2004 AMC 10A Solutions | LIVE by Po-Shen Loh

1.

You and five friends need to raise $ $1500$ in donations for a charity, dividing the fundraising equally. How many dollars will each of you need to raise?

$250$

$300$

$1500$

$7500$

$9000$

Concepts:money fraction

Difficulty rating: 450

Small Hint:

Including you, count the total number of people sharing the work

Big Hint:

Divide $ $1500$ evenly among the $6$ people

Solution:

Including you, there are $6$ people sharing the fundraising equally. Each must raise $\dfrac{1500}{6} = 250$ dollars.

Thus, the correct answer is A.

2.

For any three real numbers $a,$ $b,$ and $c,$ with $b \neq c,$ the operation $\diamond$ is defined by $\diamond(a, b, c) = \dfrac{a}{b - c}.$ What is $\diamond(\diamond(1, 2, 3), \diamond(2, 3, 1), \diamond(3, 1, 2))?$

$-\dfrac{1}{2}$

$-\dfrac{1}{4}$

$0$

$\dfrac{1}{4}$

$\dfrac{1}{2}$

Concepts:custom operation fraction

Difficulty rating: 980

Small Hint:

Evaluate the three inner expressions before combining them

Big Hint:

$\diamond(1,2,3) = -1,$ $\diamond(2,3,1) = 1,$ and $\diamond(3,1,2) = -3$

Solution:

The inner values are $\begin{aligned} \diamond(1,2,3) &= \dfrac{1}{2-3} = -1, \\ \diamond(2,3,1) &= \dfrac{2}{3-1} = 1, \\ \diamond(3,1,2) &= \dfrac{3}{1-2} = -3. \end{aligned}$

Therefore $\begin{aligned} \diamond(-1, 1, -3) &= \dfrac{-1}{1 - (-3)} \\ &= -\dfrac{1}{4}. \end{aligned}$

Thus, the correct answer is B.

3.

Alicia earns $ $20$ per hour, of which $1.45\%$ is deducted to pay local taxes. How many cents per hour of Alicia’s wages are used to pay local taxes?

$0.0029$

$0.029$

$0.29$

$2.9$

$29$

Concepts:percentage unit conversion

Difficulty rating: 870

Small Hint:

Convert $ $20$ into cents before taking the percentage

Big Hint:

Compute $1.45\%$ of $2000$ cents

Solution:

Since $ $20$ equals $2000$ cents, the local tax is $0.0145 \times 2000 = 29$ cents per hour.

Thus, the correct answer is E.

4.

What is the value of $x$ if $|x - 1| = |x - 2|?$

$-\dfrac{1}{2}$

$\dfrac{1}{2}$

$1$

$\dfrac{3}{2}$

$2$

Concepts:absolute value midpoint

Difficulty rating: 1030

Small Hint:

$|x - a|$ measures the distance from $x$ to $a$

Big Hint:

The point $x$ must be equally far from $1$ and $2$

Solution:

Since $|x - 1|$ and $|x - 2|$ are the distances from $x$ to $1$ and $2,$ the point $x$ is equidistant from $1$ and $2.$

That midpoint is $x = \dfrac{1 + 2}{2} = \dfrac{3}{2}.$

Thus, the correct answer is D.

5.

A set of three points is chosen randomly from the grid shown. Each three-point set has the same probability of being chosen. What is the probability that the points lie on the same straight line?

$\dfrac{1}{21}$

$\dfrac{1}{14}$

$\dfrac{2}{21}$

$\dfrac{1}{7}$

$\dfrac{2}{7}$

Concepts:basic probability combinations lattice point

Difficulty rating: 1240

Small Hint:

There are $\binom{9}{3}$ equally likely three-point sets

Big Hint:

Count the collinear triples: $3$ rows, $3$ columns, and $2$ diagonals

Solution:

The number of three-point sets is $\binom{9}{3} = 84.$

The collinear triples are the $3$ rows, the $3$ columns, and the $2$ main diagonals, for a total of $8.$

The probability is therefore $\dfrac{8}{84} = \dfrac{2}{21}.$

Thus, the correct answer is C.

6.

Bertha has $6$ daughters and no sons. Some of her daughters have $6$ daughters, and the rest have none. Bertha has a total of $30$ daughters and granddaughters, and no great-granddaughters. How many of Bertha’s daughters and granddaughters have no daughters?

