2002 AMC 10A Problem 6

P1-10~1 minPrint

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

Cindy was asked by her teacher to subtract 3 from a certain number and then divide the result by 9. Instead, she subtracted 9 and then divided the result by 3, giving an answer of 43. What would her answer have been had she worked the problem correctly?

Show solution without answering

Key insight

Undo Cindy's wrong steps backwards to find the starting number, 138. Then do the steps the teacher actually asked for.

Solution

We know where Cindy ended up (43) and what she did, so we can walk backwards to the number she started with.

Her last step was dividing by $3$ . To undo it, multiply by $3$ : $43 \times 3 = 129$ .

Her first step was subtracting $9$ . To undo it, add $9$ : $129 + 9 = 138$ .

So the starting number is $138$ . Check: $138 - 9 = 129$ , and $129 \div 3 = 43$ . Correct.

3. Now do what the teacher asked: subtract $3$ , then divide by $9$ :
$138 - 3 = 135, \qquad 135 \div 9 = 15.$

The answer is $\boxed{\textbf{(A)}\ 15}$ .

Why this works

A chain of steps can be undone by doing the opposite steps in the opposite order: undo the last thing first. Once the starting number is known, the correct calculation is quick. Watch out: the numbers $3$ and $9$ are swapped on purpose, and $138$ sits among the choices to catch students who stop early.

Concise solution

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

Let the starting number be $N$ . Cindy's wrong procedure gives
$\frac{N - 9}{3} = 43.$
Reverse her steps: multiply by $3$ to get $N - 9 = 129$ , then add $9$ to get $N = 138$ .

Now do what the teacher asked: subtract $3$ and divide by $9$ :
$\frac{138 - 3}{9} = \frac{135}{9} = 15.$

The answer is $\boxed{\textbf{(A)}\ 15}$ .

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)