Difference between revisions of "2015 AMC 8 Problems/Problem 9"
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===Solution 1=== | ===Solution 1=== | ||
| + | First, we have to find how many widgets she makes on Day <math>20</math>. We can write the linear equation <math>y=-1+2x</math> to represent this situation. Then, we can plug in <math>20</math> for <math>x</math>: <math>y=-1+2(20)</math> -- <math>y=-1+40</math> -- <math>y=39</math> | ||
The sum of <math>1,3,5, ........ 39</math> is <math>\dfrac{(1 + 39)(20)}{2}= \boxed{\textbf{(D)}~400}</math> | The sum of <math>1,3,5, ........ 39</math> is <math>\dfrac{(1 + 39)(20)}{2}= \boxed{\textbf{(D)}~400}</math> | ||
Revision as of 23:27, 27 October 2020
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 
days?
Solution 1
First, we have to find how many widgets she makes on Day . We can write the linear equation 
to represent this situation. Then, we can plug in for 
: 
-- 
-- 
The sum of 
is 
Solution 2
The sum is just the sum of the first odd integers, which is 
See Also
| 2015 AMC 8 (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
