Download SMLE 2009

Interesting problems from the AMATYC Student Math League Exams 2009
(February 2009, #3) The perimeter of a rectangle is 36 ft and a diagonal is 170 ft. Its area in
ft 2 is
L2  W 2  170, L  W  18  L2  2LW  W 2  324  2LW  154  LW  77 .
So the correct answer is D) 77. [See the section on Algebraic Formulas]
(February 2009, #5) For what values of k will the equation x 14  7  kx2 have exactly two
real solutions?
The equation can be rearranged into kx2  x 14  7  0 . The discriminant of the quadratic
formula can be used to determine when there will be exactly two real solutions, for k  0 . For
this quadratic equation for k  0 , is 14  28k . In order to have exactly two real solutions,
1
14  28k  0  28k  14  k   . But this includes the value k  0 , where the equation is
2
not quadratic, and in fact has only one real solution.
1
So the correct answer k   , k  0 , which is not one of the possible answers.
2
[See the section on Algebraic Formulas]
(February 2009, #6) If x and n are positive integers with x  n and xn  x n1  x n2  2009 ,
find x  n .
x n  x n1  x n2  2009  x n2  x 2  x  1  2009  x n2  x 2  x  1  72  41 . So for x  7 and
n  4 , you get that x  n  11 .
So the correct answer is B) 11.
3
of the women are matched against half of the men.
7
What fraction of all the players is matched against someone of the other gender?
(February 2009, #7) In a tournament,
W  12 M
This fraction is
, and we know that 73 W  12 M , so plugging into the first expression
W M
3
3
6
6
7W  7W
7W


yields
.
13
W  76 W
13
7W
3
7
So the correct answer is D)
6
.
13
(February 2009, #11) At one point as Elena climbs a ladder, she finds that the number of rungs
above her is twice the number below her(not counting the rung she is on). After climbing 5
more rungs, she finds that the number of rungs above and below are equal. How many more
rungs must she climb to have the number below her be four times the number above her?
A  2B, A  5  B  5  2B  5  B  5  B  10, A  20 , so
4 15  S   15  S  5S  45  S  9
So the correct answer is E) 9.
(February 2009, #12) If sin  cos  .2 and sin 2  .96 , find sin 3   cos3  .
sin 3   cos3    sin   cos   sin 2   sin  cos  cos 2     sin   cos  1  12 sin 2 
 .2 1  .48   .296
.
So the correct answer is D) .296. [See the section on Trigonometric Formulas]
[See the section on Algebraic Formulas]
(February 2009, #13) How many asymptotes does the function g  x  
It
has
vertical
 lim
1
x
of
1
10
and
10 100 x  1
2
x
1
.
10
1
have?
Since
1
x
1
 lim

and lim
,
2
2
x 
x 
x 
x 
100
100
1
1
10 100 x  1
10 100 x  1
10 100  2
10 100  2
x
x
1
1
it has horizontal asymptotes of y 
and y  
. So it has four asymptotes.
100
100
lim
x
asymptotes
x