$22$

$23$

$24$

$25$

$26$

Concepts:basic counting logical deduction

Difficulty rating: 1170

Small Hint:

Bertha has $30 - 6 = 24$ granddaughters, and none of them have daughters

Big Hint:

The granddaughters come in groups of $6,$ so find how many daughters are mothers

Solution:

Bertha has $30 - 6 = 24$ granddaughters, none of whom have daughters.

These granddaughters belong to $\frac{24}{6} = 4$ of Bertha’s daughters. So exactly $4$ women have daughters, and the number with no daughters is $30 - 4 = 26.$

Thus, the correct answer is E.

7.

A grocer stacks oranges in a pyramid-like stack whose rectangular base is $5$ oranges by $8$ oranges. Each orange above the first level rests in a pocket formed by four oranges in the level below. The stack is completed by a single row of oranges. How many oranges are in the stack?

$96$

$98$

$100$

$101$

$134$

Concepts:basic counting summation

Difficulty rating: 1100

Small Hint:

Each layer up has one fewer orange in each dimension than the layer below

Big Hint:

Sum $5\cdot 8 + 4\cdot 7 + 3\cdot 6 + 2\cdot 5 + 1\cdot 4$

Solution:

There are five layers, each one shorter and narrower than the one below. The total number of oranges is $\begin{aligned} &5\cdot 8 + 4\cdot 7 + 3\cdot 6 \\ &\quad {}+ 2\cdot 5 + 1\cdot 4 \\ &= 40 + 28 + 18 + 10 + 4 = 100. \end{aligned}$

Thus, the correct answer is C.

8.

A game is played with tokens according to the following rule. In each round, the player with the most tokens gives one token to each of the other players and also places one token into a discard pile. The game ends when some player runs out of tokens. Players $A,$ $B,$ and $C$ start with $15,$ $14,$ and $13$ tokens, respectively. How many rounds will there be in the game?

$36$

$37$

$38$

$39$

$40$

Concepts:process simulation pattern recognition

Difficulty rating: 1390

Small Hint:

Simulate a few rounds and look for a repeating pattern

Big Hint:

Every three rounds, each player’s total drops by exactly one

Solution:

After the first three rounds the counts go from $(15, 14, 13)$ to $(14, 13, 12).$ In general, every three rounds each player loses exactly one token.

After $36$ rounds the counts are $(3, 2, 1).$ On the $37$ th round the leader gives away three tokens and drops to $0,$ ending the game.

Thus, the correct answer is B.

9.

In the figure, $\angle EAB$ and $\angle ABC$ are right angles, $AB = 4,$ $BC = 6,$ $AE = 8,$ and $\overline{AC}$ and $\overline{BE}$ intersect at $D.$ What is the difference between the areas of $\triangle ADE$ and $\triangle BDC?$

$2$

$4$

$5$

$8$

$9$

Concepts:triangle area area decomposition

Difficulty rating: 1330

Small Hint:

Adding $\triangle ABD$ to each of the two triangles produces two larger triangles

Big Hint:

Subtracting the shared area makes $[ADE] - [BDC]$ equal to $[ABE] - [ABC]$

Solution:

Let $[ABD]$ be the area shared by both large triangles. Then $[ABE] = [ADE] + [ABD]$ and $[ABC] = [BDC] + [ABD].$

Subtracting, $\begin{aligned} &[ADE] - [BDC] \\ &= [ABE] - [ABC]. \end{aligned}$ Since $\angle EAB$ and $\angle ABC$ are right angles, $\begin{aligned} [ABE] &= \tfrac12(4)(8) = 16, \\ [ABC] &= \tfrac12(4)(6) = 12. \end{aligned}$

The difference is $16 - 12 = 4.$

Thus, the correct answer is B.

10.

Coin $A$ is flipped three times and coin $B$ is flipped four times. What is the probability that the number of heads obtained from flipping the two fair coins is the same?

