AMC 8 2012
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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.
Rachelle uses $3$ pounds of meat to make $8$ hamburgers for her family. How many pounds of meat does she need to make $24$ hamburgers for a neighborhood picnic?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}6\dfrac{2}3\qquad\textbf{(C)}\hspace{.05in}7\dfrac{1}2\qquad\textbf{(D)}\hspace{.05in}8\qquad\textbf{(E)}\hspace{.05in}9$
$\textbf{E}$
Since the number of hamburgers she need to make is 3 times as usual, she need $3\times3=9$ pounds of meat.
In the country of East Westmore, statisticians estimate there is a baby born every $8$ hours and a death every day. To the nearest hundred, how many people are added to the population of East Westmore each year?
$\textbf{(A)}\hspace{.05in}600\qquad\textbf{(B)}\hspace{.05in}700\qquad\textbf{(C)}\hspace{.05in}800\qquad\textbf{(D)}\hspace{.05in}900\qquad\textbf{(E)}\hspace{.05in}1000$
$\textbf{B}$
Since there are $24/8=3$ births and 1 death every day, the population will increase by $(3-1)\times365=730\approx700$ per year.
On February 13 $\textit{The Oshkosh Northwester}$ listed the length of daylight as 10 hours and 24 minutes, the sunrise was $6:57\text{ AM}$, and the sunset as $8:15\text{ PM}$. The length of daylight and sunrise were correct, but the sunset was wrong. When did the sun really set?
$\textbf{B}$
The time of sunset is $6:57\text{ AM} + 10:24 \implies 17:21 \implies 5:21\text{ PM}$.
Peter's family ordered a 12-slice pizza for dinner. Peter ate one slice and shared another slice equally with his brother Paul. What fraction of the pizza did Peter eat?
$\textbf{(A)}\hspace{.05in}\dfrac{1}{24}\qquad\textbf{(B)}\hspace{.05in}\dfrac{1}{12}\qquad\textbf{(C)}\hspace{.05in}\dfrac{1}{8}\qquad\textbf{(D)}\hspace{.05in}\dfrac{1}{6}\qquad\textbf{(E)}\hspace{.05in}\dfrac{1}{4}$
$\textbf{C}$
Peter ate 1.5 slices, which is $\dfrac{1.5}{12}=\dfrac{1}{8}$ of the pizza.
In the diagram, all angles are right angles and the lengths of the sides are given in centimeters. Note the diagram is not drawn to scale. What is the length in $X$, in centimeters?
$\textbf{(A)}\hspace{.05in}1\qquad\textbf{(B)}\hspace{.05in}2\qquad\textbf{(C)}\hspace{.05in}3\qquad\textbf{(D)}\hspace{.05in}4\qquad\textbf{(E)}\hspace{.05in}5$
$\textbf{E}$
The height of the figure is $1+1+1+2+X=X+5$ from the left side , and $1+2+1+6=10$ from the right side. Hence, we have $X+5=10\rightarrow X=5$.
A rectangular photograph is placed in a frame that forms a border two inches wide on all sides of the photograph. The photograph measures 8 inches high and 10 inches wide. What is the area of the border, in square inches?
$\textbf{(A)}\hspace{.05in}36\qquad\textbf{(B)}\hspace{.05in}40\qquad\textbf{(C)}\hspace{.05in}64\qquad\textbf{(D)}\hspace{.05in}72\qquad\textbf{(E)}\hspace{.05in}88$
$\textbf{E}$
In order to find the area of the frame, we need to subtract the area of the photograph from the area of the photograph and the frame together. The area of the photograph is $8 \times 10 = 80$ square inches. The height of the whole frame (including the photograph) would be $8+2+2 = 12$ inches, and the width of the whole frame, $10+2+2 = 14$ inches. Therefore, the area of the whole figure would be $12 \times 14 = 168$ square inches. Subtracting the area of the photograph from the area of both the frame and photograph, we find the answer to be $168-80 = 88$.
Isabella must take four 100-point tests in her math class. Her goal is to achieve an average grade of 95 on the tests. Her first two test scores were 97 and 91. After seeing her score on the third test, she realized she can still reach her goal. What is the lowest possible score she could have made on the third test?
