2006 AMC 10A — Problems & Online Timed Exam

Timed

1:15:00

1.

Sandwiches at Joe’s Fast Food cost $\$3$ each and sodas cost $\$2$ each. How many dollars will it cost to purchase $5$ sandwiches and $8$ sodas?

$31$

$32$

$33$

$34$

$35$

Answer: A

Concepts:money whole number operations

Difficulty rating: 450

Small Hint:

Find the sandwich total and the soda total separately

Big Hint:

Compute $5 \cdot 3 + 8 \cdot 2$

Solution:

Five sandwiches cost $5 \cdot 3 = 15$ dollars and eight sodas cost $8 \cdot 2 = 16$ dollars. Together they cost $15 + 16 = 31$ dollars.

Thus, the correct answer is A.

2.

Define $x \otimes y = x^3 - y.$ What is $h \otimes (h \otimes h)?$

$-h$

$0$

$h$

$2h$

$h^3$

Answer: C

Concepts:custom operation algebraic manipulation

Difficulty rating: 960

Small Hint:

First evaluate the inner $h \otimes h$

Big Hint:

$h \otimes h = h^3 - h,$ then apply the operation again

Solution:

The inner operation gives $h \otimes h = h^3 - h.$ Then $\begin{aligned} h \otimes (h^3 - h) &= h^3 - (h^3 - h) \\ &= h. \end{aligned}$

Thus, the correct answer is C.

3.

The ratio of Mary’s age to Alice’s age is $3 : 5.$ Alice is $30$ years old. How many years old is Mary?

$15$

$18$

$20$

$24$

$50$

Answer: B

Concepts:ratio and proportion ages

Difficulty rating: 560

Small Hint:

Mary’s age is $\dfrac{3}{5}$ of Alice’s age

Big Hint:

Compute $\dfrac{3}{5} \cdot 30$

Solution:

Since the ratio is $3 : 5$ and Alice is $30,$ Mary is $\dfrac{3}{5} \cdot 30 = 18$ years old.

Thus, the correct answer is B.

4.

A digital watch displays hours and minutes with am and pm. What is the largest possible sum of the digits in the display?

$17$

$19$

$21$

$22$

$23$

Answer: E

Concepts:clock digits optimization

Difficulty rating: 1030

Small Hint:

The display uses a $12$ -hour clock, so the hour is $1$ through $12$

Big Hint:

Maximize the minutes’ digit sum and the hour’s digit sum separately

Solution:

The minutes run from $00$ to $59,$ so the largest digit sum for the minutes is $5 + 9 = 14,$ at $59$ minutes.

For the hour, the single digit $9$ beats $1 + 2 = 3$ from $12.$ The largest total is $9 + 14 = 23,$ occurring at $9\!:\!59.$

Thus, the correct answer is E.

5.

Doug and Dave shared a pizza with $8$ equally-sized slices. Doug wanted a plain pizza, but Dave wanted anchovies on half of the pizza. The cost of a plain pizza was $\$8,$ and there was an additional cost of $\$2$ for putting anchovies on one half. Dave ate all the slices of anchovy pizza and one plain slice. Doug ate the remainder. Each then paid for what he had eaten. How many more dollars did Dave pay than Doug?

$1$

$2$

$3$

$4$

$5$

Answer: D

Concepts:money fraction

Difficulty rating: 1120

Small Hint:

A plain pizza costs $\$8$ for $8$ slices, so each slice is $\$1$

Big Hint:

Dave also pays the full $\$2$ for the anchovies on top of his five slices

Solution:

Each of the $8$ slices costs $\$1.$ Dave ate $5$ slices and also pays the extra $\$2$ for the anchovies, for a total of $5 + 2 = 7$ dollars.

Doug ate $3$ slices, paying $3$ dollars. So Dave paid $7 - 3 = 4$ dollars more.

Thus, the correct answer is D.

6.

