2011 AMC 10B Problem 6

P1-10~2 minPrint

Linear Equations & Word Problems Start from the end state / desired conclusion and reverse

On Halloween Casper ate $1/3$ of his candies and then gave $2$ candies to his brother. The next day he ate $1/3$ 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?

Show solution without answering

Key insight

Undo the story backwards from the end. Add back what he gave away, then remember what is left after eating a third is two thirds of what he had.

Solution

We know the end of the story (8 candies), so run the story backwards. Each step is undone by doing the opposite.

On day three he ate his last $8$ candies. So at the end of day two he had $8$ .
Just before giving $4$ to his sister, he had $8 + 4 = 12$ .
Eating $\frac{1}{3}$ leaves $\frac{2}{3}$ . So $12$ is two thirds of what he had at the start of day two. One third is $6$ , so he had $3 \times 6 = 18$ .
Just before giving $2$ to his brother, he had $18 + 2 = 20$ .
$20$ is two thirds of what he started with. One third is $10$ , so he started with $30$ .
Check forward: $30$ , eat a third to get $20$ , give $2$ to get $18$ , eat a third to get $12$ , give $4$ to get $8$ . It works.

The answer is $\boxed{\textbf{(A)}\ 30}$ .

Why this works

A story made of "eat a fraction, then give some away" steps can always be run backwards: add back what was given, then undo the fraction. Working backwards means you never have to solve a messy equation with fractions inside fractions.

Alternative approach

Test the choices. The starting number must split into thirds, and so must the number after giving $2$ away. Start with (A): $30 \to 20 \to 18 \to 12 \to 8$ . It ends at exactly $8$ , so (A) is correct.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Reverse each step, starting from the $8$ candies left on day three.

End of day two: before giving $4$ to his sister he had $8 + 4 = 12$ . Those $12$ were two thirds of what he started day two with, so day two began with $12 \cdot \frac{3}{2} = 18$ .

End of day one: before giving $2$ to his brother he had $18 + 2 = 20$ . Those $20$ were two thirds of the original amount, so he began with $20 \cdot \frac{3}{2} = 30$ .

Check forward: $30 \to 20 \to 18 \to 12 \to 8$ . The answer is $\boxed{\textbf{(A)}\ 30}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.4)
Introduction to Algebra, Chapter 7: Proportion (§7.4)