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2022 AMC 10A Problem 3

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2022 AMC 10A Problem 3AMC 10 Step by Step
Name:                             Date: September 13, 2026
Linear Equations & Word Problems · difficulty 1 of 5 · about 2 min · position P1-10

The sum of three numbers is 96.96. The first number is 66 times the third number, and the third number is 4040 less than the second number. What is the absolute value of the difference between the first and second numbers?

  1. A)

    11

  2. B)

    22

  3. C)

    33

  4. D)

    44

  5. E)

    55

Answer

E

Key insight

Write everything in terms of the third number c: 6c + (c + 40) + c = 96 gives c = 7, so the numbers are 42, 47, 7.

Solution

Both other numbers are described relative to the third, so let the third number be cc. Then the first number is 6c6c, and since the third is 4040 less than the second, the second is c+40c + 40.

The sum condition gives
6c+(c+40)+c=96⟹8c=56⟹c=7.6c + (c + 40) + c = 96 \quad\Longrightarrow\quad 8c = 56 \quad\Longrightarrow\quad c = 7.
So the first number is 4242 and the second is 4747, and ∣42−47∣=5|42 - 47| = 5.

The answer is (E) 5\boxed{\textbf{(E)}\ 5}.

Why this works

When several quantities are each described in terms of one of them, use that one as the variable; the problem collapses to a single linear equation. Translate comparison phrases carefully: "xx is 4040 less than yy" means x=y−40x = y - 40, equivalently y=x+40y = x + 40.

The trap

Writing the second number as c - 40 instead of c + 40; 'the third is 40 less than the second' means the second is the larger one.

Common mistakes

  • Writing the second number as c - 40 instead of c + 40; 'the third is 40 less than the second' means the second is the larger one.
  • Reporting an intermediate value such as c=7c = 7 or the difference 47−7=4047 - 7 = 40 instead of the requested ∣first−second∣|{\rm first} - {\rm second}|.

Techniques

Set up the equation/formula and compute; no special trick needed