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Question

Two boys enter a running escalator at the ground floor of a shopping mall . The first boy repeatedly follows a cycle of p1=1 step up and then q1=2 steps down whereas the second boy repeatedly follows a cycle of p2=2 steps up and then q2=1 step down. Both of them move relative to escalator with a speed v=50cm/s . If the boys take t1=250s and t2=50s respectively to reach the first floor in complete number of cycles, how fast is the escalator running ?

Joshi sir comment

$l e t u a n d v a r e t h e s p e e d s o f b o y a n d e s c a l a t o r a c c o r d i n g t o g i v e n c o n d i t i o n [1 (u + v) T - 2 (u - v) T] n_{1} = [2 (u + v) T - 1 (u - v) T] n_{2} \Rightarrow [1 (u + v) - 2 (u - v)] T n_{1} = [2 (u + v) - 1 (u - v)] T n_{2} G i v e n t h a t T n_{1} = 250 a n d T n_{2} = 50 \Rightarrow [1 (u + v) - 2 (u - v)] 250 = [2 (u + v) - 1 (u - v)] 50 n o w s o l v e i t f o r v, u i s g i v e n$

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