AMC 10 · 2013 · #3

Grade 7 arithmetic

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

mean-median-mode-rangelinear-equations-one-var convert-to-algebra ↑ Prerequisites: linear-equations-one-var

📏 Medium solution 💡 1 insight

Problem

On a particular January day, the high temperature in Lincoln, Nebraska, was $16$ degrees higher than the low temperature, and the average of the high and low temperatures was $3$ . In degrees, what was the low temperature in Lincoln that day?

Pick an answer.

(A)

-13

(B)

-8

(C)

-5

(D)

-3

(E)

AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: One January day the high temperature was $16$ degrees above the low temperature, and the average of the high and the low was $3$ degrees. Find the low temperature.

Givens: The high temperature is $16$ degrees higher than the low temperature; The average of the high and the low temperature is $3$ degrees; Answer choices: (A) $-13$, (B) $-8$, (C) $-5$, (D) $-3$, (E) $11$

Unknowns: The low temperature, in degrees

Understand

Restated: One January day the high temperature was $16$ degrees above the low temperature, and the average of the high and the low was $3$ degrees. Find the low temperature.

Plan

Primary tool: #4 Introduce a Variable

Secondary: #13 Convert to Algebra, #3 Eliminate Possibilities

There is one unknown, the low temperature, and every other quantity is described in terms of it, so Tool #4 (Introduce a Variable) names the low as $L$ and lets the high be written as $L+16$. Tool #13 (Convert to Algebra) then turns the sentence “the average is $3$” into a single equation to solve. Tool #3 (Eliminate Possibilities) is a fast backstop: only one listed low pairs with a high that is $16$ larger and averages $3$.

Execute — Answer: C

#4 Introduce a Variable 7.EE.B.4 Step 1

Name the low and write the high

Let $L$ stand for the low temperature.
The high is $16$ degrees higher, so the high equals $L+16$.
Now both temperatures are written with the single unknown $L$.

$$\text{low}=L,\quad \text{high}=L+16$$

💡 Once the low has a name, the high is just that name plus $16$.

#13 Convert to Algebra 6.SP.A.3 Step 2

Turn the average into an equation

The average of the high and the low is their sum divided by $2$.
Setting that equal to $3$ gives $\dfrac{L+(L+16)}{2}=3$.
This single equation captures the whole problem.

$$\dfrac{L+(L+16)}{2}=3$$

💡 The average of two numbers is just their total shared evenly between them.

#4 Introduce a Variable 7.EE.B.4 Step 3

Simplify the equation

Add the two terms in the top: $L+(L+16)=2L+16$.
Divide by $2$ to get $L+8$.
So the equation becomes $L+8=3$.
The average of the high and low is always $8$ above the low, because the low sits half of the $16$-degree gap below the middle.

$$\dfrac{2L+16}{2}=L+8=3$$

💡 Halving the gap of $16$ shows the average is exactly $8$ warmer than the low.

#3 Eliminate Possibilities 7.NS.A.1 Step 4

Solve for the low temperature

Subtract $8$ from both sides: $L=3-8$.
Since $8$ is larger than $3$, the result drops below zero to $L=-5$.
That is choice (C).
Checking the others: a low of $-8$ (B) would average $0$, and $-3$ (D) would average $5$, so only $-5$ works.

$$L=3-8=-5\;\Rightarrow\;\textbf{(C)}$$

💡 Taking away more than you have pushes the number below zero.

[1] #4 7.EE.B.4 Let $L$ stand for the low temperature. The high is $16$ degrees higher, so the h

[2] #13 6.SP.A.3 The average of the high and the low is their sum divided by $2$. Setting that eq

[3] #4 7.EE.B.4 Add the two terms in the top: $L+(L+16)=2L+16$. Divide by $2$ to get $L+8$. So t

[4] #3 7.NS.A.1 Subtract $8$ from both sides: $L=3-8$. Since $8$ is larger than $3$, the result

Review

Reasonableness: Test the low of $-5$: the high is $-5+16=11$, and the average is $\dfrac{-5+11}{2}=\dfrac{6}{2}=3$, which matches. The gap between high and low is $11-(-5)=16$, also as required. Both facts hold, so (C) is correct. The answer should be below $3$ (a low is colder than the average), which rules out (E) $11$ immediately.

Alternative: Skip algebra using the midpoint idea: the average always sits exactly halfway between the low and the high. Since the two are $16$ apart, each is $8$ from the average. The low is the colder one, so it lies $8$ below $3$, giving $3-8=-5$.

CCSS standards used (min grade 7)

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities (Naming the low as $L$, writing the high as $L+16$, and solving $L+8=3$ for $L$.)
6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number (Reading “the average of the high and low is $3$” as $\dfrac{L+(L+16)}{2}=3$.)
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers (Computing $3-8=-5$, where subtracting a larger number gives a negative temperature.)

⭐ The average of two numbers sits exactly halfway between them, so the low is the average minus half the gap.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Name the low and write the high

Turn the average into an equation

Simplify the equation

Solve for the low temperature

Review

CCSS standards used (min grade 7)

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