Download SMLE 2007

Interesting problems from the AMATYC Student Math League Exams 2007
(February 2007, #2) The teachers at Oak Tech have cars with average mileage 39,000 miles.
George buys a brand-new car, keeping his old car, and the average mileage drops to 36,400.
How many cars do the teachers now own?
Let T be the current total mileage of the cars and N be the current number of cars. So
T
T
 39,000 . We also have that
 36,400 , so 39,000 N  36,400  N  1 . This implies
N
N 1
36,400
that 2,600 N  36,400 , or N 
 14 .
2,600
So the correct answer is C) 14.
(February 2007, #3) The sequence log x,log x 2 ,log x3 ,log x 4 ,
is best described as which of
the following?
log x,log x 2 ,log x3 ,log x 4 ,  log x,2log x,3log x,4log x,
 log x,log x  log x,log x  2log x,log x  3log x,
So the correct answer is C) arithmetic with common difference log x .
[See the section on Algebraic Formulas]
(February 2007, #4) A set of seven different positive integers has mean and median both equal
to 20. What is the largest possible value this set can contain?
Let’s call the seven positive integers x1  x2  x3  x4  x5  x6  x7 . The median being 20
implies that x4  20 , and the mean being 20 implies that the sum of the other values must be
120.
20
x1
x2
x3
x4
x5
x6
x7
We want the value of x7 to be as large as possible, so we need to make the values of
x1 , x2 , x3 , x5 , x6 as small as possible. This leads to
1
2
3
20
21
22
71
x1
x2
x3
x4
x5
x6
x7
So the correct answer is C) 71. [See the section on Statistics Formulas]
(February 2007, #7) If ln s  .6 and ln t  .9 , find log st e5.4 .
s  e.6 and t  e.9 , so st  e1.5 . log st e5.4  log e1.5 e5.4 
5.4
 3.6 .
1.5
So the correct answer is A) 3.6. [See the section on Logarithmic Properties]
(February 2007, #8) A function is symmetric to the origin and periodic with period 8. If
f  2   3 , what is the value of f  4   f  6  ?
A
particular
function
that
has
these
properties
is
f  x   3sin  4 x  .
f  4   f  6   3sin   3sin 32  0  3  3 .
So the correct answer is B) 3 .
(February 2007, #9) For how many integer values of k do the graphs of x  y  k and xy  k
NOT intersect?
Intersections correspond to real solutions of the system
x yk
. Using substitution, you get
xy  k
k
 k  x 2  kx  k  0 . In order for this quadratic equation to not have real solutions, its
x
discriminant must be negative. So we want k 2  4k  0 or k  k  4   0 . This leads to
x
k  1,2,3.
So the correct answer is D) 3. [See the section on Algebraic Formulas]
(February 2007, #10) A point is chosen at random from the interior of a square of side 16.
Find the probability that the point is at least
2 units from both diagonals.
6
2
6
The point must be chosen from the four triangles outside of the central dashed cross. Each has
an area of 36 square units for a total of 144 square units. The area of the entire square is 256
144 9
square units, so the probability is
 .
256 16
So the correct answer is A)
9
. [See the section on Geometric Formulas]
16
[See the section on Probability Formulas]
(February 2007, #12) If cos  arctan  x    x (x in radians), then x 2 can be expressed in the form
a b
. Find a  b .
2
cos  arctan  x    x 
1
1  x2
 x   x2   x2  1  0  x2 
2
1  5
.
2
So the correct answer is A) 4. [See the section on Trigonometric Formulas]
(February 2007, #15) A farmer plants A acres of wheat one year. Each year thereafter, he
1
harvests(removes) of the planted acreage and then plants 1500 more acres. The number of
4
acres of wheat approaches what number?
P1  A
P2 
3
A  1500
4
2
3 3
3

3
P3   A  1500   1500    A   1500  1500
4 4
4

4
3
2
3
3
3
P4    A     1500   1500  1500
4
4
4
n
n 1
n2
3
3
3
3
Pn    A     1500     1500    1500  1500
4
4
4
4
n 1
n
 3  3 2
3  3
 1500 1            
 4    4 
 4  4 
As the number of years goes to infinity, we get
 3  3 2
Pn  1500 1     
 4  4 
3
 
