice, and all 4 AP
science classes 3 times from 157. 157-9-2(10)-3(4) =116
5. To find where the graphs intersect, substitute y=-5/4 into the equation
116
y  x 4  5 x 2  5 to get the equation
5
 x 4  5 x 2  5 . Multiply each term of the equation by 4 and simplify to 4 x 4  20 x 2  25  0 . This
4
2
5
2
equation factors to 2 x  5  0 , so x  
so there are two solutions to this equation. Therefore, the
2

50

2
graphs intersect 2 times.
6. Performing a triangular cut on ABCDEFG to get Solid X results in the subtraction of 1 vertex and the addition
of three new ones which results in 10 total vertices. Solid Y has two two addition faces than a cube (because of
the two triangular cuts that have been performed on it) resulting in 8 faces. Finally, Solid Z has 9 edges (3
triangular cuts) added to the original cube resulting in 21 edges. 10+8+21=39
7. Using the binomial theorem, the coefficient of x5 is (1)5 (4)3(8c5) = 3584.
39
3584
8. The equation x 2  y 2  z 2  16 forms a sphere with a radius of 16  4 . The region that contains all line
segments tangent to the sphere with length 1 is in the shape of a hollow sphere. The inner radius of region T = 4,
while the outer radius of T is the hypotenuse of the right triangle formed by the inner radius, the tangent line of 1,
2
2
2
and the outer radius. a 2  b 2  c 2 , so rinner
. 4 2  12  router
, so the outer radius = 17 and R = 17 .
 12  router
The surface area of T = the surface area of the inner sphere + the surface area of the outer sphere.
2
2
2
Therefore, A  4rinner
 4router
 4 (42 )  4 17  64  68  132 . The volume of T = the volume of the outer
 
3
4 3
4
4
68 17 256 .
3
sphere minus the volume of the inner sphere, so V  4 router
 rinner
  17   43 

3
3
3
3
3
3
 68 17 256 
  203456 2  45056 2 17
RAV 4  4 17 132 


3
3


 
203456 2
 45056 2 17
 
9. First, the key is to figure out what the solid is. Start by working to find the boundaries in 2 dimensions.
Let x=0, then you have 1 z 2  y 2 which becomes z   10 y (figure 1). Let y=0, then you have
10
2
z   10 x (figure 2). If you let z be some constant, c, then you have c  x 2  y 2 which is a circle (figure 3).
10
541373400
z
z
y
y
x
Figure 1
x
Figure 2
Figure 3
Together, these figures make a circular double cone. Since z is restricted to positive numbers, you only have a
single cone with vertex at the origin. Since z is given to be between 0 and 2010, the height of the cone is 2010.
plug in z=2010 and you have 1 2010 2  x 2  y 2 which is a circle with radius 201 10 . Now that we have the
10
boundaries, we must decide if the solid is located above or below the graph. Since the z squared term is less than
the sum of the squares of x and y, the solid is located below the graph of the cone. This can be pictured as a
cylinder of radius 201 10 and height 2010 with a cone of the same radius and height removed from it. The
volume of a cylinder is V  r 2 h and the volume of a cone is V  1 r 2 h , so the volume of the figure in question is


3
2
2
1
2
V  r 2 h  r 2 h  r 2 h so V   201 10 2010  541373400
3
3
3
10. Using a model of stars and bars, you can let each * represent a value of 1 and let a | represent s division
between 2 letters. You are now essentially creating “buckets” in which you are collecting *s, and the number of
stars in a bucket represents the number assigned to that letter. With the form (R,C,M,L) the solution (3,4,5,8)
would be represented by ***|****|*****|********. Now notice that the solution must have positive integers, so
there must be at least 1 * in every “bucket.” To account for this, we can redefine a * to be equivalent to each
additional unit of 1 that can be arranged between the “buckets.” Now, (3,4,5,8) would look like
**|***|****|*******. And you can calculate the number of ways to do this in a similar manner to the number of
ways to make a “word” out of 16 *s and 3 |s. So this would be equal to 19 choose 3. Since the question asks for
less than or equal to 20, you also have to account for when the sum equals 19, 18, 17….and 4. To do this, you
calculate the number of ways to arrange the stars and bars with 3 bars, and 15, 14, 13….0 stars, and add it all
together. You should get
You
19 C3 18 C3 17 C3 16 C3 15 C3 14 C3 13 C3 12 C3 19 C3 10 C3  9 C3  8 C3  7 C3  6 C3  5 C3  4 C3  3 C3
x
could solve for all of these, but there is also a rule that says that

