AMC 8 2005

Instructions

This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.
You will receive 1 point for each correct answer. There is no penalty for wrong answers.
No aids are permitted other than plain scratch paper, writing utensils, ruler, and erasers. In particular, graph paper, compass, protractor, calculators, computers, smartwatches, and smartphones are not permitted.
Figures are not necessarily drawn to scale.
You will have 40 minutes working time to complete the test.

Question 1

Connie multiplies a number by 2 and gets 60 as her answer. However, she should have divided the number by 2 to get the correct answer. What is the correct answer?

$\textbf{(A)}\ 7.5\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 240$

Solution

$\textbf{B}$

The original number is $60/2=30$. So the correct answer is $30/2=15$.

Question 2

Karl bought five folders from Pay-A-Lot at a cost of $\$2.50$ each. Pay-A-Lot had a $20\%$-off sale the following day. How much could Karl have saved on the purchase by waiting a day?

$\textbf{(A)}\ \$1.00 \qquad\textbf{(B)}\ \$2.00 \qquad\textbf{(C)}\ \$2.50\qquad\textbf{(D)}\ \$2.75 \qquad\textbf{(E)}\ \$5.00$

Solution

$\textbf{C}$

He could save $\$2.50\times20\%=\$0.50$ for each folder. Since he bought 5 folders, he could save $\$0.50\times5=\$2.50$.

Question 3

What is the minimum number of small squares that must be colored black so that a line of symmetry lies on the diagonal $\overline{BD}$ of square $ABCD$?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution

$\textbf{D}$

There are 5 black squares in the figure now. Only the top right square is in symmetry with diagonal $\overline{BD}$. So we need 4 more black squares to make the rest 4 black squares in symmetry.

Question 4

A square and a triangle have equal perimeters. The lengths of the three sides of the triangle are 6.1 cm, 8.2 cm and 9.7 cm. What is the area of the square in square centimeters?

$\textbf{(A)}\ 24\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 64$

Solution

$\textbf{C}$

The perimeter of the triangle is $6.1+8.2+9.7=24$ cm. A square's perimeter is four times its side length, since all its side lengths are equal. So the side length is $24/4=6$ cm, and the area is $6^2=36\ \text{cm}^2$.

Question 5

Soda is sold in packs of 6, 12 and 24 cans. What is the minimum number of packs needed to buy exactly 90 cans of soda?

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 15$

Solution

$\textbf{B}$

Start by trying the largest packs first. After three $24$-packs, $90-3\times24=18$ cans are left. After one $12$-pack, $18-12=6$ cans are left. Then buy one more $6$-pack. The total number of packs is $3+1+1=5$.

Question 6

Suppose $d$ is a digit. For how many values of $d$ is $2.00d5 > 2.005$?

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 10$

Solution

$\textbf{C}$

We see that $2.0055$ works but $2.0045$ does not. The digit $d$ can be from $5$ through $9$, which is $5$ values.

Question 7

Bill walks $\dfrac12$ mile south, then $\dfrac34$ mile east, and finally $\dfrac12$ mile south. How many miles is he, in a direct line, from his starting point?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 1\dfrac14\qquad\textbf{(C)}\ 1\dfrac12\qquad\textbf{(D)}\ 1\dfrac34\qquad\textbf{(E)}\ 2$

Solution

$\textbf{B}$

The length of the diagonal of the rectangle is the length of the direct line: \[\sqrt{\left(\dfrac12+\dfrac12\right)^2+\left(\dfrac34\right)^2} = \sqrt{1+\dfrac{9}{16}} = \sqrt{\dfrac{25}{16}} = \dfrac54 =1 \dfrac14\]

Question 8

Suppose $m$ and $n$ are positive odd integers. Which of the following must also be an odd integer?

$\textbf{(A)}\ m+3n\qquad\textbf{(B)}\ 3m-n\qquad\textbf{(C)}\ 3m^2 + 3n^2\qquad\textbf{(D)}\ (nm + 3)^2\qquad\textbf{(E)}\ 3mn$

Solution

$\textbf{E}$

An odd multiples by another odd is still an odd. So the answer is E.

Question 9

In quadrilateral $ABCD$, sides $\overline{AB}$ and $\overline{BC}$ both have length 10, sides $\overline{CD}$ and $\overline{DA}$ both have length 17, and the measure of angle $ADC$ is $60^\circ$. What is the length of diagonal $\overline{AC}$?

$\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 15.5\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18.5$

Solution

$\textbf{D}$

Since $\triangle ACD$ is isosceles, and $\angle ADC=60^\circ$, we know that $\triangle ACD$ is an equilateral triangle. Hence, $\overline{AC}=\overline{CD}=\overline{DA}=17$.