4
n1


1
 1500  4  6000 .
  1500 
1  34

So the correct answer is D) 6000.
Or
If the amount of wheat planted is to stabilize, it must be that
3
 P  1500  P . Solving for P
4
leads to 6000.
(February 2007, #16) Right ABC (right angle at B) has legs of length 68 and 285. If the
medians from vertex A and vertex C intersect at D, find the area of ADC to the nearest ten
square units.
A
285
D
B
68
C
The three medians of a triangle divide the triangle into 6 triangles of equal area. Since ADC
1
consists of two of these 6 triangles, its area must be of the area of the complete triangle.
3
1 1
  68  285  3230 .
3 2
So the correct answer is B) 3230.
Here’s the proof of the previous result about medians and area:
A
B
B
A
C
C
The triangular regions with the same names have the same areas since they have the same base
and height. To conclude that all 6 of the triangles have the same area, notice that the triangles
made up of AAC and BBC must have the same area as well as the triangles made up of AAB
and CCB.
(February 2007, #17) If f  x  
f 1  x  
Notice
f
1
 x
x 2  3x  4
, the inverse of f  x  can be written as
x 1
x 2  ax  b
. Find a  b  c .
xc
f  x   x  4; x  1,
that
so
f 1  x   x  4; x  5 .
This
means
that
x  4  x  5 x 2  9 x  20

.


x5
x5
So the correct answer is E) 34.
 x  y  kz  1

(February 2007, #18) Choose k so that the system  x  ky  z  2 is dependent. For which
kx  y  z  3

pair  x, y  below does there exist a z such that  x, y, z  will satisfy the resulting dependent
system?
x  y  kz  1
x  ky  z  2
subtract the first equation from the second equation, and subtract k times the
kx  y  z  3
x  y  kz  1
first equation from the third equation.  k  1 y  1  k  z  1
1  k  y  1  k 2  z  3  k
add the second equation to
x  y  kz  1
x  y  kz  1
the third equation.  k  1 y  1  k  z  1 simplify into  k  1 y   k  1 z  1 . For k  1, the
 2  k  k  z  2  k
 k  2  k  1 z  k  2
2
system has no solution, but for k  2 , you get the dependent system
solutions of y  z  13 , x  z  43 , z  z . The only pair that works is
x  y  2z  1
, which has
3 y  3z  1
 83 ,1 , and it corresponds to
z  43 .
So the correct answer is C)
 83 ,1 .
(February 2007, #19) A pentagon is circumscribed about a circle of diameter 6 cm. If the
pentagon has area 42 cm2 , find it perimeter in centimeters.
3
The area of the pentagon is
pentagon is 5 
s
15s
28
15s
, so we get that
 42  s  . So the perimeter of the
2
5
2
28
 28 .
5
So the correct answer is D) 28. [See the section on Geometric Formulas]
1
 1 
 4 
(February 2007, #20) The sum of the solutions of arctan    arctan 
  arctan 

 x
 x  2
 x  4
is
1
 1 
 4 
arctan    arctan 
  arctan 

 x
 x  2
 x  4


1
 1 
 4 
tan arctan    arctan 
   tan arctan 
 
x
x

2
x

4
 






 x1 2
4
x 1
2

 2

 x2  5x  4  2 x2  4 x  2
1
1  x x  2  x  4
x  2x  1 x  4
1
x
 x 2  x  6  0  x  3, 2
But x  2 , so the sum of the solutions is 3.
So the correct answer is E) prime. [See the section on Trigonometric Formulas]
(October 2007, #1) One can of frozen juice concentrate, when mixed with 4 13 cans of water,
makes 2 quarts(64 oz.) of juice. Assuming no volume is gained or lost by mixing, how many
oz. does a can hold?
 4 13  1 x  64  5 13  x  64 
16
3
x  64  x  64   12 .
3
16
So the correct answer is C) 12.
(October 2007, #2) Define the operation  by ab  ab  b . Find  32    23 .
32   23  89  72  9  81 .
So the correct answer is D) 81.
(October 2007, #3) A square is covered by a design made up of four identical rectangles
4
surrounding a central square, as shown. If the area of the central square is of the area of the
9
entire design, find the ration of the length of a rectangle to the side of the central square.
s + 2w
s 2  94  s  2w  
2
s  2w 3
w 1
  
s
2
s 4
sw
1 5
1 
s
4 4
So the correct answer is A)
5
.
4
s
(October 2007, #4) A radio station advertises, “Traffic every 10 minutes, 24 hours a day; 1000
reports each week.” What is the difference between the advertised number of reports and the
exact number?
The exact number is the number of 10 minute intervals in a week.
24  60  7
 1008 .
10
So the correct answer is A) 8.
(October 2007, #5) Trina has two dozen coins, all dimes and nickels, worth between $1.72 and
$2.11. What is the least number of dimes she could have?
172  10D  5 N  211
D  N  24
172  10 D  5  24  D   211  172  5D  120  211  52  5 D  91
 10.4  D  18.2
So the correct answer is B) 11.
(October 2007, #7) A bicycle travels at s feet/min. When its speed is expressed in inches/sec.,
the numerical value decreases by 16. Find s. (1 foot = 12 inches)
s
12 1