$\textbf{(A)}\hspace{.05in}90\qquad\textbf{(B)}\hspace{.05in}92\qquad\textbf{(C)}\hspace{.05in}95\qquad\textbf{(D)}\hspace{.05in}96\qquad\textbf{(E)}\hspace{.05in}97$
$\textbf{B}$
If she really gets the lowest possible score of the third test, she must get 100 score on the fourth test to achieve her goal of an average of 95. Hence, the score of the third test is $95\times4-97-91-100=92$.
A shop advertises everything is "half price in today's sale." In addition, a coupon gives a $20\%$ discount on sale prices. Using the coupon, the price today represents what percentage off the original price?
$\textbf{(A)}\hspace{.05in}10\qquad\textbf{(B)}\hspace{.05in}33\qquad\textbf{(C)}\hspace{.05in}40\qquad\textbf{(D)}\hspace{.05in}60\qquad\textbf{(E)}\hspace{.05in}70$
$\textbf{D}$
A coupon makes the sale price $1-20\%=80\%$ as usual. So the sale price on the half-price day is $80\%\times50\%=40\%$ of the original price, or we can say $60\%$ off the original price.
The Fort Worth Zoo has a number of two-legged birds and a number of four-legged mammals. On one visit to the zoo, Margie counted 200 heads and 522 legs. How many of the animals that Margie counted were two-legged birds?
$\textbf{(A)}\hspace{.05in}61\qquad\textbf{(B)}\hspace{.05in}122\qquad\textbf{(C)}\hspace{.05in}139\qquad\textbf{(D)}\hspace{.05in}150\qquad\textbf{(E)}\hspace{.05in}161$
$\textbf{C}$
Let the number of two-legged birds be $x$ and the number of four-legged mammals be $y$. Now we have
\begin{align*}
2x + 4y &= 522\\x + y &= 200
\end{align*}
Hence, the number of two-legged birds is $x=139$.
How many 4-digit numbers greater than 1000 are there that use the four digits of 2012?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}7\qquad\textbf{(C)}\hspace{.05in}8\qquad\textbf{(D)}\hspace{.05in}9\qquad\textbf{(E)}\hspace{.05in}12$
$\textbf{D}$
Since 0 can not be the first digit, we have 3 ways to arrange the position of 0. Then we have 3 ways to arrange the digit 1. When the positions of digit 0 and 1 are fixed, the rest two digits 2 are fixed either. Hence, there are $3\times3=9$ numbers for the answer.
The mean, median, and unique mode of the positive integers 3, 4, 5, 6, 6, 7, and $x$ are all equal. What is the value of $x$?
$\textbf{(A)}\hspace{.05in}5\qquad\textbf{(B)}\hspace{.05in}6\qquad\textbf{(C)}\hspace{.05in}7\qquad\textbf{(D)}\hspace{.05in}11\qquad\textbf{(E)}\hspace{.05in}12$
$\textbf{D}$
Since the mode is unique, we get that the mean, median and mode must be 6. Hence, we have $$\dfrac{3+4+5+6+6+7+x}{7}=6\rightarrow x=11$$
What is the units digit ( ones place digit ) of $13^{2012}$?
$\textbf{(A)}\hspace{.05in}1\qquad\textbf{(B)}\hspace{.05in}3\qquad\textbf{(C)}\hspace{.05in}5\qquad\textbf{(D)}\hspace{.05in}7\qquad\textbf{(E)}\hspace{.05in}9$
$\textbf{A}$
Since we only care about the units digit, we need to find the pattern of the units digit of $3^n$.
\begin{align*}
3^1 \implies 3\\
3^2 \implies 9\\
3^3 \implies 7\\
3^4 \implies 1\\
3^5 \implies 3
\end{align*}
The units digit recurs when the power increased by 4. Hence, we know that the units digit of $3^{2012}$ is the same as $3^0=1$.
Jamar bought some pencils costing more than a penny each at the school bookstore and paid $\$1.43$. Sharona bought some of the same pencils and paid $\$1.87$. How many more pencils did Sharona buy than Jamar?
$\textbf{(A)}\hspace{.05in}2\qquad\textbf{(B)}\hspace{.05in}3\qquad\textbf{(C)}\hspace{.05in}4\qquad\textbf{(D)}\hspace{.05in}5\qquad\textbf{(E)}\hspace{.05in}6$
$\textbf{C}$
By prime factorization of 143 and 187, we get $143=11\times13$ and $187=11\times17$. Therefore, 1 and 11 are the common factors of 143 and 187. Since the price of each pencil is more than 1 penny, we know that the price of a pencil is 11 pennies. Hence, Jamar bought 13 pencils while Sharona bought 17, which is 4 more than Jamar.
