2014 AMC 10B Solutions | LIVE by Po-Shen Loh

1.

Leah has $13$ coins, all of which are pennies and nickels. If she had one more nickel than she has now, then she would have the same number of pennies and nickels. In cents, how much are Leah’s coins worth?

$33$

$35$

$37$

$39$

$41$

Concepts:linear equation money

Difficulty rating: 560

Small Hint:

Translate the nickel condition into an equation for pennies and nickels

Big Hint:

Use the total of $13$ coins after finding the two counts

Solution:

Let the number of pennies be $p.$ Then, the number of nickels is $13-p$ and $p-1,$ so $13-p=p-1.$ This means we have $7$ pennies and $6$ nickels.

Therefore, the number of cents is $7+6\cdot 5=37.$

Thus, the correct answer is C .

2.

What is $\dfrac{2^3 + 2^3}{2^{-3} + 2^{-3}}?$

$16$

$24$

$32$

$48$

$64$

Concepts:exponent

Difficulty rating: 870

Small Hint:

Factor the common $2$ from the numerator and denominator

Big Hint:

Dividing powers of $2$ means subtracting exponents

Solution:

$\begin{aligned} \dfrac{2^3 + 2^3}{2^{-3} + 2^{-3}} &= \dfrac{2\cdot 2^3}{2\cdot 2^{-3}} \\&=2^6 \\&= 64 \end{aligned}$

Thus, the correct answer is E .

3.

Randy drove the first third of his trip on a gravel road, the next $20$ miles on pavement, and the remaining one-fifth on a dirt road. In miles, how long was Randy’s trip?

$30$

$\dfrac{400}{11}$

$\dfrac{75}{2}$

$40$

$\dfrac{300}{7}$

Concepts:fraction linear equation

Difficulty rating: 900

Small Hint:

The paved part is what remains after the gravel and dirt fractions

Big Hint:

Set $\frac7{15}$ of the trip equal to $20$

Solution:

Let the path distance be $t.$ Then, we get: $\begin{aligned}t &= \dfrac t5 + 20 + \dfrac t3 \\ t &= 20 + \dfrac {8t}{15} \\ \dfrac 7{15} t &= 20\\ t &= \dfrac{300}7.\end{aligned}$

Thus, the correct answer is E .

4.

Susie pays for $4$ muffins and $3$ bananas. Calvin spends twice as much paying for $2$ muffins and $16$ bananas. A muffin is how many times as expensive as a banana?

$\dfrac{3}{2}$

$\dfrac{5}{3}$

$\dfrac{7}{4}$

$2$

$\dfrac{13}{4}$

Concepts:system of equations ratio and proportion

Difficulty rating: 960

Small Hint:

Let the muffin and banana prices be variables

Big Hint:

Use Calvin spent twice Susie’s amount before solving the ratio

Solution:

Let the price for a muffin be $m$ and let the price for a banana be $b.$ Then, $\begin{aligned} 2(4m+3b) &= 16b + 2m\\ 8m+6b &= 16b + 2m \\ 10b &= 6m\\ m &= \dfrac 53 b.\end{aligned}$

Thus, the correct answer is B .

5.

Doug constructs a square window using $8$ equal-size panes of glass, as shown. The ratio of the height to width for each pane is $5 : 2 ,$ and the borders around and between the panes are $2$ inches wide. In inches, what is the side length of the square window?

$\ 26$

$\ 28$

$\ 30$

$\ 32$

$\ 34$

Concepts:ratio and proportion linear equation

Difficulty rating: 1280

Small Hint:

Write the pane dimensions as $5x$ by $2x$

Big Hint:

Count border widths horizontally and vertically and set the square sides equal

Solution:

Let each pane have height $5x$ and width $2x$ . Vertically, the window contains two pane heights and three horizontal borders, so its height is $10x+6$ . Horizontally, it contains four pane widths and five vertical borders, so its width is $8x+10$ .

Because the window is square, $10x+6=8x+10$ , so $x=2$ .

