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Difference between revisions of "1992 AJHSME Problems/Problem 1"
(→‎Solution)
Line 8: Line 8:
  
 
<cmath>\begin{align*}
 
<cmath>\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{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} \\
 
&= 1 \rightarrow \boxed{\text{B}}.
 
&= 1 \rightarrow \boxed{\text{B}}.

Revision as of 21:48, 2 November 2016

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 (Problems • Answer Key • Resources)
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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