2009 AMC 10B Solutions | LIVE by Po-Shen Loh

1.

Each morning of her five-day workweek, Jane bought either a $50$ -cent muffin or a $75$ -cent bagel. Her total cost for the week was a whole number of dollars. How many bagels did she buy?

$1$

$2$

$3$

$4$

$5$

Concepts:modular arithmetic money

Difficulty rating: 720

Small Hint:

With $b$ bagels the total is $50(5-b)+75b$ cents

Big Hint:

A whole number of dollars means that total is a multiple of $100$

Solution:

If Jane buys $b$ bagels, she buys $5-b$ muffins, for a total of $50(5-b)+75b = 250+25b$ cents. This is a whole number of dollars when $250+25b$ is a multiple of $100,$ that is, when $25b\equiv50\pmod{100},$ or $b\equiv2\pmod4.$

The only value with $0\le b\le5$ is $b=2.$

Thus, the correct answer is B.

2.

Which of the following is equal to $\dfrac{\frac13-\frac14\ }{\ \frac12-\frac13\ }?$

$\dfrac14$

$\dfrac13$

$\dfrac12$

$\dfrac23$

$\dfrac34$

Concepts:fraction

Difficulty rating: 720

Small Hint:

Multiply the numerator and denominator by $12$

Big Hint:

$\dfrac13-\dfrac14=\dfrac1{12}$ and $\dfrac12-\dfrac13=\dfrac16$

Solution:

The least common denominator of the small fractions is $12,$ so multiply top and bottom by $12:$ $\dfrac{\frac13-\frac14}{\frac12-\frac13}=\dfrac{4-3}{6-4}=\dfrac12.$

Thus, the correct answer is C.

3.

Paula the painter had just enough paint for $30$ identically sized rooms. Unfortunately, on the way to work, three cans of paint fell off her truck, so she had only enough paint for $25$ rooms. How many cans of paint did she use for the $25$ rooms?

$10$

$12$

$15$

$18$

$25$

Concepts:ratio and proportion rate

Difficulty rating: 870

Small Hint:

Losing $3$ cans cost her exactly $5$ rooms

Big Hint:

The paint-per-room rate is $\dfrac{3}{5}$ of a can

Solution:

The lost $3$ cans would have painted $30-25=5$ rooms, so each room takes $\dfrac35$ of a can.

For $25$ rooms she used $\dfrac35\cdot25=15$ cans.

Thus, the correct answer is C.

4.

A rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. The remainder of the yard has a trapezoidal shape, as shown. The parallel sides of the trapezoid have lengths $15$ and $25$ meters. What fraction of the yard is occupied by the flower beds?

$\dfrac18$

$\dfrac16$

$\dfrac15$

$\dfrac14$

$\dfrac13$

Concepts:triangle area trapezoid area ratio

Difficulty rating: 960

Small Hint:

Each triangle’s legs are half of $25-15$

Big Hint:

Compare the total triangle area to the full rectangle of dimensions $25\times5$

Solution:

The two parallel sides differ by $25-15=10,$ split evenly between the two triangles, so each isosceles right triangle has legs of length $5$ and area $\dfrac12\cdot5^2=\dfrac{25}{2}.$

Together the beds cover $25$ square meters. The rectangle has length $25$ and width $5,$ so area $125.$ The fraction is $\dfrac{25}{125}=\dfrac15.$

Thus, the correct answer is C.

5.

Twenty percent less than $60$ is one-third more than what number?

$16$

$30$

$32$

$36$

$48$

Concepts:percentage linear equation

Difficulty rating: 870

Small Hint:

Twenty percent less than $60$ is $\dfrac45\cdot60$

Big Hint:

One-third more than $n$ is $\dfrac43 n$

Solution:

Twenty percent less than $60$ is $\dfrac45\cdot60=48.$

If $n$ is the unknown number, one-third more than $n$ is $\dfrac43 n,$ so $\dfrac43 n=48$ $n=36.$

Thus, the correct answer is D.

6.

Kiana has two older twin brothers. The product of their three ages is $128.$ What is the sum of their three ages?

