2005 AMC 10A 考试题目

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1.

While eating out, Mike and Joe each tipped their server $\$2.$ Mike tipped $10\%$ of his bill and Joe tipped $20\%$ of his bill. What was the difference, in dollars, between their bills?

$2$

$4$

$5$

$10$

$20$

Answer: D

Concepts:percentage money

Difficulty rating: 900

Solution:

Mike's $\$2$ tip is $10\%$ of his bill, so his bill is $2 \times 10 = 20$ dollars. Joe's $\$2$ tip is $20\%$ of his bill, so his bill is $2 \times 5 = 10$ dollars. The difference is $20 - 10 = 10$ dollars.

Thus, the correct answer is D.

2.

For each pair of real numbers $a \neq b,$ define the operation $\star$ as $(a \star b) = \frac{a+b}{a-b}.$

What is the value of $((1 \star 2) \star 3)?$

$-\dfrac{2}{3}$

$-\dfrac{1}{5}$

$0$

$\dfrac{1}{2}$

This value is not defined.

Answer: C

Concepts:custom operation fraction order of operations

Difficulty rating: 960

Solution:

First $(1 \star 2) = \dfrac{1+2}{1-2} = \dfrac{3}{-1} = -3.$ Then $(-3 \star 3) = \dfrac{-3+3}{-3-3} = \dfrac{0}{-6} = 0.$

Thus, the correct answer is C.

3.

The equations $2x + 7 = 3$ and $bx - 10 = -2$ have the same solution $x.$ What is the value of $b?$

$-8$

$-4$

$-2$

$4$

$8$

Answer: B

Concepts:linear equation substitution

Difficulty rating: 960

Solution:

From $2x + 7 = 3$ we get $x = -2.$ Substituting, $-2b - 10 = -2,$ so $-2b = 8$ and $b = -4.$

Thus, the correct answer is B.

4.

A rectangle with a diagonal of length $x$ is twice as long as it is wide. What is the area of the rectangle?

$\dfrac{1}{4}x^2$

$\dfrac{2}{5}x^2$

$\dfrac{1}{2}x^2$

$x^2$

$\dfrac{3}{2}x^2$

Answer: B

Concepts:rectangle Pythagorean Theorem area

Difficulty rating: 1100

Solution:

Let the width be $w,$ so the length is $2w.$ Then $x^2 = w^2 + (2w)^2 = 5w^2,$ giving $w^2 = \dfrac{x^2}{5}.$ The area is $w \cdot 2w = 2w^2 = \dfrac{2}{5}x^2.$

Thus, the correct answer is B.

5.

A store normally sells windows at $\$100$ each. This week the store is offering one free window for each purchase of four. Dave needs seven windows and Doug needs eight windows. How many dollars will they save if they purchase the windows together rather than separately?

$100$

$200$

$300$

$400$

$500$

Answer: A

Concepts:money basic counting

Difficulty rating: 1170

Solution:

Alone, Dave pays for $6$ windows and receives one free to reach $7,$ costing $\$600;$ Doug pays for $7$ and receives one free to reach $8,$ costing $\$700.$ Separately they pay $\$1300.$ Together they need $15$ windows: buying $12$ yields $3$ free, for $\$1200.$ The savings are $1300 - 1200 = 100$ dollars.

Thus, the correct answer is A.

6.

The average (mean) of $20$ numbers is $30,$ and the average of $30$ other numbers is $20.$ What is the average of all $50$ numbers?

$23$

$24$

$25$

$26$

$27$

Answer: B

Concepts:mean ratio and proportion

Difficulty rating: 1020

Solution:

The combined sum is $20 \cdot 30 + 30 \cdot 20 = 600 + 600 = 1200.$ The average of all $50$ numbers is $\dfrac{1200}{50} = 24.$

Thus, the correct answer is B.

7.

Josh and Mike live $13$ miles apart. Yesterday Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met?

$4$

$5$

$6$

$7$

$8$

Answer: B

Concepts:distance rate and time linear equation

Difficulty rating: 1240

Solution:

Let Mike ride $m$ miles. Josh rides $\frac{4}{5}$ the rate for $2$ times the time, so Josh's distance is $\frac{8}{5}m.$ Together they cover $13,$ so $m + \frac{8}{5}m = \frac{13}{5}m = 13,$ giving $m = 5.$

Thus, the correct answer is B.

8.

