AMC8 2005

If $x$ is the number, then $2x=60$ and $x=30$. Dividing the number by $2$ yields $\dfrac{30}{2} = \boxed{\textbf{(B)}\ 15}$. A second way to do it is to divide the number by $4$, as you multiplied by $2$ when you were supposed to divide by $2$. So, $\dfrac{60}{4} = \boxed{\textbf{(B)}\ 15}$.

(C) Karl spent $5 \times \$2.50 = \$12.50$ on the folders. If he had purchased the folders a day later, he would have saved $20\%$ of this total, or $0.20 \times \$12.50 = \$2.50$.

(D) For diagonal $BD$ to lie on a line of symmetry in square $ABCD$, the four small squares labeled $bl$ must be colored black.

(C) The perimeter of the triangle is $6.1 + 8.2 + 9.7 = 24$ cm. The perimeter of the square is also $24$ cm. Each side of the square is $24 \div 4 = 6$ cm. The area of the square is $6^2 = 36$ square centimeters.

(B) To get the minimum number of packs, purchase as many 24-packs as possible: three 24-packs contain 72 cans, which leaves $90 - 72 = 18$ cans. To get the remaining 18 cans, purchase one 12-pack and one 6-pack. The minimum number of packs is 5.

(C) The number $2.00d5$ is greater than $2.005$ for $d = 5, 6, 7, 8$ and $9$. Therefore, there are five digits satisfying the inequality.

(B) The diagram on the left shows the path of Bill’s walk. As the diagram on the right illustrates, he could also have walked from $A$ to $B$ by first walking 1 mile south then $\frac{3}{4}$ mile east. By the Pythagorean Theorem, $(AB)^2 = 1^2 + \left(\frac{3}{4}\right)^2 = 1 + \frac{9}{16} = \frac{25}{16},$ so $AB = \frac{5}{4} = 1\frac{1}{4}.$

(E) To check the possible answers, choose the easiest odd numbers for $m$ and $n$. If $m=n=1$, then $m+3n=4,\quad 3m-n=2,\quad 3m^2+3n^2=6,\quad (mn+3)^2=16\text{ and }3mn=3.$ This shows that (A), (B), (C) and (D) can be even when $m$ and $n$ are odd. On the other hand, because the product of odd integers is always odd, $3mn$ is always odd if $m$ and $n$ are odd. Questions: Which of the expressions are always even if $m$ and $n$ are odd? What are the possibilities if $m$ and $n$ are both even? If one is even and the other odd?

(D) Triangle $ACD$ is an isosceles triangle with a $60^\circ$ angle, so it is also equilateral. Therefore, the length of $AC$ is $17$.

(D) Covering the same distance three times as fast takes one-third the time. So Joe ran for 2 minutes. His total time was 6 + 2 = 8 minutes.

(C) To add 6% sales tax to an item, multiply the price by 1.06. To calculate a 20% discount, multiply the price by 0.8. Because both actions require only multiplication, and because multiplication is commutative, the order of operations doesn’t matter. Jack and Jill will get the same total. Note: Jack’s final computation is 0.80(1.06 × \$90.00) and Jill’s is 1.06(0.80 × \$90.00). Both yield the same product, \$76.32.

(D) You can solve this problem by guessing and checking. If Big Al had eaten 10 bananas on May 1, then he would have eaten $10 + 16 + 22 + 28 + 34 = 110$ bananas. This is 10 bananas too many, so he actually ate 2 fewer bananas each day. Thus, Big Al ate 8 bananas on May 1 and 32 bananas on May 5.

Rectangle $ABCG$ has area $8 \times 9 = 72$, so rectangle $FEDG$ has area $72 - 52 = 20$. The length of $FG$ equals $DE = 9 - 5 = 4$, so the length of $EF$ is $\frac{20}{4} = 5$. Therefore, $DE + EF = 4 + 5 = 9$.

(B) Each team plays 10 games in its own division and 6 games against teams in the other division. So each of the 12 teams plays 16 conference games. Because each game involves two teams, there are $\frac{12\times16}{2}=96$ games scheduled.

(C) Because the perimeter of such a triangle is 23, and the sum of the two equal side lengths is even, the length of the base is odd. Also, the length of the base is less than the sum of the other two side lengths, so it is less than half of 23. Thus the six possible triangles have side lengths 1, 11, 11; 3, 10, 10; 5, 9, 9; 7, 8, 8; 9, 7, 7 and 11, 6, 6.

(D) It is possible for the Martian to pull out at most 4 red, 4 white and 4 blue socks without having a matched set. The next sock it pulls out must be red, white or blue, which gives a matched set. So the Martian must select $4 \times 3 + 1 = 13$ socks to be guaranteed a matched set of five socks.