$10$

$12$

$16$

$18$

$24$

Concepts:prime factorization power of 2 ages

Difficulty rating: 960

Small Hint:

Every age divides $128=2^7,$ and the twins share one age

Big Hint:

If the twins are $t$ and Kiana is $k,$ then $t^2 k=128$ with $k\lt t$

Solution:

Since $128=2^7,$ every age is a power of $2.$ Writing the twins’ common age as $t$ and Kiana’s as $k,$ we need $t^2k=128$ with $k\lt t.$

Write $t=2^j.$ Then $k=2^{7-2j}.$ For $k$ to be a positive integer age we need $7-2j\ge0,$ so $j\le3.$ Because Kiana is younger than the twins, $7-2j\lt j,$ so $j\ge3.$ Therefore $j=3,$ giving $t=8$ and $k=2.$ The sum is $8+8+2=18.$

Thus, the correct answer is D.

7.

By inserting parentheses, it is possible to give the expression $2\times3+4\times5$ several values. How many different values can be obtained?

$2$

$3$

$4$

$5$

$6$

Concepts:order of operations systematic listing

Difficulty rating: 1050

Small Hint:

Decide which operation is performed last

Big Hint:

Try groupings such as $(2\times3)+(4\times5)$ and $(2\times3+4)\times5$

Solution:

There are five full parenthesizations of four numbers joined by three operations. Evaluating all five gives

$((2\times3)+4)\times5=50$ and $(2\times(3+4))\times5=70.$

$(2\times3)+(4\times5)=26.$

$2\times((3+4)\times5)=70$ and $2\times(3+(4\times5))=46.$

Thus the distinct values are $26, 46, 50,$ and $70,$ for a total of $4.$

Thus, the correct answer is C.

8.

In a certain year the price of gasoline rose by $20\%$ during January, fell by $20\%$ during February, rose by $25\%$ during March, and fell by $x\%$ during April. The price of gasoline at the end of April was the same as it had been at the beginning of January. To the nearest integer, what is $x?$

$12$

$17$

$20$

$25$

$35$

Concepts:percentage algebraic manipulation

Difficulty rating: 1170

Small Hint:

After March the price is $(1.2)(0.8)(1.25)$ times the starting price

Big Hint:

April must divide out a factor of $1.2$

Solution:

Let $p$ be the starting price. After March the price is $(1.2)(0.8)(1.25)p=1.2p.$ The April drop must return it to $p,$ so it removes $0.2p,$ a fraction $x=100\cdot\dfrac{0.2p}{1.2p}=\dfrac{100}{6}\approx16.7.$

To the nearest integer, $x=17.$

Thus, the correct answer is B.

9.

Segment $BD$ and $AE$ intersect at $C,$ as shown, $AB=BC=CD=CE,$ and $\angle A=\dfrac52\angle B.$ What is the degree measure of $\angle D?$

$52.5$

$55$

$57.5$

$60$

$62.5$

Concepts:angle chasing isosceles triangle angle sum

Difficulty rating: 1240

Small Hint:

In $\triangle ABC,$ $AB=BC$ so $\angle A=\angle C$

Big Hint:

$\angle DCE$ equals $\angle ACB$ by vertical angles, and $\triangle CDE$ is isosceles

Solution:

Since $\triangle ABC$ is isosceles with $AB=BC,$ we have $\angle A=\angle C.$ With $\angle A=\dfrac52\angle B,$ the angle sum gives $\dfrac52\angle B+\dfrac52\angle B+\angle B=180^\circ,$ so $\angle B=30^\circ$ and $\angle ACB=75^\circ.$

By vertical angles $\angle DCE=75^\circ.$ Since $CD=CE,$ triangle $CDE$ is isosceles, so $2\angle D+75^\circ=180^\circ,$ giving $\angle D=52.5^\circ.$

Thus, the correct answer is A.

10.

A flagpole is originally $5$ meters tall. A hurricane snaps the flagpole at a point $x$ meters above the ground so that the upper part, still attached to the stump, touches the ground $1$ meter away from the base. What is $x?$

$2.0$

$2.1$

$2.2$

$2.3$

$2.4$

Concepts:Pythagorean Theorem right triangle

Difficulty rating: 1140

Small Hint:

The broken piece has length $5-x$ and forms the hypotenuse

Big Hint:

$x^2+1^2=(5-x)^2$

Solution:

The standing stump has height $x,$ and the snapped piece of length $5-x$ is the hypotenuse of a right triangle with legs $x$ and $1.$ By the Pythagorean Theorem, $\begin{aligned} x^2+1^2 &= (5-x)^2 \\ &= x^2-10x+25, \end{aligned}$ so $10x=24$ and $x=2.4.$

Thus, the correct answer is E.

11.

