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

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==Problem==
 
Let X and Y be the following sums of arithmetic sequences:  
 
Let X and Y be the following sums of arithmetic sequences:  
  
<cmath> \begin{eqnarray*}X &=& 10+12+14+\cdots+100,\\ Y &=& 12+14+16+\cdots+102.\end[eqnarray*} </cmath>
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<cmath> \begin{eqnarray*}X &=& 10+12+14+\cdots+100,\\ Y &=& 12+14+16+\cdots+102.\end{eqnarray*} </cmath>
  
What is the value of Y - X?  
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What is the value of <math>Y - X? </math>
  
 
<math> \textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112 </math>
 
<math> \textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112 </math>
  
==Solution==
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==Solution 1==
 
We see that both sequences have equal numbers of terms, so reformat the sequence to look like:  
 
We see that both sequences have equal numbers of terms, so reformat the sequence to look like:  
  
Line 14: Line 15:
 
X = 10 \ + \ &12 +  14 + \cdots + 100\\
 
X = 10 \ + \ &12 +  14 + \cdots + 100\\
 
\end{align*}</cmath>
 
\end{align*}</cmath>
From here it is obvious that Y - X = 102 - 10 = <math>\boxed{92 \ \mathbf{(A)}}</math>.
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From here it is obvious that <math>Y - X = 102 - 10 = \boxed{92 \ \mathbf{(A)}}</math>.
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===Note===
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Another way to see this is to let the sum <math>12+14+16+...+100=x.</math> So, the sequences become
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<cmath>\begin{align*}
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X = 10+x \\
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Y= x+102 \\
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\end{align*}</cmath>
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Like before, the difference between the two sequences is <math>Y-X=102-12=92.</math>
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==Solution 2==
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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\cdot 2=\boxed{92}</math>
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==Solution 3==
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<cmath>\begin{align*}
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X&=10+12+14+\cdots +100 \\
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Y&=X-10+102 = X+92 \\
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Y-X &= (X+92)-X \\
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Y-X &= X-X+92 \\
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Y-X &= 0+92 \\
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Y-X &= \boxed{92} \quad \quad \textbf{(A)}\\
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\end{align*} </cmath>
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<math>\blacksquare</math>
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- <math>\text{herobrine-india}</math>
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==Video Solution==
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https://youtu.be/L6utIF9FzPQ
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 +
~savannahsolver
  
 
== 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}}
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{{MAA Notice}}

Revision as of 20:02, 24 October 2020

Problem

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 $Y-X=102-12=92.$

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: $46\cdot 2=\boxed{92}$

Solution 3

\begin{align*} X&=10+12+14+\cdots +100 \\ Y&=X-10+102 = X+92 \\  Y-X &= (X+92)-X \\  Y-X &= X-X+92 \\  Y-X &= 0+92 \\  Y-X &= \boxed{92} \quad \quad \textbf{(A)}\\ \end{align*} $\blacksquare$

- $\text{herobrine-india}$

Video Solution

https://youtu.be/L6utIF9FzPQ

~savannahsolver

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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