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

m
(One intermediate revision by one other user not shown)
Line 13: Line 13:
 
x &= \boxed{\textbf{(A)} 30}
 
x &= \boxed{\textbf{(A)} 30}
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 +
 +
 +
== 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 <math>\boxed{\textbf{(A)} 30}</math>
 +
 +
~bobjoebilly
  
 
== See Also==
 
== See Also==
  
 
{{AMC10 box|year=2011|ab=B|num-b=5|num-a=7}}
 
{{AMC10 box|year=2011|ab=B|num-b=5|num-a=7}}
 +
{{MAA Notice}}

Revision as of 18:10, 14 April 2020

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS