David starts at the 11th floor and rides up at 57 floors/min; Albert starts at the 51st floor and rides down at 63 floors/min. At which floor do they meet?

Given data

• David: position D(t) = 11 + 57t
• Albert: position A(t) = 51 − 63t
• t in minutes, both start at t = 0.

Concept / Approach

• They meet where floors are equal: set D(t) = A(t) and solve for t, then compute the floor.

Step-by-step calculation
11 + 57t = 51 − 63t120t = 40 ⇒ t = 1/3 min = 20 sMeeting floor = 11 + 57 × (1/3) = 11 + 19 = 30

Verification
Albert: 51 − 63 × (1/3) = 51 − 21 = 30 ✔

Common pitfalls

• Adding speeds but forgetting to equate positions.
• Arithmetic slips when handling fractions of a minute.

Final Answer
30