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2015 AMC 10A Problem 4

2015 AMC 10A Problem 4AMC 10 Step by Step

Name: Date: September 13, 2026

Ratios, Percents & Averages · difficulty 1 of 5 · about 2 min · position P1-10

Pablo, Sofia, and Mia got some candy eggs at a party. Pablo had three times as many eggs as Sofia, and Sofia had twice as many eggs as Mia. Pablo decides to give some of his eggs to Sofia and Mia so that all three will have the same number of eggs. What fraction of his eggs should Pablo give to Sofia?

A)
$\frac{1}{12}$
B)
$\frac{1}{6}$
C)
$\frac{1}{4}$
D)
$\frac{1}{3}$
E)
$\frac{1}{2}$

Answer

Key insight

Give Mia 1 egg, Sofia 2, Pablo 6; the fair share is 3 each, so Sofia needs 1 of Pablo's 6 eggs.

Solution

Only ratios matter, so choose convenient numbers. Let Mia have $1$ egg. Then Sofia has $2$ and Pablo has $3 \cdot 2 = 6$ .

The total is $1 + 2 + 6 = 9$ eggs, so after the redistribution each person holds $3$ . Sofia must go from $2$ to $3$ , receiving $1$ egg from Pablo. (Mia receives $2$ .)

Pablo gives Sofia $1$ of his $6$ eggs, which is the fraction $\frac{1}{6}$ .

The answer is $\boxed{\textbf{(B)}\ \frac{1}{6}}$ .

Why this works

When a problem only asks for a fraction or ratio, the actual quantities are irrelevant, so pick the smallest numbers that satisfy the ratios. Then the redistribution becomes concrete arithmetic. Read the question twice: it asks for the share going to Sofia, as a fraction of Pablo's eggs.

The trap

Answering 1/3, the fraction Pablo gives away in total (to Sofia and Mia combined), instead of the part that goes to Sofia alone.

Common mistakes

Answering $\frac{1}{3}$ , the fraction Pablo gives away in total (to Sofia and Mia combined), instead of the part that goes to Sofia alone.
Computing the fraction relative to the total $9$ eggs ( $\frac{1}{9}$ , not a choice) or relative to Sofia's final count.

Techniques

Set up the equation/formula and compute; no special trick needed · Test small/specific values or special cases to find or verify the answer