In the figure, the length of side $AB$ of square $ABCD$ is $\sqrt{50},$ $E$ is between $B$ and $H,$ and $BE = 1.$ What is the area of the inner square $EFGH?$

$25$

$32$

$36$

$40$

$42$

Answer: C

Concepts:square (geometry)Pythagorean Theorem congruence (geometry)

Difficulty rating: 1280

Solution:

The triangles $ABH, BCE, CDF,$ and $DAG$ are congruent right triangles. In $\triangle BCE$ the hypotenuse is $BC = \sqrt{50}$ and $BE = 1,$ so $CE = \sqrt{50 - 1} = 7.$ Since $BH = CE = 7$ and $E$ lies on $BH$ with $BE = 1,$ the inner square's side is $EH = 7 - 1 = 6,$ giving area $6^2 = 36.$

Thus, the correct answer is C.

9.

Three tiles are marked X and two other tiles are marked O. The five tiles are randomly arranged in a row. What is the probability that the arrangement reads XOXOX?

$\dfrac{1}{12}$

$\dfrac{1}{10}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

Answer: B

Concepts:basic probability combinations

Difficulty rating: 1280

Solution:

The three X positions can be any of $\binom{5}{3} = 10$ equally likely choices, and exactly one of them produces XOXOX. So the probability is $\dfrac{1}{10}.$

Thus, the correct answer is B.

10.

There are two values of $a$ for which the equation $4x^2 + ax + 8x + 9 = 0$ has only one solution for $x.$ What is the sum of those values of $a?$

$-16$

$-8$

$0$

$8$

$20$

Answer: A

Concepts:quadratic algebraic manipulation

Difficulty rating: 1370

Solution:

Writing the equation as $4x^2 + (a+8)x + 9 = 0,$ there is one solution exactly when the discriminant $(a+8)^2 - 144 = 0.$ Then $a + 8 = \pm 12,$ so $a = 4$ or $a = -20,$ and their sum is $-16.$

Thus, the correct answer is A.

11.

A wooden cube $n$ units on a side is painted red on all six faces and then cut into $n^3$ unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is $n?$

$3$

$4$

$5$

$6$

$7$

Answer: B

Concepts:cube geometry surface area ratio and proportion

Difficulty rating: 1400

Solution:

The unit cubes have $6n^3$ faces total, of which the original surface accounts for $6n^2$ red faces. Then $\dfrac{6n^2}{6n^3} = \dfrac{1}{n} = \dfrac{1}{4},$ so $n = 4.$

Thus, the correct answer is B.

12.

The figure shown is called a trefoil and is constructed by drawing circular sectors about sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length $2?$

$\dfrac{1}{3}\pi + \dfrac{\sqrt{3}}{2}$

$\dfrac{2}{3}\pi$

$\dfrac{2}{3}\pi + \dfrac{\sqrt{3}}{4}$

$\dfrac{2}{3}\pi + \dfrac{\sqrt{3}}{3}$

$\dfrac{2}{3}\pi + \dfrac{\sqrt{3}}{2}$

Answer: B

Concepts:sector area decomposition equilateral triangle

Difficulty rating: 1460

Solution:

Since the base $2$ equals two radii, the radius is $1.$ The trefoil is made of four equilateral triangles and four circular segments, which reassemble into four $60^\circ$ sectors of a circle of radius $1.$ Their total area is $4 \cdot \dfrac{60}{360}\pi (1)^2 = \dfrac{2}{3}\pi.$

Thus, the correct answer is B.

13.

How many positive integers $n$ satisfy the following condition:

$(130n)^{50} \gt n^{100} \gt 2^{200}?$

$0$

$7$

$12$

$65$

$125$

Answer: E

Concepts:inequality exponent counting integers in a range

Difficulty rating: 1540

Solution:

Taking $50$ th roots, the condition becomes $130n \gt n^2 \gt 2^4 = 16.$ From $n^2 \gt 16$ we get $n \gt 4,$ and from $130n \gt n^2$ we get $n \lt 130.$ So $n$ ranges over the integers $5, 6, \ldots, 129,$ which is $125$ values.

Thus, the correct answer is E.

14.

How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?

$41$

$42$

$43$

$44$

$45$

Answer: E

Concepts:digits parity casework

Difficulty rating: 1460

Solution:

The first and last digits must have the same parity so their average is a digit. Both odd gives $5 \cdot 5 = 25$ pairs. Both even, with a nonzero leading digit, gives $4 \cdot 5 = 20$ pairs. Each pair fixes the middle digit, for a total of $25 + 20 = 45$ numbers.

Thus, the correct answer is E.

15.

