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2002 AMC 10A Problem 6

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2002 AMC 10A Problem 6AMC 10 Step by Step
Name:                             Date: September 13, 2026
Linear Equations & Word Problems · difficulty 1 of 5 · about 1 min · position P1-10

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?

  1. A)

    1515

  2. B)

    3434

  3. C)

    4343

  4. D)

    5151

  5. E)

    138138

Answer

A

Key insight

Undo Cindy's steps in reverse (multiply 43 by 3, add 9) to recover the original number 138, then apply the correct steps.

Solution

Let the starting number be NN. Cindy's wrong procedure gives
N−93=43.\frac{N - 9}{3} = 43.
Reverse her steps: multiply by 33 to get N−9=129N - 9 = 129, then add 99 to get N=138N = 138.

Now do what the teacher asked: subtract 33 and divide by 99:
138−39=1359=15.\frac{138 - 3}{9} = \frac{135}{9} = 15.

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

Why this works

A sequence of operations applied to an unknown can be peeled off in reverse order using inverse operations; this recovers the input without any guessing. Once the original number is known, the "correct" computation is a single line. Read carefully which pair of steps is wrong and which is right; the numbers 33 and 99 are deliberately swapped to cause confusion.

The trap

Stopping at 138 (the original number, choice E) or at 43 without redoing the problem correctly.

Common mistakes

  • Stopping at 138 (the original number, choice E) or at 43 without redoing the problem correctly.
  • Applying the inverse operations in the wrong order (adding 99 before multiplying by 33), which gives N=156N = 156 and a non-integer result.

Techniques

Start from the end state / desired conclusion and reverse