So the correct answer is E) 4.
(February 2009, #14) For how many solutions of the equation x2  4 x  6  y 2 are both x and y
integers?
x2  4 x  6  y 2  x2  4x  4  2  y 2   x  2   2  y 2  y 2   x  2   2
2
  y  x  2  y  x  2   2
2
.
Assuming x and y are integers and since 2 is prime There are only four possibilities:
y  x  2  1 y  x  2  2 y  x  2  1 y  x  2  2
,
,
,
y  x  2  2 y  x  2  1 y  x  2  2 y  x  2  1
None of these lead to integer solutions.
So the correct answer is A) 0.
(February 2009, #17) How many different ordered pairs of integers with y  0 are solutions
for the system of equations 6 x 2 y  y 3  10 xy  0 and 2 x2 y  2 xy  0 ?
Adding
the
equations
8x 2 y  y 3  12 xy  0 .
yields
8x 2  y 2  12 x  0  y 2  8  x  34  
2
 4 x  3  0 , the
2
 4 x  3  7 which
2
Since
y  0,
we
get
that
9
2
 2 y 2   4 x  3  9 . Since y must be an integer and
2
only possible values for y are
won’t yield an integer solution.
1, 2 .
y  1 both lead to
y  2 both lead to  4 x  3  1 , and
2
hence to the ordered pair solutions of  1,2  and  1, 2  .
So the correct answer is B) 2.
(February 2009, #18) The graph of the equation x  y  x3  y 3 is the union of a
x  y  x3  y 3   x  y   x 2  xy  y 2    x  y   0   x  y   x 2  xy  y 2  1  0 .
So x  y  0 , which is a line, or x 2  xy  y 2  1  0 , which is a hyperbola.
So the correct answer is E) line and a hyperbola. [See the section on Algebraic Formulas]
(February 2009, #19) A four-digit number each of whose digits is 1, 5, or 9 is divisible by 37.
If the digits add up to 16, find the sum of the last two digits.
The possible four-digit numbers that meet the conditions are
the arrangements of three 5’s and one 1: 1555, 5155, 5515, 5551, none of which are divisible by
37
and the arrangements of two 1’s, one 9, and one 5: 1195,1159, 9115, 5119, 5911, 9511, 1591,
1951, 1519, 1915, 9151, 5191, of which only 1591 is divisible by 37.
So the correct answer is C) 10.
(October 2009, #1) Find the sum of the solutions to the equations x 2  5x  6  0 and
x2  4 x  3  0 which DO NOT satisfy both equations at once.
The solutions of x 2  5x  6  0 are 6 and 1 from
 x  6 x  1  0 .
The solutions of
x2  4 x  3  0 are 1 and 3 from  x  1 x  3  0 . So the sum of the solutions which don’t
satisfy both is 6  3  3 .
So the correct answer is E) 3.
(October 2009, #2) Four consecutive integers are substituted in every possible order for a, b, c,
and d. Find the difference between the maximum and minimum values of ab  cd .
For the four consecutive integers n, n  1, n  2, n  3, the maximum value of ab  cd is
 n  3 n  2   n  n  1  2n2  6n  6 ,
n  n  3   n  1 n  2   2n 2  6n  2 .
and
the
minimum
value
of
ab  cd
is
So the difference between the maximum and
minimum values of ab  cd is 4 .
So the correct answer is D) 4.
(October 2009, #3) The product of a number and b more than its reciprocal is y(b>0). Express
the number in terms of b and y.
If the number is x, then y  x  1x  b  , so y  1  bx , and x 
So the correct answer is A)
y 1
. So the answer is A.
b
y 1
.
b
(October 2009, #4) If f  x   x 2  x  2 , find the sum of all x values satisfying f  x  2   22 .
If f  x  2   22 , then x  2 must be a solution of y 2  y  2  22 . The solutions come from
y 2  y  20  0 or  y  5 y  4   0 . So x  7 and x  2 . So the sum of the x values is 5 .
So the correct answer is E) 5.
(October 2009, #5) Sue bikes 2.5 times as fast as Joe runs, and in 1 hr they cover a total of 42
miles. What is their combined distance if Sue bikes for .5 hr and Joe runs for 1.5 hr?
ratesue   2.5 rate joe , and
rate joe 
 rate
sue
 rate joe   1  42 , from which we can deduce that
42
42
42
42
and ratesue 
 2.5 . So the combined distance is
 2.5  .5 
 1.5  33 .
3.5
3.5
3.5
3.5
So the correct answer is C) 33.
(October 2009, #6) The equation a 4  b3  c 2  2009 (a, b, c positive integers) has a solution in
which a and b are both perfect squares. Find a  b  c .
b
1
a4
1
1
b3
1
4
256
4
64
9
729
a
2009  1  64   1944 ,
2009  1  1  2007 ,
2009  1  729   1279 ,
2009   256  1  1752 , 2009   256  64   1689 , 2009   256  729   1024 . Of the numbers
2007, 1944, 1279, 1752, 1689, and 1024, the only perfect square is 1024. So the numbers are
a  4, b  9, c  32 , and the sum a  b  c  45 .
So the correct answer is D) 45.
(October 2009, #7) How many 3-digit numbers have one digit equal to the average of the other
2?
Average digit
Other two digits
# of 3-digit numbers with these digits
1
0 and 2 or 1 and 1
5
0 and 4 or 1 and 3
2
11
or 2 and 2
0 and 6 or 1 and 5
3
or 2 and 4 or
17
3 and 3
0 and 8 or 1 and 7
4
or 2 and 6 or
23
3 and 5 or 4 and 4
1 and 9 or 2 and 8
5
or 3 and 7 or
4 and 6 or 5 and 5
25
3 and 9 or 4 and 8
6
or 5 and 7 or
19
6 and 6
5 and 9 or 6 and 8
7
13
or 7 and 7
8
7 and 9 or 8 and 8
7
9
9 and 9
1
The total is 121.
So the correct answer is E) 121.
(October 2009, #8) A rectangular solid has integer dimensions with length  width  height
and volume 60. How many such distinct solids are there?
60  22  3  5
Height
Width
Length
1
1
60
1
2
30
1
3
20
1
4
15
1
5
12
1
6
10
2
2
15
2
3
10
2
5
6
3
4
5
So there are 10 distinct rectangular solids.
So the correct answer is C) 10.
(October 2009, #9)
2sin x