Therefore, the side length is $10(2)+6=26$ inches.

Thus, the correct answer is A .

6.

Orvin went to the store with just enough money to buy $30$ balloons. When he arrived, he discovered that the store had a special sale on balloons: buy $1$ balloon at the regular price and get a second at $\frac{1}{3}$ off the regular price. What is the greatest number of balloons Orvin could buy?

$33$

$34$

$36$

$38$

$39$

Concepts:ratio and proportion money pairing and grouping

Difficulty rating: 960

Small Hint:

Pair each full-price balloon with its discounted companion

Big Hint:

A two-balloon sale pair costs $\frac53$ regular balloons

Solution:

Each pair of balloons costs $1+\frac23=\frac53$ times the regular price of one balloon.

Thus, with the money for $30$ regular-price balloons, Orvin can buy $30\cdot\frac65=36$ balloons.

Thus, the correct answer is C .

7.

Suppose $A > B > 0$ and $A$ is $x\%$ greater than $B.$ What is $x?$

$100\left(\frac{A-B}{B}\right)$

$100\left(\frac{A+B}{B}\right)$

$100\left(\frac{A+B}{A}\right)$

$100\left(\frac{A-B}{A}\right)$

$100\left(\frac{A}{B}\right)$

Concepts:percentage algebraic manipulation

Difficulty rating: 870

Small Hint:

Percent greater is measured relative to the smaller value $B$

Big Hint:

First find the fraction $\frac{A-B}{B}$

Solution:

By definition, we know $A = \dfrac {x+ 100}{100}B$ $= B + \dfrac x{100}B.$

This implies, $\begin{aligned} (A-B) &= \dfrac x{100}B\\ 100\left(\dfrac{A-B} B\right) &=x.\end{aligned}$

Thus, the correct answer is A .

8.

A truck travels $\dfrac{b}{6}$ feet every $t$ seconds. There are $3$ feet in a yard. How many yards does the truck travel in $3$ minutes?

$\dfrac{b}{1080t}$

$\dfrac{30t}{b}$

$\dfrac{30b}{t}$

$\dfrac{10t}{b}$

$\dfrac{10b}{t}$

Concepts:unit conversion rate

Difficulty rating: 960

Small Hint:

Convert feet to yards before scaling the time

Big Hint:

Three minutes is $180$ seconds

Solution:

This means it travels $\dfrac{b}{18}$ yards in $t$ seconds since a yard is $3$ feet. Then, in one second, it travels $\dfrac{b}{18t}$ yards.

Therefore, in $3$ minutes which is $180$ seconds, it travels $\dfrac{b}{18t}\cdot 180 = \dfrac {10b}t.$

Thus, the correct answer is E .

9.

For real numbers $w$ and $z ,$ $\cfrac{\frac{1}{w} + \frac{1}{z}}{\frac{1}{w} - \frac{1}{z}} = 2014.$ What is $\frac{w+z}{w-z} ?$

$-2014$

$\dfrac{-1}{2014}$

$\dfrac{1}{2014}$

$1$

$2014$

Concepts:algebraic manipulation fraction

Difficulty rating: 1020

Small Hint:

Clear the tiny fractions by multiplying numerator and denominator by $wz$

Big Hint:

Watch the sign change between $z-w$ and $w-z$

Solution:

Observe that: $\begin{aligned} \dfrac{wz}{wz}\cdot \dfrac{\frac{1}{w} + \frac{1}{z}}{\frac{1}{w} - \frac{1}{z}} &= 2014\\ \dfrac{w+z}{z-w} &= 2014\\ \dfrac{w+z}{w-z} &= -1\cdot 2014\\ \dfrac{w+z}{w-z}&=-2014.\end{aligned}$

Thus, the correct answer is A .

10.

