2005 AMC 10B ws-25

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- A scout troop buys 1000 candy bars at a price of five for $ 2. They sell allthe 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 = 400dollars. They sold 1000/2 = 500 groups of 2 candy bars for a total of 500 1 = 500 dollars. Their profitwas $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

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- A positive number x has the property that x% of x is 4. What is x?

(A) 2 (B) 4 (C) 10 (D) 20 (E) 40

2005 AMC 10 B, Problem #2 Convert x% to a fraction.

- Solution (D) We havex

100 x = 4, 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

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- A gallon of paint is used to paint a room. One third of the paint is used onthe 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 13

=2

3

of the paint remains. On the second day,1

3 2

3=

2

9

of the paint is used. So for the third day

1 13 2

9=

4

9

of the original gallon of paint is available.

Difficulty: Medium


Mathworld.com Classification:Number Theory > Arithmetic > Fractions > Fraction

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- 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 squareroot.

-Solution (D)

It follows from the definition that

(5 12) ((12) (5)) =

52 + 122

(12)2 + (5)2=13 13 =

132 + 132 = 13

2.

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

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- Brianna is using part of the money she earned on her weekend job to buyseveral equally-priced CDs. She used one fifth of her money to buy one thirdof 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 onethird of the CDs then what fraction of her money has she spent to buy threethirds of the CDs?

- Solution (C) The number of CDs that Brianna will finally buy is three times the number she has alreadybought. The fraction of her money that will be required for all the purchases is (3)(1/5) = 3/5. Thefraction she will have left is 1 3/5 = 2/5.

Difficulty: Easy


Mathworld.com Classification:Number Theory > Arithmetic > Fractions > Fraction

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- At the beginning of the school year, Lisas goal was to earn an A on at least80% of her 50 quizzes for the year. She earned an A on 22 of the first 30quizzes. If she is to achieve her goal, on at most how many of the remainingquizzes 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 Aon?

- 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 shecan earn a grade lower than an A on at most 2 of the remaining quizzes.

Difficulty: Easy



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- A circle is inscribed in a square, then a square is inscribed in this circle, andfinally, a circle is inscribed in this square. What is the ratio of the area ofthe smaller circle to the area of the larger square?

(A)

16(B)

8(C)

3

16(D)

4(E)

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

. Hencethe larger square has sides of length 22r. The ratio of the area of the smaller circle to the area of thelarger square is therefore

r22

2r2 = 8 .

2 2

2

r

r

2r

r

Difficulty: Hard

NCTM Standard: Geometry Standard: Analyze characteristics and properties of two- and three-dimensional geometric shapesand develop mathematical arguments about geometric relationships

Mathworld.com Classification:Geometry > Line Geometry > Concurrence > Inscribed

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- 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 isshaded. How many square feet of the floor are shaded?

(A) 80 20 (B) 60 10 (C) 80 10 (D) 60 + 10 (E) 80 + 10

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, (1/2)2 = /4 square feet. So the area of the shaded portion of each tile is 1

/4 square feet.

Since there are 8 10 = 80 tiles in the entire floor, the area of the total shaded region in square feet is80

1 4

= 80 20.

Difficulty: Medium


Mathworld.com Classification:Geometry > Plane Geometry > Circles > Circle

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- 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 isthe 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 whatmust 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 firstdie is odd and the second is even. The probability is

1

3 1

3+

2

3 2

3=

1

9+

4

9=

5

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

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- In ABC, we have AC = BC = 7 and AB = 2. Suppose that D is apoint on line AB such that B lies between A and D and CD = 8. What isBD?

(A) 3 (B) 2

3 (C) 4 (D) 5 (E) 4

2

2005 AMC 10 B, Problem #10 Use the Pythagorean Theorem and create aright angle coming down from C to the line AB.

- Solution (A) Let CH be an altitude ofABC. Applying the Pythagorean Theorem to CH B and toCH D produces

82 (BD + 1)2 = CH2 = 72 12 = 48, so (BD + 1)2 = 16.

Thus BD = 3.

A H B D

C

77

11

8

Difficulty: Hard


Mathworld.com Classification:Geometry > Plane Geometry > Triangles > Special Triangles > Other Triangles > Triangle

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- The first term of a sequence is 2005. Each succeeding term is the sum ofthe cubes of the digits of the previous term. What is the 2005th term of thesequence?

(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, thesequence repeats the cycle 133, 55, 250. Because 2005 = 1 + 3 668, the 2005th term is the same as thelast term of the repeating cycle, 250.

Difficulty: Hard

NCTM Standard: Number and Operations Standard: Understand numbers, ways of representing numbers, relationships amongnumbers, and number systems

Mathworld.com Classification:Number Theory > Sequences > Sequence

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- Twelve fair dice are rolled. What is the probability that the product of thenumbers on the top faces is prime?

(A)

1

12

12(B)

1

6

12(C) 2

1

6

11(D)

5

2

1

6

11(E)

1

6

10

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 1s. Theprime 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

12

1

2

1

6

11=

1

6

10.

Difficulty: Hard



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- How many numbers between 1 and 2005 are integer multiples of 3 or 4 butnot 12?

(A) 501 (B) 668 (C) 835 (D) 1002 (E) 1169

2005 AMC 10 B, Problem #13 Find the integer multiples of 3 that arentmultiples of 12 and the integer multiples of 4 that arent multiples of 12.

- Solution (C) Between 1 and 2005, there are 668 multiples of 3, 501 multiples of 4, and 167 multiples of12. So there are

(668 167) + (501 167) = 835numbers between 1 and 2005 that are integer multiples of 3 or of 4 but not of 12.

