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

Latest revision as of 22:14, 10 June 2024

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 (Best)

First, we recognize that the number of widgets Janabel sells each day forms an arithmetic sequence. On day 1, she sells 1 widget, on day 2, she sells 3 widgets, on day 3, she sells 5 widgets, and so on. This sequence can be described by the first term $ a = 1 $ and the common difference $ d = 2 $.

Step 1: Find the number of widgets sold on day 20

The formula for the $ n $-th term of an arithmetic sequence is: $a_n = a + (n-1)d$ For $ n = 20 $: $a_{20} = 1 + (20-1) \cdot 2 = 1 + 38 = 39$

So, Janabel sells 39 widgets on day 20.

Step 2: Calculate the total number of widgets sold in 20 days}

The sum $ S_n $ of the first $ n $ terms of an arithmetic sequence is given by: $S_n = \frac{n}{2} \cdot (a + a_n)$

For $ n = 20 $: $S_{20} = \frac{20}{2} \cdot (1 + 39) = 10 \cdot 40 = 400$

Therefore, the total number of widgets sold after 20 days is: $(40\cdot10)=\boxed{\textbf{(D)}~400}$

~ GeometryMystery

Solution 2

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 3

The sum is just the first $20$ odd counting/natural numbers, which is $20^2=\boxed{\textbf{(D)}~400}$ .

Note: The sum of the first $x$ odd numbers is $x^2$ .

Solution 4

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

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/nw2v8sfT1i0

~Education, the Study of Everything

Video Solution

https://youtu.be/1xHX4p3YnFs

~savannahsolver

@@ Line 1: / Line 1: @@
+==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 <math>20</math> days?
 <math>\textbf{(A) }39\qquad\textbf{(B) }40\qquad\textbf{(C) }210\qquad\textbf{(D) }400\qquad \textbf{(E) }401</math>
+==Solutions==
+===Solution 1 (Best)===
+First, we recognize that the number of widgets Janabel sells each day forms an arithmetic sequence. On day 1, she sells 1 widget, on day 2, she sells 3 widgets, on day 3, she sells 5 widgets, and so on. This sequence can be described by the first term \( a = 1 \) and the common difference \( d = 2 \).
+Step 1: Find the number of widgets sold on day 20
+The formula for the \( n \)-th term of an arithmetic sequence is:
+<cmath> a_n = a + (n-1)d </cmath>
+For \( n = 20 \):
+<cmath> a_{20} = 1 + (20-1) \cdot 2 = 1 + 38 = 39 </cmath>
+So, Janabel sells 39 widgets on day 20.
+Step 2: Calculate the total number of widgets sold in 20 days}
+The sum \( S_n \) of the first \( n \) terms of an arithmetic sequence is given by:
+<cmath> S_n = \frac{n}{2} \cdot (a + a_n) </cmath>
+For \( n = 20 \):
+<cmath> S_{20} = \frac{20}{2} \cdot (1 + 39) = 10 \cdot 40 = 400 </cmath>
+Therefore, the total number of widgets sold after 20 days is:
+<math>(40\cdot10)=\boxed{\textbf{(D)}~400}</math>
+~ GeometryMystery
+===Solution 2===
+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>.
+===Solution 3===
+The sum is just the first <math>20</math> odd counting/natural numbers, which is <math>20^2=\boxed{\textbf{(D)}~400}</math>.
+Note: The sum of the first <math>x</math> odd numbers is <math>x^2</math>.
+===Solution 4===
+We can easily find out she makes <math>2\cdot20-1 = 39</math> widgets on Day <math>20</math>. Then, we make the sum of <math>1,3, 5, ... ,35,37,39</math> by adding in this way: <math>(1+39)+(3+37)+(5+35)+...+(19+21)</math>, which include <math>10</math> pairs of <math>40</math>. So, the sum of <math>1,3,5, ...~39</math> is <math>(40\cdot10)=\boxed{\textbf{(D)}~400}</math>.
+--LarryFlora
+==Video Solution (HOW TO THINK CRITICALLY!!!)==
+https://youtu.be/nw2v8sfT1i0
+~Education, the Study of Everything
+==Video Solution==
+https://youtu.be/1xHX4p3YnFs
+~savannahsolver
 ==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

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