In the BIG N, a middle school football conference, each team plays every other team exactly once. If a total of 21 conference games were played during the 2012 season, how many teams were members of the BIG N conference?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}7\qquad\textbf{(C)}\hspace{.05in}8\qquad\textbf{(D)}\hspace{.05in}9\qquad\textbf{(E)}\hspace{.05in}10$
$\textbf{B}$
Let the number of teams be $x$. Since each team plays with every other team once, we have $$\dbinom{x}{2}=\dfrac{x(x-1)}2=21\rightarrow x=7$$
The smallest number greater than 2 that leaves a remainder of 2 when divided by 3, 4, 5, or 6 lies between what numbers?
$\textbf{(A)}\hspace{.05in}40\text{ and }50\qquad\textbf{(B)}\hspace{.05in}51\text{ and }55\qquad\textbf{(C)}\hspace{.05in}56\text{ and }60\qquad\textbf{(D)}\hspace{.05in}\text{61 and 65}\qquad\textbf{(E)}\hspace{.05in}\text{66 and 99}$
$\textbf{D}$
If the number subtracts 2, the new number will be a multiple of 3, 4, 5, and 6. The least common multiple of 3, 4, 5, 6 is 60. Hence, the smallest number we want is $60+2=62$, which lies between 61 and 65.
Each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 is used only once to make two five-digit numbers so that they have the largest possible sum. Which of the following could be one of the numbers?
$\textbf{(A)}\hspace{.05in}76531\qquad\textbf{(B)}\hspace{.05in}86724\qquad\textbf{(C)}\hspace{.05in}87431\qquad\textbf{(D)}\hspace{.05in}96240\qquad\textbf{(E)}\hspace{.05in}97403$
$\textbf{C}$
In order to maximum the sum, 9 and 8 should be the first digits of the two numbers. Similarly, 7 and 6 should be the second digits of the two numbers, 5 and 4 be the third, 3 and 2 be the fourth, then 1 and 0 be the last. Hence, 87431 is a possible number.
A square with integer side length is cut into 10 squares, all of which have integer side length and at least 8 of which have area 1. What is the smallest possible value of the length of the side of the original square?
$\textbf{(A)}\hspace{.05in}3\qquad\textbf{(B)}\hspace{.05in}4\qquad\textbf{(C)}\hspace{.05in}5\qquad\textbf{(D)}\hspace{.05in}6\qquad\textbf{(E)}\hspace{.05in}7$
$\textbf{B}$
The sum of the areas of the 10 squares is no less than 10. Hence, the side length of the original square can not be 3. If the side length is 4, we can divide the original square into eight squares of area 1 and two squares of area 4. So the smallest value of the side length is 4.
What is the smallest positive integer that is neither prime nor square and that has no prime factor less than 50?
$\textbf{(A)}\hspace{.05in}3127\qquad\textbf{(B)}\hspace{.05in}3133\qquad\textbf{(C)}\hspace{.05in}3137\qquad\textbf{(D)}\hspace{.05in}3139\qquad\textbf{(E)}\hspace{.05in}3149$
$\textbf{A}$
The list of primes greater than 50 is $\{53,59,\dots\}$. Hence, the smallest positive integer is $53\times59=3127$.
In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}8\qquad\textbf{(C)}\hspace{.05in}9\qquad\textbf{(D)}\hspace{.05in}10\qquad\textbf{(E)}\hspace{.05in}12$
$\textbf{C}$
According to the given information, we know that the sum of blue and green is 6, the sum of blue and red is 8, and the sum of red and green is 4. By adding them together, we get twice the sum of red, green, and blue as $6+8+4=18$. So the sum of red, green, and blue is 9.
What is the correct ordering of the three numbers $\dfrac{5}{19}$, $\dfrac{7}{21}$, and $\dfrac{9}{23}$, in increasing order?
