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Difference between revisions of "2013 AMC 8 Problems"

(Problem 13)
(Problem 14)
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<math> \textbf{(A)}\hspace{.05in}\qquad\textbf{(B)}\hspace{.05in}\qquad\textbf{(C)}\hspace{.05in}\qquad\textbf{(D)}\hspace{.05in}\qquad\textbf{(E)}\hspace{.05in} </math>
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The probability that both show a green bean is <math>\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}</math>. The probability that both show a red bean is <math>\frac{1}{2}\cdot\frac{2}{4}=\frac{1}{4}</math>. Therefore the probability is <math>\frac{1}{4}+\frac{1}{8}=\boxed{\textbf{(C)}\ \frac{3}{8}}</math>
  
 
[[2013 AMC 8 Problems/Problem 14|Solution]]
 
[[2013 AMC 8 Problems/Problem 14|Solution]]

Revision as of 13:20, 27 November 2013

Problem 1

Danica wants to arrange her model cars in rows with exactly 6 cars in each row. She now has 23 model cars. What is the smallest number of additional cars she must buy in order to be able to arrange all her cars this way?

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

Solution

Problem 2

A sign at the fish market says, "50% off, today only: half-pound packages for just $3 per package." What is the regular price for a full pound of fish, in dollars?

$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15$

Solution

Problem 3

What is the value of $4 \cdot (-1+2-3+4-5+6-7+\cdots+1000)$?

$\textbf{(A)}\ -10 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 500 \qquad \textbf{(E)}\ 2000$

Solution

Problem 4

Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill?

$\textbf{(A)}\ &#036;120\qquad\textbf{(B)}\ &#036;128\qquad\textbf{(C)}\ &#036;140\qquad\textbf{(D)}\ &#036;144\qquad\textbf{(E)}\ &#036;160$ (Error compiling LaTeX. Unknown error_msg)

Solution

Problem 5

Hammie is in the $6^\text{th}$ grade and weighs 106 pounds. His quadruplet sisters are tiny babies and weigh 5, 5, 6, and 8 pounds. Which is greater, the average (mean) weight of these five children or the median weight, and by how many pounds?

$\textbf{(A)}\ \text{median, by 60} \qquad \textbf{(B)}\ \text{median, by 20} \qquad \textbf{(C)}\ \text{average, by 5} \qquad \textbf{(D)}\ \text{average, by 15} \qquad \textbf{(E)}\ \text{average, by 20}$

Solution

Problem 6

The number in each box below is the product of the numbers in the two boxes that touch it in the row above. For example, $30 = 6\times5$. What is the missing number in the top row?

[asy] unitsize(0.8cm); draw((-1,0)--(1,0)--(1,-2)--(-1,-2)--cycle); draw((-2,0)--(0,0)--(0,2)--(-2,2)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((-3,2)--(-1,2)--(-1,4)--(-3,4)--cycle); draw((-1,2)--(1,2)--(1,4)--(-1,4)--cycle); draw((1,2)--(1,4)--(3,4)--(3,2)--cycle); label("600",(0,-1)); label("30",(-1,1)); label("6",(-2,3)); label("5",(0,3)); [/asy]

$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$

Solution

Problem 7

Trey and his mom stopped at a railroad crossing to let a train pass. As the train began to pass, Trey counted 6 cars in the first 10 seconds. It took the train 2 minutes and 45 seconds to clear the crossing at a constant speed. Which of the following was the most likely number of cars in the train?

$\textbf{(A)}\ 60 \qquad \textbf{(B)}\ 80 \qquad \textbf{(C)}\ 100 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 140$

Solution

Problem 8

A fair coin is tossed 3 times. What is the probability of at least two consecutive heads?

$\textbf{(A)}\ \frac18 \qquad \textbf{(B)}\ \frac14 \qquad \textbf{(C)}\ \frac38 \qquad \textbf{(D)}\ \frac12 \qquad \textbf{(E)}\ \frac34$

Solution

Problem 9

$\textbf{(A)}\hspace{.05in}\qquad\textbf{(B)}\hspace{.05in}\qquad\textbf{(C)}\hspace{.05in}\qquad\textbf{(D)}\hspace{.05in}\qquad\textbf{(E)}\hspace{.05in}$

Solution

Problem 10

$\textbf{(A)}\hspace{.05in}\qquad\textbf{(B)}\hspace{.05in}\qquad\textbf{(C)}\hspace{.05in}\qquad\textbf{(D)}\hspace{.05in}\qquad\textbf{(E)}\hspace{.05in}$

