Help with time and distance questions please!!

1) The distance between Bob and Jane is 40 Floors.

The rate at which they are travelling towards each other is 57 + 63 (Since they are travelling towards each other, I am adding up their rates) = 120 floors/min

Time taken = \(\frac{40}{120}\) = \(\frac{1}{3}\) mins

So at 120 floors/min, they will cover the 40 floors in a third of a min.

In a third of a min, Bob would have travelled \(\frac{57}{3}\) = 19 Floors

So, the floor at which they would meet is 21+19 = 40th Floor

2) The distance between X and Y is 100 miles.

The rate at which they are travelling towards each other is \(\frac{100}{3}\) + \(\frac{100}{5}\) (Since they are travelling towards each other, I am adding up their rates) = \(\frac{800}{15}\) miles/hour

Time taken = \(\frac{100}{800/15}\) = \(\frac{15}{8}\) hours

So at \(\frac{800}{15}\) miles/hour, they will cover the 100 miles in \(\frac{15}{8}\) hours

In \(\frac{15}{8}\) hours, A would have travelled \(\frac{100}{3}\)*\(\frac{15}{8}\) = 62.5 miles from X

1. Bob enters a lift on the 21st floor, which is going up @ 57 floors/min. At the same time Jane enters
another lift on the 61st floor which is going down at 63 floors/min. At what floor will they be together?

Bob and Jane will cover 61 - 21 = 40 floors.
They are moving in opposite directions.So we need to add up their speed to find their relative speed.

They cover 57 + 63 = 120 floors/min.So, they will take 40/120 = 1/3 min.
Now,we need to calculate no of floors anyone has covered in that period.
Bob covers 57/3 = 19 floors.
Hence they will meet at floor = 21+19 = 40.


2. Train A can travel from X to Y, a distance of 100 miles, in 3hrs while train B can travel the same distance
in 5hrs. If A starts from X to Y at the same time as B starts from Y to X, then at what distance from X will they meet?

We'll apply the same logic as above.

Speed of A = 100/3 miles/hr.
Speed of B = 100/5 = 20 miles/hr.

Relative speed of A and B = 20 + 100/3 = 160/3 miles/hr.
Time taken to cover up the distance = (100 * 3 )/160 = 300/160 hrs.
So,the required distance from X = distance travelled by A = 100/3 * 300/160 = 62.5 miles.

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Distance between the two = 61-21 = 40
Speed = 57+63 = 120 / min (As both are coming towards each other) = 120 / 60 sec = 2 / sec

Therefore, time taken = 40/2 = 20 sec = 20/60min = 1/3 min

So, In 1/3 min, bob will be at 21 + (57/3) = 40th floor
same for Jane, 61 - (63/3) = 40th floor

Thank you so much, guys!!!