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2011 AMC 10B Problem 4

2011 AMC 10B Problem 4AMC 10 Step by Step

Name: Date: September 13, 2026

Linear Equations & Word Problems · difficulty 1 of 5 · about 1 min · position P1-10

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?

A)
$\frac{A + B}{2}$
B)
$\dfrac{A - B}{2}$
C)
$\dfrac{B - A}{2}$
D)
$B - A$
E)
$A + B$

Answer

Key insight

Each person's fair share is half the total, (A+B)/2, so LeRoy must hand over the gap between that share and what he already paid.

Solution

The trip cost $A + B$ dollars in total, so each person's fair share is $\dfrac{A + B}{2}$ .

LeRoy has paid only $A$ , so he owes
$\frac{A + B}{2} - A = \frac{B - A}{2}.$
This is positive because $A < B$ , and after the transfer Bernardo's net payment is $B - \frac{B - A}{2} = \frac{A + B}{2}$ as well.

The answer is $\boxed{\textbf{(C)}\ \dfrac{B - A}{2}}$ .

Why this works

Equalizing two amounts means moving half of their difference from the larger to the smaller. Sanity-check with numbers: if $A = 10$ and $B = 30$ , each should end at $20$ , so LeRoy pays $10 = (30 - 10)/2$ .

Alternative approach

Plug in $A = 10$ , $B = 30$ : the fair share is $20$ , so LeRoy gives $10$ . Only choice (C) evaluates to $10$ ; (D) gives $20$ and (B) is negative.

The trap

Giving the whole difference B - A, which would overshoot and leave Bernardo paying less than LeRoy.

Common mistakes

Giving the whole difference $B - A$ , which would overshoot and leave Bernardo paying less than LeRoy.
Choosing $\frac{A - B}{2}$ , which has the right size but the wrong sign since $A < B$ .

Techniques

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