2026 AMC 10A Problem 1

We are never told how many minutes a record lasts, and we do not need to know. Measure everything in "EP-lengths": the playing time of one EP record.

1. Change the LPs into EP-lengths. One LP plays as long as $3$ EPs. So $33$ LPs play as long as
$$33 \times 3 = 99 \text{ EPs}.$$

1. Count the EP-lengths she already has. She owns $45$ EP records, which is $45$ EP-lengths.

3. Find the gap. For the EP total to match the LP total, she needs $99$ EP records altogether. She has $45$, so she must buy
$$99 - 45 = 54$$
more.

1. Check. After buying, she has $45 + 54 = 99$ EPs. Since $99 = 3 \times 33$, that is exactly as long as her $33$ LPs.

The answer is $\boxed{\textbf{(B)}\ 54}$.

When two kinds of things are compared by a ratio ("three times as long"), pick the smaller one as your unit and rewrite the other in that unit. Then both totals are plain counts of the same thing and can be compared by subtracting. The actual length of a record cancels out, which is why the problem never gives it.

Use a letter for the time. Say one EP plays for $t$ minutes, so one LP plays for $3t$ minutes, and say Helena buys $n$ more EPs. Then
$$(45 + n)\,t = 33 \times 3t = 99t.$$
Divide both sides by $t$: $45 + n = 99$, so $n = 54$.

In units of one EP's playing time, the $33$ LPs last $33 \cdot 3 = 99$. Helena needs $99$ EPs in total and owns $45$, so she must buy $99 - 45 = 54$ more.

The answer is $\boxed{\textbf{(B)}\ 54}$.