How many positive cubes divide $3! \cdot 5! \cdot 7!\,?$

$2$

$3$

$4$

$5$

$6$

Answer: E

Concepts:prime factorization factor counting factorial

Difficulty rating: 1580

Solution:

As a product of primes, $3! \cdot 5! \cdot 7! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7.$ A cube divisor uses exponents that are multiples of $3:$ the exponent of $2$ can be $0, 3,$ or $6$ ( $3$ choices), the exponent of $3$ can be $0$ or $3$ ( $2$ choices), and the exponents of $5$ and $7$ must be $0.$ That gives $3 \cdot 2 \cdot 1 \cdot 1 = 6$ cubes.

Thus, the correct answer is E.

16.

The sum of the digits of a two-digit number is subtracted from the number. The units digit of the result is $6.$ How many two-digit numbers have this property?

$5$

$7$

$9$

$10$

$19$

Answer: D

Concepts:digits place value units digit

Difficulty rating: 1510

Solution:

If the number is $10a + b,$ then $(10a + b) - (a + b) = 9a.$ The units digit of $9a$ is $6$ only when $a = 4,$ since $9 \cdot 4 = 36.$ The digit $b$ can then be anything from $0$ to $9,$ giving the ten numbers $40$ through $49.$

Thus, the correct answer is D.

17.

In the five-sided star shown, the letters $A, B, C, D,$ and $E$ are replaced by the numbers $3, 5, 6, 7,$ and $9,$ although not necessarily in this order. The sums of the numbers at the ends of the line segments $AB, BC, CD, DE,$ and $EA$ form an arithmetic sequence, although not necessarily in this order. What is the middle term of the arithmetic sequence?

$9$

$10$

$11$

$12$

$13$

Answer: D

Concepts:arithmetic sequence double counting mean

Difficulty rating: 1660

Solution:

Every number is an endpoint of two segments, so the five segment sums total $2(3 + 5 + 6 + 7 + 9) = 60.$ The middle term of a five-term arithmetic sequence equals its mean, which is $\dfrac{60}{5} = 12.$

Thus, the correct answer is D.

18.

Team A and team B play a series. The first team to win three games wins the series. Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If team B wins the second game and team A wins the series, what is the probability that team B wins the first game?

$\dfrac{1}{5}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

Answer: A

Concepts:conditional probability systematic listing

Difficulty rating: 1860

Solution:

Suppose all five games are played, so every sequence of five results is equally likely. Requiring that B wins game $2$ and A ends up with the series (three wins) leaves the equally likely sequences

$\text{BBAAA}, \quad \text{ABBAA}, \quad \text{ABABA}, \quad \text{ABAAB}, \quad \text{ABAAA}.$

Only in BBAAA does team B win the first game, so the probability is $\dfrac{1}{5}.$

Thus, the correct answer is A.

19.

Three one-inch squares are placed with their bases on a line. The center square is lifted out and rotated $45^\circ,$ as shown. Then it is centered and lowered into its original location until it touches both of the adjoining squares. How many inches is the point $B$ from the line on which the bases of the original squares were placed?

$1$

$\sqrt{2}$

$\dfrac{3}{2}$

$\sqrt{2} + \dfrac{1}{2}$

$2$

Answer: D

Concepts:transformation square (geometry)special right triangle

Difficulty rating: 1760

Solution:

When lowered, the rotated square's two lower edges rest on the inner top corners of the adjoining squares, which are at height $1.$ Working out the geometry, the square's bottom vertex settles at height $\frac{1}{2}.$ The point $B$ is the opposite vertex, a full vertical diagonal of length $\sqrt{2}$ higher, so its height is $\frac{1}{2} + \sqrt{2}.$

Thus, the correct answer is D.

20.

An equiangular octagon has four sides of length $1$ and four sides of length $\dfrac{\sqrt{2}}{2},$ arranged so that no two consecutive sides have the same length. What is the area of the octagon?

$\dfrac{7}{2}$

$\dfrac{7\sqrt{2}}{2}$

$\dfrac{5 + 4\sqrt{2}}{2}$

$\dfrac{4 + 5\sqrt{2}}{2}$

$7$

Answer: A

Concepts:equiangular polygon area decomposition special right triangle

Difficulty rating: 1760

Solution:

Extend the four sides of length $1$ to form a square. Each short side $\dfrac{\sqrt{2}}{2}$ is the hypotenuse of an isosceles right triangle with legs $\dfrac{1}{2},$ and cutting these four corners from a square of side $1 + 2 \cdot \frac{1}{2} = 2$ gives the octagon. Its area is $2^2 - 4 \cdot \frac{1}{2}\left(\frac{1}{2}\right)^2 = 4 - \frac{1}{2} = \dfrac{7}{2}.$

Thus, the correct answer is A.

21.