How many $7$ -digit palindromes (numbers that read the same backward as forward) can be formed using the digits $2,$ $2,$ $3,$ $3,$ $5,$ $5,$ $5?$

$6$

$12$

$24$

$36$

$48$

Concepts:palindrome permutations parity

Difficulty rating: 1170

Small Hint:

In a $7$ -digit palindrome the middle digit appears an odd number of times

Big Hint:

The first three digits determine the last three; count their orderings

Solution:

A $7$ -digit palindrome has the form with the middle digit used once and the outer three digits each used twice. Only $5$ appears an odd number of times, so $5$ must be the middle digit.

The remaining digits $2,3,5$ fill the first three positions in some order and mirror to the last three. There are $3!=6$ such orderings.

Thus, the correct answer is A.

12.

Distinct points $A,$ $B,$ $C,$ and $D$ lie on a line, with $AB=BC=CD=1.$ Points $E$ and $F$ lie on a second line, parallel to the first, with $EF=1.$ A triangle with positive area has three of the six points as its vertices. How many possible values are there for the area of the triangle?

$3$

$4$

$5$

$6$

$7$

Concepts:triangle area parallel lines systematic listing

Difficulty rating: 1310

Small Hint:

The height is the fixed distance between the two parallel lines

Big Hint:

Count the distinct base lengths available on either line

Solution:

A positive-area triangle uses two points on one line as its base and one point on the other line as its apex. The height is always the fixed distance between the lines, so the area depends only on the base length.

Bases on the first line can be $1, 2,$ or $3;$ a base on the second line is $1.$ So the distinct base lengths are $1, 2, 3,$ giving three possible areas.

Thus, the correct answer is A.

13.

As shown below, convex pentagon $ABCDE$ has sides $AB=3,$ $BC=4,$ $CD=6,$ $DE=3,$ and $EA=7.$ The pentagon is originally positioned in the plane with vertex $A$ at the origin and vertex $B$ on the positive $x$ -axis. The pentagon is then rolled clockwise to the right along the $x$ -axis. Which side will touch the point $x=2009$ on the $x$ -axis?

$\overline{AB}$

$\overline{BC}$

$\overline{CD}$

$\overline{DE}$

$\overline{EA}$

Concepts:perimeter modular arithmetic process simulation

Difficulty rating: 1480

Small Hint:

The perimeter is $23,$ and $2009=23\cdot87+8$

Big Hint:

After $87$ full turns, track where vertices $A, B, C, D$ land

Solution:

The pentagon has perimeter $3+4+6+3+7=23.$ One full roll advances the contact point by $23,$ and $2009=23\cdot87+8.$

After $87$ rolls, vertex $A$ sits at $x=23\cdot87=2001$ and $B$ at $2004.$ Rolling further, $C$ touches at $2004+4=2008$ and $D$ at $2008+6=2014.$

Since $2009$ lies between $2008$ and $2014,$ side $\overline{CD}$ touches that point.

Thus, the correct answer is C.

14.

On Monday, Millie puts a quart of seeds, $25\%$ of which are millet, into a bird feeder. On each successive day she adds another quart of the same mix of seeds without removing any seeds that are left. Each day the birds eat only $25\%$ of the millet in the feeder, but they eat all of the other seeds. On which day, just after Millie has placed the seeds, will the birds find that more than half the seeds in the feeder are millet?

Tuesday

Wednesday

Thursday

Friday

Saturday

Concepts:geometric sequence recursion inequality

Difficulty rating: 1660

Small Hint:

Each day the standing millet is multiplied by $\dfrac34,$ then $\dfrac14$ quart is added

Big Hint:

Other seeds always total $\dfrac34$ quart; find when millet first exceeds that

Solution:

Each quart adds $\dfrac14$ quart of millet, and the birds leave $\dfrac34$ of the standing millet. On day $n$ the millet present is $\begin{gathered} \dfrac14\left(1+\dfrac34+\cdots+\Big(\dfrac34\Big)^{n-1}\right) \\ = 1-\Big(\dfrac34\Big)^{n}. \end{gathered}$

The other seeds always total $\dfrac34$ quart. Millet exceeds half when $1-\Big(\dfrac34\Big)^n\gt\dfrac34,$ i.e. $\Big(\dfrac34\Big)^n\lt\dfrac14.$

Since $\Big(\dfrac34\Big)^4=\dfrac{81}{256}\gt\dfrac14$ but $\Big(\dfrac34\Big)^5=\dfrac{243}{1024}\lt\dfrac14,$ this first happens on day $5,$ which is Friday.