 s  16  4s  80  s  20 .
1 60
So the correct answer is D) 20.
(October 2007, #8) The average of A and 2B is 7, and the average of A and 2C is 8. What is the
average of A, B, and C?
Adding the two equations A  2B  14, A  2C  16 leads to 2 A  2B  2C  30 .
A BC
A  B  C  15 
5.
3
So the correct answer is C) 5.
So
(October 2007, #11) A class is exactly 40% female. When 3 male students are replaced by
female students, the class becomes exactly 44% female. How many more males than females
are in the original class?
F
 .4  2M  3F  0
FM
F 3
 .44  11M  14 F  75
FM
 M  45, F  30
So the correct answer is C) 15.
(October 2007, #12) A piece has 2 saxophone parts, 3 trumpet parts, and 3 trombone parts. If a
band has 2 saxophonists, 3 trumpeters, and 3 trombonists, in how many ways can different parts
be assigned to each player?
2
1
3
2
1
3
2
1
1st sax
2nd sax
1st trump
2nd trump
3rd trump
1st trom
2nd trom
3rd trom
So the correct answer is B) 72.
(October 2007, #13) Add any integer N to the square of 2N to produce an integer M. For how
many values of N is M prime?
M  4 N 2  N  N  4 N  1 . In order for M to be prime, its only factors must be 1 and M. For
N  1, we get the prime numbers 5 and 3. Other choices of N produce 0 or factors different
from 1.
So the correct answer is C) 2.
(October 2007, #14) Sixteen students in a dance contest have numbers 1 to 16. When they are
paired up, they discover that each couple’s numbers add to a perfect square. What is the largest
difference between the two numbers for any couple?
The pairings leading to perfect square sums are
4
9
1 + 8, 6 + 3, 7 + 2, 4 + 5
16
25
16 + 9, 15 + 10, 12 + 13, 11 + 14
So the correct answer is B) 7.
(October 2007, #16) What is the smallest positive integer that cannot be the degree measure of
an exterior angle of a regular polygon?
360
n
360
n
The exterior angle of a regular polygon with n sides is
360
, so we need to find a value that is
n
not a factor of 360.
So the correct answer is E) 7.
(October 2007, #18) Let r, s, and t be nonnegative integers. How many such triples  r , s, t 
rs  t  14
satisfy the system 
?
r

st

13

Subtracting the second equation from the first equation leads to rs  r  t  st  1.
rs  r  t  st  1  r  s  1  t  s  1  1   s  1 r  t   1 .
So s  1  1  s  0,2 .
For
s  0 , we get t  14 and r  13 , and hence the triple 13,0,14  . For s  2 , we get r  5 and
t  4 , and hence the triple  5,2,4  .
So the correct answer is A) 2.
(October 2007, #19) The average of any 17 consecutive perfect square integers is always k
greater than a perfect square. If k  2r m , where m is odd, find r.
0  1  4  9  16  25  36  49  64  81  100  121  144  169  196  225  256
 88
17
88  7  81,7  2r m
88  24  64,24  23  3
88  39  49,39  2r m
88  52  36,52  22  13
88  63  25,63  2r m
88  72  16,72  23  9
88  79  9,79  2r m
88  84  4,84  22  21
88  87  1,87  2r m
At this point the evidence is not conclusive.
i 2   i  1    i  16  17i 2   2  4  6 
In general,

17
2
2
 32  i  1  4  9 
17
 256 
.
 i 2  16i  88   i  8  24 . 24  23  3 .
2
So the correct answer is D) 3.
(October 2007, #20) In SML , SM  7 and ML  9 . If M is exactly twice as large as S ,
find SL .
From the Law of Sines and Double Angle Identity, we get that
sin S sin 2S

 SL  18cos S . From the Law of Cosines and Double Angle Identity , we get
9
SL
SL  49  81  126cos 2 S  256  252cos 2 S .
Putting these together, we get
256
2
2
324cos2 S  256  252cos2 S  cos2 S 
 cos S   SL  18   12 .
576
3
3
that
So the correct answer is C) 12.