Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (Added the section title)
(→‎Solution 3)
 
(18 intermediate revisions by 5 users not shown)
Line 4: Line 4:
 
<cmath> \begin{eqnarray*}X &=& 10+12+14+\cdots+100,\\ Y &=& 12+14+16+\cdots+102.\end{eqnarray*} </cmath>
 
<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?  
+
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>
Line 15: 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>.
+
From here it is obvious that <math>Y - X = 102 - 10 = \boxed{92 \ \mathbf{(A)}}</math>.
  
 
===Note===
 
===Note===
Line 28: Line 28:
 
==Solution 2==
 
==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:
+
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
 
<math>46\cdot 2=\boxed{92}</math>
 
<math>46\cdot 2=\boxed{92}</math>
 +
 +
==Solution 3==
 +
 +
<cmath>\begin{align*}
 +
X&=10+12+14+\cdots +100 \\
 +
Y&=X-10+102 = X+92 \\
 +
Y-X &= (X+92)-X \\
 +
    &= \boxed{92} \quad \quad \textbf{(A)}\\
 +
\end{align*} </cmath>
 +
 +
 +
- <math>\text{herobrine-india}</math>
 +
 +
==Solution 4==
 +
 +
In an actual contest, this would probably take too much time but is nevertheless a solution.
 +
The general formula for computing sums of any arithmetic sequence where <math>x</math> is the number of terms, <math>f</math> is the first term and <math>l</math> is the last term is <math>\frac{(f+l)x}{2}</math>. If one uses that formula for both sequences, they will get <math>2530</math> as the sum for <math>X</math> and <math>2622</math> as the sum for <math>Y</math>.
 +
Subtracting <math>X</math> from <math>Y</math>, one will get the answer <math>\boxed{92 \text{\textbf{ (A)}}}</math>.
 +
- danfan
 +
 +
==Video Solution==
 +
https://youtu.be/L6utIF9FzPQ
 +
 +
~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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 10:02, 17 June 2024

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 \\ &= \boxed{92} \quad \quad \textbf{(A)}\\ \end{align*}


- $\text{herobrine-india}$

Solution 4

In an actual contest, this would probably take too much time but is nevertheless a solution. The general formula for computing sums of any arithmetic sequence where $x$ is the number of terms, $f$ is the first term and $l$ is the last term is $\frac{(f+l)x}{2}$. If one uses that formula for both sequences, they will get $2530$ as the sum for $X$ and $2622$ as the sum for $Y$. Subtracting $X$ from $Y$, one will get the answer $\boxed{92 \text{\textbf{ (A)}}}$. - danfan

Video Solution

https://youtu.be/L6utIF9FzPQ

~savannahsolver

See Also

2011 AMC 10A (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_4&oldid=220655"