(E) Evelyn covered more distance in less time than Briana, Debra and Angela, so her average speed is greater than any of their average speeds. Evelyn went almost as far as Carla in less than half the time that it took Carla, so Evelyn’s average speed is also greater than Carla’s.

(C) The smallest three-digit number divisible by 13 is $13\times 8=104$, so there are seven two-digit multiples of 13. The greatest three-digit number divisible by 13 is $13\times 76=988$. Therefore, there are $76-7=69$ three-digit numbers divisible by 13.

By the Pythagorean Theorem, $AE=\sqrt{30^2-24^2}=\sqrt{324}=18$. (Or note that triangle $AEB$ is similar to a 3-4-5 right triangle, so $AE=3\times 6=18$.) Also $CF=24$ and $FD=\sqrt{25^2-24^2}=\sqrt{49}=7$. The perimeter of the trapezoid is $50+30+18+50+7+25=180$.

(A) Write the points where Alice and Bob will stop after each move and compare points. \[ \begin{array}{c|ccccccc} \text{Move} & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text{Alice:} & 12 & 5 & 10 & 3 & 8 & 1 & 6 \\ \text{Bob:} & 12 & 3 & 6 & 9 & 12 & 3 & 6 \end{array} \] So Alice and Bob will be together again after six moves.

(C) To make a triangle, select as vertices two dots from one row and one from the other row. To select two dots in the top row, decide which dot is not used. This can be done in three ways. There are also three ways to choose one dot to use from the bottom row. So there are $3 \times 3 = 9$ triangles with two vertices in the top row and one in the bottom. Similarly, there are nine triangles with one vertex in the top row and two in the bottom. This gives a total of $9 + 9 = 18$ triangles. Note: Can you find the four noncongruent triangles?

(E) Neither the units of size nor the cost are important in this problem. So for convenience, suppose the small size costs \$1 and weighs 10 ounces. To determine the relative value, we compare the cost per unit weight. $S:\ \dfrac{\$1.00}{10}=10\text{¢ per oz.}$ $M:\ \dfrac{\$1.50}{0.8\times 20}=9.375\text{¢ per oz.}$ $L:\ \dfrac{1.3\times \$1.50}{20}=9.75\text{¢ per oz.}$ So the value, or buy, from best to worst is medium, large and small, that is MLS.

(B) Reflect the triangle and the semicircle across the hypotenuse $AB$ to obtain a circle inscribed in a square. The circle has area $4\pi$. The radius of a circle with area $4\pi$ is $2$. The side length of the square is $4$ and the area of the square is $16$. So the area of the triangle is $8$.

(B) One way to solve the problem is to work backward, either dividing by 2 if the number is even or subtracting 1 if the number is odd. $200/2 \rightarrow 100/2 \rightarrow 50/2 \rightarrow 25-1 \rightarrow 24/2 \rightarrow 12/2 \rightarrow 6/2 \rightarrow 3-1 \rightarrow 2/2 \rightarrow 1$ So if you press $[\times 2]\ [+1]\ [\times 2]\ [\times 2]\ [\times 2]\ [+1]\ [\times 2]\ [\times 2]\ [\times 2]$ or 9 keystrokes, you can reach “200” from “1.” To see that no sequence of eight keystrokes works, begin by noting that of the four possible sequences of two keystrokes, $[\times 2]\ [\times 2]$ produces the maximum result. Furthermore, $[+1]\ [\times 2]$ produces a result larger than either $[\times 2]\ [+1]$ or $[+1]\ [+1]$. So the largest possible result of a sequence of eight keystrokes is “256,” produced by either $[\times 2]\ [\times 2]\ [\times 2]\ [\times 2]\ [\times 2]\ [\times 2]\ [\times 2]\ [\times 2]$ or $[+1]\ [\times 2]\ [\times 2]\ [\times 2]\ [\times 2]\ [\times 2]\ [\times 2]\ [\times 2].$ The second largest result is “192,” produced by $[\times 2]\ [+1]\ [\times 2]\ [\times 2]\ [\times 2]\ [\times 2]\ [\times 2]\ [\times 2].$ Thus no sequence of eight keystrokes produces a result of “200.”

(A) Because the circle and square share the same interior region and the area of the two exterior regions indicated are equal, the square and the circle must have equal area. The area of the square is $2^2$ or $4$. Because the area of both the circle and the square is $4$, $4=\pi r^2$. Solving for $r$, the radius of the circle, yields $r^2=\frac{4}{\pi}$, so $r=\sqrt{\frac{4}{\pi}}=\frac{2}{\sqrt{\pi}}$. Note: It is not necessary that the circle and square have the same center.