Thus, the correct answer is D.

15.

When a bucket is two-thirds full of water, the bucket and water weigh $a$ kilograms. When the bucket is one-half full of water the total weight is $b$ kilograms. In terms of $a$ and $b,$ what is the total weight in kilograms when the bucket is full of water?

$\dfrac23 a+\dfrac13 b$

$\dfrac32 a-\dfrac12 b$

$\dfrac32 a+b$

$\dfrac32 a+2b$

$3a-2b$

Concepts:system of equations linear equation

Difficulty rating: 1310

Small Hint:

Let $x$ be the empty bucket’s weight and $y$ a full load of water

Big Hint:

Subtract $x+\dfrac12 y=b$ from $x+\dfrac23 y=a$ to find $y$

Solution:

Let $x$ be the bucket’s weight and $y$ the weight of a full load of water. Then $x+\dfrac23 y=a$ and $x+\dfrac12 y=b.$

Subtracting gives $\dfrac16 y=a-b,$ so $y=6a-6b,$ and $x=b-\dfrac12 y=4b-3a.$ The full bucket weighs $\begin{aligned} x+y &= (4b-3a)+(6a-6b) \\ &= 3a-2b. \end{aligned}$

Thus, the correct answer is E.

16.

Points $A$ and $C$ lie on a circle centered at $O,$ each of $\overline{BA}$ and $\overline{BC}$ are tangent to the circle, and $\triangle ABC$ is equilateral. The circle intersects $\overline{BO}$ at $D.$ What is $\dfrac{BD}{BO}?$

$\dfrac{\sqrt2}{3}$

$\dfrac12$

$\dfrac{\sqrt3}{3}$

$\dfrac{\sqrt2}{2}$

$\dfrac{\sqrt3}{2}$

Concepts:tangent line special right triangle circle

Difficulty rating: 1420

Small Hint:

Equilateral $\triangle ABC$ gives $\angle OBC=30^\circ,$ and $OC\perp BC$

Big Hint:

In a $30$ - $60$ - $90$ triangle the hypotenuse $BO=2\,OC$

Solution:

Let the radius be $r.$ By symmetry $BO$ bisects the $60^\circ$ angle $ABC,$ so $\angle OBC=30^\circ.$ Since $OC\perp BC,$ triangle $BCO$ is a $30$ - $60$ - $90$ triangle with hypotenuse $BO=2\,OC=2r.$

Then $BD=BO-OD=2r-r=r,$ so $\dfrac{BD}{BO}=\dfrac{r}{2r}=\dfrac12.$

Thus, the correct answer is B.

17.

Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from $(a,0)$ to $(3,3),$ divides the entire region into two regions of equal area. What is $a?$

$\dfrac12$

$\dfrac35$

$\dfrac23$

$\dfrac34$

$\dfrac45$

Concepts:triangle area area decomposition coordinate geometry

Difficulty rating: 1540

Small Hint:

The whole region has area $5,$ so each part must be $\dfrac52$

Big Hint:

The lower-right part is a triangle of base $3-a$ and height $3,$ with one unit square removed

Solution:

The five unit squares have total area $5,$ so each region must have area $\dfrac52.$

The region to the lower right of the line is a right triangle with legs $3-a$ and $3,$ minus the one unit square it does not cover. Setting its area to $\dfrac52$ gives $\dfrac{3(3-a)}{2}-1=\dfrac52,$ so $3(3-a)=7$ and $a=\dfrac23.$

Thus, the correct answer is C.

18.

Rectangle $ABCD$ has $AB=8$ and $BC=6.$ Point $M$ is the midpoint of diagonal $\overline{AC},$ and $E$ is on $\overline{AB}$ with $\overline{ME}\perp\overline{AC}.$ What is the area of $\triangle AME?$

$\dfrac{65}{8}$

$\dfrac{25}{3}$

$9$

$\dfrac{75}{8}$

$\dfrac{85}{8}$

Difficulty rating: 1370

Small Hint:

$AC=10,$ so $AM=5$

Big Hint:

$\triangle AME\sim\triangle ABC,$ giving $\dfrac{ME}{AM}=\dfrac{BC}{AB}$

Solution:

By the Pythagorean Theorem, $AC=\sqrt{8^2+6^2}=10,$ so $AM=5.$ Right triangles $AME$ and $ABC$ share angle $A,$ so they are similar with $\dfrac{ME}{AM}=\dfrac{BC}{AB}=\dfrac68,$ giving $ME=\dfrac{15}{4}.$

Then $\begin{gathered} [\triangle AME]=\dfrac12\cdot AM\cdot ME \\ = \dfrac12\cdot5\cdot\dfrac{15}{4}=\dfrac{75}{8}. \end{gathered}$

Thus, the correct answer is D.

