Difference between revisions of "2015 AMC 8 Problems/Problem 9"

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<math>\textbf{(A) }39\qquad\textbf{(B) }40\qquad\textbf{(C) }210\qquad\textbf{(D) }400\qquad \textbf{(E) }401</math>
 
<math>\textbf{(A) }39\qquad\textbf{(B) }40\qquad\textbf{(C) }210\qquad\textbf{(D) }400\qquad \textbf{(E) }401</math>
  
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===Solution 1===
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The sum of <math>1,3,5, ........ 39</math> is <math>\dfrac{(1 + 39)(20)}{2}= \boxed{\textbf{(D)}~400}</math>
  
  

Revision as of 17:56, 25 November 2015

On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working $20$ days?

$\textbf{(A) }39\qquad\textbf{(B) }40\qquad\textbf{(C) }210\qquad\textbf{(D) }400\qquad \textbf{(E) }401$

Solution 1

The sum of $1,3,5, ........ 39$ is $\dfrac{(1 + 39)(20)}{2}= \boxed{\textbf{(D)}~400}$


See Also

2015 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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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