For how many positive integers $n$ does $1 + 2 + \cdots + n$ evenly divide $6n?$

$3$

$5$

$7$

$9$

$11$

Answer: B

Concepts:triangular number divisibility factor

Difficulty rating: 1790

Solution:

Since $1 + 2 + \cdots + n = \dfrac{n(n+1)}{2},$ the quotient is $\dfrac{6n}{n(n+1)/2} = \dfrac{12}{n+1},$ which is an integer exactly when $n + 1$ divides $12.$ The divisors of $12$ that are at least $2$ are $2, 3, 4, 6, 12,$ giving $n = 1, 2, 3, 5, 11$ — five values.

Thus, the correct answer is B.

22.

Let $S$ be the set of the $2005$ smallest positive multiples of $4,$ and let $T$ be the set of the $2005$ smallest positive multiples of $6.$ How many elements are common to $S$ and $T?$

$166$

$333$

$500$

$668$

$1001$

Answer: D

Concepts:least common multiple multiple counting integers in a range

Difficulty rating: 1690

Solution:

The elements common to $S$ and $T$ are the multiples of $\operatorname{lcm}(4,6) = 12.$ Now $S$ contains multiples of $4$ up to $4 \cdot 2005 = 8020,$ while $T$ reaches up to $6 \cdot 2005 = 12{,}030,$ so the common elements are the multiples of $12$ not exceeding $8020.$ There are $\left\lfloor \dfrac{8020}{12} \right\rfloor = 668$ of them.

Thus, the correct answer is D.

23.

Let $AB$ be a diameter of a circle and $C$ be a point on $AB$ with $2 \cdot AC = BC.$ Let $D$ and $E$ be points on the circle such that $DC \perp AB$ and $DE$ is a second diameter. What is the ratio of the area of $\triangle DCE$ to the area of $\triangle ABD?$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

Answer: C

Concepts:area ratio circle triangle area

Difficulty rating: 2010

Solution:

Let $O$ be the center. From $2 \cdot AC = BC$ and $AC + BC = AB,$ we get $AC = \dfrac{AB}{3},$ so $CO = \dfrac{AB}{2} - \dfrac{AB}{3} = \dfrac{AB}{6}.$ Triangles $DCO$ and $DAB$ share the apex $D$ with bases $CO$ and $AB$ on the same line, so $[\triangle DCO] = \dfrac{CO}{AB}[\triangle DAB] = \dfrac{1}{6}[\triangle DAB].$ Because $O$ is the midpoint of $DE,$ $[\triangle DCE] = 2\,[\triangle DCO] = \dfrac{1}{3}[\triangle DAB].$

Thus, the correct answer is C.

24.

For each positive integer $m \gt 1,$ let $P(m)$ denote the greatest prime factor of $m.$ For how many positive integers $n$ is it true that both $P(n) = \sqrt{n}$ and $P(n + 48) = \sqrt{n + 48}?$

$0$

$1$

$3$

$4$

$5$

Answer: B

Concepts:prime perfect square difference of squares

Difficulty rating: 2120

Solution:

The condition $P(n) = \sqrt{n}$ means $n$ is the square of a prime $q,$ and likewise $n + 48 = p^2$ for a prime $p.$ Then $48 = p^2 - q^2 = (p - q)(p + q).$ Checking the same-parity factorizations of $48,$ only $p - q = 2,\ p + q = 24$ yields primes, giving $(p, q) = (13, 11)$ and $n = 121.$ So there is exactly one such $n.$

Thus, the correct answer is B.

25.

In $\triangle ABC$ we have $AB = 25,$ $BC = 39,$ and $AC = 42.$ Points $D$ and $E$ are on $AB$ and $AC$ respectively, with $AD = 19$ and $AE = 14.$ What is the ratio of the area of triangle $ADE$ to the area of the quadrilateral $BCED?$

$\dfrac{266}{1521}$

$\dfrac{19}{75}$

$\dfrac{1}{3}$

$\dfrac{19}{56}$

$1$

Answer: D

Concepts:area ratio triangle area

Difficulty rating: 1760

Solution:

Triangles $ADE$ and $ABC$ share angle $A,$ so $\dfrac{[ADE]}{[ABC]} = \dfrac{AD \cdot AE}{AB \cdot AC} = \dfrac{19 \cdot 14}{25 \cdot 42} = \dfrac{266}{1050} = \dfrac{19}{75}.$ Since $[BCED] = [ABC] - [ADE],$ we get $\dfrac{[ADE]}{[BCED]} = \dfrac{19}{75 - 19} = \dfrac{19}{56}.$

Thus, the correct answer is D.