2010 AMC 8 — Problems & Online Timed Exam

Timed

40:00

1.

At Euclid Middle School the mathematics teachers are Miss Germain, Mr. Newton, and Mrs. Young. There are $11$ students in Miss Germain’s class, $8$ students in Mr. Newton’s class, and $9$ students in Mrs. Young’s class taking the AMC $8$ Contest this year. How many mathematics students at Euclid Middle School are taking the contest?

$\ 26$

$\ 27$

$\ 28$

$\ 29$

$\ 30$

Answer: C

Concepts:basic counting

Difficulty rating: 370

Small Hint:

No class overlap is described, so the total is a direct sum.

Big Hint:

Add the three class counts.

Solution:

There are $11+8+9 = 28$ students.

Therefore, the answer is C .

2.

If $a \ast b = \dfrac{a\times b}{a+b}$ for $a,$ $b$ positive integers, then what is $5 \ast 10?$

$\ \dfrac{3}{10}$

$\ 1$

$\ 2$

$\ \dfrac{10}{3}$

$\ 50$

Answer: D

Concepts:custom operation fraction

Difficulty rating: 610

Small Hint:

Reduce $\frac{50}{15}$ after adding the denominator.

Big Hint:

Substitute the two given numbers into the definition of $a\ast b$ .

Solution:

Given our definition of $a\ast b,$ we have $5 \ast 10 = \dfrac{5 \times 10}{5 + 10} = \dfrac{50}{15} = \dfrac{10}{3}.$

Thus, the answer is D .

3.

The graph shows the price of five gallons of gasoline during the first ten months of the year. By what percent is the highest price more than the lowest price?

$\ 50$

$\ 62$

$\ 70$

$\ 89$

$\ 100$

Answer: C

Concepts:data and graph interpretation percentage

Difficulty rating: 790

Small Hint:

Percent more means the increase divided by the lowest price.

Big Hint:

Read the highest and lowest bar heights from the graph.

Solution:

The highest price is $17$ and the lowest price is $10.$ This means the percent is $(\dfrac{17}{10}-1)\cdot 100 = 70.$

Thus, the answer is C .

4.

What is the sum of the mean, median, and mode of the numbers $2,$ $3,$ $0,$ $3,$ $1,$ $4,$ $0,$ $3?$

$\ 6.5$

$\ 7$

$\ 7.5$

$\ 8.5$

$\ 9$

Answer: C

Concepts:mean median (data)mode

Difficulty rating: 720

Small Hint:

The mean uses all eight numbers, while the median uses the two middle numbers.

Big Hint:

Put the list in increasing order before finding the median and mode.

Solution:

The list reordered is $0,0,1,2,3,3,3,4.$

The median is the mean of the middle two numbers, which would be $\dfrac{2+3}2 = 2.5.$ The mode is $3$ since $3$ appears the most. The mean is $\dfrac{0+0+1+2+3+3+3+4}{8}$ $= \dfrac{16}{8} = 2.$

Their sum is $2.5+3+2 = 7.5.$

Thus, the answer is C .

5.

Alice needs to replace a light bulb located $10$ centimeters below the ceiling in her kitchen. The ceiling is $2.4$ meters above the floor. Alice is $1.5$ meters tall and can reach $46$ centimeters above the top of her head. Standing on a stool, she can just reach the light bulb. What is the height of the stool, in centimeters?

$\ 32$

$\ 34$

$\ 36$

$\ 38$

$\ 40$

Answer: B

Concepts:unit conversion

Difficulty rating: 770

Small Hint:

The stool covers the remaining gap after Alice’s height, reach, and the bulb offset.

Big Hint:

Convert the ceiling and Alice’s height to centimeters first.

Solution:

We know the height of the ceiling is $2.4 \cdot 100 = 240$ cm. Subtracting out all the given values, we get $240 - 150 - 46 - 10 = 34.$ Therefore, the height of the stool is $34$ cm. Thus, B is the correct answer.

6.

Which of the following figures has the greatest number of lines of symmetry?

equilateral triangle

non-square rhombus

non-square rectangle

isosceles trapezoid

square

Answer: E

Concepts:transformation symmetry

Difficulty rating: 560

Small Hint:

A square has both diagonal and midpoint-to-midpoint symmetry lines.