Question 10

Joe had walked half way from home to school when he realized he was late. He ran the rest of the way to school. He ran 3 times as fast as he walked. Joe took 6 minutes to walk half way to school. How many minutes did it take Joe to get from home to school?

$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 7.3\qquad\textbf{(C)}\ 7.7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 8.3$

Solution

$\textbf{D}$

Since he ran 3 times as fast as he walked, it takes hims $6/3=2$ minutes to finish the rest half way. So the total time is $6+2=8$ minutes.

Question 11

The sales tax rate in Bergville is $6\%$. During a sale at the Bergville Coat Closet, the price of a coat is discounted $20\%$ from its $\$90.00$ price. Two clerks, Jack and Jill, calculate the bill independently. Jack brings up $\$90.00$ and adds $6\%$ sales tax, then subtracts $20\%$ from this total. Jill brings up $\$90.00$, subtracts $20\%$ of the price, then adds $6\%$ of the discounted price for sales tax. What is Jack's total minus Jill's total?

$\textbf{(A)}\ -\$1.06\qquad\textbf{(B)}\ -\$0.53\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ \$0.53\qquad\textbf{(E)}\ \$1.06$

Solution

$\textbf{C}$

The price Jack rings up is $\$90.00\times(1+6\%)(1-20\%)$. The price Jill rings up is $\$90.00\times(1-20\%)(1+6\%)$. By the commutative property of multiplication, these quantities are the same, and the difference is $0$.

Question 12

Big Al the ape ate 100 delicious yellow bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many delicious bananas did Big Al eat on May 5?

$\textbf{(A)}\ 20\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34$

Solution

$\textbf{D}$

The average number of bananas he ate per day is 20, which is also the number of bananas he ate on May 3. So he ate $20+6+6=32$ bananas on May 5.

Question 13

The area of polygon $ABCDEF$ is 52 with $AB=8$, $BC=9$ and $FA=5$. What is $DE+EF$?

$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$

Solution

$\textbf{C}$

Notice that $AF + DE = BC$, so $DE=4$. Let $O$ be the intersection of the extensions of $AF$ and $CD$, which makes rectangle $ABCO$. The area of the polygon is the area of $FEDO$ subtracted from the area of $ABCO$. \[52 = 8 \times 9- EF \times 4\] Solving for the unknown, $EF=5$, therefore $DE+EF=4+5=9$.

Question 14

The Little Twelve Basketball League has two divisions, with six teams in each division. Each team plays each of the other teams in its own division twice and every team in the other division once. How many games are scheduled?

$\textbf{(A)}\ 80\qquad\textbf{(B)}\ 96\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 108\qquad\textbf{(E)}\ 192$

Solution

$\textbf{B}$

We have $\dbinom{6}{2}=15$ ways to choose 2 teams in a division. Since there are 2 divisions, and each team plays each of the other teams twice in its own division, we have $2\times2\times15=60$ games within divisions. The number of games between the two divisions is $6\times6=36$. Together there are $60+36=96$ games.

Question 15

How many different isosceles triangles have integer side lengths and perimeter 23?

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11$

Solution

$\textbf{C}$

Let $b$ be the base of the isosceles triangles, and $a$ be the lengths of the two equal sides. Hence, the perimeter is $2a+b=23$. According to the triangle inequality, we have $0<b<2a$. So $2a<2a+b=23<2a+2a=4a$, which can be simplified as $5.75<a<11.5$. So the value of $a$ could be $6,7,8,9,10,11$. The answer is C.

Question 16

A five-legged Martian has a drawer full of socks, each of which is red, white or blue, and there are at least five socks of each color. The Martian pulls out one sock at a time without looking. How many socks must the Martian remove from the drawer to be certain there will be 5 socks of the same color?

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 15$

Solution

$\textbf{D}$

The Martian can pull out $12$ socks, $4$ of each color, without having $5$ of the same kind yet. However, the next one he pulls out must be the fifth of one of the colors. So he must remove $13$ socks.

Question 17

The results of a cross-country team's training run are graphed below. Which student has the greatest average speed?

$\textbf{(A)}\ \text{Angela}\qquad\textbf{(B)}\ \text{Briana}\qquad\textbf{(C)}\ \text{Carla}\qquad\textbf{(D)}\ \text{Debra}\qquad\textbf{(E)}\ \text{Evelyn}$

Solution

$\textbf{E}$

The average speed is distance over time, or the slope of the line through the point and the origin. Evelyn has the steepest line, so she has the greatest average speed.

Question 18

How many three-digit numbers are divisible by 13?

$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 67\qquad\textbf{(C)}\ 69\qquad\textbf{(D)}\ 76\qquad\textbf{(E)}\ 77$

Solution

$\textbf{C}$

Let $k$ be any positive integer so that $13k$ is a multiple of $13$. When $100\le 13k \le999$, we have $7\dfrac{9}{13}\le k\le 76\dfrac{11}{13}$. So the number $k$ can range from $8$ to $76$. Hence, there are $76-8+1=69$ three-digit numbers.

