2025 AMC 10A Problem 2

The $10$ pounds already in the box consist of $5$ pounds of peanuts, $2$ pounds of cashews and $3$ pounds of almonds.

Let $x$ be the number of pounds of the second mix added. It brings $0.2x$ pounds of peanuts and $0.4x$ pounds of cashews, and the box now holds $10 + x$ pounds in total. The peanut condition says
$$\frac{5 + 0.2x}{10 + x} = 0.4 \implies 5 + 0.2x = 4 + 0.4x \implies 1 = 0.2x \implies x = 5.$$

Now return to the cashews: the box holds
$$2 + 0.4(5) = 2 + 2 = 4$$
pounds of them.

The answer is $\boxed{\textbf{(B)}\ 4}$.

Only one ingredient is constrained, so only that ingredient needs an equation; the cashews are dead weight until the amount added is known. The general form of every mixture problem is the same: amount of the ingredient equals the fraction times the total, with both sides linear in the unknown weight, which is why a single equation settles it. The almonds never enter at all, and their percentages are there only to make the two mixes look complicated.

Use alligation on the peanut percentages. The result, $40$ percent, sits $10$ points below the first mix's $50$ percent and $20$ points above the second mix's $20$ percent, so the two mixes are used in the ratio $20 : 10 = 2 : 1$. With $10$ pounds of the first mix, that means $5$ pounds of the second, and the cashews total $2 + 0.4(5) = 4$.

• Stopping at x = 5, the number of pounds added, and marking (D) instead of going on to the cashews.
• Averaging the two peanut percentages to $35$ percent, which silently assumes equal weights of the two mixes.
• Adding the cashew percentages $20$ and $40$ and applying the result to the final $15$ pounds.