Difference between revisions of "2015 AMC 8 Problems/Problem 9"
Larryflora (talk | contribs) |
Larryflora (talk | contribs) |
||
Line 15: | Line 15: | ||
===Solution 3=== | ===Solution 3=== | ||
− | We can easily find out she makes <math>2*20-1 = 39</math> widgets on Day <math>20</math>. Then, we make the sum of <math>1,3, 5,......,35,37,39</math> by adding <math>(1+39)+(3+37)+(5+35)+...+(19+21)</math> which have 10 pairs <math>40</math>. So The sum of <math>1,3,5, ........ 39</math> is <math>(40*10) | + | We can easily find out she makes <math>2*20-1 = 39</math> widgets on Day <math>20</math>. Then, we make the sum of <math>1,3, 5,......,35,37,39</math> by adding <math>(1+39)+(3+37)+(5+35)+...+(19+21)</math> which have 10 pairs <math>40</math>. So The sum of <math>1,3,5, ........ 39</math> is is <math>(40*10)= \boxed{\textbf{(D)}~400}</math> ----LarryFlora |
==See Also== | ==See Also== |
Revision as of 13:37, 5 July 2021
Problem
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?
Solutions
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
Solution 3
We can easily find out she makes widgets on Day . Then, we make the sum of by adding which have 10 pairs . So The sum of is is ----LarryFlora
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.