In the addition shown below $A,$ $B,$ $C,$ and $D$ are distinct digits. How many different values are possible for $D?$ $\begin{array}{r} ABBCB \\ + \ BCADA \\ \hline DBDDD \end{array}$

$2$

$4$

$7$

$8$

$9$

Concepts:cryptarithm digits casework

Difficulty rating: 1370

Small Hint:

Column addition forces $C=0$

Big Hint:

After that, identify which digits can appear as $D=A+B$

Solution:

From the leftmost column, there is no carry into a sixth digit, so $A+B=D\le 9$ .

The units column is $B+A=D$ , so it also has no carry. The tens column then gives $C+D=D$ , hence $C=0$ .

Since $A$ and $B$ are distinct nonzero digits, $D=A+B$ can be any digit from $3$ through $9$ . For example, $(A,B)=(1,2),$ $(1,3),$ $(2,3),$ $(2,4),$ $(2,5),$ $(2,6),$ $(2,7)$ give $D=3,4,5,6,7,8,9$ .

Thus there are $7$ possible values of $D$ , and the correct answer is C .

11.

For the consumer, a single discount of $n\%$ is more advantageous than any of the following discounts:

$(1)$ Two successive $15\%$ discounts.

$(2)$ Three successive $10\%$ discounts.

$(3)$ A $25\%$ discount followed by a $5\%$ discount.

What is the smallest possible positive integer value of $n?$

$\ \ 27$

$\ 28$

$\ 29$

$\ 31$

$\ 33$

Concepts:percentage inequality

Difficulty rating: 1540

Small Hint:

Convert each stacked discount into the final fraction of the price paid

Big Hint:

The single discount must exceed the largest of the three effective discounts

Solution:

The three successive-discount plans leave the customer paying $\begin{aligned} 0.85^2&=0.7225,\\ 0.9^3&=0.729,\\ 0.75(0.95)&=0.7125 \end{aligned}$ times the listed price. Their effective discounts are therefore $27.75\%$ , $27.1\%$ , and $28.75\%$ .

A single $n\%$ discount must exceed all three, so $n>28.75$ . The smallest positive integer that works is $29$ .

Thus, the correct answer is C .

12.

The largest divisor of $2{,}014{,}000{,}000$ is itself. What is its fifth-largest divisor?

$125{,}875{,}000$

$201{,}400{,}000$

$251{,}750{,}000$

$402{,}800{,}000$

$503{,}500{,}000$

Concepts:factor prime factorization

Difficulty rating: 1280

Small Hint:

The fifth largest divisor pairs with the fifth smallest divisor

Big Hint:

List the smallest divisors from the prime factorization

Solution:

The fifth-largest divisor is $2,014,000,000$ divided by the fifth smallest divisor.

The prime factorization of $2,014,000,000$ is: $2^7\cdot 5^6 \cdot 19\cdot 53.$ This makes the first $5$ smallest divisors $1,2,4,5,8.$ Therefore, the fifth smallest divisor is $8,$ and the fifth largest divisor must be: $\dfrac{2014000000}8 =251750000.$

Thus, the correct answer is C .

13.

Six regular hexagons surround a regular hexagon of side length $1$ as shown. What is the area of $\triangle{ABC}?$

$2\sqrt{3}$

$3\sqrt{3}$

$1+3\sqrt{2}$

$2+2\sqrt{3}$

$3+2\sqrt{3}$

Concepts:regular polygon equilateral triangle triangle area

Difficulty rating: 1480

Small Hint:

Use the regular hexagon angles to find the side of $\triangle ABC$

Big Hint:

The resulting triangle is equilateral

Solution:

Follow the $60^\circ$ grid formed by the unit hexagons and place $A=(0,0)$ . The marked vertices may then be written as $B=(3,\sqrt3)$ and $C=(3,-\sqrt3)$ .

Thus $BC=2\sqrt3$ , and $\begin{aligned} AB=AC&=\sqrt{3^2+(\sqrt3)^2}\\ &=2\sqrt3. \end{aligned}$ Hence $\triangle ABC$ is equilateral, with area $\frac{(2\sqrt3)^2\sqrt3}{4}=3\sqrt3.$

Thus, the correct answer is B .

