Difference between revisions of "1992 AJHSME Problems/Problem 1"

(Created page with '==Problem== <math>\dfrac{10-9+8-7+6-5+4-3+2-1}{1-2+3-4+5-6+7-8+9}=</math> <math>\text{(A)}\ -1 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 9 \qquad \text{(E)}\…')
 
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\dfrac{10-9+8-7+6-5+4-3+2-1}{1-2+3-4+5-6+7-8+9} &= \dfrac{(10-9)+(8-7)+(6-5)+(4-3)+(2-1)}{1+(-2+3)+(-4+5)+(-6+7)+(-8+9)} \\
 
\dfrac{10-9+8-7+6-5+4-3+2-1}{1-2+3-4+5-6+7-8+9} &= \dfrac{(10-9)+(8-7)+(6-5)+(4-3)+(2-1)}{1+(-2+3)+(-4+5)+(-6+7)+(-8+9)} \\
 
&= \dfrac{1+1+1+1+1}{1+1+1+1+1} \\
 
&= \dfrac{1+1+1+1+1}{1+1+1+1+1} \\
&= \boxed{1}.
+
&= 1 \rightarrow \boxed{\text{B}}.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
  
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{{AJHSME box|year=1992|before=First<br />Problem|num-a=2}}
 
{{AJHSME box|year=1992|before=First<br />Problem|num-a=2}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 16:00, 9 November 2019

Problem

$\dfrac{10-9+8-7+6-5+4-3+2-1}{1-2+3-4+5-6+7-8+9}=$

$\text{(A)}\ -1 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 10$

Solution

\begin{align*} \dfrac{10-9+8-7+6-5+4-3+2-1}{1-2+3-4+5-6+7-8+9} &= \dfrac{(10-9)+(8-7)+(6-5)+(4-3)+(2-1)}{1+(-2+3)+(-4+5)+(-6+7)+(-8+9)} \\ &= \dfrac{1+1+1+1+1}{1+1+1+1+1} \\ &= 1 \rightarrow \boxed{\text{B}}. \end{align*}

See Also

1992 AJHSME (ProblemsAnswer KeyResources)
Preceded by
First
Problem
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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