19.

A particular $12$ -hour digital clock displays the hour and minute of a day. Unfortunately, whenever it is supposed to display a $1,$ it mistakenly displays a $9.$ For example, when it is $1{:}16$ PM the clock incorrectly shows $9{:}96$ PM. What fraction of the day will the clock show the correct time?

$\dfrac12$

$\dfrac58$

$\dfrac34$

$\dfrac56$

$\dfrac{9}{10}$

Concepts:clock digits basic counting

Difficulty rating: 1540

Small Hint:

The hour is wrong only for $1,$ $10,$ $11,$ $12$

Big Hint:

A minute is wrong when either of its digits is a $1;$ count those out of $60$

Solution:

Among the hours $1$ through $12,$ exactly $1,$ $10,$ $11,$ $12$ contain a $1,$ so the hour is correct $\dfrac{8}{12}=\dfrac23$ of the time.

A minute is displayed wrong when its tens digit is $1$ (minutes $10$ – $19$ ) or its units digit is $1$ ( $01,11,\dots,51$ ), which is $15$ of the $60$ minutes. So the minute is correct $\dfrac{45}{60}=\dfrac34$ of the time.

The clock is correct $\dfrac23\cdot\dfrac34=\dfrac12$ of the day.

Thus, the correct answer is A.

20.

Triangle $ABC$ has a right angle at $B,$ $AB=1,$ and $BC=2.$ The bisector of $\angle BAC$ meets $\overline{BC}$ at $D.$ What is $BD?$

$\dfrac{\sqrt3-1}{2}$

$\dfrac{\sqrt5-1}{2}$

$\dfrac{\sqrt5+1}{2}$

$\dfrac{\sqrt6+\sqrt2}{2}$

$2\sqrt3-1$

Concepts:angle bisector theorem Pythagorean Theorem

Difficulty rating: 1600

Small Hint:

By the Angle Bisector Theorem, $\dfrac{BD}{DC}=\dfrac{AB}{AC}$

Big Hint:

$AC=\sqrt5$ and $BD+DC=2$

Solution:

By the Pythagorean Theorem, $AC=\sqrt{1^2+2^2}=\sqrt5.$ The Angle Bisector Theorem gives $\dfrac{BD}{DC}=\dfrac{AB}{AC}=\dfrac{1}{\sqrt5},$ so $DC=\sqrt5\,BD.$

Since $BD+DC=2,$ we have $BD(1+\sqrt5)=2,$ so $BD=\dfrac{2}{1+\sqrt5}=\dfrac{\sqrt5-1}{2}.$

Thus, the correct answer is B.

21.

What is the remainder when $3^0+3^1+3^2+\cdots+3^{2009}$ is divided by $8?$

$0$

$1$

$2$

$4$

$6$

Concepts:modular arithmetic geometric sequence pairing and grouping

Difficulty rating: 1420

Small Hint:

$3^0+3^1+3^2+3^3=40$ is a multiple of $8$

Big Hint:

Group the $2010$ terms into blocks of four and handle the leftover

Solution:

Any four consecutive powers of $3$ sum to a multiple of $3^0+3^1+3^2+3^3=40,$ which is divisible by $8.$

The terms from $3^2$ to $3^{2009}$ split into such blocks and contribute remainder $0.$ What remains is $3^0+3^1=4.$

Thus, the correct answer is D.

22.

A cubical cake with edge length $2$ inches is iced on the sides and the top. It is cut vertically into three pieces as shown in this top view, where $M$ is the midpoint of a top edge. The piece whose top is triangle $B$ contains $c$ cubic inches of cake and $s$ square inches of icing. What is $c+s?$

$\dfrac{24}{5}$

$\dfrac{32}{5}$

$8+\sqrt5$

$5+\dfrac{16\sqrt5}{5}$

$10+5\sqrt5$

Concepts:3D geometry similarity volume surface area

Difficulty rating: 1750

Small Hint:

Triangle $B$ is similar to the triangle whose legs are $1$ and $2$

Big Hint:

The icing on $B$ covers its top face and the entire cube face it borders

Solution:

