Difference between revisions of "2005 AMC 8 Problems/Problem 12"

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n&=\boxed{\textbf{(D)}\ 32}
 
n&=\boxed{\textbf{(D)}\ 32}
 
\end{align*}</cmath>
 
\end{align*}</cmath>
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==Solution 2==
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There are <math>5</math> days from May 1 to May 5. If we set the first day as <math>n</math>, then the second day can be expressed as <math>n+6</math>
  
 
==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}}

Revision as of 04:12, 23 July 2018

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 $n$, then the second day can be expressed as $n+6$

See Also

2005 AMC 8 (ProblemsAnswer KeyResources)
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

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