2005 AMC 8 Problems/Problem 12


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?

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


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.


See Also

2005 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AJHSME/AMC 8 Problems and Solutions

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