*Problem 2.* In any circle, the circumference is proportional to
the radius (i.e., C=2pi· r), and the area is proportional to the square
of the radius (i.e., A=pi· r^{2}). The first statement says that
the radius of Batman's pizza is 1.2 times the radius of Robin's, hence
the area of Batman's pizza is (1.2)^{2}=1.44 times the area of Robin's.
Answer: e.

*Problem 3.* There are 12 large dogs, 5 of which are mutts,
so 7 are pure breeds.
Answer: b.

*Problem 4. * If x satisfies the inequality, then 3x is at least
13 units away from -4, hence 3x is either less than -17 or greater than 9.
Answer: b.

*Problem 5.* If one considers all triangles with one side of length
a and another of length b, and one varies that angle between these two sides,
the length of the third side (i.e., the side opposite the angle increases).
Since we know that a^{2}+b^{2}=c^{2} whenever the
angle is a right angle, it follows that the angle must be obtuse whenever
a^{2}+b^{2} < c^{2}.
Note: The numerical relationship is given by the Law of Cosines,
a^{2}+b^{2} -2ab cos(theta)= c^{2}.
Answer: b.

*Problem 6.* log_{2}(8)=3 and log_{3}(9)=2, so x
should satisfy log_{5}(x)=3-2=1. Since 5^{1}=5, x=5.
Answer: d.

*Problem 7*.
In the smallest such class there would be exactly one boy. So, if n represents
the size of such a class, then 1/n < 7%. Thus, we want the smallest n so that
1/n < 7/100, hence n=15.
Answer: c.

*Problem 8.* Let Q be the number of questions on the exam. Since all the numbers
are integers, Q is divisible by 2, 3, 4 and 5. 60 is the only number < 100
divisible by all of these. Now each of the sentences say that each Beatle
got exactly 37 questions correct.
Answer: e.

*Problem 9.* Let s denote the length of a side of the cube.
Then the `bottom diagonal' has length s · 2^{1/2}
and the main diagonal is the hypotenuse of a right triangle with
edges s and
s · 2^{1/2}. By the Pythagorean theorem, the length
of the main diagonal is
s · 3^{1/2}. But the volume of the cube is s^{3}.
Setting these equal, cancelling out s, and taking a square root yields
s=3^{1/4}.
Answer: e.

*Problem 10.*
All right triangles with one angle of size x are similar, so choose one
where the side opposite x has length 1 and the side adjacent to x has
length 3. Clearly, tan x = 1/3. But now the hypotenuse has length
10^{1/2}, so sin x =1/10^{1/2}.
Answer: d.

*Problem 11*.
Let x=Jack's age now and y=Jill's age now. So r=x/y and we want
to find the ratio y/(y-(x-y)), or y/(2y-x). Dividing numerator and denominator
by y yields 1/(2-r).
Answer: d.

*Problem 12.*
Note that 25=10^{log(25)}, so

25^{1/log(25)}=(10^{log(25)})^{1/log(25)}=10^{1}=10.
Answer: c.

*Problem 13*.
The statements give us two equations, s^{2}=pi· r^{2} and
4s=2pi· r. The first implies s/r=pi^{1/2}, while the second implies
s/r=pi/2. Taken together, these would imply pi=0 or 4, both of which are false.
Hence Jim is wrong.
Answer: e.

*Problem 14*.
The parabola y=x^{2}+bx+c intersects the line y=dx in exactly one point
precisely when the equation x^{2}+bx+c=dx has exactly one solution.
The equation x^{2}+(b-d)x+c=0 has one solution when
(b-d)^{2}- 4· 1· c=0. Since b,c,d are all integers, this implies
d-b is even.
Answer: b.

*Problem 15.*
The first statement translates to (LF) implies NOT(GE), the second to
(W) implies (LF) and the third (T) implies (W), where (LF) abbreviates "loves fish"
(GE) abbreviates green eyes, (W) abbreviates "has whiskers" and (T)
abbreviates "has a tail". Putting these together yields
(T) implies NOT(GE), which is the translation of sentence (a).
Answer: a.

*Problem 16*.
For x > 1, the value of y^{y} (when y=x^{x}) is strictly increasing.
When x=4, y=4^{4}=256 and

y^{y}=256^{256} < 10^{768} < 10^{2003}
(since 256 < 10^{3}).
When x=5, y=3125 so y^{y} > 10^{2003},
so the value of x making y^{y}=10^{2003} is between 4 and 5.
Answer: c.

