Difference between revisions of "2011 AMC 10B Problems/Problem 6"

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x &= \boxed{\textbf{(A)} 30}
 
x &= \boxed{\textbf{(A)} 30}
 
\end{align*}</cmath>
 
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== See Also==
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{{AMC10 box|year=2011|ab=B|num-b=5|num-a=7}}

Revision as of 15:43, 4 June 2011

Problem 6

On Halloween Casper ate $\frac{1}{3}$ of his candies and then gave $2$ candies to his brother. The next day he ate $\frac{1}{3}$ of his remaining candies and then gave $4$ candies to his sister. On the third day he ate his final $8$ candies. How many candies did Casper have at the beginning?

$\textbf{(A)}\ 30 \qquad\textbf{(B)}\ 39 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 57 \qquad\textbf{(E)}\ 66$

Solution

Let $x$ represent the amount of candies Casper had at the beginning. \begin{align*} \frac{2}{3} \left(\frac{2}{3} x - 2\right) - 4 - 8 &= 0\\ \frac{2}{3} x - 2 &= 18\\ \frac{2}{3} x &= 20\\ x &= \boxed{\textbf{(A)} 30} \end{align*}

See Also

2011 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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 AMC 10 Problems and Solutions
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