2014 AMC 10B — Problems & Online Timed Exam

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

Answer: C

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

Answer: E

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}

Answer: E

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}

Answer: B

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

Answer: A

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

Answer: C

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)

Answer: A

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}

Answer: E

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

Answer: A

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

Answer: C

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

Answer: C

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

Answer: C

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}

Answer: B

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

Answer: D

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}

Answer: A

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}

Answer: B

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}

Answer: D

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

Answer: E

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}

Answer: D

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

Answer: C

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

Answer: B

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

Answer: B

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

Answer: E

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 .

Answer: B

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}

Answer: C

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}

.

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 .

2014 AMC 10B Problems

Answer: C

Difficulty rating: 560

Small hint:

Big hint:

Solution:

Answer: E

Difficulty rating: 870

Small hint:

Big hint:

Solution:

Answer: E

Difficulty rating: 900

Small hint:

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Solution:

Answer: B

Difficulty rating: 960

Small hint:

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Solution:

Answer: A

Difficulty rating: 1280

Small hint:

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Solution:

Answer: C

Difficulty rating: 960

Small hint:

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Solution:

Answer: A

Difficulty rating: 870

Small hint:

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Solution:

Answer: E

Difficulty rating: 960

Small hint:

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Solution:

Answer: A

Difficulty rating: 1020

Small hint:

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Solution:

Answer: C

Difficulty rating: 1370

Small hint:

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Solution:

Answer: C

Difficulty rating: 1540

Small hint:

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Solution:

Answer: C

Difficulty rating: 1280

Small hint:

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Solution:

Answer: B

Difficulty rating: 1480

Small hint:

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Solution:

Answer: D

Difficulty rating: 1540

Small hint:

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Solution:

Answer: A

Difficulty rating: 1660

Small hint:

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Solution:

Answer: B

Difficulty rating: 1420

Small hint:

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