14.

Danica drove her new car on a trip for a whole number of hours, averaging $55$ miles per hour. At the beginning of the trip, $abc$ miles was displayed on the odometer, where $abc$ is a $3$ -digit number with $a\ge1$ and $a+b+c\le7.$ At the end of the trip, the odometer showed $cba$ miles. What is $a^2+b^2+c^2?$

$26$

$27$

$36$

$37$

$41$

Concepts:digits divisibility

Difficulty rating: 1540

Small Hint:

The odometer change is $99(c-a)$

Big Hint:

It must also be a multiple of $55$

Solution:

We know that the difference of the numbers $cba$ and $abc$ is equal to: $100c + 10b+a - 100a - 10b-c$ $= 99(c-a)$ We know that this number also must be a multiple of $55.$ As $\gcd(55,99)$ is $11,$ we know that $c-a$ is a multiple of $5,$ and $c > a.$

This makes $a = 1, b = 0, c = 6$ the only possible value with $a+ b+c \leq 7$ as every other combination has $a+b+c > 7.$ As such, $a^2+b^2+c^2 = 37.$

Thus, the correct answer is D .

15.

In rectangle $ABCD,$ $DC = 2 \cdot CB$ and points $E$ and $F$ lie on $\overline{AB}$ so that $\overline{ED}$ and $\overline{FD}$ trisect $\angle ADC$ as shown. What is the ratio of the area of $\triangle DEF$ to the area of rectangle $ABCD?$

$\ \ \dfrac{\sqrt{3}}{6}$

$\ \dfrac{\sqrt{6}}{8}$

$\ \dfrac{3\sqrt{3}}{16}$

$\ \dfrac{1}{3}$

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

Concepts:trigonometry area ratio rectangle

Difficulty rating: 1660

Small Hint:

Use the $30^\circ$ and $60^\circ$ rays from $D$

Big Hint:

Find $EF$ , then compare triangle and rectangle areas

Solution:

Let $AD=h.$ Since $DC=AB=2h,$ the area of rectangle $ABCD$ is $2h^2.$

The rays $DE$ and $DF$ make angles of $60^\circ$ and $30^\circ,$ respectively, with $DC.$

Therefore, $AE=h\tan30^\circ=\dfrac{h}{\sqrt3}$ and $AF=h\tan60^\circ=h\sqrt3.$

Hence $\begin{aligned}EF&=AF-AE\\&=h\sqrt3-\dfrac{h}{\sqrt3}\\&=\dfrac{2h\sqrt3}{3}.\end{aligned}$ The area of $\triangle DEF$ is $\dfrac12\cdot EF\cdot h=\dfrac{h^2\sqrt3}{3}.$

The desired ratio is $\dfrac{\frac{h^2\sqrt3}{3}}{2h^2}=\dfrac{\sqrt3}{6}.$ Thus, the correct answer is A.

16.

Four fair six-sided dice are rolled. What is the probability that at least three of the four dice show the same value?

$\dfrac{1}{36}$

$\dfrac{7}{72}$

$\dfrac{1}{9}$

$\dfrac{5}{36}$

$\dfrac{1}{6}$

Concepts:dice (probability)casework

Difficulty rating: 1420

Small Hint:

Count exactly three equal dice and four equal dice separately

Big Hint:

For exactly three equal dice, choose the odd die position

Solution:

There are $6^4$ equally likely ordered outcomes.

If exactly three dice show the same value, choose the repeated value in $6$ ways, the different value in $5$ ways, and the position of the different die in $4$ ways. This gives $6\cdot5\cdot4=120$ outcomes.

If all four dice match, there are $6$ outcomes.

The probability is $\frac{120+6}{6^4}=\frac{126}{1296}=\frac7{72}$ .

Thus, the correct answer is B .

17.

