AMC 10 · 2011 · #4

Grade 7 arithmetic

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

linear-equations-one-varfraction-arithmetic convert-to-algebra ↑ Prerequisites: linear-equations-one-var

📏 Short solution 💡 2 insights

Problem

LeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, each of them paid for various joint expenses such as gasoline and car rental. At the end of the trip it turned out that LeRoy had paid $A$ dollars and Bernardo had paid $B$ dollars, where $A < B$ . How many dollars must LeRoy give to Bernardo so that they share the costs equally?

Pick an answer.

(A)

$\frac{A + B}{2}$

(B)

$\dfrac{A - B}{2}$

(C)

$\dfrac{B - A}{2}$

(D)

B - A

(E)

A + B

AMC 10 2011 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.

View mode:

Toolkit + CCSS Solution

Understand

Restated: LeRoy and Bernardo split a trip's costs equally. LeRoy paid $A$ dollars and Bernardo paid $B$ dollars, with $A < B$. Find how many dollars LeRoy must hand to Bernardo so each ends up having paid the same amount.

Givens: LeRoy paid $A$ dollars; Bernardo paid $B$ dollars; The two agreed to share all costs equally; $A < B$, so LeRoy paid less than Bernardo; Answer choices: (A) $\dfrac{A+B}{2}$, (B) $\dfrac{A-B}{2}$, (C) $\dfrac{B-A}{2}$, (D) $B-A$, (E) $A+B$

Unknowns: The number of dollars LeRoy must give Bernardo to even out their payments

Understand

Plan

Primary tool: #13 Convert to Algebra

Secondary: #4 Introduce a Variable, #6 Guess and Check

The answer choices are all expressions in $A$ and $B$, so Tool #13 (Convert to Algebra) turns the words into an equation. The fair share is half the total, $\dfrac{A+B}{2}$. Tool #4 (Introduce a Variable) names the transfer amount $x$ and writes $A+x=\dfrac{A+B}{2}$, and Tool #6 (Guess and Check) confirms the simplified answer against real numbers like $A=30,\;B=60$.

Execute — Answer: C

#13 Convert to Algebra 6.EE.A.2 Step 1

Write the fair share

Together they spent $A+B$ dollars.
Sharing equally means each person should end up having paid exactly half of that total.
So each person's fair share is $\dfrac{A+B}{2}$ dollars.

$$\text{total}=A+B \;\Rightarrow\; \text{fair share}=\dfrac{A+B}{2}$$

💡 Splitting one shared cost evenly means each person owes exactly half of the whole.

#4 Introduce a Variable 7.EE.B.4 Step 2

Set up the transfer

LeRoy paid $A$, which is the smaller amount, so he still owes some money.
Let $x$ be the dollars he gives Bernardo.
After paying $x$, LeRoy has paid $A+x$ toward his share, and that should equal the fair share.
This gives the equation $A+x=\dfrac{A+B}{2}$.

$$A + x = \dfrac{A+B}{2}$$

💡 The person who underpaid just needs to add enough to reach the halfway amount.

#13 Convert to Algebra 7.EE.A.1 Step 3

Solve for the transfer

Subtract $A$ from both sides to isolate $x$: $x=\dfrac{A+B}{2}-A$.
Write $A$ as $\dfrac{2A}{2}$ so both terms share the denominator $2$, then combine: $\dfrac{A+B-2A}{2}=\dfrac{B-A}{2}$.
Because $A<B$, this amount is positive, as it should be.
This matches choice (C).

$$x=\dfrac{A+B}{2}-A=\dfrac{A+B-2A}{2}=\dfrac{B-A}{2}\;\Rightarrow\;\textbf{(C)}$$

💡 The extra LeRoy owes is exactly half of the gap between what the two of them paid.

[1] #13 6.EE.A.2 Together they spent $A+B$ dollars. Sharing equally means each person should end

[2] #4 7.EE.B.4 LeRoy paid $A$, which is the smaller amount, so he still owes some money. Let $x

[3] #13 7.EE.A.1 Subtract $A$ from both sides to isolate $x$: $x=\dfrac{A+B}{2}-A$. Write $A$ as

Review

Reasonableness: The transfer should be positive (LeRoy underpaid) and smaller than the whole gap $B-A$, since handing over the full gap would overshoot and flip who paid more. The answer $\dfrac{B-A}{2}$ is half the gap, so it lands exactly in the middle — positive and less than $B-A$. Check with numbers: if $A=30$ and $B=60$, the total is $90$ and each owes $45$; LeRoy gives $\dfrac{60-30}{2}=15$, making his total $45$ and Bernardo's $60-15=45$. Equal, as required.

Alternative: Reason about the gap directly instead of solving an equation. Bernardo paid $B-A$ more than LeRoy. Moving one dollar from LeRoy to Bernardo widens the difference, while moving it the other way narrows it; to make the two equal, LeRoy must cover half of that $B-A$ gap, which is $\dfrac{B-A}{2}$ — again choice (C).

CCSS standards used (min grade 7)

6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Writing the total $A+B$ and the fair share $\dfrac{A+B}{2}$ with the letters $A$ and $B$.)
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities (Naming the transfer $x$ and building the equation $A+x=\dfrac{A+B}{2}$.)
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions (Simplifying $\dfrac{A+B}{2}-A$ into $\dfrac{B-A}{2}$ over a common denominator.)

⭐ To even up who paid what, the person who paid less hands over half of the gap between the two payments.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Write the fair share

Set up the transfer

Solve for the transfer

Review

CCSS standards used (min grade 7)

More like this