2011 AMC 10B Problems/Problem 6

Problem

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.

$\begin{align*} \frac{2}{3} \left(\frac{2}{3} x - 2\right) - 4 - 8 &= 0\\ \frac{2}{3} x - 2 &= 18\\ \frac{2}{3} x &= 20\\ x &= \boxed{\textbf{(A)} 30} \end{align*}$

Solution 2

We work backwards. If he had 8 candies at the end, then before he gave candies to his sister he had 12 candies. This means that at the end of Halloween he had 18 candies, so before he gave candies to his brother he had 20 candies. Therefore, at the start he had $\boxed{\textbf{(A)} 30}$

~bobjoebilly

Solution 3 (answer choices)

A solve by algebra is more secure and safe (and usually faster), but you can also test the answer choices. We are lucky and option A works $\Longrightarrow \boxed{\textbf{(A) } 30}$ .

~JH. L

2011 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2011 AMC 10B Problems/Problem 6

Contents

Problem

Solution

Solution 2

Solution 3 (answer choices)

See Also