CAT 2025 Lesson : Time & Speed - Past Questions

Observation/Strategy

1) As the cat travels a shorter distance to meet the train at A, the train will first pass through A and go to B.
2) Ratio of speeds is the same as ratio of distance covered.
Let the length of the tunnel be $x$.

Scenario 1: When the cat covers $\dfrac{3}{8}x$, it meets the train at the start of the tunnel, A.
Scenario 2: When the cat covers the remaining $\dfrac{5x}{8}$, it meets the train at the end of the tunnel, B.

As compared to scenario 1, in scenario 2, the additional distance covered by the train is the length of the tunnel $(x)$ and by the cat is $\dfrac{5x}{8} - \dfrac{3x}{8} = \dfrac{x}{4}$

Ratio of Speeds of train and cat $=$ Ratio of Distance covered in a given time $= x : \dfrac{x}{4} = 4 : 1$

Alternatively

Let the distance between the train and City A be y and City A and City B be 8x.

Let the speeds of the train and cat be $s$ and $c$ respectively.






Scenario 1: Time taken $= \dfrac{y}{s} = \dfrac{3x}{c}$ ⇒ $\dfrac{y}{3x} = \dfrac{s}{c} \longrightarrow(1)$

Scenario 2: Time taken $= \dfrac{y + 8x}{s} = \dfrac{5x}{c}$ ⇒ $\dfrac{y + 8x}{5x} = \dfrac{s}{c} \longrightarrow(2)$

Equating $(1)$ and $(2)$,

$\dfrac{y}{3x} = \dfrac{y + 8x}{5x}$

⇒ $5y = 3y + 24x$

⇒ $y = 12x$

Substituting this in (1),

$\dfrac{s}{c} = \dfrac{y}{3x} = \dfrac{12x}{3x} = 4 : 1$

Answer: $(1) 4 : 1$

Observation/Strategy

1) Ratio of speeds is reciprocal of ratio of time taken.
2) We can form the equations as the time taken in the two scenarios are equal.

Ratio of time taken while walking and driving = 8 : 1
Ratio of walking and driving speed $= 1 : 8$

Let the walking and driving speeds be $s$ and $8s$ respectively. Let the distance between temple and office be $x$ km.






Time taken to walk from to office = Time taken to walk to house and then drive to office

$\dfrac{x}{s} = \dfrac{1}{s} + \dfrac{1 + x}{8s}$

⇒ $x - 1 = \dfrac{x + 1}{8}$

⇒ $8x - 8 = x + 1$

⇒ $x = \dfrac{9}{7}$

Answer: $(3) \dfrac{9}{7}$

Observation/Strategy

1) Time taken by the three is minimized if all of them are in movement (never idle) and when all three reach Mathura at the same time.
2) We can draw the path and form the equations as the time taken should be equal for all three.

Let D and M be Delhi and Mathura respectively.

Let Mukesh and Suresh start on the bike while Dinesh starts by walk. Mukesh should drop Suresh at Point A, from where Suresh will walk to Mathura. Mukesh from point A should come back to pick up Dinesh at point B and then ride to Mathura, such that all three reach at the same time.

Time taken by Mukesh $= \dfrac{x + y + y + y + z}{60} = \dfrac{x}{60} + \dfrac{3y}{60} + \dfrac{z}{60} \longrightarrow(1)$

Time taken by Suresh $\dfrac{x}{15} + \dfrac{y + z}{60} = \dfrac{x}{15} + \dfrac{y}{60} + \dfrac{z}{60} \longrightarrow(2)$

Time taken by Dinesh $= \dfrac{x + y}{60} + \dfrac{z}{15} = \dfrac{x}{60} + \dfrac{y}{60} + \dfrac{z}{15} \longrightarrow(3)$

$(1), (2)$ and $(3)$ are equal. Let's start with equating $(2)$ and $(3)$,

$\dfrac{x}{15} + \dfrac{y}{60} + \dfrac{z}{60} = \dfrac{x}{60} + \dfrac{y}{60} + \dfrac{z}{15}$

$\dfrac{x}{15} - \dfrac{x}{60} = \dfrac{z}{15} - \dfrac{z}{60}$

⇒ $x = z \longrightarrow(4)$

Equating $(1)$ and $(2)$ and substituting $(4)$,

$\dfrac{x}{60} + \dfrac{3y}{60} + \dfrac{z}{60} = \dfrac{x}{15} + \dfrac{y}{60} + \dfrac{z}{60}$

⇒ $\dfrac{3y}{60} - \dfrac{y}{60} = \dfrac{x}{15} - \dfrac{x}{60}$

⇒ $\dfrac{2y}{60} = \dfrac{3x}{60}$ ⇒ $y = \dfrac{3x}{2}$

Total distance = $x + y + z = 300$ km

⇒ $x + \dfrac{3x}{2} + x = 300$ ⇒ $x = \dfrac{600}{7}$

As all of them take the same time, we can substitute in any of $(1), (2)$ or $(3)$ to find the time taken.
Substituting in $(1)$, $\dfrac{x}{60} + \dfrac{3y}{60} + \dfrac{z}{60} = \dfrac{1}{60} \times \left( x + \dfrac{9x}{2} + x \right) = \dfrac{1}{60} \times \dfrac{13x}{2}$

$= \dfrac{13}{120} \times \dfrac{600}{7} = \dfrac{65}{7} = 9\dfrac{2}{7}$ hours

Answer: $(2) 9\dfrac{2}{7}$ hours