$\dfrac{19}{128}$

$\dfrac{23}{128}$

$\dfrac{1}{4}$

$\dfrac{35}{128}$

$\dfrac{1}{2}$

Concepts:binomial probability casework

Difficulty rating: 1470

Small Hint:

Add up $P(A = k)\,P(B = k)$ over each possible common count $k$

Big Hint:

The counts for $A$ are weighted $1, 3, 3, 1$ and for $B$ are $1, 4, 6, 4, 1$

Solution:

The two coins match when both show $0,$ $1,$ $2,$ or $3$ heads. Coin $A$ has weights $1, 3, 3, 1$ out of $8$ and coin $B$ has weights $1, 4, 6, 4, 1$ out of $16.$

The probability is $\begin{aligned} &\dfrac{1\cdot 1 + 3\cdot 4 + 3\cdot 6 + 1\cdot 4}{8 \cdot 16} \\ &= \dfrac{35}{128}. \end{aligned}$

Thus, the correct answer is D.

11.

A company sells peanut butter in cylindrical jars. Marketing research suggests that using wider jars will increase sales. If the diameter of the jars is increased by $25\%$ without altering the volume, by what percent must the height be decreased?

$10$

$25$

$36$

$50$

$60$

Concepts:cylinder volume percentage

Difficulty rating: 1310

Small Hint:

The volume $\pi r^2 h$ is unchanged, so $r^2 h$ stays fixed

Big Hint:

Multiplying the radius by $1.25$ multiplies $r^2$ by $1.25^2$

Solution:

Keeping $\pi r^2 h$ constant while multiplying the radius by $1.25$ requires the height to be multiplied by $\dfrac{1}{1.25^2} = \dfrac{1}{1.5625} = 0.64.$

So the height becomes $64\%$ of the original, a decrease of $36\%.$

Thus, the correct answer is C.

12.

Henry’s Hamburger Heaven offers its hamburgers with the following condiments: ketchup, mustard, mayonnaise, tomato, lettuce, pickles, cheese, and onions. A customer can choose one, two, or three meat patties, and any collection of condiments. How many different kinds of hamburgers can be ordered?

$24$

$256$

$768$

$40{,}320$

$120{,}960$

Concepts:multiplication principle subsets

Difficulty rating: 1190

Small Hint:

Each of the $8$ condiments is independently included or left out

Big Hint:

Multiply the $2^8$ condiment choices by the $3$ choices of patty count

Solution:

Each of the $8$ condiments is independently in or out, giving $2^8 = 256$ condiment combinations.

For each of these there are $3$ choices of patty count, so the number of hamburgers is $3 \times 256 = 768.$

Thus, the correct answer is C.

13.

At a party, each man danced with exactly three women and each woman danced with exactly two men. Twelve men attended the party. How many women attended the party?

$8$

$12$

$16$

$18$

$24$

Concepts:double counting counting pairs

Difficulty rating: 1190

Small Hint:

Count the man-woman dancing pairs in two ways

Big Hint:

There are $12 \cdot 3$ pairs, and each woman accounts for $2$ of them

Solution:

The number of dancing pairs is $12 \cdot 3 = 36,$ counting from the men’s side. Each woman was in exactly $2$ pairs, so the number of women is $\dfrac{36}{2} = 18.$

Thus, the correct answer is D.

14.

The average value of all the pennies, nickels, dimes, and quarters in Paula’s purse is $20$ cents. If she had one more quarter, the average value would be $21$ cents. How many dimes does she have in her purse?

$0$

$1$

$2$

$3$

$4$

Concepts:mean money linear equation

Difficulty rating: 1450

Small Hint:

If she has $n$ coins, their total value is $20n$ cents

Big Hint:

Adding a quarter gives $20n + 25 = 21(n + 1)$

Solution:

With $n$ coins the total value is $20n$ cents. Adding a quarter gives $20n + 25 = 21(n + 1),$ so $n = 4.$

If there were at most two quarters, the other coins would be worth at most $10$ cents each, so four coins would total at most $2(25)+2(10)=70$ cents. Thus there must be three quarters, leaving $5$ cents for the fourth coin. The purse contains three quarters and one nickel, so it has $0$ dimes.

Thus, the correct answer is A.

15.

