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

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In an actual contest, this would probably take too much time but is nevertheless a solution.
 
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 ((f+l)/2)*x. If one uses that formula for both sequences, they will get 2530 as the sum for X and 2622 as the sum for Y.
 
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 ((f+l)/2)*x. 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 Y-X, one will get the answer, 92.
+
Subtracting Y-X, one will get the answer, 92 (A).
 
- danfan
 
- danfan
  

Revision as of 11:01, 19 October 2023

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*} $\blacksquare$

- $\text{herobrine-india}$

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 ((f+l)/2)*x. 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 Y-X, one will get the answer, 92 (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

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