2005 AMC 8 Problems/Problem 12

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

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.

Page

Toolbox

Search

2005 AMC 8 Problems/Problem 12

Contents

Problem

Solution

Solution 2

Solution 3 (Lightning Quick)

See Also