Big Hint:

Count reflection lines for each named shape.

Solution:

First, each line of symmetry must go through the center of the shape. This would ensure that the center of the shape isn’t on only one side, as that would make it asymmetric.

Next, if a line goes through any side, it must go through its midpoint, to ensure that after a reflection over this line, the same amount of the line is on both sides. Moreover, it must be perpendicular, to ensure that the line when reflected stays on itself.

Similarly, if a line goes through any corner, it must go through its angle bisector, to ensure that after a reflection over this line, the angle of the line is the same after reflection. Moreover, it must have the same side length on both sides.

Now, let’s look at each of the shapes. An equilateral triangle has $3$ lines that intersect a corner or a midpoint, so it has at most $3$ symmetry lines. A non-square rhombus only has two symmetry lines, as only the lines that go through the corners work, but not the ones through the midpoints as they would not intersect perpendicularly. A non-square rectangle only has two symmetry lines as it has symmetry lines through the midpoints of opposite sides, but not through the corners since it doesn’t bisect the angle. An isosceles trapezoid only has $1$ symmetry line, that goes through the midpoints of opposite sides. A square has $4$ symmetry lines, which go through opposite corners and opposite midpoints.

A square therefore has the most symmetry lines.

Thus, the answer is E .

7.

Using only pennies, nickels, dimes, and quarters, what is the smallest number of coins Freddie would need so he could pay any amount of money less than a dollar?

$\ 6$

$\ 10$

$\ 15$

$\ 25$

$\ 99$

Answer: B

Concepts:money optimization

Difficulty rating: 1350

Small Hint:

After making every amount through $24$ cents, quarters can extend the range to $99$ cents.

Big Hint:

First force the coins needed to make small amounts like $4$ , $9$ , and $24$ cents.

Solution:

At least four pennies are necessary to pay $4$ cents. After those four pennies, the next coin must be worth at most $5$ cents so that $5$ cents can be paid; a nickel is the best choice. Those five coins total only $9$ cents. The next coin must be worth at most $10$ cents, and even a dime brings the total to only $19$ cents. One more coin worth at most $20$ cents is therefore necessary, so at least seven coins are needed before quarters can be used without leaving a gap.

With only nine coins total, at most two coins could remain after those required seven. The first seven can total at most $4+5+10+10=29$ cents, and two more quarters would bring the total to only $79$ cents, so nine coins cannot pay every amount through $99$ cents.

Ten coins do work: four pennies, one nickel, two dimes, and three quarters can make every amount from $1$ through $99$ cents. Therefore the minimum is $10$ .

Therefore, the answer is B .

8.

As Emily is riding her bicycle on a long straight road, she spots Emerson skating in the same direction $\frac{1}{2}$ mile in front of her. After she passes him, she can see him in her rear view mirror until he is $\frac{1}{2}$ mile behind her. Emily rides at a constant rate of $12$ miles per hour, and Emerson skates at a constant rate of $8$ miles per hour. For how many minutes can Emily see Emerson?

$\ 6$

$\ 8$

$\ 12$

$\ 15$

$\ 16$

Answer: D

Concepts:relative speed distance rate and time unit conversion

Difficulty rating: 1100

Small Hint:

During the visible interval Emily gains a total of $1$ mile on Emerson.

Big Hint:

Use Emily’s speed relative to Emerson’s speed.

Solution:

Let $d$ be how far Emily is ahead of Emerson. Emily sees Emerson if $-\dfrac 12 \leq d \leq \frac 12.$ Suppose at $t= 0,$ where $t$ is in hours, that $d = -\frac 12.$ Then, $d = (12-8)t - \frac 12 .$ Since we must find where $d = \frac 12,$ we find the time where $\frac 12 = 4t - \frac 12 \implies 1 = 4t$ $\implies t = 0.25.$ Since $0.25$ hours passed, we know that $60\cdot0.25 = 15$ minutes passed.

Therefore, the answer is D .

9.

Ryan got $80\%$ of the problems correct on a $25$ -problem test, $90\%$ on a $40$ -problem test, and $70\%$ on a $10$ -problem test. What percent of all the problems did Ryan answer correctly?

$\ 63$

$\ 75$

$\ 80$

$\ 84$

$\ 86$

Answer: D

Concepts:percentage

Difficulty rating: 960

Small Hint:

Divide the total correct answers by the total number of problems.

