Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2005 AMC 8 Problems/Problem 12"
(→‎Solution 3 (Lightning Quick))
 
(10 intermediate revisions by 2 users not shown)
Line 12: Line 12:
 
n&=\boxed{\textbf{(D)}\ 32}
 
n&=\boxed{\textbf{(D)}\ 32}
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 +
 +
==Solution 2==
 +
There are <math>5</math> days from May 1 to May 5. If we set the first day as <math>n</math>, the second day can be expressed as <math>n+6</math>, the third as <math>n+12</math>, and so on, for five days.
 +
The sum <math>n+(n+6)+(n+12)+(n+18)+(n+24)</math> is equal to <math>100</math>, as stated in the problem. We can write a very simple equation, that is: <math>5n+60=100</math>.
 +
Now all we do is just solve. <math>5n=40</math>, so Big Al eats <math>8</math> bananas on the first day. The fifth day, <math>n+24</math>, is then <math>32</math>, which is your answer.
 +
 +
==Solution 3 (Lightning Quick)==
 +
 +
Simply realize that the middle term of the arithmetic sequence is the arithmetic mean of all terms, which is simply <math>\frac{100}{5}=20</math>. This means that the number of bananas the ape ate on May 5th is just <math>20+6*2=32</math>. Select <math>\boxed{D}</math>.
 +
 +
You can do this quicker than you read it if you truly master arithmetic sequences.
 +
 +
~hastapasta
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2005|num-b=11|num-a=13}}
 
{{AMC8 box|year=2005|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 10:48, 4 May 2022

Problem

Big Al, the ape, ate 100 bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many bananas did Big Al eat on May 5?

$\textbf{(A)}\ 20\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34$

Solution

There are $5$ days from May 1 to May 5. The number of bananas he eats each day is an arithmetic sequence. He eats $n$ bananas on May 5, and $n-4(6)=n-24$ bananas on May 1. The sum of this arithmetic sequence is equal to $100$.

\begin{align*} \frac{n+n-24}{2} \cdot 5 &= 100\\ n-12&=20\\ n&=\boxed{\textbf{(D)}\ 32} \end{align*}

Solution 2

There are $5$ days from May 1 to May 5. If we set the first day as $n$, the second day can be expressed as $n+6$, the third as $n+12$, and so on, for five days. The sum $n+(n+6)+(n+12)+(n+18)+(n+24)$ is equal to $100$, as stated in the problem. We can write a very simple equation, that is: $5n+60=100$. Now all we do is just solve. $5n=40$, so Big Al eats $8$ bananas on the first day. The fifth day, $n+24$, is then $32$, which is your answer.

Solution 3 (Lightning Quick)

Simply realize that the middle term of the arithmetic sequence is the arithmetic mean of all terms, which is simply $\frac{100}{5}=20$. This means that the number of bananas the ape ate on May 5th is just $20+6*2=32$. Select $\boxed{D}$.

You can do this quicker than you read it if you truly master arithmetic sequences.

~hastapasta

See Also

2005 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_8_Problems/Problem_12&oldid=173771"