What non-zero real value for $x$ satisfies $(7x)^{14} = (14x)^7?$

$\dfrac{1}{7}$

$\dfrac{2}{7}$

$1$

$7$

$14$

Answer: B

Concepts:exponent algebraic manipulation

Difficulty rating: 1190

Small Hint:

Take the seventh root of both sides

Big Hint:

$(7x)^2 = 14x,$ then divide by $x$

Solution:

Taking the seventh root of both sides gives $(7x)^2 = 14x,$ so $49x^2 = 14x.$ Since $x \neq 0,$ divide by $x$ to get $49x = 14,$ hence $x = \dfrac{2}{7}.$

Thus, the correct answer is B.

7.

The $8 \times 18$ rectangle $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $y?$

$6$

$7$

$8$

$9$

$10$

Answer: A

Concepts:area area decomposition

Difficulty rating: 1190

Small Hint:

The two hexagons form a square with the same area as the rectangle

Big Hint:

The rectangle has area $8 \cdot 18 = 144,$ so the square has side $12$

Solution:

The rectangle’s area is $8 \cdot 18 = 144,$ so the square formed has side $\sqrt{144} = 12.$

Let $E$ and $F$ be the upper and lower endpoints of the stair-step cut. For the two congruent pieces to fit into a square, the three horizontal segments $DE,$ the middle segment of length $y,$ and $FB$ must be equal. Together they span the rectangle’s width, so $DE+y+FB=18$ gives $3y=18$ and $y=6.$ Indeed, the resulting square has side $2y=12,$ agreeing with the area calculation.

Thus, the correct answer is A.

8.

A parabola with equation $y = x^2 + bx + c$ passes through the points $(2, 3)$ and $(4, 3).$ What is $c?$

$2$

$5$

$7$

$10$

$11$

Answer: E

Concepts:parabola system of equations

Difficulty rating: 1270

Small Hint:

The two points have the same $y$ -value, so they are symmetric about the vertex

Big Hint:

The axis of symmetry is $x = 3,$ giving $b = -6$

Solution:

Substituting the points gives $3 = 4 + 2b + c$ and $3 = 16 + 4b + c.$ Subtracting yields $0 = 12 + 2b,$ so $b = -6.$

Then $c = 3 - 4 - 2(-6) = 11.$

Thus, the correct answer is E.

9.

How many sets of two or more consecutive positive integers have a sum of $15?$

$1$

$2$

$3$

$4$

$5$

Answer: C

Concepts:arithmetic sequence casework

Difficulty rating: 1170

Small Hint:

The sum of $n$ consecutive integers is $n$ times the middle value

Big Hint:

Check $n = 2, 3, 4, 5$

Solution:

The sum of $n$ consecutive integers equals $n$ times their median. For a sum of $15$ : $n = 2$ gives $7 + 8,$ $n = 3$ gives $4 + 5 + 6,$ and $n = 5$ gives $1 + 2 + 3 + 4 + 5.$

No set of $4$ works (their sum is even), and $6$ or more consecutive positive integers already exceed $15.$ There are $3$ such sets.

Thus, the correct answer is C.

10.

For how many real values of $x$ is $\sqrt{120 - \sqrt{x}}$ an integer?

$3$

$6$

$9$

$10$

$11$

Answer: E

Concepts:radical counting integers in a range

Difficulty rating: 1390

Small Hint:

Let $k = \sqrt{120 - \sqrt{x}}$ and bound $k$

Big Hint:

$k$ is an integer with $0 \le k \le 10$

Solution:

Let $k = \sqrt{120 - \sqrt{x}}.$ Since $\sqrt{x} \ge 0,$ we need $0 \le k \le \sqrt{120},$ so $k \in \{0, 1, \ldots, 10\},$ giving $11$ values.

Each $k$ yields $\sqrt{x} = 120 - k^2,$ and since $120 - k^2$ is positive and strictly decreasing, the resulting values $x = (120 - k^2)^2$ are distinct.

Thus, the correct answer is E.

11.

Which of the following describes the graph of the equation $(x + y)^2 = x^2 + y^2?$

the empty set

one point

two lines

a circle

the entire plane

Answer: C

Concepts:algebraic manipulation coordinate geometry

Difficulty rating: 1270

Small Hint:

Expand $(x + y)^2$ and simplify

Big Hint:

The equation reduces to $xy = 0$

Solution:

Expanding, $x^2 + 2xy + y^2 = x^2 + y^2,$ which reduces to $2xy = 0,$ i.e. $xy = 0.$

This holds exactly when $x = 0$ or $y = 0,$ the two coordinate axes, so the graph is two lines.