a b
a
4845
Cb  ( x1) C(b1) . Thus, the above is equal
20! 20 * 19 * 18 * 17 * 16!
 4845
16!4! 16!*4 * 3 * 2 * 1
11. One useful way to model this question is known as stars and bars. Let each star(*) represent 1 inch of the
cake and let each bar (|) represent a division between layers of the cake. Thus, ***|***|************** would
represent the cake 3 – 3 – 14. You are currently using 20 *s and 2 |s. However, since each layer of the cake must
be at least 3 inches (have at least 3 *s) you can redefine each star to be any additional inch of cake beyond the
required 3 inches that is being added to that level, meaning you take away 3*s from each layer. Now the cake
with 3 – 3 – 14 would be ||*********** (no stars before the first bar and between the 2 bars means that there are
no additional inches in either of the first 2 layers). When you divide the cake up into 4 layers, you will now have
3 dividers (3 bars). Since each of the 4 layers must have at least 3 inches of cake, that means that there are 8
inches of additional cake (8 stars) remaining that must be arranged. Similarly, with 5 layers, there are 4 bars, and
5 stars. And with 6 layers there are 5 bars and 2 stars. For each of these layers you can use combinations to
arrange the “word” formed by the stars and bars.
to
20
C4 
406
# of layers
# of bars
# of stars
1
0
20
# of stars +
# of bars
20
2
1
14
15
3
2
11
13
4
3
8
11
# of arrangements
20
C0
15
C1
13
C2
11
C3
20!
20!0!
15!
14!1!
13!
11!2!
11!
8!3!
1
15
78
165
5
4
5
9
6
5
2
7
9
C4
7
C5
126
9!
5!4!
7!
2!5!
21
Total=406
12.
First, it is important to determine how the ant can get through
square theta. You can write out the number of ways to get to
each point (which follows the pattern of Pascal’s triangle) as has
been done below with each number written up and to the left of
the point it corresponds to.
.
1
6
21
56
126
252
1
5
15
35
70
126
1
4
10
20
35
56
1
3
6
10
15
21
1
2
3
4
5
6
1
1
1
1
1
X
Or you can consider the question in terms of combinations. There is only one way to get to square M, by moving
straight up, but to get to square A, the ant must make some combination of 5 moves to the right and 5 moves up.
Thus, his movements can be modeled by the arrangements of RRRRRUUUUU, so 10 choose
5= 10 C5 
10!
 252 .
5!5!
Y
Z
V
5363
W
X
Now consider the movements through square M. Following similar logic, there are C  8!  70 ways for the
8 4
4!4!
ant to get to point Y, starting at point V.
To get from point W to point Z, there are 6 C3  6!  20 ways. However, for each way to move from point W to
3!3!
point Z, there are 252 ways to move from point X to point W, so the number of ways for the ant to go through
square theta and out square A is (252)(20)=5040.
5040+70=5110 ways for the ant to get out through point Y or point Z by going through 2 squares.
However, the ant can also travel through 3 squares by going from point X to point V to point Z, or from point X
to point W to point Y. Because of the limited directions of travel (right and up in square theta, and moving up in
squares M and A), there is only one way for the ant to travel from point X to point V to point Z. There is also
only 1 way for the ant to travel from point W to point Y, but since there are 252 ways for the ant to travel from
point X to point W, there are (252)(1)(1)=252 ways for the ant to get out point Y by going through point W.
Thus, there are 1+252=253 ways for the ant to get out by going through 3 squares.
This means that the total number of ways for the ant to get out of the Mu Alpha Theta insignia is 5110+253=5363.
13. Let e represent only the even numbers and o represent only the odd numbers. You can now rewrite your
expression as
  1 (e )    1 o .
30
e
e2
30
o
3
Since -1 to an even power is 1 and -1 to an odd power is -1, this is
o 1
29
e  o
3
e2
29
3
3
o 1
To solve this question, you take advantage of the difference of cubes formula: a 3  b3  a  ba 2  ab  b 2 . Let