Big Hint:

Convert each test percentage into a number of correct answers.

Solution:

There were a total of $25+40+10 = 75$ problems.

On the first test, he solved $0.8\cdot25 = 20$ problems.

On the second test, he solved $0.9\cdot40 = 36$ problems.

On the third test, he solved $0.7\cdot10 = 7$ problems.

Therefore, he solved a total of $20+36+7=63.$ This means the fraction he solved is $\dfrac{63}{75} = 84\%.$

Therefore, the answer is D .

10.

Six pepperoni circles will exactly fit across the diameter of a $12$ -inch pizza when placed as shown. If a total of $24$ circles of pepperoni are placed on this pizza without overlap, what fraction of the pizza is covered by pepperoni?

$\ \dfrac 12$

$\ \dfrac 23$

$\ \dfrac 34$

$\ \dfrac 56$

$\ \dfrac 78$

Answer: B

Concepts:circle area area ratio power scaling of length, area, and volume

Difficulty rating: 1030

Small Hint:

Area scales with the square of the diameter, then multiply by $24$ pepperoni circles.

Big Hint:

Six small diameters equal the pizza diameter.

Solution:

Each circle has $\dfrac{1}{6}$ the diameter of the large circle, so it has $(\frac 16)^2 = \frac 1{36}$ of the total area.

Since there are $24$ pepperoni, they take up $24 \cdot \dfrac 1{36} = \frac 23$ of the area.

Therefore, the answer is B .

11.

The top of one tree is $16$ feet higher than the top of another tree. The heights of the two trees are in the ratio $3:4.$ In feet, how tall is the taller tree?

$\ 48$

$\ 64$

$\ 80$

$\ 96$

$\ 112$

Answer: B

Concepts:ratio and proportion linear equation

Difficulty rating: 900

Small Hint:

The taller tree has four of those equal parts.

Big Hint:

The ratio $3:4$ means the height difference is one ratio part.

Solution:

Let $b,$ $s$ be the heights of bigger and smaller trees respectively. Then, $b= s+16$ and $s = 0.75b.$ If we substitute, we get $b = 0.75b + 16 \implies 0.25b = 16$ $\implies b = 64.$

Thus, the answer is B .

12.

Of the $500$ balls in a large bag, $80\%$ are red and the rest are blue. How many of the red balls must be removed so that $75\%$ of the remaining balls are red?

$\ 25$

$\ 50$

$\ 75$

$\ 100$

$\ 150$

Answer: D

Concepts:percentage

Difficulty rating: 1120

Small Hint:

If red balls are $75\%$ , then blue balls are $25\%$ of the remaining bag.

Big Hint:

The number of blue balls stays fixed.

Solution:

If $r$ is the number of red balls, then $r = 0.8\cdot500 = 400.$ Therefore, if $b$ is the number of blue balls, then $b = 500-400 = 100.$

If there are $75\%$ red balls after removing balls, then there are $25\%$ blue balls. This means the total number of balls is $\dfrac{100}{0.25} = 400.$ This means the total number of balls decreased by $500-400 = 100.$

Thus, the answer is D .

13.

The lengths of the sides of a triangle in inches are three consecutive integers. The length of the shortest side is $30\%$ of the perimeter. What is the length of the longest side?

$\ 7$

$\ 8$

$\ 9$

$\ 10$

$\ 11$

Answer: E

Concepts:perimeter linear equation percentage

Difficulty rating: 1060

Small Hint:

Use the condition that the shortest side is $30\%$ of the perimeter.

Big Hint:

Let the side lengths be three consecutive integers starting with the shortest.

Solution:

Let $s$ be the smallest length. Then, all the side lengths are $s,$ $s+1,$ $s+2.$ This would make the perimeter equal to $3s+3.$ Since $s = 0.3(3s+3),$ then $s = 0.9s + 0.9 .$ This makes $0.1s = 0.9,$ so $s = 9$ which makes the longest side length $s + 2 = 11.$

Thus, the answer is E .

14.

What is the sum of the prime factors of $2010?$

$\ 67$

$\ 75$

$\ 77$

$\ 201$

$\ 210$

Answer: C

Concepts:prime factorization

Difficulty rating: 960

Small Hint:

After removing $2$ , $5$ , and $3$ , the remaining factor is prime.