Solution

Problem 11

$\textbf{(A)}\hspace{.05in}\qquad\textbf{(B)}\hspace{.05in}\qquad\textbf{(C)}\hspace{.05in}\qquad\textbf{(D)}\hspace{.05in}\qquad\textbf{(E)}\hspace{.05in}$

Solution

Problem 12

$\textbf{(A)}\hspace{.05in}\qquad\textbf{(B)}\hspace{.05in}\qquad\textbf{(C)}\hspace{.05in}\qquad\textbf{(D)}\hspace{.05in}\qquad\textbf{(E)}\hspace{.05in}$

Solution

Problem 13

When Clara totaled her scores, she inadvertently reversed the units digit and the tens digit of one score. By which of the following might her incorrect sum have differed from the correct one?

$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 46 \qquad \textbf{(C)}\ 47 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 49$

Solution

Problem 14

The probability that both show a green bean is $\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}$. The probability that both show a red bean is $\frac{1}{2}\cdot\frac{2}{4}=\frac{1}{4}$. Therefore the probability is $\frac{1}{4}+\frac{1}{8}=\boxed{\textbf{(C)}\ \frac{3}{8}}$

Solution

Problem 15

$\textbf{(A)}\hspace{.05in}\qquad\textbf{(B)}\hspace{.05in}\qquad\textbf{(C)}\hspace{.05in}\qquad\textbf{(D)}\hspace{.05in}\qquad\textbf{(E)}\hspace{.05in}$

Solution

Problem 16

$\textbf{(A)}\hspace{.05in}\qquad\textbf{(B)}\hspace{.05in}\qquad\textbf{(C)}\hspace{.05in}\qquad\textbf{(D)}\hspace{.05in}\qquad\textbf{(E)}\hspace{.05in}$

Solution

Problem 17

$\textbf{(A)}\hspace{.05in}\qquad\textbf{(B)}\hspace{.05in}\qquad\textbf{(C)}\hspace{.05in}\qquad\textbf{(D)}\hspace{.05in}\qquad\textbf{(E)}\hspace{.05in}$

Solution

Problem 18

$\textbf{(A)}\hspace{.05in}\qquad\textbf{(B)}\hspace{.05in}\qquad\textbf{(C)}\hspace{.05in}\qquad\textbf{(D)}\hspace{.05in}\qquad\textbf{(E)}\hspace{.05in}$

Solution

Problem 19

$\textbf{(A)}\hspace{.05in}\qquad\textbf{(B)}\hspace{.05in}\qquad\textbf{(C)}\hspace{.05in}\qquad\textbf{(D)}\hspace{.05in}\qquad\textbf{(E)}\hspace{.05in}$

Solution

Problem 20

$\textbf{(A)}\hspace{.05in}\qquad\textbf{(B)}\hspace{.05in}\qquad\textbf{(C)}\hspace{.05in}\qquad\textbf{(D)}\hspace{.05in}\qquad\textbf{(E)}\hspace{.05in}$

Solution

Problem 21

$\textbf{(A)}\hspace{.05in}\qquad\textbf{(B)}\hspace{.05in}\qquad\textbf{(C)}\hspace{.05in}\qquad\textbf{(D)}\hspace{.05in}\qquad\textbf{(E)}\hspace{.05in}$

Solution

Problem 22

$\textbf{(A)}\hspace{.05in}\qquad\textbf{(B)}\hspace{.05in}\qquad\textbf{(C)}\hspace{.05in}\qquad\textbf{(D)}\hspace{.05in}\qquad\textbf{(E)}\hspace{.05in}$

Solution

Problem 23

$\textbf{(A)}\hspace{.05in}\qquad\textbf{(B)}\hspace{.05in}\qquad\textbf{(C)}\hspace{.05in}\qquad\textbf{(D)}\hspace{.05in}\qquad\textbf{(E)}\hspace{.05in}$

Solution

Problem 24

$\textbf{(A)}\hspace{.05in}\qquad\textbf{(B)}\hspace{.05in}\qquad\textbf{(C)}\hspace{.05in}\qquad\textbf{(D)}\hspace{.05in}\qquad\textbf{(E)}\hspace{.05in}$

Solution

Problem 25

$\textbf{(A)}\hspace{.05in}\qquad\textbf{(B)}\hspace{.05in}\qquad\textbf{(C)}\hspace{.05in}\qquad\textbf{(D)}\hspace{.05in}\qquad\textbf{(E)}\hspace{.05in}$

Solution The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png