3
2
2
2
2
b=a-1, then you have a  a  1  a  a  1 a  aa  1  a  1  1a  a  a  a  2a  1
 3a 2  3a  1
e 30
30
e 30
30
Using this, you can again re-write the expression as
 3e 2  3e  1   3e 2   3e  1
3
2
e2
e2
e2
e2
Now considering you are taking the summation through the first 15 even numbers, you can divide out powers of 2
from the first 2 summations. Let m represent positive integers, then you have:
 32m    32m  15 , so this
15
m 1
2
15
m 1
simplifies to 15  12m 2  6m . Using that the sum of the first m squares is equal to m2m  1m  1 ** and that

6
m 1
15
the sum of an arithmetic sequence is equal to the average of the sequence times the number of terms in the
14175
sequence, you can evaluate the expression: 15  12153116  15 61  615   15  14880  720  14175


6
2


**note: a proof of the sum of the first n squares is included in the solution for number 4 of the junior interschool
test from this year
14. At the beginning you have 420 pages. First you want to subtract all pages that are multiples of 6 and all pages
that are multiples of 7. There are 420/6=70 pages that are multiples of 6 and 420/7=60 pages that are multiples of
7. However, when you subtract both of these from 420, you have subtracted the pages that are multiples of both 6
and 7 (multiples of 42) twice. Thus you must add one set of pages that are multiples of 42 back into your set of
pages that are in the book because you cannot take these pages out twice. Now you must also consider the pages
where you have a multiple of 6 on one side of the page and a multiple of 7 on the other side, for example pages 35
and 36. Since 6 is even, the multiple of 6 can only occur on the back side of a page, so this only occurs when
there is a multiple of 7 on the front side of a page and a multiple of 6 on the back side of a page. If you subtract
the multiples of 6 and 7 independently, then you have also subtracted these multiples twice, so you must add one
set of these back. After you find the first place where this occurs (pages 35 and 36) you can see that this will
occur every 42 pages after that because when it happens again the first page must still be a multiple of 7 and the
back side must still be a multiple of 6.
310
Mathematically:
420-(multiples of 6)-(multiples of 7)+(multiples of 42)=420-70-60+10=300
Now add back the front/back combos that have been subtracted twice.
300+10=310
15. First, it is necessary to solve for A. The first thing you do is plug x=0 into the expression and see what you
get.
Since you get 0 in the denominador, you have to factor the expressions.
 x( x  2)
 x
cos 2 ( x) 
( x  2)
cos 2 ( x) 
(0  2)
12
1 1
A  lim 
 ln(x2) x   lim 


 1

 1  
x
2
sin(
x
)
cos(
x
)
e
e
sin
x
2
cos(
x
)
(
x

2
)
e
2

1
(
0

2)  1
2 2
x0 
 x0 

If A= -1/2, then you have
 (24x 5 - 20x 4  18x 3 - 15x 2 - 6x  5) 
B  lim 

(12x 3  4x 2 - 11x - 5)
1 

x
First you try substituting x= -1/2 into the equation and
2
you get B=0/0, so this doesn’t work. This means you must factor the top and bottom of the polynomial and then
cancel any like terms. After that you substitute x= -1/2 again.
 (24x 5 - 20x 4  18x 3 - 15x 2 - 6x  5) 
 (2 x  1)( 2 x  1)(6 x  5)( x 2  1) 
 (2 x  1)(6 x  5)( x 2  1) 
B  lim 
  lim 
  lim 

3
2
(12x

4x
11x
5)
(
x

1
)(
2
x

1
)(
6
x

5
)
( x  1)(6 x  5)
1 

1

1
 x 
 x 

x
2
2
2
2 6
4
   1 2 
(1  1)( 3  5)    1 (2)( 8) 5 
 2 



 4   20   20

3
3
 1 
  3
 1 3  5


(2)
 2

 2 
If A= -1/2 and B= -20/3, then C= 9(-1/2)(-20/3)+45=9(10/3)+45=30+45=75
Now evaluate sin(75o). Since sin(75o)=sin(30º+45º), then you use the formula
sin(A+B)=sin(A)cos(B)+sin(B)cos(A).
sin( 75)  sin( 30) cos( 45)  sin( 45) cos(30) 
1 2
2 3


2 2
2 2
2 6
4
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