AMC8 1999

Simplifying the given expression, we get: $(6?3)=2.$ At this point, it becomes clear that it should be $(\textrm{A})$ $\div$.

At $10:00$, the hour hand will be on the $10$ while the minute hand on the $12$. This makes them $\frac{1}{6}$th of a circle apart, and $\frac{1}{6}\cdot360^{\circ}=\boxed{\textbf{(C) } 60}$.

By adding each triplet, we can see that $\boxed{(D)}$ gives us $0$, not $1$, as our sum.

After 4 hours, we see that Bjorn biked 45 miles, and Alberto biked 60. Thus the answer is $60-45=15$ $\boxed{\text{(A)}}$.

We need the same perimeter as a $60$ by $20$ rectangle, but the greatest area we can get. right now the perimeter is $160$. To get the greatest area while keeping a perimeter of $160$, the sides should all be $40$. that means an area of $1600$. Right now, the area is $20 \times 60$ which is $1200$. $1600-1200=400$ which is $\boxed{D}$.

Use logic to solve this problem. You don't actually need to use any equations. Neither Jo nor Bo has as much money as Flo. So Flo clearly does not have the least amount of money. Rule out Flo. Both Bo and Coe have more than Moe. Rule out Bo and Coe; they clearly do not have the least amount of money. Jo has more than Moe. Rule out Jo. The only person who has not been ruled out is Moe. So $\boxed{\text{(E)}}$ is the answer.

There are $160-40=120$ miles between the third and tenth exits, so the service center is at milepost $40+(3/4)(120) = 40+90=\boxed{\text{(E)}\ 130}$.

When G is arranged to be the base, B is the back face and W is the front face. Thus, $\boxed{\text{(A)}\ B}$ is opposite W.

Plants shared by two beds have been counted twice, so the total is $500 + 450 + 350 - 50 - 100 = \boxed{\text{(C)}\ 1150}$. Bed A has $350$ plants it doesn't share with B or C. Bed B has $400$ plants it doesn't share with A or C. And C has $250$ it doesn't share with A or B. The total is $350 + 400 + 250 + 50 + 100 = \boxed{\text{(C)}\ 1150}$ plants.

\[\frac{\text{time not green}}{\text{total time}} = \frac{R + Y}{R + Y + G} = \frac{35}{60} = \boxed{\text{(E)}\ \frac{7}{12}}\] The probability of green is $\frac{25}{60} = \frac{5}{12}$, so the probability of not green is $1- \frac{5}{12} = \boxed{\text{(E)}\ \frac{7}{12}}$ .

The largest sum occurs when $13$ is placed in the center. This sum is $13 + 10 + 1 = 13 + 7 + 4 = \boxed{\text{(D)}\ 24}$. Note: Two other common sums, $18$ and $21$, are also possible. Since the horizontal sum equals the vertical sum, twice this sum will be the sum of the five numbers plus the number in the center. When the center number is $13$, the sum is the largest, \[[10 + 4 + 1 + 7 + 2(13)]=2S\\ 48=2S\\ S=\boxed{\text{(D)}\ 24}\] The other four numbers are divided into two pairs with equal sums.

The ratio means that for every $11$ games won, $4$ are lost, so the team has won $11x$ games, lost $4x$ games, and played $15x$ games for some positive integer $x$. The percentage of games lost is just $\dfrac{4x}{15x}\times100=\dfrac{4}{15}\times 100=26.\overline{6}\%\approx\boxed{\text{(B)}\ 27\%}$

First, find the total amount of the girl's ages and add it to the total amount of the boy's ages. It equals $(20)(15)+(15)(16)=540$. The total amount of everyone's ages can be found from the average age, $17\cdot40=680$. Then you do $680-540=140$ to find the sum of the adult's ages. The average age of an adult is divided among the five of them, $140\div5=\boxed{\text{(C)}\ 28}$.

There is a rectangle present, with both horizontal bases being $8$ units in length. The excess units on the bottom base must then be $16-8=8$. The fact that $AB$ and $CD$ are equal in length indicate, by the Pythagorean Theorem, that these excess lengths are equal. There are two with a total length of $8$ units, so each is $4$ units. The triangle has a hypotenuse of $5$, because the triangles are $3-4-5$ right triangles. So, the sides of the trapezoid are $8$, $5$, $16$, and $5$. Adding those up gives us the perimeter, $8 + 5 + 16 + 5 = 13 + 21 = \boxed{\text{(D)}\ 34}$ units.