Thus, the correct answer is C.

12.

Rolly wishes to secure his dog with an $8$ -foot rope to a square shed that is $16$ feet on each side. His preliminary drawings are shown.

Which of these arrangements gives the dog the greater area to roam, and by how many square feet?

$\mathrm{I},$ by $8\pi$

$\mathrm{I},$ by $6\pi$

$\mathrm{II},$ by $4\pi$

$\mathrm{II},$ by $8\pi$

$\mathrm{II},$ by $10\pi$

Answer: C

Concepts:sector circle area

Difficulty rating: 1420

Small Hint:

In each case the dog sweeps a half-disk of radius $8$ on the open side

Big Hint:

In arrangement II the rope also wraps around the near corner

Solution:

In arrangement I the dog is tied at the middle of a side and sweeps a half-disk of radius $8$ : area $\frac12 \pi \cdot 8^2 = 32\pi.$ The rope reaches exactly to the corners, so nothing wraps.

In arrangement II the dog is tied $4$ feet from a corner. It sweeps the same $32\pi$ half-disk, and after the rope reaches the corner, $4$ feet remain to sweep a quarter-disk of radius $4$ : $\frac14 \pi \cdot 4^2 = 4\pi.$

So II gives $36\pi,$ exceeding I by $4\pi.$

Thus, the correct answer is C.

13.

A player pays $\$5$ to play a game. A die is rolled. If the number on the die is odd, the game is lost. If the number on the die is even, the die is rolled again. In this case the player wins if the second number matches the first and loses otherwise. How much should the player win if the game is fair? (In a fair game the probability of winning times the amount won is what the player should pay.)

$\$12$

$\$30$

$\$50$

$\$60$

$\$100$

Answer: D

Concepts:expected value dice (probability)

Difficulty rating: 1390

Small Hint:

Find the probability of winning: an even roll, then a matching roll

Big Hint:

The win probability is $\frac12 \cdot \frac16 = \frac{1}{12};$ set (probability)(prize) equal to $5$

Solution:

The player wins only if the first roll is even (probability $\frac12$ ) and the second roll matches it (probability $\frac16$ ), so the win probability is $\frac12 \cdot \frac16 = \frac{1}{12}.$

For a fair game, $\frac{1}{12} x = 5,$ so $x = 60.$

Thus, the correct answer is D.

14.

A number of linked rings, each $1$ cm thick, are hanging on a peg. The top ring has an outside diameter of $20$ cm. The outside diameter of each of the other rings is $1$ cm less than that of the ring above it. The bottom ring has an outside diameter of $3$ cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring?

$171$

$173$

$182$

$188$

$210$

Answer: B

Concepts:arithmetic sequence summation

Difficulty rating: 1330

Small Hint:

The exposed part of each lower ring is its outside diameter minus the $2$ cm of overlap with the ring above

Big Hint:

Add $20$ to $17 + 16 + \cdots + 1$

Solution:

The top ring contributes its full outside diameter, $20$ cm. Because the rings are $1$ cm thick, each ring hangs $2$ cm below the top of the ring above it, so each lower ring adds its outside diameter minus $2.$

The outside diameters run $20, 19, \ldots, 3,$ so the added distances are $17, 16, \ldots, 1.$ The total is $\begin{gathered} 20 + (17 + 16 + \cdots + 1) \\ = 20 + \frac{17 \cdot 18}{2} \\ = 20 + 153 \\ = 173. \end{gathered}$

Thus, the correct answer is B.

15.

Odell and Kershaw run for $30$ minutes on a circular track. Odell runs clockwise at $250$ m/min and uses the inner lane with a radius of $50$ meters. Kershaw runs counterclockwise at $300$ m/min and uses the outer lane with a radius of $60$ meters, starting on the same radial line as Odell. How many times after the start do they pass each other?

