A straight road connects points $A$ and $B.$ Car $1$ travels from $A$ to $B$ and Car $2$ travels from $B$ to $A$ , both leaving at the same time. After meeting each other, they take $45$ minutes and $20$ minutes, respectively, to complete their journeys. If Car $1$ travels at the speed of $60 km/hr$ , then the speed of Car $2,$ in km/hr, is

100

✅ Correct Option: 3

Looking at this problem, I need to find Car 2's speed when two cars meet on a road and then take different times to finish their journeys.

Let me set up what we know:

Car 1: goes $A \to B$ at $60$ km/hr, takes $45$ min after meeting to reach $B$
Car 2: goes $B \to A$ at unknown speed, takes $20$ min after meeting to reach $A$
Both cars start at the same time

Key insight: When the cars meet, they've been traveling for the same amount of time!

Let me call the meeting point $M$ , and let $t =$ time (in hours) from start until they meet.

At the meeting point $M$ :

Car 1 has traveled: $60t$ km (distance $AM$ )
Car 2 has traveled: $v_2 t$ km (distance $BM$ ), where $v_2$ is Car 2's speed

After meeting:

Car 1 takes $45$ min $= 0.75$ hr to go from $M$ to $B$

So distance $MB = 60 \times 0.75 = 45$ km

Car 2 takes $20$ min $= \frac{1}{3}$ hr to go from $M$ to $A$

So distance $MA = v_2 \times \frac{1}{3}$ km

The crucial connection:

Distance $AM$ (what Car 1 already traveled) $=$ Distance $MA$ (what Car 2 needs to travel)
Distance $BM$ (what Car 2 already traveled) $=$ Distance $MB$ (what Car 1 needs to travel)

This gives us:

$60t = \frac{v_2}{3}$ ... (equation 1)

$v_2 t = 45$ ... (equation 2)

From equation 2: $t = \frac{45}{v_2}$

Substituting into equation 1:

$60 \times \frac{45}{v_2} = \frac{v_2}{3}$

$\frac{2700}{v_2} = \frac{v_2}{3}$

$2700 \times 3 = v_2^2$

$8100 = v_2^2$

$v_2 = 90$

Therefore, Car 2's speed is $90$ km/hr.

CAT 2021 Slot 2

Two trains $A$ and $B$ were moving in opposite directions, their speeds being in the ratio $5: 3$ . The front end of $A$ crossed the rear end of $B$ $46$ seconds after the front ends of the trains had crossed each other. It took another $69$ seconds for the rear ends of the trains to cross each other. The ratio of length of train $A$ to that of train $B$ is

CAT 2019 Slot 1

The wheels of bicycles $A$ and $B$ have radii $30$ cm and $40$ cm, respectively. While traveling a certain distance, each wheel of $A$ required $5000$ more revolutions than each wheel of $B$ . If bicycle $B$ traveled this distance in $45$ minutes, then its speed, in km per hour, was

CAT 2017 Slot 2

A motorbike leaves point $A$ at $1$ pm and moves towards point $B$ at a uniform speed. A car leaves point $B$ at $2$ pm and moves towards point $A$ at a uniform speed which is double that of the motorbike. They meet at $3:40$ pm at a point which is $168 \mathrm{~km}$ away from $A$ . What is the distance, in km, between $A$ and $B$ ?

2025 Slot 3

Rahul starts on his journey at 5 pm at a constant speed so that he reaches his destination at 11 pm the same day. However, on his way, he stops for 20 minutes, and after that, increases his speed by 3 km per hour to reach on time. If he had stopped for 10 minutes more, he would have had to increase his speed by 5 km per hour to reach on time. His initial speed, in km per hour, was

CAT 2018 Slot 2

On a long stretch of east-west road, $A$ and $B$ are two points such that $B$ is $350$ km west of $A$ . One car starts from $A$ and another from $B$ at the same time. If they move towards each other, then they meet after $1$ hour. If they both move towards east, then they meet in $7 \mathrm{hrs}$ . The difference between their speeds, in km per hour, is

CAT 2021 Slot 1

Two trains cross each other in $14$ seconds when running in opposite directions along parallel tracks. The faster train is $160$ m long and crosses a lamp post in $12$ seconds. If the speed of the other train is $6$ km/hr less than the faster one, its length, in m, is

Solution

Related questions:

Keyboard Shortcuts

CAT Past Year Questions (Topic-Wise)