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1989 AJHSME Problems/Problem 1

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Problem

$(1+11+21+31+41)+(9+19+29+39+49)=$

$\text{(A)}\ 150 \qquad \text{(B)}\ 199 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 249 \qquad \text{(E)}\ 250$

Solution

We make use of the associative and commutative properties of addition to rearrange the sum as $\begin{align*}(1+49)+(11+39)+(21+29)+(31+19)+(41+9) &= 50+50+50+50+50 \\&= 250 \Longrightarrow \boxed{\text{E}}\end{a...$

See Also

1989 AJHSME (Problems • Resources)
Preceded by First Problem	Followed by Problem 2
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

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/1989_AJHSME_Problems/Problem_1"

Category: Introductory Algebra Problems

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