Question

David gets on the elevator at the 11th floor of a building and rides up at the rate of 57 floors per minute. At the same time Albert gets on an elevator at the 51 it floor of the same building and rides down at the rate of 63 floors per minute. If they continue traveling at these rates, then at which floor will their elevators meet?

• A. 19
• B. 30
• C. 28
• D. 37

Answer 1

The Correct Option is B

To find the floor where the elevators meet, let the time elapsed be \( t \) minutes. In these \( t \) minutes, the movement in floors for David's elevator and Albert's elevator can be described as follows:

David starts at the 11th floor and moves upwards at 57 floors per minute. Therefore, after \( t \) minutes, he reaches:

\( \text{David's floor} = 11 + 57t \)

Albert starts at the 51st floor and moves downwards at 63 floors per minute. Therefore, after \( t \) minutes, he reaches:

\( \text{Albert's floor} = 51 - 63t \)

Since the elevators meet on the same floor, we set the two floor expressions equal:

\[\begin{aligned} 11 + 57t &= 51 - 63t \end{aligned}\]

Adding \( 63t \) to both sides gives:

\[\begin{aligned} 11 + 57t + 63t &= 51 \end{aligned}\]

Which simplifies to:

\[\begin{aligned} 11 + 120t &= 51 \end{aligned}\]

Subtracting 11 from both sides gives:

\[\begin{aligned} 120t &= 40 \end{aligned}\]

Dividing both sides by 120 gives:

\[\begin{aligned} t &= \frac{1}{3} \end{aligned}\]

Substituting \( t = \frac{1}{3} \) back into the expression for David's floor, we find:

\[\begin{aligned} 11 + 57 \times \frac{1}{3} &= 11 + 19 = 30 \end{aligned}\]

Therefore, the elevators meet on the 30th floor.