Download A scout troop buys 1000 candy bars at a price of five for $2. They sell

-
A scout troop buys 1000 candy bars at a price of five for $2. They sell all
the candy bars at a price of two for $1. What was their profit, in dollars?
(A) 100
(B) 200
(C) 300
(D) 400
(E) 500
2005 AMC 10 B, Problem #1— “How much does it cost to buy one candy bar?
For how much do they sell one candy bar?”
- Solution (A) The scouts bought 1000/5 = 200 groups of 5 candy bars at a total cost of 200 · 2 = 400
dollars. They sold 1000/2 = 500 groups of 2 candy bars for a total of 500 · 1 = 500 dollars. Their profit
was $500 − $400 = $100.
Difficulty: Easy
NCTM Standard: Problem Solving Standard: Solve problems that arise in mathematics and in other contexts
Mathworld.com Classification:
Applied Mathematics > Business > Economics > Marginal Analysis
26
-
A positive number x has the property that x% of x is 4. What is x?
(A) 2
(B) 4
(C) 10
(D) 20
2005 AMC 10 B, Problem #2— “Convert x% to a fraction.”
- Solution (D) We have
x
· x = 4,
100
so
x2 = 400.
Because x > 0, it follows that x = 20.
Difficulty: Easy
NCTM Standard: Algebra Standard: Understand patterns, relations, and functions
Mathworld.com Classification:
Number Theory > Arithmetic > Fractions > Percent
27
(E) 40
-
Brianna is using part of the money she earned on her weekend job to buy
several equally-priced CDs. She used one fifth of her money to buy one third
of the CDs. What fraction of her money will she have left after she buys all
the CDs?
(A)
1
5
(B)
1
3
(C)
2
5
(D)
2
3
(E)
4
5
2005 AMC 10 B, Problem #5— “If she used one fifth of her money to buy one
third of the CDs then what fraction of her money has she spent to buy three
thirds of the CDs?”
- Solution (C) The number of CDs that Brianna will finally buy is three times the number she has already
bought. The fraction of her money that will be required for all the purchases is (3)(1/5) = 3/5. The
fraction she will have left is 1 − 3/5 = 2/5.
Difficulty: Easy
NCTM Standard: Problem Solving Standard: Solve problems that arise in mathematics and in other contexts
Mathworld.com Classification:
Number Theory > Arithmetic > Fractions > Fraction
30
-
At the beginning of the school year, Lisa’s goal was to earn an A on at least
80% of her 50 quizzes for the year. She earned an A on 22 of the first 30
quizzes. If she is to achieve her goal, on at most how many of the remaining
quizzes can she earn a grade lower than an A?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
2005 AMC 10 B, Problem #6— “How many quizzes does she need to earn an A
on?”
- Solution (B) To earn an A on at least 80% of her quizzes, Lisa needs to receive an A on at least
(0.8)(50) = 40 quizzes. Thus she must earn an A on at least 40 − 22 = 18 of the remaining 20. So she
can earn a grade lower than an A on at most 2 of the remaining quizzes.
Difficulty: Easy
NCTM Standard: Problem Solving Standard: Solve problems that arise in mathematics and in other contexts
Mathworld.com Classification:
Number Theory > Arithmetic > Fractions > Percent
31
-
An 8-foot by 10-foot floor is tiled with square tiles of size 1 foot by 1 foot.
Each tile has a pattern consisting of four white quarter circles of radius 1/2
foot centered at each corner of the tile. The remaining portion of the tile is
shaded. How many square feet of the floor are shaded?
(A) 80 − 20π
(B) 60 − 10π
(C) 80 − 10π
(D) 60 + 10π
(E) 80 + 10π
2005 AMC 10 B, Problem #8— “How much of each tile is shaded?”
- Solution (A) The four white quarter circles in each tile have the same area as a whole circle of radius 1/2,
that is, π(1/2)2 = π/4 square feet. So the area of the shaded portion of each tile is 1 − π/4 square feet.
Since there are 8 · 10 = 80 tiles in the entire floor, the area of the total shaded region in square feet is
³
π´
= 80 − 20π.
