Aptitude - Simplification - Discussion

First-person is now at the 11th floor while the second person now at the 51th floor.

First person is moving upward and the second person is moving downward staying in two different lift.

The question is at which floor their lift meets i.e they meet or they cross each other. Their lift speeds are 57 floor/ minute and 63 floor/minute respectively.

Solution:
Let, at the x floor they meet, such that 11<= x<=51.
Therefore, to reach the meeting floor first person will cover a distance of (x-11) floors
and the second person will cover a distance of (51-x) floor.

As their speeds are different but they start the journey at the same time, to reach the meeting floor they need equal time i,e t1=t2

Therefore, t1=t2.
>> (x-11)/57=(51-x)/63 [t=s/v formula].

>> x=30
Therefore, at 30th floor, their lift will cross each other or they will meet.

Since, both X and Y are traveling in opposite directions, their relative speed = (57+63) floor/minute = 120f/m.

Total distance between X an Y in terms of floors = 51-11=40 floors.

Therefore, time taken to cross each other = Distance/Speed = 40/120= 1/3 minutes.
Distance (floors) traveled by X in 1/3 minutes = Speed x Time = 57 x 1/3 = 19 floors.
So, X would have travelled to 11 + 19 = 30 floors when the meet/cross each other.

David's starting position = 11th floor.
Albert's = 51st,

So, distance covered by david till the point they meet= x
and for Albert= (40-x)

Speed of David = 57 floors/minute
and for Albert = 63 floors/minute.

As time taken by both will be same equation becomes:
(x/57) = (40-x)/63
x = 19.

Thus, answer is (11+19) or (51-21) => 30th Floor.

We can also predict it though an train problem, two people who are travelling opposite direction:

Here total speed will be 63+57 (f/m) =120 f/m.

Total distance will be subtracted 51 -11.

So the time they will meet will be 40/120=1/3 min.

So the distance will be for their meet will be:

11+57*1/3= 30.

or

51-63*1/3=30.

@Geeni.

Just implemented the concept of relative velocity.

1/3 is time at which david and albert meet. So he multiply that time to the velocity of david and calculate the distance travelled by david and finally he added that distance to david's initial position.

Consider david to be stationary then difference speed is
= 57+63 floor/minute
= 120f/m.

Total difference of floor is 51-11=40 floors.

Therefore minutes is (120f/m) /40f=1/3m.

So 11 + (1/3) * 57 = 51-(1/3)*63 = 30.

Taking the ratios 57:63
We get 19:21.

David is in 11th floor. David is going upward direction. So 11+19=30th floor.
Otherwise, Albert is in the 51st floor and he goes downward. So 51-21=30th floor.

s = d/t.
In this problem time travel is same for both=> d1/s1 = d2/s2.

Lets assume they cross the paths at yth floor.
(y-11)/57=(51-y)/63.

Solve this,
y=30.

@Kumar

Because at Yth position both of them meet so the distance from David is (y-11) and that of Albert is (51-y).

If David stationary then how we can add 57+63.

57 is David speed per min.