What is the greatest power of $2$ that is a factor of $10^{1002} - 4^{501}?$

$2^{1002}$

$2^{1003}$

$2^{1004}$

$2^{1005}$

$2^{1006}$

Concepts:power of 2 factoring difference of squares

Difficulty rating: 1660

Small Hint:

Factor out $2^{1002}$

Big Hint:

Determine the exact power of $2$ in $5^{1002}-1$

Solution:

Factor out the obvious power of $2$ : $10^{1002}-4^{501}=2^{1002}(5^{1002}-1)$ .

Since $5^{1002}-1=(5^{501}-1)(5^{501}+1)$ , and $501$ is odd, $5^{501}-1$ is divisible by $4$ but not by $8$ , while $5^{501}+1$ is divisible by $2$ but not by $4$ .

Thus $5^{1002}-1$ contributes exactly $2^3$ , so the whole expression is divisible by $2^{1005}$ but not $2^{1006}$ .

Thus, the correct answer is D .

18.

A list of $11$ positive integers has a mean of $10,$ a median of $9,$ and a unique mode of $8.$ What is the largest possible value of an integer in the list?

$24$

$30$

$31$

$33$

$35$

Concepts:mean median (data)mode optimization

Difficulty rating: 1790

Small Hint:

The total sum of the list is $110$

Big Hint:

Minimize the first ten entries while keeping $8$ the unique mode and $9$ the median

Solution:

The list has total sum $11\cdot10=110$ . To maximize the largest entry, minimize the sum of the other ten entries.

In nondecreasing order, the sixth entry is $9$ , and $8$ must be the unique mode. If $8$ appears twice, the least possible first ten entries sum to $1+2+3+8+8+9$ $+10+11+12+13$ $=77$ , giving largest entry $33$ .

If $8$ appears three times, the least possible first ten entries are $1,1,8,8,8,9,9,10,10,11$ , with sum $75$ , giving largest entry $35$ .

If $8$ appears four or five times, the least possible sum of the first ten entries is at least $80$ , so the largest entry is at most $30$ .

Therefore the largest possible entry is $35$ , and the correct answer is E .

19.

Two concentric circles have radii $1$ and $2.$ Two points on the outer circle are chosen independently and uniformly at random. What is the probability that the chord joining the two points intersects the inner circle?

$\ \dfrac{1}{6}$

$\ \dfrac{1}{4}$

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

$\ \dfrac{1}{3}$

$\ \dfrac{1}{2}$

Concepts:geometric probability chord tangent line

Difficulty rating: 1600

Small Hint:

Fix one endpoint on the outer circle

Big Hint:

The limiting chords are tangent to the inner circle

Solution:

Fix the first endpoint $A$ on the outer circle. Draw the two chords from $A$ that are tangent to the inner circle, and call their other endpoints $B$ and $C$ . A chord from $A$ meets the inner circle exactly when its second endpoint lies on the minor arc $BC$ .

If $O$ is the common center and $D$ is a tangency point, then $\triangle AOD$ is right, with $OA=2$ and $OD=1$ . Hence $\angle OAD=30^\circ$ . The two tangent chords therefore make a $60^\circ$ angle at $A$ , so the intercepted minor arc $BC$ measures $120^\circ$ .

Therefore, the probability is $\dfrac {120^\circ}{360^\circ} = \dfrac 13 .$

Thus, the correct answer is D .

20.

For how many integers $x$ is the number $x^4-51x^2+50$ negative?

$8$

$10$

$12$

$14$

$16$

Concepts:factoring inequality counting integers in a range

Difficulty rating: 1280

Small Hint:

Factor the expression as a product in $x^2$

Big Hint:

Find where the two factors have opposite signs

Solution:

First, note that $x^4-51x^2+50$ $= (x^2-50)(x^2-1).$

The product is negative exactly when its two factors have opposite signs. Since $x^2-50<x^2-1$ , this requires $x^2-50<0<x^2-1$ . Thus $1<x^2<50$ , or $2\le |x|\le7$ . There are $6$ positive and $6$ negative integer solutions, for a total of $12$ .

Thus, the correct answer is C .

21.