$29$

$42$

$45$

$47$

$50$

Answer: D

Concepts:relative speed distance rate and time circumference

Difficulty rating: 1630

Small Hint:

Compute each runner’s time for one lap; they turn out equal

Big Hint:

Running in opposite directions, they meet twice per lap; count how many laps fit in $30$ minutes

Solution:

Odell’s lap is $2\pi(50) = 100\pi$ m at $250$ m/min, taking $\frac{100\pi}{250} = 0.4\pi$ min. Kershaw’s lap is $2\pi(60) = 120\pi$ m at $300$ m/min, also $\frac{120\pi}{300} = 0.4\pi$ min.

Their periods are equal. Running in opposite directions, they meet at times $t = \frac{k}{2}(0.4\pi)$ for $k = 1, 2, \ldots$ Requiring $t \le 30$ gives $k \le \frac{60}{0.4\pi} = \frac{150}{\pi} \approx 47.7,$ so they pass $47$ times.

Thus, the correct answer is D.

16.

A circle of radius $1$ is tangent to a circle of radius $2.$ The sides of $\triangle ABC$ are tangent to the circles as shown, and the sides $AB$ and $AC$ are congruent. What is the area of $\triangle ABC?$

$\dfrac{35}{2}$

$15\sqrt{2}$

$\dfrac{64}{3}$

$16\sqrt{2}$

$24$

Answer: D

Concepts:tangent circles similarity isosceles triangle

Difficulty rating: 1720

Small Hint:

Let $O$ and $O'$ be the centers; the right triangles cut off along $AC$ are similar

Big Hint:

The centers lie $3$ and $6$ from $A,$ and the base is $8$ below $A$

Solution:

Let $O, O'$ be the centers of the small and large circles, and let $D$ be the point where the small circle touches $AC.$ The right triangles cut off along $AC$ are similar, so $\frac{AO}{1} = \frac{AO + 3}{2},$ giving $AO = 3$ and $AO' = 6.$

The tangent length is $AD = \sqrt{AO^2 - 1^2}$ $= \sqrt{3^2 - 1^2}$ $= 2\sqrt2.$ Let $F$ be the midpoint of $BC$ ; then $AF = AO' + 2 = 8.$

Since $\triangle ADO \sim \triangle AFC,$ we get $\frac{FC}{1} = \frac{AF}{2\sqrt2} = \frac{8}{2\sqrt2} = 2\sqrt2.$ Thus $BC = 4\sqrt2,$ and the area is $\frac12 \cdot BC \cdot AF$ $= \frac12 \cdot 4\sqrt2 \cdot 8$ $= 16\sqrt2.$

Thus, the correct answer is D.

17.

In rectangle $ADEH,$ points $B$ and $C$ trisect $\overline{AD},$ and points $G$ and $F$ trisect $\overline{HE}.$ In addition, $AH = AC = 2.$ What is the area of quadrilateral $WXYZ$ shown in the figure?

$\dfrac{1}{2}$

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

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

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

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

Answer: A

Concepts:coordinate geometry area square (geometry)

Difficulty rating: 1540

Small Hint:

Assign coordinates using $AB = BC = CD = 1$ and $AH = 2$

Big Hint:

$WXYZ$ is a square; find its diagonals

Solution:

Set $A = (0, 0),$ $D = (3, 0),$ $H = (0, 2),$ so $B = (1, 0),$ $C = (2, 0),$ $G = (1, 2),$ $F = (2, 2),$ and $E = (3, 2).$

The drawn segments meet at $W = (1.5, 1.5),$ $X = (1, 1),$ $Y = (1.5, 0.5),$ and $Z = (2, 1).$ These form a square whose perpendicular diagonals $WY$ and $XZ$ each have length $1.$

Its area is $\frac12 \cdot 1 \cdot 1 = \frac12.$

Thus, the correct answer is A.

18.

A license plate in a certain state consists of $4$ digits, not necessarily distinct, and $2$ letters, also not necessarily distinct. These six characters may appear in any order, except that the two letters must appear next to each other. How many distinct license plates are possible?

