Difference between revisions of "1996 AHSME Problems"

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==Problem 3==
 
==Problem 3==
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<math> \frac{(3!)!}{3!}= </math>
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<math> \text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 6\qquad\text{(D)}\ 40\qquad\text{(E)}\ 120 </math>
  
 
[[1996 AHSME Problems/Problem 3|Solution]]
 
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==Problem 4==
 
==Problem 4==
  
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Six numbers from a list of nine integers are <math>7,8,3,5,</math> and <math>9</math>. The largest possible value of the median of all nine numbers in this list is
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<math> \text{(A)}\ 5\qquad\text{(B)}\6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9 </math>
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[[1996 AHSME Problems/Problem 4|Solution]]
  
 
==Problem 5==
 
==Problem 5==
  
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Given that <math> 0 < a < b < c < d </math>, which of the following is the largest?
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<math> \text{(A)}\  \frac{a+b}{c+d} \qquad\text{(B)}\ \frac{a+d}{b+c} \qquad\text{(C)}\  \frac{b+c}{a+d} \qquad\text{(D)}\  \frac{b+d}{a+c} \qquad\text{(E)}\ \frac{c+d}{a+b} </math>
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[[1996 AHSME Problems/Problem 5|Solution]]
  
 
==Problem 6==
 
==Problem 6==
  
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If <math> f(x) = x^{(x+1)}(x+2)^{(x+3)} </math>, then <math> f(0)+f(-1)+f(-2)+f(-3) = </math>
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<math> \text{(A)}\ -\frac{8}{9}\qquad\text{(B)}\ 0\qquad\text{(C)}\ \frac{8}{9}\qquad\text{(D)}\ 1\qquad\text{(E)}\ \frac{10}{9} </math>
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==Problem 7==
 
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==Problem 9==
 
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==Problem 29==
 
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==Problem 30==
 
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Revision as of 19:16, 18 August 2011

Problem 1

The addition below is incorrect. What is the largest digit that can be changed to make the addition correct?

$\begin{tabular}{r}&\ \texttt{6 4 1}\\ \texttt{8 5 2} &+\texttt{9 7 3}\\ \hline \texttt{2 4 5 6}\end{tabular}$ (Error compiling LaTeX. Unknown error_msg)


$\text{(A)}\ 4\qquad\text{(B)}\ 5\qquad\text{(C)}\ 6\qquad\text{(D)}\ 7\qquad\text{(E)}\ 8$


Solution

Problem 2

Each day Walter gets $3$ dollars for doing his chores or $5$ dollars for doing them exceptionally well. After $10$ days of doing his chores daily, Walter has received a total of $36$ dollars. On how many days did Walter do them exceptionally well?

$\text{(A)}\ 3\qquad\text{(B)}\ 4\qquad\text{(C)}\ 5\qquad\text{(D)}\ 6\qquad\text{(E)}\ 7$

Solution

Problem 3

$\frac{(3!)!}{3!}=$

$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 6\qquad\text{(D)}\ 40\qquad\text{(E)}\ 120$

Solution

Problem 4

Six numbers from a list of nine integers are $7,8,3,5,$ and $9$. The largest possible value of the median of all nine numbers in this list is

$\text{(A)}\ 5\qquad\text{(B)}\6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$ (Error compiling LaTeX. Unknown error_msg)

Solution

Problem 5

Given that $0 < a < b < c < d$, which of the following is the largest?

$\text{(A)}\  \frac{a+b}{c+d} \qquad\text{(B)}\ \frac{a+d}{b+c} \qquad\text{(C)}\  \frac{b+c}{a+d} \qquad\text{(D)}\  \frac{b+d}{a+c} \qquad\text{(E)}\ \frac{c+d}{a+b}$

Solution

Problem 6

If $f(x) = x^{(x+1)}(x+2)^{(x+3)}$, then $f(0)+f(-1)+f(-2)+f(-3) =$

$\text{(A)}\ -\frac{8}{9}\qquad\text{(B)}\ 0\qquad\text{(C)}\ \frac{8}{9}\qquad\text{(D)}\ 1\qquad\text{(E)}\ \frac{10}{9}$

Solution

Problem 7

Solution

Problem 8

Solution

Problem 9

Solution

Problem 10

Solution

Problem 11

Solution

Problem 12

Solution

Problem 13

Solution

Problem 14

Solution

Problem 15

Solution

Problem 16

Solution

Problem 17

Solution

Problem 18

Solution

Problem 19

Solution

Problem 20

Solution

Problem 21

Solution

Problem 22

Solution

Problem 23

Solution

Problem 24

Solution

Problem 25

Solution

Problem 26

Solution

Problem 27

Solution

Problem 28

Solution

Problem 29

Solution

Problem 30

Solution