A building has 2 elevators. X gets on one of them and Y on another. X takes the elevator from the 2 nd floor which goes up at 15 floors a minute. Y takes the elevator from 107 th floor which comes down at 20 floor a minute. At which floor do X and Y meet?
47
Finding the Elevator Meeting Floor
This problem involves two objects (elevators) moving towards each other from different starting points and at different speeds. We need to find the floor where they meet and the time it takes for them to meet.
Let's denote:
- \(F_X\) as the floor number of elevator X at time \(t\).
- \(F_Y\) as the floor number of elevator Y at time \(t\).
- \(t\) as the time in minutes from when they start moving.
We are given the following information:
- Elevator X starts at floor 2 and moves up at a speed of 15 floors per minute.
- Elevator Y starts at floor 107 and moves down at a speed of 20 floors per minute.
Formulating Equations for Elevator Positions
The position of elevator X at time \(t\) can be represented by its starting floor plus the distance it travels upwards:
\[F_X(t) = \text{Starting Floor of X} + (\text{Speed of X} \times t)\]
\[F_X(t) = 2 + 15t\]
The position of elevator Y at time \(t\) can be represented by its starting floor minus the distance it travels downwards:
\[F_Y(t) = \text{Starting Floor of Y} - (\text{Speed of Y} \times t)\]
\[F_Y(t) = 107 - 20t\]
Solving for the Meeting Time
The elevators meet when their floor numbers are the same. So, we set their position equations equal to each other:
\[F_X(t) = F_Y(t)\]
\[2 + 15t = 107 - 20t\]
Now, we solve this linear equation for \(t\):
Add \(20t\) to both sides:
\[2 + 15t + 20t = 107 - 20t + 20t\]
\[2 + 35t = 107\]
Subtract 2 from both sides:
\[2 + 35t - 2 = 107 - 2\]
\[35t = 105\]
Divide by 35:
\[t = \frac{105}{35}\]
\[t = 3 \text{ minutes}\]
So, the elevators meet after 3 minutes.
Calculating the Meeting Floor
To find the floor where they meet, we can substitute the value of \(t = 3\) minutes into either the equation for \(F_X(t)\) or \(F_Y(t)\).
Using \(F_X(t)\):
\[F_X(3) = 2 + 15 \times 3\]
\[F_X(3) = 2 + 45\]
\[F_X(3) = 47\]
Using \(F_Y(t)\):
\[F_Y(3) = 107 - 20 \times 3\]
\[F_Y(3) = 107 - 60\]
\[F_Y(3) = 47\]
Both equations give the same result, confirming that the meeting floor is 47.
Summary of Results
| Elevator | Starting Floor | Speed (Floors/minute) | Direction |
|---|---|---|---|
| X | 2 | 15 | Up |
| Y | 107 | 20 | Down |
Meeting Time: 3 minutes
Meeting Floor: 47
Therefore, X and Y meet at the 47th floor.
Revision Table: Elevator Meeting Point Calculations
| Concept | Formula/Explanation | Application |
|---|---|---|
| Position (Upward) | \(P(t) = P_0 + vt\) | \(F_X(t) = 2 + 15t\) |
| Position (Downward) | \(P(t) = P_0 - vt\) | \(F_Y(t) = 107 - 20t\) |
| Meeting Point Condition | Positions are equal at time \(t\) | \(F_X(t) = F_Y(t)\) |
| Solving for Time | Algebraically solve the equation | \(35t = 105 \implies t=3\) |
| Finding Meeting Floor | Substitute \(t\) into position equation | \(F_X(3)=47\) or \(F_Y(3)=47\) |
Additional Information: Relative Speed Concepts
Another way to think about this problem is using the concept of relative speed. Since the elevators are moving towards each other, their speeds add up to find the rate at which the distance between them is closing.
- Initial distance between X and Y = Floor 107 - Floor 2 = 105 floors.
- Speed of X (up) = 15 floors/min.
- Speed of Y (down) = 20 floors/min.
- Relative speed = Speed of X + Speed of Y (since they move in opposite directions towards each other)
- Relative speed = 15 + 20 = 35 floors/min.
The time it takes for them to meet is the initial distance divided by their relative speed:
\[\text{Time} = \frac{\text{Initial Distance}}{\text{Relative Speed}}\]
\[t = \frac{105 \text{ floors}}{35 \text{ floors/min}}\]
\[t = 3 \text{ minutes}\]
This matches the time calculated earlier. Once the time is known, you can find the meeting floor as before by substituting \(t=3\) into either elevator's position equation.
This relative speed method is useful for problems where objects move towards or away from each other.
Important Questions from Relative Speed
In a circular race of 2500 m, a man and a woman start from a point towards opposite directions with speeds of 37 km/h and 35 km/h, respectively. After how much time from the start of the race will they meet for the first time?
- X and Yrun a 3 km race along a circular course of length 300 m. Their speeds are in the ratio 3 : 2. If they start together in the same direction, how many times would the first one pass the other (the start-off is not counted as passing)?
A train of length 600 meters passes another train of length 1000 meters moving in the opposite direction in 96 seconds. If the speed of both the trains is the same, then what is the speed of each train?
A train, running at a speed of 36 kmph, takes 60 seconds to cross an electric pole. How much time it will take to cross a platform of length 250 meter?
A truck covered a distance of 780 km in 12 hours. What distance would a car cover in 11 hours travelling at a speed which is seven kmph more than that of the truck?