$10^4 \cdot 26^2$

$10^3 \cdot 26^3$

$5 \cdot 10^4 \cdot 26^2$

$10^2 \cdot 26^4$

$5 \cdot 10^3 \cdot 26^3$

Answer: C

Concepts:arrangements with restrictions multiplication principle

Difficulty rating: 1450

Small Hint:

Glue the two letters into a single block that moves as a unit

Big Hint:

Count digit choices, letter choices, and positions for the block

Solution:

Since the two letters must be adjacent, treat them as one block. A plate is then $4$ digits plus this block— $5$ objects—and the block can occupy $5$ positions.

There are $10^4$ choices for the digits and $26^2$ for the two letters, so the total is $5 \cdot 10^4 \cdot 26^2.$

Thus, the correct answer is C.

19.

How many non-similar triangles have angles whose degree measures are distinct positive integers in arithmetic progression?

$0$

$1$

$59$

$89$

$178$

Answer: C

Concepts:angle sum arithmetic sequence counting integers in a range

Difficulty rating: 1630

Small Hint:

If the angles are in arithmetic progression, the middle angle is $60^\circ$

Big Hint:

The common difference $d$ satisfies $1 \le d \le 59$

Solution:

Let the angles be $n - d,$ $n,$ $n + d.$ Their sum is $3n = 180,$ so $n = 60.$

The measures are distinct positive integers, so $d \ge 1,$ and $n - d \gt 0$ forces $d \lt 60.$ Thus $d \in \{1, 2, \ldots, 59\},$ giving $59$ non-similar triangles.

Thus, the correct answer is C.

20.

Six distinct positive integers are randomly chosen between $1$ and $2006,$ inclusive. What is the probability that some pair of these integers has a difference that is a multiple of $5?$

$\dfrac{1}{2}$

$\dfrac{3}{5}$

$\dfrac{2}{3}$

$\dfrac{4}{5}$

$1$

Answer: E

Concepts:pigeonhole principle modular arithmetic

Difficulty rating: 1510

Small Hint:

Consider the remainders when the integers are divided by $5$

Big Hint:

There are only $5$ possible remainders for $6$ integers

Solution:

Group the integers by their remainder modulo $5.$ There are only $5$ possible remainders but $6$ integers, so by the Pigeonhole Principle two share a remainder.

Their difference is then a multiple of $5.$ This always happens, so the probability is $1.$

Thus, the correct answer is E.

21.

How many four-digit positive integers have at least one digit that is a $2$ or a $3?$

$2439$

$4096$

$4903$

$4904$

$5416$

Answer: E

Concepts:complementary counting digits

Difficulty rating: 1450

Small Hint:

Count the complement: four-digit numbers with no $2$ and no $3$

Big Hint:

The leading digit has $7$ choices; each other digit has $8$

Solution:

There are $9000$ four-digit integers. For those avoiding $2$ and $3,$ the leading digit is one of $\{1, 4, 5, 6, 7, 8, 9\}$ ( $7$ choices) and each remaining digit is one of $\{0, 1, 4, 5, 6, 7, 8, 9\}$ ( $8$ choices): $7 \cdot 8^3 = 3584.$

So $9000 - 3584 = 5416$ have at least one $2$ or $3.$

Thus, the correct answer is E.

22.

Two farmers agree that pigs are worth $\$300$ and that goats are worth $\$210.$ When one farmer owes the other money, he pays the debt in pigs or goats, with “change” received in the form of goats or pigs as necessary. (For example, a $\$390$ debt could be paid with two pigs, with one goat received in change.) What is the amount of the smallest positive debt that can be resolved in this way?

$\$5$

$\$10$

$\$30$

$\$90$

$\$210$

Answer: C

Concepts:greatest common divisor Diophantine Equation

Difficulty rating: 1630

Small Hint:

A resolvable debt is $300p + 210g$ for integers $p,$ $g$ (possibly negative)

Big Hint:

This equals $30(10p + 7g),$ and $\gcd(10, 7) = 1$

Solution:

A resolvable debt is $D = 300p + 210g$ for integers $p,$ $g,$ where a negative value means change received. Since $D = 30(10p + 7g)$ and $\gcd(10, 7) = 1,$ the value $10p + 7g$ can be any integer, so $D$ is any multiple of $30.$

The smallest positive one is $30,$ achieved by $30 = 300(-2) + 210(3)$ (give $3$ goats, receive $2$ pigs).