Answer (D): Before new letters were added, five different letters could have been chosen for the first position, three for the second, and four for the third. This means that $5 \cdot 3 \cdot 4 = 60$ plates could have been made. If two letters are added to the second set, then $5 \cdot 5 \cdot 4 = 100$ plates can be made. If one letter is added to each of the second and third sets, then $5 \cdot 4 \cdot 5 = 100$ plates can be made. None of the other four ways to place the two letters will create as many plates. So, $100 - 60 = 40$ ADDITIONAL plates can be made. Note: Optimum results can usually be obtained in such problems by making the factors as nearly equal as possible.

Answer (B): Since 70%(10) + 40%(30) + 60%(35) = 7 + 12 + 21 = 40, she answered 40 questions correctly. She needed 60%(75) = 45 to pass, so she needed 5 more correct answers.

Answer (C): One recipe makes $15$ cookies, so $216 \div 15 = 14.4$ recipes are needed, but this must be rounded up to $15$ recipes to make enough cookies. Each recipe requires $2$ eggs. So $30$ eggs are needed. This is $5$ half-dozens.

Answer (E): The 108(0.75) = 81 students need 2 cookies each so 162 cookies are to be baked. Since 162 ÷ 15 = 10.8, Walter and Gretel must bake 11 recipes. A few leftovers are a good thing!

Answer (B): Since $216 \div 15 = 14.4$, they will have to bake 15 recipes. This requires $15 \times 3 = 45$ tablespoons of butter. So, $45 \div 8 = 5.625$, and 6 sticks are needed.

Answer (B): The front view shows the larger of the numbers of cubes in the front or back stack in each column. Therefore the desired front view will have, from left to right, 2, 3, and 4 cubes. This is choice B.

Answer (B): Since $\angle 1$ forms a straight line with angle $100^\circ$, $\angle 1 = 80^\circ$. Since $\angle 2$ forms a straight line with angle $110^\circ$, $\angle 2 = 70^\circ$. Angle $3$ is the third angle in a triangle with $\angle E = 40^\circ$ and $\angle 2 = 70^\circ$, so $\angle 3 = 180^\circ - 40^\circ - 70^\circ = 70^\circ$. Angle $4 = 110^\circ$ since it forms a straight angle with $\angle 3$. Then $\angle 5$ forms a straight angle with $\angle 4$, so $\angle 5 = 70^\circ$. (Or $\angle 3 = \angle 5$ because they are vertical angles.) Therefore, $\angle A = 180^\circ - \angle 1 - \angle 5 = 180^\circ - 80^\circ - 70^\circ = 30^\circ$.

Answer (D): One fish is worth $\frac{2}{3}$ of a loaf of bread and $\frac{2}{3}$ of a loaf of bread is worth $\frac{2}{3}\cdot 4=\frac{8}{3}=2\frac{2}{3}$ bags of rice. OR $3F=2B$ $\frac{3}{2}F=B=4R$ $\left(\frac{2}{3}\right)\left(\frac{3}{2}\right)F=\frac{2}{3}(4R)$ $F=\frac{8}{3}R=2\frac{2}{3}R.$

Answer (C): One-third of the square's area is 3, so triangle $MBC$ has area $3=\frac{1}{2}(MB)(BC)$. Since side $BC$ is 3, side $MB$ must be 2. The hypotenuse $CM$ of this right triangle is $\sqrt{2^2+3^2}=\sqrt{13}$.

Answer (D): Since any positive integer (expressed in base ten) is some multiple of 5 plus its last digit, its remainder when divided by 5 can be obtained by knowing its last digit. Note that $1999^1$ ends in 9, $1999^2$ ends in 1, $1999^3$ ends in 9, $1999^4$ ends in 1, and this alternation of 9 and 1 endings continues with all even powers ending in 1. Therefore, the remainder when $1999^{2000}$ is divided by 5 is 1.

Answer (A): At each stage the area of the shaded triangle is one-third of the trapezoidal region not containing the smaller triangle being divided in the next step. Thus, the total area of the shaded triangles comes closer and closer to one-third of the area of the triangular region $ACG$ and this is $\frac{1}{3}\cdot\frac{1}{2}\cdot 6\cdot 6=6$. The shaded areas for the first six stages are: 4.5, 5.625, 5.906, 5.976, 5.994, and 5.998. These are the calculations for the first three steps. $\frac{1}{2}\cdot\frac{6}{2}\cdot\frac{6}{2}=4.5$ $\frac{1}{2}\cdot\frac{6}{2}\cdot\frac{6}{2}+\frac{1}{2}\cdot\frac{6}{4}\cdot\frac{6}{4}=4.5+1.125=5.625$ $\frac{1}{2}\cdot\frac{6}{2}\cdot\frac{6}{2}+\frac{1}{2}\cdot\frac{6}{4}\cdot\frac{6}{4}+\frac{1}{2}\cdot\frac{6}{8}\cdot\frac{6}{8}=5.625+0.281=5.906$