Difference between revisions of "2011 AMC 10A Problems/Problem 4"

m (\mathbf{Solution})
(Solution)
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X = 10 \ + \ &12 +  14 + \cdots + 100\\
 
X = 10 \ + \ &12 +  14 + \cdots + 100\\
 
\end{align*}</cmath>
 
\end{align*}</cmath>
From here it is obvious that <math>Y - X = 102 - 10 = 92</math>
+
From here it is obvious that Y - X = 102 - 10 = <math>\boxed{92 \ \mathbf{(A)}}</math>.
 +
 
 
== See Also ==
 
== See Also ==
 
{{AMC10 box|year=2011|ab=A|num-b=3|num-a=5}}
 
{{AMC10 box|year=2011|ab=A|num-b=3|num-a=5}}

Revision as of 18:35, 9 September 2011

Let X and Y be the following sums of arithmetic sequences:

\begin{eqnarray*}X &=& 10+12+14+\cdots+100,\\ Y &=& 12+14+16+\cdots+102.\end[eqnarray*} (Error compiling LaTeX. Unknown error_msg)

What is the value of Y - X?

$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112$

Solution

We see that both sequences have equal numbers of terms, so reformat the sequence to look like:

\begin{align*} Y = \ &12 + 14 + \cdots + 100 + 102\\ X = 10 \ + \ &12 +  14 + \cdots + 100\\ \end{align*} From here it is obvious that Y - X = 102 - 10 = $\boxed{92 \ \mathbf{(A)}}$.

See Also

2011 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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 AMC 10 Problems and Solutions