Given that $-4 \le x \le -2$ and $2 \le y \le 4,$ what is the largest possible value of $\dfrac{x + y}{x}?$

$-1$

$-\dfrac{1}{2}$

$0$

$\dfrac{1}{2}$

$1$

Concepts:inequality optimization algebraic manipulation

Difficulty rating: 1420

Small Hint:

Rewrite $\dfrac{x + y}{x} = 1 + \dfrac{y}{x}$

Big Hint:

Since $\dfrac{y}{x} \lt 0,$ make its absolute value as small as possible

Solution:

Write $\dfrac{x + y}{x} = 1 + \dfrac{y}{x}.$ Here $\dfrac{y}{x} \lt 0,$ so the expression is largest when $\left|\dfrac{y}{x}\right|$ is smallest.

That happens with $y = 2$ and $x = -4,$ giving $1 + \dfrac{2}{-4} = 1 - \dfrac{1}{2} = \dfrac{1}{2}.$

Thus, the correct answer is D.

16.

The $5 \times 5$ grid shown contains a collection of squares with sizes from $1 \times 1$ to $5 \times 5.$ How many of these squares contain the shaded center square?

$12$

$15$

$17$

$19$

$20$

Concepts:counting shapes in figures casework

Difficulty rating: 1480

Small Hint:

Every $3\times3,$ $4\times4,$ and $5\times5$ square contains the center

Big Hint:

Then count how many $2\times2$ and $1\times1$ squares cover the center cell

Solution:

Every $5\times5,$ $4\times4,$ and $3\times3$ square contains the center cell, and there are $1^2 + 2^2 + 3^2 = 14$ of them.

Among the smaller squares, $4$ of the $2\times2$ squares and $1$ of the $1\times1$ squares cover the center, giving $14 + 4 + 1 = 19.$

Thus, the correct answer is D.

17.

Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run $100$ meters. They next meet after Sally has run $150$ meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters?

$250$

$300$

$350$

$400$

$500$

Concepts:relative speed distance rate and time ratio and proportion

Difficulty rating: 1540

Small Hint:

Starting from opposite points, together they cover half the track before their first meeting

Big Hint:

Use the constant speed ratio to compare Brenda’s first-meeting distance with her distance between meetings

Solution:

Before the first meeting the two together cover half the track. Between the first and second meetings they together cover a full track, which is twice as far, so Brenda runs $2 \cdot 100 = 200$ meters in that stretch.

Sally runs $150$ meters in the same stretch, so the full track length is $200 + 150 = 350.$

Thus, the correct answer is C.

18.

A sequence of three real numbers forms an arithmetic progression with a first term of $9.$ If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term of the geometric progression?

$1$

$4$

$36$

$49$

$81$

Concepts:arithmetic sequence geometric sequence quadratic

Difficulty rating: 1630

Small Hint:

Write the progression as $9,$ $9 + d,$ $9 + 2d,$ so the geometric one is $9,$ $11 + d,$ $29 + 2d$

Big Hint:

Use $(11 + d)^2 = 9(29 + 2d)$ and take the value of $d$ giving the smaller third term

Solution:

The arithmetic progression is $9,$ $9 + d,$ $9 + 2d,$ so the geometric progression is $9,$ $11 + d,$ $29 + 2d.$

The geometric condition gives $(11 + d)^2 = 9(29 + 2d),$ which simplifies to $d^2 + 4d - 140 = 0,$ so $d = 10$ or $d = -14.$

The third terms $29 + 2d$ are $49$ and $1.$ The smallest is $1.$

Thus, the correct answer is A.

19.

A white cylindrical silo has a diameter of $30$ feet and a height of $80$ feet. A red stripe with a horizontal width of $3$ feet is painted on the silo, as shown, making two complete revolutions around it. What is the area of the stripe in square feet?

$120$

$180$

$240$

$360$

$480$

Concepts:cylinder net (3D geometry)parallelogram

Difficulty rating: 1600

Small Hint:

Imagine cutting the stripe from the silo and unrolling it flat

Big Hint:

It becomes a parallelogram with horizontal width $3$ and height $80$

Solution:

Unrolling the stripe flattens it into a parallelogram. Its base (the horizontal width) is $3$ feet and its height spans the full $80$ feet of the silo.

The area is therefore $3 \times 80 = 240$ square feet.

Thus, the correct answer is C.

20.