*Problem 17.*
Let c=number of questions answered correctly and w=number wrong
(answered incorrectly). Then 8c-5w=13 and c+w < = 20.
The possibilities can be checked by hand, but we can restrict the search
by looking at the first equation modulo 5 and 8.
Modulo 5, we have 3c is congruent to 3 (mod 5), hence c is congruent to
1 (mod 5), hence c=1,6,11,or 16. Similarly, -5w is congruent to 5 (mod 8)
hence w is congruent to -1 (mod 8), hence w=7 or 15.
The only solution is c=6, w=7.
Answer: c.

*Problem 18*.
(a+(1/a))^{2}=a^{2}+2+1/a^{2}=3^{2}=9, so

(a-(1/a))^{2}=a^{2}-2+1/a^{2}=9-4=5.
Answer: a.

*Problem 19.*
Let ABC be any right triangle, where angle A is the right angle. Let D be
the point on BC where the altitude from A intersects the segment BC and let
h=the length of the altitude AD.
Let x=the length of BD and y=length of DC.
Then the length of AB=(h^{2}+x^{2})^{1/2}
and the length of AB=(h^{2}+y^{2})^{1/2}
Since ABC is similar to DAC, h/y=(length of AB)/(length of AC),
so

h^{2}/y^{2}=(h^{2}+x^{2})/(h^{2}+y^{2}),
hence h^{2}=xy. So the area of the triangle is
given by 1/2(x+y)h=1/2(x+y)(xy)^{1/2}. In this problem, just plug
in x=25, y=2003.
Answer: a.

*Problem 20.*
Let A,B,C denote the part of the job done by each pig in one hour.
Then we get three equations, namely 2A+2B=1, (6/5)A+(6/5)C=1, and (3/2)B+(3/2)C=1.
Solving these yields A=1/3, B=1/6, C=1/2, so A+B+C=1. That is, working together
it takes the three pigs exactly one hour to dig the moat.
Answer: e.

*Problem 21.*
Let D=distance between the ports, C=rate of current, and B=rate of the boat in
still water. Then D/(B+C)=5, while D/(B-C)=7.
Taking reciprocals yields C/D + B/D=1/5 and B/D-C/D=1/7.
Subtracting the first equation from the second yields 2(C/D)=2/35, hence D/C=35.
Answer: b.

*Problem 22.*
Let a_{n} be the number of shares held by the n^{th} shareholder
and let T denote the total number of shares held.
Without loss assume a_{n} < = a_{m} whenever n < = m.
Then a_{1}+...a_{1100} > = T/2. If we are trying to maximize
a_{2003} without increasing T, we should have a_{t}=a_{1100}
for all t between 1101 and 2002. Also, to keep a_{1100} as small as possible
we should have a_{t}=a_{1100}
for all t between 1 and 1099. Thus, letting a=a_{1100}
and c=a_{2003}, we have 1100a=T/2 and 2002a+c=T. Hence c=198/2200 T, so
c/T=9/100.
Answer: a.

*Problem 23.*
The number of 4-digit numbers whose digits are strictly
decreasing is in 1-1 correspondence with the number of
4-element subsets of {0,1,...,9}. Thus, the number of
such sequences is (10 choose 4)=10· 9· 8· 7/
(4· 3· 2· 1)=210.
Answer: d.

*Problem 24.*
There are 8 sequences of the form 0001aaa, 4 of the form 10001bb,
4 of the form c10001c, 4 of the form dd10001, and 8 of the form eee1000.
This gives 28 sequences, but it double-counts the sequence 0001000,
so there are 27 in all.
Answer: b.

*Problem 25.*
Any handshake will preserve the relative parity of the 3 animals.
That is, after any number of moves, the number of Trolls (T's)
and the number of Dragons (D's) will either both be even, or they will both be odd.
As well, the number of Griffins (G's) will have the opposite parity.
At the end of any play of the game,
the number of G's cannot be zero, since it's parity always differs from the other two.
So there are only G's left at the end. Furthermore, since there are an even
number of T's and D's, there must be an odd number of G's at the end of any
play of the game.
It is easy to see that there are plays of the game that result in any odd number
less than 4003 G's being left. For example, to see that 25 Griffins is possible,
first have a G and a D shake hands, resulting in 2002 T's and D's and 2001 G's.
Then, after 13 T+D handshakes there are 1989 T's and D's, and 2014 G's.
Now do 1989 rounds of triple handshakes, first T+G, then D+G, then T+D.
Answer: b.