AMC 10 Step by Step

2013 AMC 10B Problem 3

P1-10~2 minPrint

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?

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Key insight

If two numbers average 3, they add up to 6. The high is 16 more than the low, so the low is 8 below the average.

Solution

1. Call the low temperature LL. The high is 1616 degrees higher, so it is L+16L + 16.
2. The average of two numbers is their sum divided by 22. The average is 33, so the sum is 2×3=62 \times 3 = 6:
L+(L+16)=6.L + (L + 16) = 6.
3. That says 2L+16=62L + 16 = 6. Take 1616 away from both sides: 2L=−102L = -10. Divide by 22: L=−5L = -5.
4. Check: the high is −5+16=11-5 + 16 = 11, and the average is 11+(−5)2=62=3\frac{11 + (-5)}{2} = \frac{6}{2} = 3. That matches.

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

Why this works

"The average of two numbers is 33" means exactly the same thing as "the two numbers add up to 66". Once you know the sum and the difference of two numbers, both numbers are fixed. Turning an average into a sum is a move you will use again and again.

Alternative approach

The average of two numbers sits exactly halfway between them. The high and low are 1616 degrees apart, so each one is half of that, 88 degrees, away from the average. The low is 3−8=−53 - 8 = -5 and the high is 3+8=113 + 8 = 11.

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 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}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.