Points $E$ and $F$ are located on square $ABCD$ so that $\triangle BEF$ is equilateral. What is the ratio of the area of $\triangle DEF$ to that of $\triangle ABE?$

$\dfrac{4}{3}$

$\dfrac{3}{2}$

$\sqrt{3}$

$2$

$1 + \sqrt{3}$

Concepts:equilateral triangle area ratio Pythagorean Theorem

Difficulty rating: 1790

Small Hint:

Let the square have side $1$ and set $ED = DF = x$

Big Hint:

From $EF^2 = EB^2,$ get $2x^2 = 1 + (1 - x)^2$

Solution:

Let the square have side $1,$ and by symmetry let $ED = DF = x,$ so $AE = 1 - x.$

Since $\triangle BEF$ is equilateral, $EF^2 = EB^2,$ giving $2x^2 = 1 + (1 - x)^2,$ which simplifies to $x^2 = 2(1 - x).$

The right triangles have areas $[DEF] = \tfrac12 x^2$ and $[ABE] = \tfrac12(1 - x),$ so $\begin{aligned} \dfrac{[DEF]}{[ABE]} &= \dfrac{x^2}{1 - x} \\ &= \dfrac{2(1 - x)}{1 - x} = 2. \end{aligned}$

Thus, the correct answer is D.

21.

Two distinct lines pass through the center of three concentric circles of radii $3,$ $2,$ and $1.$ The area of the shaded region in the diagram is $\dfrac{8}{13}$ of the area of the unshaded region. What is the radian measure of the acute angle formed by the two lines? (Note: $\pi$ radians is $180$ degrees.)

$\dfrac{\pi}{8}$

$\dfrac{\pi}{7}$

$\dfrac{\pi}{6}$

$\dfrac{\pi}{5}$

$\dfrac{\pi}{4}$

Concepts:sector circle area annulus linear equation

Difficulty rating: 1880

Small Hint:

Let $\theta$ be the acute angle and express each shaded piece as a sector area

Big Hint:

The shaded pieces total $3\pi + 3\theta,$ out of the whole area $9\pi$

Solution:

Let $\theta$ be the acute angle. The shaded region has three parts: two acute sectors of the unit disk with total area $\theta,$ two obtuse sectors of the ring between radii $1$ and $2$ with total area $3(\pi - \theta),$ and two acute sectors of the ring between radii $2$ and $3$ with total area $5\theta.$

Adding these gives a shaded area of $\theta + 3(\pi - \theta) + 5\theta = 3\pi + 3\theta.$

The shaded region is $\dfrac{8}{13}$ of the unshaded region, so it is $\dfrac{8}{21}$ of the total area $9\pi.$ Then $3\pi + 3\theta = \dfrac{8}{21}(9\pi) = \dfrac{24\pi}{7},$ which gives $\theta = \dfrac{\pi}{7}.$

Thus, the correct answer is B.

22.

Square $ABCD$ has side length $2.$ A semicircle with diameter $\overline{AB}$ is constructed inside the square, and the tangent to the semicircle from $C$ intersects side $\overline{AD}$ at $E.$ What is the length of $\overline{CE}?$

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

$\sqrt{5}$

$\sqrt{6}$

$\dfrac{5}{2}$

$5 - \sqrt{5}$

Concepts:tangent line Pythagorean Theorem

Difficulty rating: 1790

Small Hint:

Tangents from a point have equal length: $CF = CB$ and $EF = EA$

Big Hint:

With $AE = x,$ apply the Pythagorean theorem to $\triangle CDE:$ $(2 - x)^2 + 2^2 = (2 + x)^2$

Solution:

Let $F$ be the point where $CE$ touches the semicircle and let $x = AE.$ Since tangents from a point are equal, $CF = CB = 2$ and $EF = EA = x,$ so $CE = 2 + x.$

In right triangle $CDE,$ we have $DE = 2 - x$ and $DC = 2,$ so $(2 - x)^2 + 2^2 = (2 + x)^2.$ This gives $x = \dfrac{1}{2},$ hence $CE = 2 + \dfrac{1}{2} = \dfrac{5}{2}.$

Thus, the correct answer is D.

23.

