Difference between revisions of "2011 AMC 10A Problems/Problem 4"
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\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 Y - X = 102 - 10 = <math>\boxed{92 \ \mathbf{(A)}}</math>. | ||
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| + | OR | ||
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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: | ||
| + | <math>46*2=\boxed{92}</math> | ||
== 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}} | ||
Revision as of 16:32, 4 February 2012
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*} (Error compiling LaTeX. Unknown error_msg)
What is the value of Y - X?
Solution
We see that both sequences have equal numbers of terms, so reformat the sequence to look like:
From here it is obvious that Y - X = 102 - 10 = 
.
OR
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:
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 | ||
