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

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\end{align*}</cmath>
 
\end{align*}</cmath>
 
From here it is obvious that Y - X = 102 - 10 = <math>\boxed{92 \ \mathbf{(A)}}</math>.
 
From here it is obvious that Y - X = 102 - 10 = <math>\boxed{92 \ \mathbf{(A)}}</math>.
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OR
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We see that every number in Y's sequence is two more than every corresponding number in X's sequence. Since there are 46 numbers in each sequence, the difference must be:
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<math>46*2=\boxed{92}</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 17:32, 4 February 2012

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 = $\boxed{92 \ \mathbf{(A)}}$.


OR


We see that every number in Y's sequence is two more than every corresponding number in X's sequence. Since there are 46 numbers in each sequence, the difference must be: $46*2=\boxed{92}$

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