Difference between revisions of "1989 AJHSME Problems/Problem 1"

(New page: ==Problem== <math>(1+11+21+31+41)+(9+19+29+39+49)=</math> <math>\text{(A)}\ 150 \qquad \text{(B)}\ 199 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 249 \qquad \text{(E)}\ 250</math> ==Solu...)
 
(Solution 2)
 
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&= 250 \Longrightarrow \boxed{\text{E}}
 
&= 250 \Longrightarrow \boxed{\text{E}}
 
\end{align*}</cmath>
 
\end{align*}</cmath>
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==Solution 2==
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Or, we can just combine the first, second, third, fourth, and last terms in each parentheses. This gives us:
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<cmath>(1+9)+(11+19)+(21+29)+(31+39)+(41+49),</cmath>
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which gives us
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<cmath>10+30+50+70+90 = 250 = \boxed{E}.</cmath>
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~DuoDuoling0
  
 
==See Also==
 
==See Also==
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{{AJHSME box|year=1989|before=First<br>Problem|num-a=2}}
 
{{AJHSME box|year=1989|before=First<br>Problem|num-a=2}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 13:21, 28 December 2021

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

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