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

-
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
-
A gallon of paint is used to paint a room. One third of the paint is used on
the first day. On the second day, one third of the remaining paint is used.
What fraction of the original amount of paint is available to use on the third
day?
(A)
1
10
(B)
1
9
(C)
1
3
(D)
4
9
(E)
5
9
2005 AMC 10 B, Problem #3— “How much paint is used on the second day?”
- Solution (D) After the first day,
1−
of the paint remains. On the second day,
1
2
=
3
3
1 2
2
· =
3 3
9
of the paint is used. So for the third day
1−
1 2
4
− =
3 9
9
of the original gallon of paint is available.
Difficulty: Medium
NCTM Standard: Problem Solving Standard: Solve problems that arise in mathematics and in other contexts
Mathworld.com Classification:
Number Theory > Arithmetic > Fractions > Fraction
28
-
For real numbers a and b, define a ¦ b =
√
a2 + b2 . What is the value of
(5 ¦ 12) ¦ ((−12) ¦ (−5)) ?
(A) 0
(B)
17
2
(C) 13
√
(D) 13 2
(E) 26
2005 AMC 10 B, Problem #4— “Set up the equation in terms of the square
root.”
- Solution (D) It follows from the definition that
p
p
52 + 122 ¦ (−12)2 + (−5)2
p
√
=13 ¦ 13 = 132 + 132 = 13 2.
(5 ¦ 12) ¦ ((−12) ¦ (−5)) =
Difficulty: Medium
NCTM Standard: Algebra Standard: Represent and analyze mathematical situations and structures using algebraic symbols
Mathworld.com Classification:
Calculus and Analysis > Operator Theory > Operator
29
-
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
-
A circle is inscribed in a square, then a square is inscribed in this circle, and
finally, a circle is inscribed in this square. What is the ratio of the area of
the smaller circle to the area of the larger square?
(A)
π
16
(B)
π
8
(C)
3π
16
(D)
π
4
(E)
π
2
2005 AMC 10 B, Problem #7— “Set the radius of the larger circle to something.
What is the side length of each square?”
- Solution (B) Let the radius of the smaller circle be r. Then the side length of the smaller square
is 2r.
√
The radius of the larger circle is half the
length
of
the
diagonal
of
the
smaller
square,
so
it
is
2r.
Hence
√
the larger square has sides of length 2 2r. The ratio of the area of the smaller circle to the area of the
larger square is therefore
πr2
π
¡ √ ¢2 = 8 .
2 2r
2r
2 √2 r
r
√2 r
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 > Line Geometry > Concurrence > Inscribed
32
-
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
-
One fair die has faces 1, 1, 2, 2, 3, 3 and another has faces 4, 4, 5, 5, 6, 6.
The dice are rolled and the numbers on the top faces are added. What is
the probability that the sum will be odd?
(A)
1
3
(B)
4
9
(C)
1
2
(D)
5
9
(E)
2
3
2005 AMC 10 B, Problem #9— “If the first die rolls an {1, 1, 2, 2, 3, 3} then what
must the second die roll so their sum is odd?”
- Solution (D) An odd sum requires either that the first die is even and the second is odd or that the first
die is odd and the second is even. The probability is
1 4
5
1 1 2 2
· + · = + = .
3 3 3 3
9 9
9
Difficulty: Medium
NCTM Standard: Data Analysis and Probability Standard: Understand and apply basic concepts of probability
Mathworld.com Classification:
Probability and Statistics > Probability > Probability
34
-
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
-
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
-
Twelve fair dice are rolled. What is the probability that the product of the
numbers on the top faces is prime?
µ
(A)
1
12
¶12
µ ¶12
1
(B)
6
µ ¶11
1
(C) 2
6
5
(D)
2
µ ¶11
1
6
µ ¶10
1
(E)
6
2005 AMC 10 B, Problem #12— “What are the divisors of a prime number?”
- Solution (E) Exactly one die must have a prime face on top, and the other eleven must have 1’s. The
prime die can be any one of the twelve, and the prime can be 2, 3, or 5. Thus the probability of a prime
face on any one die is 1/2, and the probability of a prime product is
µ ¶ µ ¶11 µ ¶10
1
1
1
=
.
12
2
6
6
Difficulty: Hard
NCTM Standard: Data Analysis and Probability Standard: Understand and apply basic concepts of probability
Mathworld.com Classification:
Probability and Statistics > Probability > Probability
37
-
How many numbers between 1 and 2005 are integer multiples of 3 or 4 but
not 12?