Difficulty: Hard


Mathworld.com Classification:Number Theory > Sequences > Sequence

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- Equilateral ABC has side length 2, M is the midpoint of AC, and C isthe midpoint of BD. What is the area ofCDM?

A

B C D

M

(A)

2

2(B)

3

4(C)

3

2(D) 1 (E)

2

2005 AMC 10 B, Problem #14 Make right triangles.

- Solution (C) Drop MQ perpendicular to BC. Then MQC is a 306090 triangle, so MQ = 3/2,and the area ofCDM is

1

2

2

3

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 ofCDM is half of the area ofABC. SinceArea(ABC) =

3

4 22 =

3, we have Area(CDM) =

3

2.

Difficulty: Medium-hard


Mathworld.com Classification:Geometry > Plane Geometry > Triangles > Special Triangles > Other Triangles > Equilateral Triangle

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- An envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 twenties. Twobills are drawn at random without replacement. What is the probability thattheir 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 8

2

=

8!

6! 2! = 28

ways to choose the bills. A sum of at least$

20 is obtained by choosing both$

20 bills, one of the$

20 billsand one of the six smaller bills, or both $ 10 bills. Hence the probability is

1 + 2 6 + 128

=14

28=

1

2.

Difficulty: Medium

NCTM Standard: Data Analysis and Probability Standard: Understand and apply basic concepts of probability.


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- The quadratic equation x2 + mx + n = 0 has roots that are twice those ofx2 + 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 2r1and 2r2, we have the following relationships:

m = r1r2, n = 4r1r2, p = (r1 + r2), and m = 2(r1 + r2).

So

n = 4m, p =1

2m, and

n

p=

4m12

m= 8.

OR

The roots of

x

22

+px

2 + m = 0are 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, son

p= 8.

Difficulty: Hard


Mathworld.com Classification:Algebra > Algebraic Equations > Quadratic Equation

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- Suppose that 4a = 5, 5b = 6, 6c = 7, and 7d = 8. What is a b c d?

(A) 1 (B) 32

(C) 2 (D) 52

(E) 3

2005 AMC 10 B, Problem #17 What is 4abcd equal to?

- Solution (B) Because

4abcd =

(4a)bcd

=

5bcd

= (6c)d

= 7d = 8 = 43/2, we have a b c d = 3/2.



Mathworld.com Classification:Calculus and Analysis > Special Functions > Powers > Power

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- All of Davids telephone numbers have the form 555abcdefg, where a, b,c, d, e, f, and g are distinct digits and in increasing order, and none is either0 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 sevendigits are then placed in increasing order to obtain a possible phone number. Thus there are 8 possiblephone numbers.



Mathworld.com Classification:Number Theory > Special Numbers > Digit-Related Numbers > Digit

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- On a certain math exam, 10% of the students got 70 points, 25% got 80points, 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 thisexam?

(A) 0 (B) 1 (C) 2 (D) 4 (E) 5

2005 AMC 10 B, Problem #19 For what score were fewer than half of thescores 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 8685 = 1.

Difficulty: Medium

NCTM Standard: Data Analysis and Probability Standard: Develop and evaluate inferences and predictions that are based ondata


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- What is the average (mean) of all 5-digit numbers that can be formed byusing 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 inthe 1s place, the 10s place, . . ., and the 10,000s place?

- Solution (C) Each digit appears the same number of times in the 1s place, the 10s place, . . ., and the10,000s place. The average of the digits in each place is

1

5(1 + 3 + 5 + 7 + 8) =

24

5= 4.8.

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 ondata

Mathworld.com Classification:Calculus and Analysis > Special Functions > Means > Arithmetic Mean

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- 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 drawnfrom 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 twoof the slips bear a number a and the other two bear a number b = a. Whatis 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 twodistinct 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

404

. Exactly 10 of these

possibilities result in the four slips having the same number.

Now we need to determine the number of ways that two slips can have a number a and the other two slipshave a number b, with b = a. There are 10

2

ways to choose the distinct numbers a and b. For each value

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 havesome distinct number b is

10

2

4

2

4

2

= 45 6 6 = 1620.

So the probabilities q and p are

10404

and

1620404

, respectively, which implies that

p

q=

1620

10= 162.

Difficulty: Hard



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- For how many positive integers n less than or equal to 24 is n! evenly divisibleby 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

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- Let x and y be two-digit integers such that y is obtained by reversing thedigits of x. The integers x and y satisfy x2 y2 = m2 for some positiveinteger 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 y2?

- Solution (E) By the given conditions, it follows that x > y. Let x = 10a + b and y = 10b + a, wherea > b. Then

m2 = x2 y2 = (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) = 11k2,

for some positive integer k. Because a and b are distinct digits, we have a b 9 1 = 8 anda + b 9 + 8 = 17. It follows that a + b = 11, a b = k2, and k is either 1 or 2.Ifk = 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

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- A subset B of the set of integers from 1 to 100, inclusive, has the propertythat no two elements of B sum to 125. What is the maximum possiblenumber 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 6225+ 1 = 38 pairs. In addition, B can containall of the numbers 1, 2, . . . , 24 since these cannot be paired with any of the available numbers to sumto 125. So B has at most 38 + 24 = 62 numbers. The set containing the first 62 positive integers, forexample, is one of these maximum sets.



Mathworld.com Classification:Foundations of Mathematics > Set Theory > Sets > Set

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2005 AMC 10B ws-25