2015 AMC 8 Problems/Problem 9

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 $20$ days?

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

Solutions

Solution 1

First, we have to find how many widgets she makes on Day $20$ . We can write the linear equation $y=-1+2x$ to represent this situation. Then, we can plug in $20$ for $x$ : $y=-1+2(20)$ -- $y=-1+40$ -- $y=39$ The sum of $1,3,5, ........ 39$ is $\dfrac{(1 + 39)(20)}{2}= \boxed{\textbf{(D)}~400}$

Solution 2

The sum is just the sum of the first $20$ odd integers, which is $20^2=\boxed{\textbf{(D)}~400}$

Solution 3

We can easily find out she makes $2\cdot20-1 = 39$ widgets on Day $20$ . Then, we make the sum of $1,3, 5,......,35,37,39$ by adding in this way: $(1+39)+(3+37)+(5+35)+...+(19+21)$ , which include $10$ pairs of $40$ . So the sum of $1,3,5, ........~39$ is $(40\cdot10)=\boxed{\textbf{(D)}~400}$ ----LarryFlora

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.

Page

Toolbox

Search