(A) 501
(B) 668
(C) 835
(D) 1002
(E) 1169
2005 AMC 10 B, Problem #13— “Find the integer multiples of 3 that aren’t
multiples of 12 and the integer multiples of 4 that aren’t multiples of 12.”
- Solution (C) Between 1 and 2005, there are 668 multiples of 3, 501 multiples of 4, and 167 multiples of
12. So there are
(668 − 167) + (501 − 167) = 835
numbers between 1 and 2005 that are integer multiples of 3 or of 4 but not of 12.
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
38
-
Equilateral 4ABC has side length 2, M is the midpoint of AC, and C is
the midpoint of BD. What is the area of 4CDM ?
A
M
B
√
2
(A)
2
3
(B)
4
D
C
√
(C)
3
2
(D) 1
(E)
√
2
2005 AMC 10 B, Problem #14— “Make right triangles.”
- Solution (C) Drop M Q perpendicular to BC. Then 4M QC is a 30–60–90◦ triangle, so M Q =
and the area of 4CDM is
à √ ! √
1
3
3
2·
=
.
2
2
2
√
3/2,
OR
Triangles ABC and CDM have equal bases. Because M is the midpoint of AC, the ratio of the altitudes
from M and from A is 1/2. So the area of 4CDM is half of the area of 4ABC. Since
√
√
3 2 √
3
· 2 = 3, we have Area(4CDM ) =
.
Area(4ABC) =
4
2
Difficulty: Medium-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 > Equilateral Triangle
39
-
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
-
Suppose that 4a = 5, 5b = 6, 6c = 7, and 7d = 8. What is a · b · c · d?
(A) 1
(B)
3
2
(C) 2
(D)
5
2
2005 AMC 10 B, Problem #17— “What is 4a·b·c·d equal to?”
- Solution (B) Because
³³
´c ´d ¡¡ ¢ ¢d
c
d
b
= (6c ) = 7d = 8 = 43/2 , we have a · b · c · d = 3/2.
4a·b·c·d =
(4a )
= 5b
Difficulty: Medium-hard
NCTM Standard: Algebra Standard: Understand patterns, relations, and functions
Mathworld.com Classification:
Calculus and Analysis > Special Functions > Powers > Power
42
(E) 3
-
All of David’s telephone numbers have the form 555–abc–defg, where a, b,
c, d, e, f, and g are distinct digits and in increasing order, and none is either
0 or 1. How many different telephone numbers can David have?
(A) 1
(B) 2
(C) 7
(D) 8
(E) 9
2005 AMC 10 B, Problem #18— “How many digits are unused?”
- Solution (D) The last seven digits of the phone number use seven of the eight digits {2, 3, 4, 5, 6, 7, 8, 9},
so all but one of these digits is used. The unused digit can be chosen in eight ways. The remaining seven
digits are then placed in increasing order to obtain a possible phone number. Thus there are 8 possible
phone numbers.
Difficulty: Medium-hard
NCTM Standard: Number and Operations Standard: Understand numbers, ways of representing numbers, relationships among
numbers, and number systems
Mathworld.com Classification:
Number Theory > Special Numbers > Digit-Related Numbers > Digit
43
-
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
-
What is the average (mean) of all 5-digit numbers that can be formed by
using each of the digits 1, 3, 5, 7, and 8 exactly once?
(A) 48000
(B) 49999.5
(C) 53332.8
(D) 55555
(E) 56432.8
2005 AMC 10 B, Problem #20— “How many times does each digit appear in
the 1’s place, the 10’s place, . . ., and the 10,000’s place?”
- Solution (C) Each digit appears the same number of times in the 1’s place, the 10’s place, . . ., and the
10,000’s place. The average of the digits in each place is
24
1
(1 + 3 + 5 + 7 + 8) =
= 4.8.
5
5
Hence the average of all the numbers is
4.8(1 + 10 + 100 + 1000 + 10000) = 4.8(11111) = 53332.8.
Difficulty: Hard
NCTM Standard: Data Analysis and Probability Standard: Develop and evaluate inferences and predictions that are based on
data
Mathworld.com Classification:
Calculus and Analysis > Special Functions > Means > Arithmetic Mean
45
-
Forty slips are placed into a hat, each bearing a number 1, 2, 3, 4, 5, 6, 7,
8, 9, or 10, with each number entered on four slips. Four slips are drawn
from the hat at random and without replacement. Let p be the probability
that all four slips bear the same number. Let q be the probability that two
of the slips bear a number a and the other two bear a number b 6= a. What
is the value of q/p?