$\textbf{(A)}\hspace{.05in}\dfrac{9}{23}<\dfrac{7}{21}<\dfrac{5}{19}\quad\newline$
$\textbf{(B)}\hspace{.05in}\dfrac{5}{19}<\dfrac{7}{21}<\dfrac{9}{23}\quad\newline$
$\textbf{(C)}\hspace{.05in}\dfrac{9}{23}<\dfrac{5}{19}<\dfrac{7}{21}\newline$
$\textbf{(D)}\hspace{.05in}\dfrac{5}{19}<\dfrac{9}{23}<\dfrac{7}{21}\quad\newline$
$\textbf{(E)}\hspace{.05in}\dfrac{7}{21}<\dfrac{5}{19}<\dfrac{9}{23}$
$\textbf{B}$
We can rewrite these numbers as $\dfrac{5}{19}=1-\dfrac{14}{19}$, $\dfrac{7}{21}=1-\dfrac{14}{21}$, $\dfrac{9}{23}=1-\dfrac{14}{23}$. Since $$\dfrac{14}{19}>\dfrac{14}{21}>\dfrac{14}{23}$$ we have $$1-\dfrac{14}{19}<1-\dfrac{14}{21}<1-\dfrac{14}{23}$$ or $$\dfrac{5}{19}<\dfrac{7}{21}<\dfrac{9}{23}$$
Marla has a large white cube that has an edge of 10 feet. She also has enough green paint to cover 300 square feet. Marla uses all the paint to create a white square centered on each face, surrounded by a green border. What is the area of one of the white squares, in square feet?
$\textbf{(A)}\hspace{.05in}5\sqrt2\qquad\textbf{(B)}\hspace{.05in}10\qquad\textbf{(C)}\hspace{.05in}10\sqrt2\qquad\textbf{(D)}\hspace{.05in}50\qquad\textbf{(E)}\hspace{.05in}50\sqrt2$
$\textbf{D}$
The surface area of the cube is $6\times10\times10=600$ square feet. So the area of the white square on each face is $(600-300)/6=50$ square feet.
Let $R$ be a set of nine distinct integers. Six of the elements are 2, 3, 4, 6, 9, and 14. What is the number of possible values of the median of $R$ ?
$\textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}7\qquad\textbf{(E)}\hspace{.05in}8$
$\textbf{D}$
The median of 9 integers is the 5th number. In extreme case when the rest 3 numbers are all less than 2, the median is 3. Similarly, when the rest 3 numbers are all greater than 14, the median is 9. The number of integers between 3 and 9 (including 3 and 9 )is 7. Hence, the median has 7 possible values.
An equilateral triangle and a regular hexagon have equal perimeters. If the area of the triangle is 4, what is the area of the hexagon?
$\textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}4\sqrt3\qquad\textbf{(E)}\hspace{.05in}6\sqrt3$
$\textbf{C}$
Six equilateral triangles of the same side length can form a regular hexagon of twice the perimeter and 6 times the area of the triangle. Now we halve the perimeter of the hexagon, the area will be 1/4 as much as before, which is still 1.5 times of the area of the triangle. Since the area of the triangle is 4, the area of the hexagon is $4\times1.5=6$.
A circle of radius 2 is cut into four congruent arcs. The four arcs are joined to form the star figure shown. What is the ratio of the area of the star figure to the area of the original circle?
$\textbf{(A)}\hspace{.05in}\dfrac{4-\pi}{\pi}\qquad\textbf{(B)}\hspace{.05in}\dfrac{1}\pi\qquad\textbf{(C)}\hspace{.05in}\dfrac{\sqrt2}{\pi}\qquad\textbf{(D)}\hspace{.05in}\dfrac{\pi-1}{\pi}\qquad\textbf{(E)}\hspace{.05in}\dfrac{3}\pi$
$\textbf{A}$
Let the radius of the circle be 1. Then the area of the circle is $\pi$. The area of the star is the area of the red square subtracts the area of the circle as $2^2-\pi=4-\pi$. So the ratio of the star to the circle is $\dfrac{4-\pi}{\pi}$.
A square with area 4 is inscribed in a square with area 5, with one vertex of the smaller square on each side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length $a$, and the other of length $b$. What is the value of $ab$?
$\textbf{(A)}\hspace{.05in}\dfrac{1}5\qquad\textbf{(B)}\hspace{.05in}\dfrac{2}5\qquad\textbf{(C)}\hspace{.05in}\dfrac{1}2\qquad\textbf{(D)}\hspace{.05in}1\qquad\textbf{(E)}\hspace{.05in}4$
$\textbf{C}$
The area of the larger square is the sum of the smaller square and 4 congruent triangles. So the area of each of the 4 congruent triangles is $$\dfrac{5-1}4=\dfrac14=\dfrac12ab\rightarrow ab=\dfrac12$$