Trapezoid $ABCD$ has parallel sides $\overline{AB}$ of length $33$ and $\overline {CD}$ of length $21 .$ The other two sides are of lengths $10$ and $14 .$ The angles $A$ and $B$ are acute. What is the length of the shorter diagonal of $ABCD ?$

$10\sqrt{6}$

$25$

$8\sqrt{10}$

$18\sqrt{2}$

$26$

Concepts:trapezoid Pythagorean Theorem

Difficulty rating: 1790

Small Hint:

Drop perpendiculars from the shorter base to the longer base

Big Hint:

Use the two leg lengths to find the horizontal offsets

Solution:

Let the feet of the perpendiculars from $D$ and $C$ to $AB$ be $E$ and $F$ , respectively. As drawn, take $AD=10$ and $BC=14$ ; interchanging the two legs only reflects the trapezoid.

Let $AE=x$ and let the altitude be $h$ . Since $EF=CD=21$ and $AB=33$ , we have $FB=12-x$ .

The two right triangles give $10^2=x^2+h^2$ and $14^2=(12-x)^2+h^2.$ Subtracting yields $96=144-24x$ , so $x=2$ and $h^2=96$ .

The shorter diagonal is $AC$ , whose horizontal displacement is $AE+EF=2+21=23$ . Therefore $AC=\sqrt{23^2+96}=\sqrt{625}=25.$

Thus, the correct answer is B .

22.

Eight semicircles line the inside of a square with side length $2$ as shown. What is the radius of the circle tangent to all of these semicircles?

$\dfrac{1+\sqrt2}4$

$\dfrac{\sqrt5-1}2$

$\dfrac{\sqrt3+1}4$

$\dfrac{2\sqrt3}5$

$\dfrac{\sqrt5}3$

Concepts:tangent circles Pythagorean Theorem

Difficulty rating: 1660

Small Hint:

Connect the center of the small circle to a semicircle center

Big Hint:

The needed radius is a center distance minus $\frac12$

Solution:

The distance from the center of the square to the center of the semicircles can be found as a hypotenuse of a right triangle.

One of the legs is from the center of the square to the center of one of the sides which is of distance $1.$

The other leg is from the center of the side to the center of one of the semicircles which is of distance $\dfrac 12.$ This also shows that the radius of the semicircles is $\dfrac 12.$

Therefore, the distance from the center of the square to the center of the semicircle is $\sqrt{1^2 + \left(\dfrac 12\right)^2} = \dfrac {\sqrt 5}2.$ Then we subtract $\dfrac 12$ for the radius of the semicircle. This makes the radius of the circle $\dfrac{\sqrt 5 -1}2 .$

Thus, the correct answer is B .

23.

A sphere is inscribed in a truncated right circular cone as shown. The volume of the truncated cone is twice that of the sphere. What is the ratio of the radius of the bottom base of the truncated cone to the radius of the top base of the truncated cone?

$\dfrac32$

$\dfrac{1+\sqrt5}2$

$\sqrt3$

$2$

$\dfrac{3+\sqrt5}2$

Concepts:cone sphere volume

Difficulty rating: 2300

Small Hint:

Scale the top radius to $1$

Big Hint:

Use a cross-section to relate the sphere radius to the bottom radius

Solution:

Let the top radius be $1$ , the bottom radius be $R$ , and the inscribed sphere radius be $a$ .

In the cross-section, the sphere is tangent to the two bases, so the frustum height is $2a$ . A slanted side joins $(1,a)$ to $(R,-a)$ , if the sphere’s center is the origin. Its equation is $2ax+(R-1)y-a(R+1)=0.$ Because this line is tangent to the circle of radius $a$ , its distance from the origin is $a$ . Thus $\frac{a(R+1)}{\sqrt{4a^2+(R-1)^2}}=a,$ which simplifies to $R=a^2$ .

The frustum volume is $\frac13\pi(R^2+R+1)(2a)$ $=\frac{2a\pi}{3}(a^4+a^2+1)$ .