cos x  sinx tan x
2sin x
2sin x
2sin x cos x
sin 2 x



 tan 2 x .
2
2
2
x
cos x  sinx tan x cos x  sin
cos
x

sin
x
cos
2
x
cos x
So the correct answer is A) tan 2x . [See the section on Trigonometric Formulas]
(October 2009, #10) If x 
1
1 3
 12 and y   , find the largest value of xy .
x 8
y
1 36
1 5
  xy 
 . Let
xy 8
xy 2
1 5
53
1
z  xy to get z   . Solving for z leads to 2 z 2  5 z  2  0 ,
 2, . So the largest
z 2
4
2
value is 2.
xy  1  1 
Multiplying the two equations together leads to
So the correct answer is D) 2.
(October 2009, #12) The sum of the squares of the three roots of P  x   2 x3  6 x 2  3x  5 is
 x  a  x  b  x  c   x3   a  b  c  x 2   ab  ac  bc  x  abc .
The
roots
of
2 x3  6 x 2  3x  5 are the same as the roots of x3  3x 2  32 x  52 . So it must be that the roots a,
b,
and
a  b  c
c
2
satisfy
the
a  b  c  3, ab  ac  bc  32 , abc   52 .
equations
 9  a 2  b 2  c 2  2ab  2ac  2bc  9  a 2  b 2  c 2  2 ab  ac  bc   9
 a 2  b2  c 2  2  32  9  a 2  b2  c 2  6
So the correct answer is B) 6. [See the section on Polynomial Properties]
(October 2009, #13) The value of 4
 1 1 1
log 2  2 4 28 216


4



 4log2
1
2 4 log
2
1
2 8 log
2
1
216 
 1 1 1
log 2  2 4 28 216





is
1
4
1
 44
1
 18  16

1 12
4
1
 42  2 .
So the correct answer is C) 2. [See the section on Algebraic Formulas]
(October 2009, #14) The figure shows a circle of radius 4 inscribed in a trapezoid whose longer
base is three times the radius of the circle. Find the area of the trapezoid.
y  m  x  12 
 x  4
2
  y  4   16
2
We need to find the value of m that will make the line tangent to the circle, so we want the
 x  4   y  4  16 to have just one solution.
system
This means
y  m  x  12 
2
2
2
 x  4  m  x  12  4  16  x 2  8x  16  m2  x  12   8m  x  12   16  16
2
2
that
 1  m2  x 2   24m2  8m  8 x  144m2  96m  16   0
4
must have a double root, so its discriminant must be zero. This implies that m   , and the
3
upper base measurement must be 6. So the area of the trapezoid is 12  6  12   8  72 .
So the correct answer is A) 72. [See the section on Geometric Formulas]
(October 2009, #16) The integer r  1 is both the common ratio of an integer geometric
sequence and the common difference of an integer arithmetic sequence. Summing the
corresponding terms of the sequences yields 7, 26, 90, … . The value of r is
b, br , br 2 ,
a , a  r , a  2r ,
Summing leads to a  b, a  r  br, a  2r  br 2 ,
, so we get the system
ab7
a  r  br  26
a  2r  br 2  90
Subtracting the first equation from the other two yields
r  br  b  19
2r  br 2  b  83
Subtracting twice the first equation from the second equation yields
br 2  2br  b  45  b  r  1  45  b  r  1  5  32 .
2
2
So r  1  3  r  4 .
So the correct answer is B) 4. [See the section on Algebraic Formulas]
(October 2009, #17) A hallway has 8 offices on one side and 5 offices on the other side. A
worker randomly starts in one office and randomly goes to a second and then a third office(all
different). Find the probability that the worker crosses the hallway at least once.
The probability of crossing the hallway at least once is equal to one minus the probability of not
crossing the hallway. The probability of not crossing the hallway is
8  7  6  5  4  3 396
3
3 10

 . So the probability that we want is 1   .
13  12  11
1716 13
13 13
So the correct answer is D)
10
. [See the section on Probability Formulas]
13
(October 2009, #19) In square ABCD, AB=10. The square is rotated 45 around point P, the
intersection of AC and BD . Find the area of the union of ABCD and the rotated square to the
nearest square inch.
5 2 5
C
D
P
B
A
So the area of the union is the area of square ABCD along with the area of the four small



triangles. This gives 100  4  12  2 5 2  5  5 2  5  100  100


2
2  1  117.157  117 .
So the correct answer is A) 117. [See the section on Geometric Formulas]
(October 2009, #20) The sum of the 100 consecutive perfect squares starting with a 2  a  0 
equals the sum of the next 99 consecutive perfect squares. Find a.
a 2   a  1   a  2  
2
2
  a  99    a  100    a  101   a  102  
2
2
2
  a  198
2
2
Subtracting the left side from the right side leads to
 a  100 2  a 2    a  1012   a  12    a  102 2   a  2 2  

 
 

2
2
  a  198    a  98 


  a  99   0
2
This leads to
100  2a  100   100  2a  102   100  2a  104  
 100  2a  296    a  99   0 .
2
This
leads to
100 198a  100  102  104  106 
 296    a  99   0 . This leads to
100 198a  9900  2 1  2  3  4 
 98     a  99   0 . This leads to
2
2
100198a  9900  98  99   a  99   0 . So we get the equation
2
19800a  1960200  a 2  198a  9801  0  a 2  19602a  1950399  0 . The quadratic formula
19602  196022  4  1950399 19602  19800

 19701.
yields
2
2
So the correct answer is a  19,701. [See the section on Algebraic Formulas]