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Greater New London County Mathematics League
Meet 1
Tuesday, October 5, 2010
Round 1
Arithmetic
1 Point
What is the 2010th odd natural number?
Position
1
2
3
n
Number
1
3
5
2n-1
2(2010) – 1 = 4020 – 1 = 4019
2 Points
The sum of two positive integers is nine. What is the least possible sum of their
reciprocals?
x y 9
1 1 x  y 9 In order for that fraction to be as small as possible, the product xy
 

x y
xy
xy
has to be as large as possible. xy = 4 x 5 = 20
Thus,
3
9
20
Points
The number 2 48  1 is divisible by two numbers between 60 and 70. Find the two
numbers.

 1  2
 1  2


 12  1
 12  1  65  63
2 48  1  2 24  1 2 24  1
2 24
212
12
6
12
6
Greater New London County Mathematics League
Meet 1
Tuesday, October 5, 2010
Round 2
Verbal Problems- Linear
1 Point
A father’s age is three times the sum of the ages of his two children. In four years, he
will be twice as old as the sum of their ages. How old is the father?
3S  4  2S  16
F  3S
F  4  2S  8 
2
S  12
F  36
Points
A cup of coffee contains 20 more milligrams (mg) of caffeine than a cup of tea and
85 mg more of caffeine than the average soda. If one cup of tea and four sodas
contain the same amount of caffeine as one cup of coffee, how many mg of caffeine
are there in one cup of coffee?
C  20  4(C  85)  C
T  C  20
C  20  4C  340  C
S  C  85
5C  360  C
C  T  4S
4C  360
C  90
3
Points
Television rating showed that at nine o’clock Monday night, 37% more of the city’s
television audience watched a new comedy over that of an old movie. The percentage
watching the movie was 5% more than twice the percent watching a Penn State football
game. The percent watching the comedy was 2% more than six times the percent
watching the football game. What percentage of the entire TV audience watched the
football game? Comedy?
6 F  2  M  37
C  M  37
6 F  2  2 F  5  37
M  2F  5
6 F  2  2 F  42
C  6F  2
4 F  40
F  10
Greater New London County Mathematics League
Meet 1
Tuesday, October 5, 2010
Round 3
Parallels & Perpendiculars
1
Point
Rectangle ABCD lies in the xy-coordinate plane so that its sides are not parallel to the
axes. What is the product of the slopes of all four sides of rectangle ABCD?
m AB  mCD  1
m AD  m BC  1
 1  1  1
2
Points
EFGH is a parallelogram. Find the value of x.
Area  b  h
15  9  10 x
135  10 x
x  13.5
3
Points
When two parallel lines are cut by a transversal, the measures of a pair of alternate
exterior angles are x 2  4 and 11x  22 . What are the measures of a pair of consecutive
interior angles.
x 2  4  11x  22
x2  4
2
x  11x  26  0
x2  4
( x  13)( x  2)  0
11x  22
x  13,2
11x  22
If x = -2, then 11x+22 = (non existent angle)
If x = 13, then 11x+22 = 165
Thus, 165 and 15
Greater New London County Mathematics League
Meet 1
Tuesday, October 5, 2010
Round 4
Inequalities & Absolute Values
1
Point
Solve the following inequality for x. Write your solution in interval notation.
4  x  2  ( x  3)
4 x  2 x3
4  3
x 
2 Points
Solve the following inequality for x. Write your solution in interval notation.
4
x x2
y
2
3
x2  x  2  0
2
1
( x  2)( x  1)  0
x
−6
−4
−2
2
4
6
−1
−2
---- ----x–2
x+1
++++
2
---- ++++ ++++
-1
+
+
−3
−4
x   ,1  2, 
3 Points
Beyonce’s telephone number is (ABC) DEF-GHIJ, where each letter represents a
different digit.
If A>B>C, D>E>F, G>H>I>J, D, E, F are consecutive even integers, G, H, I, J are
consecutive odd numbers and A+B+C = 9.
Find Beyonce’s phone number.
GHIJ can be either 9753 or 7531. Thus, A, B, C must have a 1 or 9. However, A+B+C=9.
So A, B, C cannot be 9. Therefore GHIJ is 9753.
Since B or C is 1, then the remaining digits of ABC must sum 8.
DEF can be 864, 642, or 420. Since the remaining two digits must equal 8, that can only
happen if DEF = 642.
Therefore, call Beyonce at (810) 642-9753
Greater New London County Mathematics League
Meet 1
Tuesday, October 5, 2010
Round 5
Quadratics over the Real Numbers
1
Point
What is the vertex of the parabola given by y  3( x  5) 2  8 ?
The vertex is (-5, 8).
2
Points
Find all the value(s) of k for which 8 x 2  kx  2  0 has no real solutions.
b 2  4ac  0
k 2  4(8)( 2)  0
k 2  64  0
3
Therefore, k   8,8
Points
A farmer has 5000 ft of fencing material with which he builds a rectangular pen for
his ewes and rams. The pen has interior partitions which also require fencing, as seen
in the diagram. Find the dimensions that will yield the maximum area for the animals
to graze?
x
x
 5000  5 x 


