1989 AJHSME Problems/Problem 1

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{align*}

Solution 2

Or, we can just combine the first, second, third, fourth, and last terms in each parentheses. This gives us:

\[(1+9)+(11+19)+(21+29)+(31+39)+(41+49),\] which gives us \[10+30+50+70+90 = 250 = \boxed{E}.\]

~DuoDuoling0

See Also

1989 AJHSME (ProblemsAnswer KeyResources)
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
All AJHSME/AMC 8 Problems and Solutions

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