The speedy problem IV

• Let the speed of the escalator be $k\mbox{ steps}\cdot\mbox{s}^{-1}$ .
• Let the speed of Ganga be $G\mbox{ steps}\cdot\mbox{s}^{-1}$ .
• Let the speed of Malik be $M\mbox{ steps}\cdot\mbox{s}^{-1}$ .
• Let there be $n\mbox{ steps}$ .

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• Time for Ganga to reach the top = $\frac{270}G$ .
• Speed of Ganga relative to the ground = $(G-k)$ .
• Distance travelled by Ganga = $n$ . $\therefore\frac{270(G-k)}G=n$
• Time for Malik to reach the top = $\frac{108}M$ .
• Speed of Malik relative to the ground = $(M+k)$ .
• Distance travelled by Malik = $n$ . $\therefore\frac{108(M+k)}M=n$

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• Time for Malik to reach the bottom = $\frac{108}M$
• Time for Ganga to take 216 steps = $\frac{216}G$ $\therefore\frac{108}M=\frac{216}G$ $G=2M$

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$\begin{array}{rcl} \frac{270(G-k)}G&=&\frac{108(M+k)}M\\ \frac{270(2M-k)}{2M}&=&\frac{108(M+k)}M\\ 270(2M-k)&=&216(M+k)\\ 324M&=&486k\\ M&=&1.5k \end{array}$

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$\begin{array}{rcl} n&=&\frac{108(M+k)}M\\ &=&\frac{108(2.5k)}{1.5k}\\ &=&\frac{108\times2.5}{1.5}\\ &=&180 \end{array}$