A bus can hold a maximum of 38 students. What is the minimum number of buses needed to transport 411 students?
Since $n$ buses will carry $38n$ students, we need $38n>411$.

Dividing both sides of this inequality by $38$, we have $n>\dfrac{411}{38}$. We can convert $\dfrac{411}{38}$ to a mixed number: $$\frac{411}{38} = \frac{380}{38}+\frac{31}{38} = 10\frac{31}{38}.$$ Since the number of buses must be an integer, the smallest possible number of buses is $\boxed{11}$.
11
Level 1
Prealgebra
Problem: A bus can hold a maximum of 38 students. What is the minimum number of buses needed to transport 411 students?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the minimum number of buses required, first, divide the total number of students by the capacity of one bus. This gives the equation: 411 students ÷ 38 students/bus. Performing this division yields approximately 10.81578947. Since the number of buses must be a whole number, round up to the nearest whole number. This rounding is necessary because even a fraction of a bus indicates that an additional bus is needed to accommodate the remaining students. Therefore, rounding 10.81578947 up results in 11. Thus, the minimum number of buses needed to transport all 411 students is 11.<|END_OF_LATENT|> The final answer is $11$.