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2013 AMC 10B Problem 3

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2013 AMC 10B 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

On a particular January day, the high temperature in Lincoln, Nebraska, was 1616 degrees higher than the low temperature, and the average of the high and the low temperatures was 3 ∘3\,^\circ . In degrees, what was the low temperature in Lincoln that day?

  1. A)

    −13-13

  2. B)

    −8-8

  3. C)

    −5-5

  4. D)

    −3-3

  5. E)

    1111

Answer

C

Key insight

An average of 3 means the high and low add to 6, and the high is the low plus 16, so twice the low is 6 minus 16.

Solution

Let the low temperature be LL, so the high is L+16L + 16.

The average of the two is 33, so their sum is 66:
L+(L+16)=6⟹2L=−10⟹L=−5.L + (L + 16) = 6 \quad\Longrightarrow\quad 2L = -10 \quad\Longrightarrow\quad L = -5.

Check: the high is 1111 and 11+(−5)2=3\frac{11 + (-5)}{2} = 3.

The answer is (C) −5\boxed{\textbf{(C)}\ -5}.

Why this works

"Average of two numbers is mm" is the same as "the two numbers add to 2m2m." Combined with a known difference, the sum-and-difference pair determines both numbers: the smaller is sum−difference2\frac{\text{sum} - \text{difference}}{2}.

Alternative approach

The low is 88 degrees below the average and the high is 88 degrees above it (half of 1616 each way). So the low is 3−8=−53 - 8 = -5.

The trap

Using 3 as the sum of the two temperatures instead of 6, or mishandling the negative result.

Common mistakes

  • Using 3 as the sum of the two temperatures instead of 6, or mishandling the negative result.
  • Reporting the high temperature 1111 (choice (E)) instead of the low.

Techniques

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