Big Hint:

Factor $2010$ using divisibility by $10$ and by $3$ .

Solution:

First, the primes $2,$ $5$ are factors of $2010$ since it is a multiple of $10.$ Dividing $2010$ by $10$ is $201.$ Then, $3$ is a factor of $201$ since the digit sum of $201$ is a multiple of $3.$ Dividing this by $3$ yields the prime number $67.$ This means the prime factors are $2,$ $3,$ $5,$ $67,$ which makes their sum $77.$

Thus, the correct answer is C .

15.

A jar contains $5$ different colors of gum drops. $30\%$ are blue, $20\%$ are brown, $15\%$ are red, $10\%$ are yellow, and the other $30$ gum drops are green. If half of the blue gum drops are replaced by brown gum drops, how many gum drops will be brown?

$\ 35$

$\ 36$

$\ 42$

$\ 48$

$\ 64$

Answer: C

Concepts:percentage

Difficulty rating: 1200

Small Hint:

Use the green count to find the total, then move half the blue count to brown.

Big Hint:

Find what percent of the jar is green.

Solution:

Since we have percentages for every color except green, the percent of green is $100\%$ minus the sum of the other colors. This would make the percent of green equal to $100\%-30\%-20\%-$ $15\%-10\% = 25\%$

Since we know $30$ gum drops is $25\%,$ we know that the total number of gum drops is $\dfrac { 30}{0.25} = 120.$

This means there are $0.2\cdot120 = 24$ brown gum drops to start and $0.3\cdot120 = 36$ blue gum drops. If half of the blue gum drops are turned to brown, then $\dfrac{36}{2} = 18$ more brown gum drops are added. Therefore, we have $24+18= 42$ brown gum drops.

Thus, the answer is C .

16.

A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

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

$\ \sqrt{\pi}$

$\ \pi$

$\ 2\pi$

$\ \pi^{2}$

Answer: B

Concepts:circle area square (geometry)

Difficulty rating: 1000

Small Hint:

Set $s^2=\pi r^2$ and solve for $\frac{s}{r}$ .

Big Hint:

Write the square area and circle area in terms of side length and radius.

Solution:

Let $s$ be the side length of the square and let $r$ be the radius of the circle. Then, since they have the same areas, $s^2 = \pi r^2 .$ This means $(\dfrac sr)^2 = \pi,$ so $\frac sr = \sqrt{ \pi} .$

Thus, the answer is B .

17.

The diagram shows an octagon consisting of $10$ unit squares. The portion below $\overline{PQ}$ is a unit square and a triangle with base $5.$ If $\overline{PQ}$ bisects the area of the octagon, what is the ratio $\dfrac{XQ}{QY}?$

$\dfrac{2}{5}$

$\dfrac{1}{2}$

$\dfrac{3}{5}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

Answer: D

Concepts:area decomposition triangle area

Difficulty rating: 1670

Small Hint:

Below $PQ$ , subtract the unit square to get a triangle area equation.

Big Hint:

The line $PQ$ cuts the octagon into two equal areas.

Solution:

Since $PQ$ bisects the area, the area under the line is $5.$ Removing the square on the right makes the bottom a triangle of base $5$ with area $4.$ Let the base of this triangle be $PZ.$

The area being $4$ means $\dfrac{(PZ)(ZQ)}{2}=\frac{5(ZQ)}{2} = 4$ $\implies ZQ = 1.6.$

Therefore, $QY = QZ-1 = 0.6$ and $XQ = 2-QZ = 0.4.$ This would make $\dfrac{XQ}{QY} = \dfrac {0.4}{0.6} = \dfrac 23.$

Thus, the answer is D .

18.

A decorative window is made up of a rectangle with semicircles on either end. The ratio of $AD$ to $AB$ is $3:2,$ and $AB=30$ inches. What is the ratio of the area of the rectangle to the combined areas of the semicircles?

$\ 2:3$

$\ 3:2$

$\ 6:\pi$

$\ 9:\pi$

$\ 30 :\pi$

Answer: C

Concepts:circle area rectangle area ratio

Difficulty rating: 1240

Small Hint:

Use the ratio $AD:AB=3:2$ to express the rectangle height in terms of $AB$ .

Big Hint:

The two semicircles combine to one full circle.

Solution:

Combining the semicircles would make a circle of diameter $d=30.$ This would make the radius equal to $\dfrac{d}{2}.$ Therefore, the combined area of the semicircles is $(\dfrac{d}{2})^2 \cdot \pi = \dfrac{\pi d^2}{4}.$

Side $AD = \dfrac 32 d$ because $AD:AB=3:2$ and $AB=d.$ The area of the rectangle is therefore $\dfrac 32 d^2.$ The ratio of the area of the rectangle to the area of the semicircles is $\dfrac{\dfrac 32 d^2}{\dfrac{\pi d^2}{4}} = \dfrac{6}{\pi}.$

Thus, the answer is C .

19.

The two circles pictured have the same center $C.$ Chord $\overline{AD}$ is tangent to the inner circle at $B,$ $AC$ is $10,$ and chord $\overline{AD}$ has length $16.$ What is the area between the two circles?

$\ 36 \pi$

$\ 49 \pi$

$\ 64 \pi$

$\ 81 \pi$

$\ 100 \pi$

Answer: C

Concepts:circle area chord Pythagorean Theorem

Difficulty rating: 1390

Small Hint:

Half the chord, the inner radius, and the outer radius form a right triangle.

Big Hint:

The radius to a tangent point is perpendicular to the chord.

Solution:

The area between the two circles is the area of the larger circle minus the area of the smaller circle. This would be $(AC)^2\pi - (CB)^2 \pi$ $= \pi(AC^2 - CB^2).$ By the Pythagorean Theorem, we can get $AC^2 - CB^2 = AB^2.$ Therefore, we need to find $AB^2 \pi.$

Since $AB$ is half of $AD,$ we get $AB = 8.$ This makes $AB^2 \pi = 64 \pi.$

Thus, the answer is C .

20.

In a room, $\frac{2}{5}$ of the people are wearing gloves, and $\frac{3}{4}$ of the people are wearing hats. What is the minimum number of people in the room wearing both a hat and gloves?

$\ 3$

$\ 5$

$\ 8$

$\ 15$

$\ 20$

Answer: A

Concepts:inclusion-exclusion least common multiple

Difficulty rating: 1610

Small Hint:

Minimize overlap by making everyone counted by at least one of the two groups if possible.

Big Hint:

The total number of people must make both fractions whole numbers.

Solution:

Since our room has $\dfrac 25$ of the people wearing gloves, the number of people must be a multiple of $5.$ Since our room has $\dfrac 34$ of the people wearing hats, the number of people must be a multiple of $4.$ Therefore, the people in the room must be a multiple of $20.$

Now, we can also use the following formula by the principle of inclusion exclusion: Fraction of people wearing both = Fraction of people wearing gloves + Fraction of people wearing hats - Fraction of people wearing either.

This makes our desired fraction equal to $\dfrac{2}{5} + \dfrac 34 -$ Fraction of people who wear either. If we wish to minimize the number who wear both, we maximize the fraction of people who wear either, up to $1.$ Therefore, the fraction of people that wear both is $\dfrac{2}{5} + \dfrac 34- 1 = \dfrac 3{20}.$

Since our number is a (positive) multiple of $20,$ we have the number of people wearing both as $3$ if we choose to have just $20$ people.

Therefore, A is the correct answer.

21.

Hui is an avid reader. She bought a copy of the bestseller Math is Beautiful. On the first day, Hui read $\frac{1}{5}$ of the pages plus $12$ more, and on the second day she read $\frac{1}{4}$ of the remaining pages plus $15$ pages. On the third day she read $\frac{1}{3}$ of the remaining pages plus $18$ pages. She then realized that there were only $62$ pages left to read, which she read the next day. How many pages are in this book?

$\ 120$

$\ 180$

$\ 240$

$\ 300$

$\ 360$

Answer: C

Concepts:fraction work backwards

Difficulty rating: 1520

Small Hint:

Undo the extra pages first, then undo the fraction read each day.

Big Hint:

Work backward from the $62$ pages left after the third day.

Solution:

The pages left after the third day is $62.$ Before reading the last $18$ pages, she had $80$ pages left. This is $\dfrac 23$ of the pages remaining, so she had $120$ pages left before the third day.

