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

(Solution 1)
(Note)
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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>.
  
=Note=
+
===Note===
 
Another way to see this is to let the sum <math>12+14+16+...+100=x.</math> So, the sequences become
 
Another way to see this is to let the sum <math>12+14+16+...+100=x.</math> So, the sequences become
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}

Revision as of 22:25, 30 January 2018

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*}

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 1

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

Note

Another way to see this is to let the sum $12+14+16+...+100=x.$ So, the sequences become \begin{align*} X = 10+x \\ Y= x+102 \\ \end{align*}

Like before, the difference between the two sequences is $Y-X=102-12=90.$

Solution 2

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
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All AMC 10 Problems and Solutions

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