Difference between revisions of "2011 AMC 10B Problems/Problem 6"
m (→Solution) |
|||
| (4 intermediate revisions by 3 users not shown) | |||
| Line 5: | Line 5: | ||
<math> \textbf{(A)}\ 30 \qquad\textbf{(B)}\ 39 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 57 \qquad\textbf{(E)}\ 66 </math> | <math> \textbf{(A)}\ 30 \qquad\textbf{(B)}\ 39 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 57 \qquad\textbf{(E)}\ 66 </math> | ||
| − | == Solution == | + | == Solution 1 == |
Let <math>x</math> represent the amount of candies Casper had at the beginning. | Let <math>x</math> represent the amount of candies Casper had at the beginning. | ||
| + | |||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\frac{2}{3} \left(\frac{2}{3} x - 2\right) - 4 - 8 &= 0\\ | \frac{2}{3} \left(\frac{2}{3} x - 2\right) - 4 - 8 &= 0\\ | ||
| Line 13: | Line 14: | ||
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 | ||
| + | |||
| + | == 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 <math>\Longrightarrow \boxed{\textbf{(A) } 30}</math>. | ||
| + | |||
| + | ~JH. L | ||
== See Also== | == See Also== | ||
Latest revision as of 20:49, 19 July 2024
Problem
On Halloween Casper ate 
of his candies and then gave 
candies to his brother. The next day he ate of his remaining candies and then gave 
candies to his sister. On the third day he ate his final 
candies. How many candies did Casper have at the beginning?
Solution 1
Let 
represent the amount of candies Casper had at the beginning.
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 
~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 
.
~JH. L
See Also
| 2011 AMC 10B (Problems • Answer Key • Resources) | ||
| 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. 