Question 19

What is the perimeter of trapezoid $ABCD$?

$\textbf{(A)}\ 180\qquad\textbf{(B)}\ 188\qquad\textbf{(C)}\ 196\qquad\textbf{(D)}\ 200\qquad\textbf{(E)}\ 204$

Solution

$\textbf{A}$

Draw altitudes from $B$ and $C$ to base $AD$ to create a rectangle and two right triangles. The side opposite $BC$ is equal to $50$. The bases of the right triangles can be found using Pythagorean or special triangles to be $18$ and $7$. Add it together to get $AD=18+50+7=75$. The perimeter is $75+30+50+25=180$.

Question 20

Alice and Bob play a game involving a circle whose circumference is divided by 12 equally-spaced points. The points are numbered clockwise, from 1 to 12. Both start on point 12. Alice moves clockwise and Bob, counterclockwise. In a turn of the game, Alice moves 5 points clockwise and Bob moves 9 points counterclockwise. The game ends when they stop on the same point. How many turns will this take?

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 24$

Solution

$\textbf{A}$

Together they walked $5+9=14$ points in a turn. To end up on the same point, the total number of points should be a multiple of 12. The least common multiple of 14 and 12 is 84. So they need to walk $84/14=6$ turns.

Question 21

How many distinct triangles can be drawn using three of the dots below as vertices?

$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24$

Solution

$\textbf{C}$

The number of ways to choose three points from six is $\dbinom63 = 20$. However, two of these are straight lines so we subtract $2$ to get $18$.

Question 22

A company sells detergent in three different sized boxes: small (S), medium (M) and large (L). The medium size costs $50\%$ more than the small size and contains $20\%$ less detergent than the large size. The large size contains twice as much detergent as the small size and costs $30\%$ more than the medium size. Rank the three sizes from best to worst buy.

$\textbf{(A)}\ \text{SML}\qquad\textbf{(B)}\ \text{LMS}\qquad\textbf{(C)}\ \text{MSL}\qquad\textbf{(D)}\ \text{LSM}\qquad\textbf{(E)}\ \text{MLS}$

Solution

$\textbf{E}$

Suppose the small size costs $\$1$ and the large size has $10$ oz. The medium size then costs $\$1.50$ and has $8$ oz. The small size has $5$ oz and the large size costs $\$1.95$. The small, medium, and large size cost respectively, $0.200, 0.188, 0.195$ dollars per oz. The sizes from best to worst buy are $\text{MLS}$.

Question 23

Isosceles right triangle $ABC$ encloses a semicircle of area $2\pi$. The circle has its center $O$ on hypotenuse $\overline{AB}$ and is tangent to sides $\overline{AC}$ and $\overline{BC}$. What is the area of triangle $ABC$?

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 3\pi\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 4\pi$

Solution

$\textbf{B}$

Draw another half of the semicircle to make it a full circle, and another half of the triangle to make it a full square. The area of the circle is $4\pi$, so the radius is 2. Hence, the side length of the square is 4, and the area of the square is 16. Therefore, the area of $\triangle ABC$ is 8.

Question 24

A certain calculator has only two keys [+1] and [x2]. When you press one of the keys, the calculator automatically displays the result. For instance, if the calculator originally displayed "9" and you pressed [+1], it would display "10." If you then pressed [x2], it would display "20." Starting with the display "1," what is the fewest number of keystrokes you would need to reach "200"?

$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$

Solution

$\textbf{B}$

The fastest way to make a number greater is [x2] but not [+1]. Hence, we start at $200$ and work our way down to $1$.
\begin{align*}
200\div2&=100\\100\div2&=50\\50\div2&=25
\end{align*}Here we come across an odd number. So we will subtract that number by $1$ and keep on dividing it by 2
\begin{align*}
25-1 &= 24\\
24 \div 2 &= 12\\
12 \div 2 &= 6\\
6 \div 2 &= 3\\
3-1 &= 2\\
2 \div 2 &= 1
\end{align*}Since we've reached $1$, it's clear that the answer should be $9$.

Question 25

A square with side length 2 and a circle share the same center. The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square. What is the radius of the circle?

$\textbf{(A)}\ \dfrac{2}{\sqrt{\pi}} \qquad \textbf{(B)}\ \dfrac{1+\sqrt{2}}{2} \qquad \textbf{(C)}\ \dfrac{3}{2} \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ \sqrt{\pi}$

Solution

$\textbf{A}$

Since the areas of the regions outside of the circle and the square are equal to each other, the area of the circle must be equal to the area of the square. Hence, we get $\pi r^2=4\rightarrow r=\dfrac{2}{\sqrt{\pi}}$.