80 1 −
4
Difficulty: Medium
NCTM Standard: Geometry Standard: Analyze characteristics and properties of two- and three-dimensional geometric shapes
and develop mathematical arguments about geometric relationships
Mathworld.com Classification:
Geometry > Plane Geometry > Circles > Circle
33
-
In 4ABC, we have AC = BC = 7 and AB = 2. Suppose that D is a
point on line AB such that B lies between A and D and CD = 8. What is
BD?
(A) 3
√
(B) 2 3
(C) 4
(D) 5
√
(E) 4 2
2005 AMC 10 B, Problem #10— “Use the Pythagorean Theorem and create a
right angle coming down from C to the line AB.”
- Solution (A) Let CH be an altitude of 4ABC. Applying the Pythagorean Theorem to 4CHB and to
4CHD produces
82 − (BD + 1)2 = CH 2 = 72 − 12 = 48,
so
(BD + 1)2 = 16.
Thus BD = 3.
C
7
7
8
1 1
A H B
D
Difficulty: Hard
NCTM Standard: Geometry Standard: Analyze characteristics and properties of two- and three-dimensional geometric shapes
and develop mathematical arguments about geometric relationships
Mathworld.com Classification:
Geometry > Plane Geometry > Triangles > Special Triangles > Other Triangles > Triangle
35
-
What is the area enclosed by the graph of |3x| + |4y| = 12?
(A) 6
(B) 12
(C) 16
(D) 24
(E) 25
2005 AMC 12 B, Problem #7— “What is the shape of the region?”
- Solution (D) The graph is symmetric with respect to both coordinate axes, and in the first quadrant it
coincides with the graph of the line 3x + 4y = 12. Therefore the region is a rhombus, and the area is
µ
¶
1
Area = 4
(4 · 3) = 24.
2
y
3
-4
4
x
-3
Difficulty: Medium-hard
NCTM Standard: Geometry Standard: Specify locations and describe spatial relationships using coordinate geometry and other
representational systems
Mathworld.com Classification:
Geometry > Plane Geometry > Quadrilaterals > Rhombus
66
-
For how many values of a is it true that the line y = x + a passes through
the vertex of the parabola y = x2 + a2 ?
(A) 0
(B) 1
(C) 2
(D) 10
(E) infinitely many
2005 AMC 12 B, Problem #8— “The vertex of the parabola is (0, a2 ).”
- Solution (C) The vertex of the parabola is (0, a2 ). The line passes through the vertex if and only if
a2 = 0 + a. There are two solutions to this equation, namely a = 0 and a = 1.
Difficulty: Medium
NCTM Standard: Geometry Standard: Analyze characteristics and properties of two- and three-dimensional geometric shapes
and develop mathematical arguments about geometric relationships
Mathworld.com Classification:
Geometry > Curves > Plane Curves > Conic Sections > Parabola Vertex
67
-
On a certain math exam, 10% of the students got 70 points, 25% got 80
points, 20% got 85 points, 15% got 90 points, and the rest got 95 points.
What is the difference between the mean and the median score on this
exam?
(A) 0
(B) 1
(C) 2
(D) 4
(E) 5
2005 AMC 10 B, Problem #19— “For what score were fewer than half of the
scores were less than it, and fewer than half of the scores were greater than it?”
- Solution (B) The percentage of students getting 95 points is
100 − 10 − 25 − 20 − 15 = 30,
so the mean score on the exam is
0.10(70) + 0.25(80) + 0.20(85) + 0.15(90) + 0.30(95) = 86.
Since fewer than half of the scores were less than 85, and fewer than half of the scores were greater than 85,
the median score is 85. The difference between the mean and the median score on this exam is 86 − 85 = 1.
Difficulty: Medium
NCTM Standard: Data Analysis and Probability Standard: Develop and evaluate inferences and predictions that are based on
data
Mathworld.com Classification:
Number Theory > Arithmetic > Fractions > Percent
44
-
The first term of a sequence is 2005. Each succeeding term is the sum of
the cubes of the digits of the previous term. What is the 2005th term of the
sequence?