(A) 162
(B) 180
(C) 324
(D) 360
(E) 720
2005 AMC 10 B, Problem #21— “How many ways are there to choose two
distinct numbers? How many ways are there to choose each individual number?”
- Solution (A) The total number of ways that the numbers can be chosen is
possibilities result in the four slips having the same number.
¡40¢
4
. Exactly 10 of these
Now we need to determine the number of¡ ways
¢ that two slips can have a number a and the other two slips
have a number b, with b 6= a. There are 10
ways to choose the distinct numbers a and b. For each value
2
¡¢
¡¢
of a there are 42 to choose the two slips with a and for each value of b there are 42 to choose the two
slips with b. Hence the number of ways that two slips have some number a and the other two slips have
some distinct number b is
µ ¶ µ ¶ µ ¶
10
4
4
·
·
= 45 · 6 · 6 = 1620.
2
2
2
10
1620
So the probabilities q and p are ¡40¢ and ¡40¢ , respectively, which implies that
4
4
p
1620
=
= 162.
q
10
Difficulty: Hard
NCTM Standard: Data Analysis and Probability Standard: Understand and apply basic concepts of probability
Mathworld.com Classification:
Probability and Statistics > Probability > Probability
46
-
For how many positive integers n less than or equal to 24 is n! evenly divisible
by 1 + 2 + · · · + n?
(A) 8
(B) 12
(C) 16
(D) 17
(E) 21
2005 AMC 10 B, Problem #22— “What does 1 + 2 + · · · + n sum to?”
- Solution (C) Since
1 + 2 + ··· + n =
n(n + 1)
,
2
the condition is equivalent to having an integer value for
n!
.
n(n + 1)/2
This reduces, when n ≥ 1, to having an integer value for
2(n − 1)!
.
n+1
This fraction is an integer unless n + 1 is an odd prime. There are 8 odd primes less than or equal to 25,
so there are 24 − 8 = 16 numbers less than or equal to 24 that satisfy the condition.
Difficulty: Hard
NCTM Standard: Number and Operations Standard: Understand meanings of operations and how they relate to one another
Mathworld.com Classification:
Number Theory > Divisors > Divisible
47
-
In trapezoid ABCD we have AB parallel to DC, E as the midpoint of BC,
and F as the midpoint of DA. The area of ABEF is twice the area of
F ECD. What is AB/DC?
(A) 2
(B) 3
(C) 5
(D) 6
(E) 8
2005 AMC 10 B, Problem #23— “Note that F E = (AB + DC)/2.”
- Solution (C) First note that F E = (AB + DC)/2. Because trapezoids ABEF and F ECD have the
same height, the ratio of their areas is equal to the ratio of the averages of their parallel sides. Since
AB +
and
AB + DC
3AB + DC
=
2
2
AB + DC
AB + 3DC
+ DC =
,
2
2
we have
3AB + DC = 2(AB + 3DC) = 2AB + 6DC,
and
AB
= 5.
DC
C
D
E
F
A
B
Difficulty: Hard
NCTM Standard: Geometry Standard: Apply transformations and use symmetry to analyze mathematical situations
Mathworld.com Classification:
Geometry > Plane Geometry > Quadrilaterals > Trapezoid
48
-
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
-
A subset B of the set of integers from 1 to 100, inclusive, has the property
that no two elements of B sum to 125. What is the maximum possible
number of elements in B?
(A) 50
(B) 51
(C) 62
(D) 65
(E) 68
2005 AMC 10 B, Problem #25— “How many pairs sum to 125?”
- Solution (C) Several pairs of numbers from 1 to 100 sum to 125. These pairs are (25, 100), (26, 99), . . . , (62, 63).
Set B can have at most one number from each of these 62 − 25 + 1 = 38 pairs. In addition, B can contain
all of the numbers 1, 2, . . . , 24 since these cannot be paired with any of the available numbers to sum
to 125. So B has at most 38 + 24 = 62 numbers. The set containing the first 62 positive integers, for
example, is one of these maximum sets.
Difficulty: Medium-hard
NCTM Standard: Number and Operations Standard: Understand numbers, ways of representing numbers, relationships among
numbers, and number systems
Mathworld.com Classification:
Foundations of Mathematics > Set Theory > Sets > Set
50