2


x
 5000  5 x 
A  x

2


5
A   x 2  2500 x
2
Since this graphs a reflected parabola,
the maximum area occurs at the
vertex.
x
 b  2500

 500
2a
5
x
Therefore, the dimensions are 500 ft x 1250 ft
x
Greater New London County Mathematics League
Meet 1
Tuesday, October 5, 2010
Team Round
4 Points
Line m is perpendicular to line n at point (x ,2). The point (-4, 6) is on line m and the
point (2x, 3x) is on line n. Find the value of x.
 3 x  2  6  2 


  1
 2 x  x   4  x 
8  64  4(1)(8) 8  64  32
x

2
x  4 x  12 x  8
 3 x  2  4 
2
2


1



2
 x   4  x 
x  8x  8  0
8  32 8  4 2

 42 2
 12 x  8 
2
2

  1
2
  x  4x 
4 Points
A carpenter wants to put four shelves on an 8-foot wall so that the five spaces created
decrease by 6 inches as we move up the wall. If the thickness of each shelf is ½ inch,
how far will the bottom shelf be from the floor?
8 feet = 96 inches
x  24
x 18
x 12
x6
x
1
1
1
1
  x  6     x  12     x  18    x  24   96
2
2
2
2
5 x  2  60  96
x
5 x  58  96
5 x  154
x
154
4
 30  30.8
5
5
Greater New London County Mathematics League
Meet 1
Tuesday, October 5, 2010
4
Points
Solve the following absolute value inequality for x. Write your solution in interval
notation.
x 2  2x  2  x 2  2x  2
Personally, I thought about the two graphs without the absolute values.
y  x 2  2 x  2 has a y-intercept of (0, -2) and a vertex at (1, -3) while
y  x 2  2 x  2 has a y-intercept of (0, 2) and a vertex at (1, 1).
I also know that the absolute value function will make all the “negative” parts of the
second graph reflect above the x-axis.
My sketch would look like this.
4
y
3
2
1
x
−6
−4
−2
2
4
6
−1
−2
−3
−4
Based on symmetry of sa parabola, since they intersect at (0, 2) and have an axis of
symmetry at x = 1, then they must intersect at x = 2.
Thus x  (0,2)
Greater New London County Mathematics League
Meet 1
Tuesday, October 5, 2010
4
Points
The length of two opposite sides of a square with side length s is increased by 5 inches
while the lengths of the other two sides is decreased by 2 inches. If the area of the
2
rectangle is s  1 square inches less than the area of the original square. Find the area
of the original square.
Area of square = s 2 and the area of the rectangle = s  5( s  2)  s 2  3s  10
s 2  3s  10  s  1  s 2
2
s 2  3s  10  s 2  2 s  1  s 2
s 2  5s  9  0
s
 5  25  4(1)( 9)  5  25  36  5  61


2
2
2
 5  61
. Thus the area of the square is
2
  5  61   5  61 






2
2



Therefore s 

25  10 61  61 86  10 61 43  5 61


4
4
2
Greater New London County Mathematics League
Meet 1
Tuesday, October 5, 2010
Answers
Round 1
Round 4
1.
4019
1.

2.
9
20
2.
(,1)  (2, )
3.
63, 65
3.
(810) 642-9753
or no solution
Round 5
Round 2
1.
36
2.
90
3.
% F = 10
%C = 62
1.
(-5, 8)
2.
(8,8)
3.
500 x 1250
Team
Round 3
1.
4 2
1.
1
2.
154
4
 30  30.8
5
5
2.
27
2
3.
(0, 2)
3.
15, 165
or 13.5
4.
43  5 61
2