Download 2008_2009 meet 4

DUSO MATHEMATICS LEAGUE
INDIVIDUAL QUESTIONS - MEET #4
JANUARY 7, 2009
1.
EL. ALG (MATH A)
5 MINUTES
One cup of yellow candies (bought in groups of 3 for 16¢) and twice as many red
candies (bought in groups of 4 for 21¢) were mixed together. For what integer k
will selling the mixture in groups for 3 for k¢ earn the seller a 20% profit based
on his cost?
2.
GEOMETRY
(MATH A)
5 MINUTES
There’s only one point P inside ∆ABC for which ∆PAB, ∆PBC, and ∆PAC all
have equal areas. How many such points P lie in the plane of and exterior to
∆ABC?
3.
IntALG./TRIG
If
(MATH B)
7 MINUTES
f ( x)  3 x 6  x 5  4 x 4  2 x 3  x 2  5 x  4
 3 .
compute the value of f 2
Express your answer in simplest form.
DUSO MATHEMATICS LEAGUE
INDIVIDUAL QUESTIONS - MEET #4
JANUARY 7, 2009
4.
GEOMETRY
(MATH B)
5 MINUTES
Assuming that the Earth is a sphere with a radius of 4000 miles, how far is it, in
miles, from an astronaut seated in a horizontal spacecraft 100 miles from the
Earth’s surface, to the Earth’s visible horizon?
5.
EL. ALG (MATH A)
6 MINUTES
What are all ordered pairs of integers ( k, n) which satisfy the following equation.
n 2  (k  1) 2  5 2
Hint: There are six such pairs.
6.
IntALG./TRIG
(MATH B)
8 MINUTES
If a and c are rational, and if
a



x  cx  5 x  a  ( x  c)( x  c) x  2 
c 



where c ≠0, what are all possible values of a?
3
2
DUSO MATHEMATICS LEAGUE
GROUP TEAM QUESTION- MEET #4
JANUARY 7, 2009
1. Of all the whole numbers that are smaller than one million, what is the greatest
whole number, which when written in English, does not contain the letter “s” ,
nor the letter “e”.?
2.
 
In how many zeroes does 510 ! terminate?
(Express your answer in simplest form.)
3. Sea Biscuit, Secretariat, and Citation are in a 3 horse race in which ties are not
possible. The probability that Sea Biscuit will win is
1
.
The probability that
4
Secretariat will win is
3
. The “odds against” Citation winning are p to 3.
5
Find the value of p.
( Hint: to clarify what is meant by “odds against”, if the “odds against” a
horse winning are said to be “a to b”, that means that the probability that the
b
horse will win is
.)
ab
DUSO MATHEMATICS LEAGUE
SOLUTIONS - MEET #4
JANUARY 7, 2009
Answers for GROUP TEAM QUESTION
1). Fifty-four or 54
2). 2441406
3). 17
(Solutions for the Relay question)
1). To find the largest number, eliminate numbers from 999,999 on down.
The numbers from one thousand to 999,999 all contain at least one “s”.
The numbers from one hundred to 999 all contain at least one “e”.
The numbers in the nineties, eighties, seventies and sixties all contain at least one “e” or one “s”.
Fifty-nine, fifty- eight, fifty-seven, fifty-six, fifty-five, each have an “e” or an “s”, but
fifty-four does not contain an “e” nor an “s”.
Thus the number is fifty-four or 54
 510   510   510 
 510 


  ...  

2). Let T = the number of zeroes = 
 5   5 2   53 
 510 

 
 



So
T = 5 9  58  5 7  ...  1
Now
5T = 510  5 9  58  ...  5
qq
Thus 4T = 510  1
10
QSo
= T = (5  1) / 4 = 2441406
3). In a 3 horse race with no ties, P(Citation wins) + P(Sea Biscuit wins) + P(Secretariat wins) = 1
1
3
P(Citation wins) +
+
= 1
or
4
5
P(Citation wins) + 25% + 60%
= 100%
15
So
P(Citation wins)
= 15%
=
100
Thus the “odds against” Citation are
Therefore
a to b
3

20
or
3

17  3
b

ab
17 to 3.
p = 17
Individual Questions written by Steve Conrad www.mathleague.com; edited by Dan Flegler & J.S.
Relay team question written for DUSO by J.A.; DUSO Editor J.S.
DUSO MATHEMATICS LEAGUE
SOLUTIONS - MEET #4
JANUARY 7, 2009
Answers for INDIVIDUAL QUESTIONS
1). 19
2). 3
3). 2
4). 900
5). (11,13), (11,-13), (-13,13), (-13,-13), (-1,5), (-1,-5)
6). 3, -3
Solutions for the Individual questions)
1).
 16
 21  




2



120% of the cost per 3 =  6  3
k
4 

 

 



3
3
5


 






 16
 21  



 32 63 
 95 

2



 4    6 

6
19
 6  3


 

     6
6     6  

  


 5 


3
3
3
  5  3 
  5 
 














k = 19
2) . For P exterior to ∆ABC, P may be any of the 3 points that
are vertices of a parallelograms whose other 3 vertices are
those of ∆ABC.
P
2
A
B
P
1
C
P
3
This is so because a diagonal divides a parallelogram into two
equal area regions.
4)
3) Use synthetic division to evaluate f(2/3) which is the
remainder when dividing by x – 2/3
3 1 4 2 -1 -5 4
2/3
2 2 4
4
2 -2
3 3 6 6
3
-3 2
Or
100
5
4
3
2
f (2 )  3 2  2  4 2  2 2  2  5 2  4
3
3
3
3
3
3
3
 3  32 3  42 3  2 3  4 9 10 3  4 
6
 2
4
5
5
4
4
5
 144
4
81
 34
9
 4  16
9
 34
34
9
5). Rewrite: n 2  5 2  ( k  1) 2
If (k  1) 2  0 , then k = -1 and n 2  5 2  25
So n = ±5 giving the two solutions: (-1, 5) and (-1, -5)
If (k  1) 2  0 , then ( k+1) = ± 12 and n = ± 13
(Pythag. Triple: 5, 12, 13).
So k = 11 or -13
This gives four more solutions:
(11,13),(11,-13), (-13,13) , (-13, -13)
The six solutions are:
{(-1,5),(-1,-5),(11,13),(11,-13),(-13,13),(-13,-13)}
2
2
2
x2 = (4000+4000+100)(100)
x2 = (8100)(100) = 810000
x = 900
 34  4
9
42
Method 1
2
2
+ 4000 = 4100
x
2
x
2
x
4000
32  64  48
3
2
Method 2
3
96 3  416 3  16 3  34 9  4 
x
= 4100 - 4000
= (4100+4000)(4100-4000)
4000
= (8100)(100) =810000
x = 900
Note: A Pythag. Triple 9, 40, 41
4000
           
6
x
6)
x3
+
cx2
-5x
+
a =
a 
a 






( x  c)( x  c) x  2   ( x 2  2cx  c 2 ) x  2 
c 
c 






2a 
 a






 x 3  x 2  2  2c   x c 2 
a
c 
c







Equating coefficients of like powers :
a
c
 2c  c
2
2a
and
c2 
c
c  5c
3
So
a  3c 3
and
a
2
Thus 5c  5c , so c =1, c = ±1 and a = ±3
3
2
 5