Circles $A,$ $B,$ and $C$ are externally tangent to each other and internally tangent to circle $D.$ Circles $B$ and $C$ are congruent. Circle $A$ has radius $1$ and passes through the center of $D.$ What is the radius of circle $B?$

$\dfrac{2}{3}$

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

$\dfrac{7}{8}$

$\dfrac{8}{9}$

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

Concepts:tangent circles coordinate geometry Pythagorean Theorem

Difficulty rating: 1990

Small Hint:

Since $A$ passes through $D$ ’s center and is internally tangent, circle $D$ has radius $2$

Big Hint:

Place the centers on coordinates; circle $B$ of radius $r$ has center at distance $1 + r$ from $A$ and $2 - r$ from $D$

Solution:

Because circle $A$ passes through $D$ ’s center and is internally tangent to $D,$ circle $D$ has radius $2.$ Place $D$ ’s center at the origin and $A$ ’s center at $(-1, 0).$

Let circle $B$ have radius $r$ and center $(x, r),$ using the symmetry of $B$ and $C$ about the horizontal axis. Tangency gives $\begin{aligned} (x + 1)^2 + r^2 &= (1 + r)^2, \\ x^2 + r^2 &= (2 - r)^2. \end{aligned}$

Subtracting yields $x = 3r - 2.$ Substituting into the second equation gives $9r^2 - 8r = 0,$ so $r = \dfrac{8}{9}.$

Thus, the correct answer is D.

24.

Let $a_1,$ $a_2,$ $\ldots$ be a sequence with the following properties: $a_1 = 1,$ and $a_{2n} = n \cdot a_n$ for any positive integer $n.$ What is the value of $a_{2^{100}}?$

$1$

$2^{99}$

$2^{100}$

$2^{4950}$

$2^{9999}$

Concepts:recursion triangular number pattern recognition

Difficulty rating: 2010

Small Hint:

Compute $a_{2^1},$ $a_{2^2},$ $a_{2^3},$ $a_{2^4}$ and track the exponent of $2$

Big Hint:

The exponents are $0,$ $1,$ $1 + 2,$ $1 + 2 + 3,$ $\ldots$

Solution:

Applying the rule repeatedly, $\begin{aligned} a_{2^1} &= 2^0, \\ a_{2^2} &= 2^1, \\ a_{2^3} &= 2^{1+2}, \\ a_{2^4} &= 2^{1+2+3}, \ldots \end{aligned}$ so in general $a_{2^n} = 2^{1 + 2 + \cdots + (n - 1)} = 2^{\frac{n(n-1)}{2}}.$

For $n = 100,$ the exponent is $\dfrac{100 \cdot 99}{2} = 4950,$ so $a_{2^{100}} = 2^{4950}.$

Thus, the correct answer is D.

25.

Three mutually tangent spheres of radius $1$ rest on a horizontal plane. A sphere of radius $2$ rests on them. What is the distance from the plane to the top of the larger sphere?

$3 + \dfrac{\sqrt{30}}{2}$

$3 + \dfrac{\sqrt{69}}{3}$

$3 + \dfrac{\sqrt{123}}{4}$

$\dfrac{52}{9}$

$3 + 2\sqrt{2}$

Concepts:3D geometry sphere centroid Pythagorean Theorem

Difficulty rating: 2180

Small Hint:

The three small centers form an equilateral triangle of side $2$ at height $1$

Big Hint:

The big center sits above the triangle’s centroid; the centroid is $\dfrac{2\sqrt{3}}{3}$ from each small center, and the slant distance between centers is $1 + 2 = 3$

Solution:

The three small centers form an equilateral triangle of side $2,$ each $1$ unit above the plane. Its centroid $D$ is at distance $\dfrac{2\sqrt{3}}{3}$ from each vertex.

The large sphere’s center $E$ sits directly above $D,$ and the distance between $E$ and a small center is $1 + 2 = 3.$ Thus $\begin{aligned} DE &= \sqrt{3^2 - \left(\dfrac{2\sqrt{3}}{3}\right)^2} \\ &= \sqrt{9 - \dfrac{4}{3}} = \dfrac{\sqrt{69}}{3}. \end{aligned}$

Adding the $1$ unit from the plane to $D$ and the $2$ units from $E$ to the top of the large sphere gives $1 + \dfrac{\sqrt{69}}{3} + 2 = 3 + \dfrac{\sqrt{69}}{3}.$

Thus, the correct answer is B.