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> | <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 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 = | + | From here it is obvious that <math>Y - X = 102 - 10 = \boxed{92 \ \mathbf{(A)}}</math>. |
===Note=== | ===Note=== | ||
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\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Like before, the difference between the two sequences is <math>Y-X=102-12= | + | Like before, the difference between the two sequences is <math>Y-X=102-12=92.</math> |
==Solution 2== | ==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: | 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 \\ | ||
+ | Y-X &= X-X+92 \\ | ||
+ | Y-X &= 0+92 \\ | ||
+ | Y-X &= \boxed{92} \quad \quad \textbf{(A)}\\ | ||
+ | \end{align*} </cmath> | ||
+ | <math>\blacksquare</math> | ||
+ | |||
+ | - <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 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. | ||
+ | - 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}} |
Revision as of 12:36, 11 September 2021
Contents
Problem
Let X and Y be the following sums of arithmetic sequences:
What is the value of
Solution 1
We see that both sequences have equal numbers of terms, so reformat the sequence to look like:
From here it is obvious that .
Note
Another way to see this is to let the sum So, the sequences become
Like before, the difference between the two sequences is
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:
Solution 3
-
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. - danfan
Video Solution
~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.