(A) 29
(B) 55
(C) 85
(D) 133
(E) 250
2005 AMC 10 B, Problem #11— “Look for a pattern in the sequence.”
- Solution (E) The sequence begins 2005, 133, 55, 250, 133, . . . . Thus after the initial term 2005, the
sequence repeats the cycle 133, 55, 250. Because 2005 = 1 + 3 · 668, the 2005th term is the same as the
last term of the repeating cycle, 250.
Difficulty: Hard
NCTM Standard: Number and Operations Standard: Understand numbers, ways of representing numbers, relationships among
numbers, and number systems
Mathworld.com Classification:
Number Theory > Sequences > Sequence
36
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An envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 twenties. Two
bills are drawn at random without replacement. What is the probability that
their sum is $20 or more?
(A)
1
4
(B)
2
5
(C)
3
7
(D)
1
2
(E)
2
3
2005 AMC 10 B, Problem #15— “How many ways are there to choose two bills?
What ways is the sum greater than or equal to $20?”
- Solution (D) There are
µ ¶
8
8!
=
= 28
2
6! · 2!
ways to choose the bills. A sum of at least $20 is obtained by choosing both $20 bills, one of the $20 bills
and one of the six smaller bills, or both $10 bills. Hence the probability is
14
1
1+2·6+1
=
= .
28
28
2
Difficulty: Medium
NCTM Standard: Data Analysis and Probability Standard: Understand and apply basic concepts of probability.
Mathworld.com Classification:
Probability and Statistics > Probability > Probability
40
-
The quadratic equation x2 + mx + n = 0 has roots that are twice those of
x2 + px + m = 0, and none of m, n, and p is zero. What is the value of
n/p?
(A) 1
(B) 2
(C) 4
(D) 8
(E) 16
2005 AMC 10 B, Problem #16— “Set r1 and r2 to be the roots of x2 + px + m = 0”
- Solution (D) Let r1 and r2 be the roots of x2 + px + m = 0. Since the roots of x2 + mx + n = 0 are 2r1
and 2r2 , we have the following relationships:
m = r1 r2 ,
n = 4r1 r2 ,
p = −(r1 + r2 ),
So
n = 4m,
1
m,
2
p=
and
and m = −2(r1 + r2 ).
n
4m
= 1 = 8.
p
2m
OR
The roots of
³ x ´2
2
+p
³x´
2
+m=0
are twice those of x2 + px + m = 0. Since the first equation is equivalent to x2 + 2px + 4m = 0, we have
m = 2p and
n = 4m,
so
n
= 8.
p
Difficulty: Hard
NCTM Standard: Algebra Standard: Understand patterns, relations, and functions
Mathworld.com Classification:
Algebra > Algebraic Equations > Quadratic Equation
41
-
Let x and y be two-digit integers such that y is obtained by reversing the
digits of x. The integers x and y satisfy x2 − y 2 = m2 for some positive
integer m. What is x + y + m?
(A) 88
(B) 112
(C) 116
(D) 144
(E) 154
2005 AMC 10 B, Problem #24— “Let x = 10a + b. What is y? What is x2 − y 2 ?”
- Solution (E) By the given conditions, it follows that x > y. Let x = 10a + b and y = 10b + a, where
a > b. Then
m2 = x2 − y 2 = (10a + b)2 − (10b + a)2 = 99a2 − 99b2 = 99(a2 − b2 ).
Since 99(a2 − b2 ) must be a perfect square,
a2 − b2 = (a + b)(a − b) = 11k 2 ,
for some positive integer k. Because a and b are distinct digits, we have a − b ≤ 9 − 1 = 8 and
a + b ≤ 9 + 8 = 17. It follows that a + b = 11, a − b = k 2 , and k is either 1 or 2.
If k = 2, then (a, b) = (15/2, 7/2), which is impossible. Thus k = 1 and (a, b) = (6, 5). This gives x = 65,
y = 56, m = 33, and x + y + m = 154.
Difficulty: Hard
NCTM Standard: Algebra Standard: Represent and analyze mathematical situations and structures using algebraic symbols
Mathworld.com Classification:
Algebra > General Algebra > Algebra
49