Question

A person walks up a stalled escalator in $90s$ . When just standing on the same moving escalator, he is carried in $60s$. The time it would take him to walk up the moving escalator will be

• A. $27s$
• B. $50s$
• C. $18s$
• D. $36s$

Answer 1

The correct option is D $36s$

Let $S$ be the distance moved by the person on the escalator.
Let $v_1$ be the velocity of the person when he/she walks up a stalled escalator
$\Rightarrow v_1=\frac{S}{90}$
Let $v_2$ be the velocity of the moving escalator
$\Rightarrow v_2=\frac{S}{60}$
Time taken by the person to walk up the moving escalator is given by
$t=\frac{S}{v_1+v_2}=\frac{S}{\frac{S}{90}+\frac{S}{60}}$
$=\frac{90\times 60}{90+60}=36s$
Note: General method to solve such types of problems:
To understand such types of problems, let us take a case of man swimming in the river.
Let $AB$ be the distance covered by the man while swimming.
$AB=V_m{t}_1=V_r{t}_2=(V_m+V_r)t_3=(V_m-V_r)t_4$
where $V_m=$ Velocity of man is still water
$V_r=$Velocity of river
$t_1=$ time taken by man to travel a distance AB when swimming in still water
$t_2=$ time taken by man to travel a distance AB when flowing with the river
$t_3=$ time taken by man to swim downstream in flowing river
$t_4=$ Time taken by man to swim upstream in flowing river
$\Rightarrow t_3=\frac{t_1{t}_2}{t_1+t_2}$
& $t_4=\frac{t_1{t}_2}{t_2-t_1}$