This is twice the sphere volume, $\frac{8a^3\pi}{3}$ . Cancelling gives $a^4-3a^2+1=0$ , so $R^2-3R+1=0$ .

Since the bottom radius is larger than the top radius, $R>1$ . Thus $R=\frac{3+\sqrt5}{2}$ , and the correct answer is E .

24.

The numbers $1,$ $2,$ $3,$ $4,$ $5$ are to be arranged in a circle. An arrangement is $\textit{bad}$ if it is not true that for every $n$ from $1$ to $15$ one can find a subset of the numbers that appear consecutively on the circle that sum to $n.$ Arrangements that differ only by a rotation or a reflection are considered the same. How many different bad arrangements are there?

$1$

$2$

$3$

$4$

$5 .$

Concepts:circular arrangements casework

Difficulty rating: 2390

Small Hint:

It is enough to make consecutive sums $6$ and $7$

Big Hint:

Classify arrangements where $6$ or $7$ is impossible

Solution:

Single numbers give sums $1$ through $5$ , complements give sums $10$ through $14$ , and all five numbers give $15$ . So an arrangement is good exactly when consecutive blocks can make sums $6$ and $7$ .

If sum $6$ is impossible, then $1$ is not adjacent to $5$ . By rotating and reflecting, write the arrangement as $1bc5e$ . The adjacent pair $bc$ cannot be $\{2,3\}$ or $\{2,4\}$ , since $1+2+3=6$ and $2+4=6$ . Thus $e=2$ , and avoiding the consecutive block $2,1,3$ forces the bad arrangement $14352$ .

If sum $7$ is impossible, then $2$ is not adjacent to $5$ . Similarly write the arrangement as $2bc5e$ . Now $bc$ cannot be $\{3,4\}$ or $\{1,4\}$ , so $e=4$ . To avoid the consecutive block $4,2,1$ , the remaining order must be $b=3,\ c=1$ , giving $23154$ .

These two arrangements are indeed bad, one missing sum $6$ and the other missing sum $7$ . Hence there are $2$ bad arrangements.

Thus, the correct answer is B .

25.

In a small pond there are eleven lily pads in a row labeled $0$ through $10.$ A frog is sitting on pad $1.$ When the frog is on pad $N,$ where $0 < N < 10,$ it will jump to pad $N-1$ with probability $\frac{N}{10}$ and to pad $N+1$ with probability $1-\frac{N}{10}.$ Each jump is independent of the previous jumps.

If the frog reaches pad $0$ it will be eaten by a patiently waiting snake. If the frog reaches pad $10$ it will exit the pond, never to return. What is the probability that the frog will escape without being eaten by the snake?

$\dfrac{32}{79}$

$\dfrac{161}{384}$

$\dfrac{63}{146}$

$\dfrac{7}{16}$

$\dfrac{1}{2}$

Concepts:random walk recursive probability symmetry

Difficulty rating: 2440

Small Hint:

Let $p_i$ be the escape probability starting on pad $i$

Big Hint:

Use symmetry at pad $5$ and solve backward to $p_1$

Solution:

Let $p_i$ be the probability that the frog eventually escapes starting from pad $i$ . Then $p_0=0$ , $p_{10}=1$ , and by symmetry $p_5=\frac12$ .

For $1\le i\le 4$ , $p_i=\frac{i}{10}p_{i-1}+\frac{10-i}{10}p_{i+1}$ .

Working downward from $p_5=\frac12$ , we get $p_4=\frac25p_3+\frac3{10}$ , then $p_3=\frac3{10}p_2+\frac7{10}p_4=\frac5{12}p_2+\frac7{24}$ .

Next $p_2=\frac15p_1+\frac45p_3=\frac3{10}p_1+\frac7{20}$ . Finally $p_1=\frac9{10}p_2$ .

Substituting the expression for $p_2$ gives $p_1=\frac9{10}\left(\frac3{10}p_1+\frac7{20}\right)$ , so $p_1=\frac{63}{146}$ .

Thus, the correct answer is C .