Thus, the correct answer is C.

23.

Circles with centers $A$ and $B$ have radii $3$ and $8,$ respectively. A common internal tangent touches the circles at $C$ and $D,$ as shown. Lines $AB$ and $CD$ intersect at $E,$ and $AE = 5.$ What is $CD?$

$13$

$\dfrac{44}{3}$

$\sqrt{221}$

$\sqrt{255}$

$\dfrac{55}{3}$

Answer: B

Concepts:tangent line similarity Pythagorean Theorem

Difficulty rating: 1720

Small Hint:

Radii $AC$ and $BD$ are perpendicular to the tangent $CD$

Big Hint:

$\triangle ACE \sim \triangle BDE$ ; first find $CE$

Solution:

Since $AC \perp CD,$ we have $CE = \sqrt{AE^2 - AC^2}$ $= \sqrt{25 - 9}$ $= 4.$

Because $\triangle ACE \sim \triangle BDE,$ $\frac{DE}{CE} = \frac{BD}{AC},$ so $DE = 4 \cdot \frac{8}{3} = \frac{32}{3}.$

Then $CD = CE + DE$ $= 4 + \frac{32}{3}$ $= \frac{44}{3}.$

Thus, the correct answer is B.

24.

Centers of adjacent faces of a unit cube are joined to form a regular octahedron. What is the volume of this octahedron?

$\dfrac{1}{8}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

Answer: B

Concepts:3D geometry pyramid volume

Difficulty rating: 1760

Small Hint:

The octahedron is two square pyramids glued at their bases

Big Hint:

Adjacent face-centers are $\frac{\sqrt2}{2}$ apart; find the base area and height

Solution:

The six face centers form a regular octahedron, viewed as two congruent square pyramids sharing a base. Adjacent face centers are $\frac{\sqrt2}{2}$ apart, so the square base has area $\left(\frac{\sqrt2}{2}\right)^2 = \frac12.$

Each pyramid has height $\frac12,$ so its volume is $\frac13 \cdot \frac12 \cdot \frac12 = \frac{1}{12}.$ The octahedron has volume $2 \cdot \frac{1}{12} = \frac16.$

Thus, the correct answer is B.

25.

A bug starts at one vertex of a cube and moves along the edges of the cube according to the following rule. At each vertex the bug will choose to travel along one of the three edges emanating from that vertex. Each edge has equal probability of being chosen, and all choices are independent. What is the probability that after seven moves the bug will have visited every vertex exactly once?

$\dfrac{1}{2187}$

$\dfrac{1}{729}$

$\dfrac{2}{243}$

$\dfrac{1}{81}$

$\dfrac{5}{243}$

Answer: C

Concepts:graph theory basic probability casework

Difficulty rating: 2120

Small Hint:

The bug must trace a path visiting all $8$ vertices, i.e. $7$ moves with no repeated vertices

Big Hint:

Count such paths from the start; there are $3^7$ equally likely walks

Solution:

After $7$ moves there are $3^7 = 2187$ equally likely walks. A successful walk visits every vertex exactly once.

Label the cube’s vertices by binary triples, with adjacent vertices differing in one coordinate. There are $3$ choices for the first move and $2$ for the second move if the bug is not to return to its starting point. By symmetry, fix these first moves as $000\to100\to110.$

The successful continuations are exactly $\begin{aligned} 110&\to111\to101\\ &\to001\to011\to010, \end{aligned}$ $\begin{aligned} 110&\to010\to011\\ &\to001\to101\to111, \end{aligned}$ and $\begin{aligned} 110&\to010\to011\\ &\to111\to101\to001. \end{aligned}$ Thus each allowed pair of first moves has $3$ successful continuations, giving $3\cdot2\cdot3=18$ successful walks.

The probability is $\frac{18}{2187} = \frac{2}{243}.$

Thus, the correct answer is C.