CAT 2025 Lesson : Time & Speed - Escalators

In a second, John can take 3 steps while Mary can take 2 steps. John's steps are 20% longer than that of Mary. If John gives Mary a head start of 80 steps in a race, how long does it take John to catch-up with Mary?

Solution

Let a step of Mary be $1$ feet. Therefore, John's step would measure $1.2$ feet.

As Mary take $2$ steps every second, speed of Mary $= 2$ ft/s

As John takes 3 steps every second, speed of John $= 3 \times 1.2 = 3.6$ ft/s

John gives Mary a head start of $80$ steps, which is $80$ ft.

As they move in the same direction, relative speed $= 3.6 - 2 = 1.6$ ft/s

Time taken to catch-up $= \dfrac{80}{1.6} = 50$ seconds

Answer: $50$ seconds

Michael climbs up or down at a constant speed of 8 steps per second. He takes $15$ seconds to climb up the down-escalator and $5$ seconds to climb down the down-escalator. How long would it take for him to climb up if the escalator is not functioning (moving)?

Solution

Let the speed of the escalator be x steps per second. Relative going-up and going-down speeds are $8 - x$ and $8 + x$ respectively.

Ratio of time taken for going up and going down $= 15 : 5 = 3 : 1$

Ratio of speed for going up and going down $= \dfrac{1}{3} : \dfrac{1}{1} = 1 : 3$

$\dfrac{8 - x}{8 + x} = \dfrac{1}{3}$

⇒ $x = 4$ steps/second

Distance or Visible steps $15 \times (8 - 4) = 60$ steps

Time taken when escalator is not moving $= \dfrac{60}{8} = 7.5$ seconds

Alternatively

Let the speed of the escalator be $\bm{x}$ steps per second.

Michael takes 15 seconds to climb up the down-escalator.

Visible steps = Michael's steps up $-$ Escalator down movement
$= (15 \times 8) - (15 \times x)$ = 120 - 15x

Michael takes 5 seconds to climb down the down-escalator.

Visible steps $=$ Michael's steps down $+$ Escalator down movement
$= (5 \times 8) + (5 \times x) =$ 40 + 5x

Visible steps $= 120 - 15x = 40 + 5x$
⇒ $x = 4$

$\therefore$ Visible steps $= 40 + (5 \times 4) = 60$ steps

Time taken to go up a still escalator $= \dfrac{60}{8} = 7.5$ seconds

Answer: $7.5$ seconds

Kunal and Jinal start travelling at the same time, in a straight line, from points A and B towards points B and A respectively. After they meet each other on the way, Kunal and Jinal take 3 hours and 12 hours more to reach B and A respectively. How long did it take for them to meet each other?
Let $k$ km/hr and $j$ km/hr be the speeds of Kunal and Jinal respectively and $t$ hours be the time taken for them to meet.

Distances covered by Kunal and Jinal to meet are $kt$ km and $jt$ km.



After meeting, Kunal covers the rest of the distance of $jt$ km in 3 hours, i.e. $\dfrac{jt}{k} = 3$

After meeting, Jinal covers the rest of the distance of $kt$ km in 12 hours, i.e.$\dfrac{kt}{j} = 12$

Multiplying the two equations formed,

$\dfrac{jt}{k} \times \dfrac{kt}{j} = 3 \times 12$

⇒ $t^2 = 36$
⇒ $t = 6$ hours

Alternatively (Recommended Method)

Time taken to meet $= \sqrt{3 \times 12} = 6$ hours

Answer: $6$ hours