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2010 AMC 10A Problem 3

2010 AMC 10A Problem 3AMC 10 Step by Step

Name: Date: September 13, 2026

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

Tyrone had $97$ marbles and Eric had $11$ marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?

A)
$3$
B)
$13$
C)
$18$
D)
$25$
E)
$29$

Answer

Key insight

The total 108 marbles is fixed, so Tyrone ends with 72 and Eric with 36; Tyrone gave away 97 - 72 = 25.

Solution

Giving marbles back and forth does not change the total, which is $97 + 11 = 108$ .

At the end Tyrone has twice as many as Eric, so the $108$ marbles split in the ratio $2 : 1$ : Eric has $\dfrac{108}{3} = 36$ and Tyrone has $72$ .

Tyrone therefore gave away $97 - 72 = 25$ marbles (equivalently, Eric gained $36 - 11 = 25$ ).

The answer is $\boxed{\textbf{(D)}\ 25}$ .

Why this works

When objects are transferred between two people, the total is an invariant. A final ratio then determines both final amounts directly, and the transfer is just the difference from the start. This avoids setting up and solving an equation in the unknown transfer, though $97 - x = 2(11 + x)$ works just as well.

The trap

Solving 97 - x = 2 * 11 (ignoring that Eric also gains the marbles) and getting 75, or answering Eric's final count.

Common mistakes

Solving 97 - x = 2 * 11 (ignoring that Eric also gains the marbles) and getting 75, or answering Eric's final count.
Sign slip in $97 - x = 22 + 2x$ , e.g. writing $3x = 119$ instead of $3x = 75$ .

Techniques

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