Before reading the $15$ pages, she had $120+15=135$ pages left. This is $\dfrac 34$ of the pages remaining, so she had $180$ pages left before the second day.

Before reading the $12$ pages, she had $180+12=192$ pages left. This is $\dfrac 45$ of the pages remaining, so she had $240$ pages left before the first day, making the book $240$ pages.

Thus, the answer is C .

22.

The hundreds digit of a three-digit number is $2$ more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result?

$0$

$2$

$4$

$6$

$8$

Answer: E

Concepts:place value digits

Difficulty rating: 1260

Small Hint:

The tens digit cancels when the reversed number is subtracted.

Big Hint:

Represent the original number with hundreds digit two more than the units digit.

Solution:

Let the units digit be $u,$ and let the tens digit be $t.$ This makes the hundreds digit be $u+2.$ This makes the number equal to $100(u+2) +10t+u$ $=101u+200+10t$ and the reversed number is $100u +10t+(u+2)$ $=101u+2+10t.$ This makes the difference equal to $(101u+10t+200)-$ $(101u+10t+2)=198.$ This makes the units digit $8.$

Therefore, the answer is E .

23.

Semicircles $POQ$ and $ROS$ pass through the center of circle $O.$ What is the ratio of the combined areas of the two semicircles to the area of circle $O?$

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

$\dfrac{1}{2}$

$\dfrac{2}{\pi}$

$\dfrac{2}{3}$

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

Answer: B

Concepts:circle area Pythagorean Theorem area ratio

Difficulty rating: 1540

Small Hint:

The two semicircles together have the area of one circle of radius $1$ .

Big Hint:

Find the radius of circle $O$ from the coordinates.

Solution:

The area of each of the semicircles is $\pi \dfrac {r^2}{2} .$ Each of them has a radius of $1,$ so their combined area is $\pi \frac{1}{2} + \pi \frac 12 = \pi.$

Next, the radius of the larger circle is equal to the length of $OQ,$ which is equal to $\sqrt{1^2+1^2} = \sqrt 2.$ Its area is $\pi r^2 = \pi(\sqrt{2})^2 = 2\pi.$

This means the ratio is $\dfrac{\pi}{2\pi} = \frac 12.$

Thus, the answer is B .

24.

What is the correct ordering of the three numbers, $10^8,$ $5^{12},$ and $2^{24}?$

2^{24} < 10^8 < 5^{12}

2^{24} < 5^{12} < 10^8

5^{12} < 2^{24} < 10^8

10^8 < 5^{12} < 2^{24}

10^8 < 2^{24} < 5^{12}

Answer: A

Concepts:exponent

Difficulty rating: 1480

Small Hint:

Compare $2^{24}$ , $10^8=(2\cdot5)^8$ , and $5^{12}$ .

Big Hint:

Rewrite the powers so they have comparable eighth-power factors.

Solution:

First, we get $2^{24} = (2^8)(4^8) < (2^8)(5^8) = 10^8.$

Next, we get $10^8 = (2^8)(5^8) =$ $(4^4)(5^8) < (5^4)(5^8) = 5^{12}.$ This means $2^{24} < 10^8 < 5^{12} .$

Thus, the answer is A .

25.

Every day at school, Jo climbs a flight of $6$ stairs. Jo can take the stairs $1,$ $2,$ or $3$ at a time. For example, Jo could climb $3,$ then $1,$ then $2.$ In how many ways can Jo climb the stairs?

$\ 13$

$\ 18$

$\ 20$

$\ 22$

$\ 24$

Answer: E

Concepts:partitions and compositions recursive counting

Difficulty rating: 1540

Small Hint:

The first step can be $1$ , $2$ , or $3$ stairs, giving a recurrence.

Big Hint:

Let $w_n$ be the number of ways to climb $n$ stairs.

Solution:

Let $w_n$ be the number of ways to climb $n$ stairs. For $n\ge4$ , the first step can be $1$ , $2$ , or $3$ stairs, so $w_n=w_{n-1}+w_{n-2}+w_{n-3}$ .

We have $w_1=1$ , $w_2=2$ , and $w_3=4$ . Therefore $w_4=4+2+1=7$ , $w_5=7+4+2=13$ , and $w_6=13+7+4=24$ .

Thus, the answer is E .