aishu4 wrote:

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?

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?

Please help...if possible with explanation...

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

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