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

Revision as of 13:09, 25 May 2012

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

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

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

Page

Toolbox

Search

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

Revision as of 13:09, 25 May 2012

Problem

Solution

See Also

@@ Line 10: / Line 10: @@
 \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{A}}.
+&= 1 \rightarrow \boxed{\text{B}}.
 \end{align*}</cmath>