Set the top face as a $2\times2$ square. The cut from $M$ toward the far corner creates the top triangle $A$ with legs $1$ and $2,$ so area $1$ and hypotenuse $\sqrt5.$

Triangle $B$ is similar to $A$ but with hypotenuse $2,$ so its area is $\left(\dfrac{2}{\sqrt5}\right)^2\cdot1=\dfrac45.$ Since the cake has height $2,$ the volume is $c=\dfrac45\cdot2=\dfrac85.$

The icing on this piece is its top ( $\dfrac45$ ) plus the full cube side face it borders ( $2\times2=4$ ), so $s=\dfrac45+4=\dfrac{24}{5}.$ Therefore $c+s=\dfrac85+\dfrac{24}{5}=\dfrac{32}{5}.$

Thus, the correct answer is B.

23.

Rachel and Robert run on a circular track. Rachel runs counterclockwise and completes a lap every $90$ seconds, and Robert runs clockwise and completes a lap every $80$ seconds. Both start from the start line at the same time. At some random time between $10$ minutes and $11$ minutes after they begin to run, a photographer standing inside the track takes a picture that shows one-fourth of the track, centered on the starting line. What is the probability that both Rachel and Robert are in the picture?

$\dfrac{1}{16}$

$\dfrac18$

$\dfrac{3}{16}$

$\dfrac14$

$\dfrac{5}{16}$

Concepts:geometric probability distance rate and time

Difficulty rating: 1920

Small Hint:

The picture covers $\dfrac18$ of a lap on each side of the start line

Big Hint:

Find each runner’s window of time inside that region during the $10$ th minute, then overlap them

Solution:

The picture spans $\dfrac18$ lap on each side of the start. After $600$ seconds Rachel is $30$ seconds short of the line; running $\dfrac14$ lap in $22.5$ seconds, she is in view between $30-11.25=18.75$ and $30+11.25=41.25$ seconds of the $10$ th minute.

After $600$ seconds Robert is $40$ seconds from the line; running $\dfrac14$ lap in $20$ seconds, he is in view between $30$ and $50$ seconds.

Both appear between $30$ and $41.25$ seconds, a window of length $11.25$ out of $60,$ so the probability is $\dfrac{11.25}{60}=\dfrac{3}{16}.$

Thus, the correct answer is C.

24.

The keystone arch is an ancient architectural feature. It is composed of congruent isosceles trapezoids fitted together along the non-parallel sides, as shown. The bottom sides of the two end trapezoids are horizontal. In an arch made with $9$ trapezoids, let $x$ be the angle measure in degrees of the larger interior angle of the trapezoid. What is $x?$

$100$

$102$

$104$

$106$

$108$

Concepts:regular polygon angle sum symmetry

Difficulty rating: 1600

Small Hint:

Reflect the arch to close it into a full loop of $18$ trapezoids

Big Hint:

The inner vertices form a regular $18$ -gon; use its interior angle

Solution:

Adding a mirror image completes the arch into a symmetric closed loop of $18$ trapezoids. Their inner edges form a regular $18$ -gon, each interior angle of which is $\dfrac{(18-2)\cdot180^\circ}{18}=160^\circ.$

At each inner vertex, two of the trapezoids’ larger angles $x$ meet the $160^\circ$ angle around a full turn: $x+x+160^\circ=360^\circ,$ so $x=100.$

Thus, the correct answer is A.

25.

Each face of a cube is given a single narrow stripe painted from the center of one edge to the center of its opposite edge. The choice of the edge pairing is made at random and independently for each face. What is the probability that there is a continuous stripe encircling the cube?

$\dfrac18$

$\dfrac{3}{16}$

$\dfrac14$

$\dfrac38$

$\dfrac12$

Concepts:basic probability independent events symmetry

Difficulty rating: 2090

Small Hint:

Each face has $2$ equally likely stripe orientations, so $2^6$ total configurations

Big Hint:

There are $3$ possible encircling bands, each fixing the orientation of $4$ faces

Solution:

Each face’s stripe has $2$ orientations, giving $2^6=64$ equally likely configurations.

An encircling stripe runs around one of the $3$ pairs of opposite faces. For a given band, the $4$ faces it crosses must each be oriented to continue it, a probability of $\left(\dfrac12\right)^4=\dfrac{1}{16}.$

The three bands are mutually exclusive, so the probability is $3\cdot\dfrac{1}{16}=\dfrac{3}{16}.$

Thus, the correct answer is B.