Time, Speed & Distance Questions for CAT 2025 with Solutions

The speed of Emily is 12 km/hr, and she starts 30 minutes before Jack, which implies the distance traveled by Emily in these 30 minutes is 6 km.

Now the effective distance is (66-6) = 60 km when Jack started moving from the opposite end. The ratio of the speed of Emily to the speed of Jack is 2: 3.

Hence, the distance, traveled by Emily after Jack started moving is 60*(2/5) = 24 km.

Hence, the total distance traveled by Emily when she meets Jack is (6+24) = 30 km.

The correct option is B

Speed of train = 72kmph = 20m/s
Total distance travelled by train = 150 + 850 =1000 m
Time taken = 1000/20 = 50 seconds
Since the boy waits for 5 seconds , the time taken by him to reach the last compartment = 50 -5 = 45 seconds
Total distance = 300m => Speed = $$\frac{300}{45\ }=\frac{20}{3}=6\ \frac{2}{3}\ $$ m/s

In 3 hours, the bus travelled 120 km.
Relative speed of the car with respect to bus = 60 - 40 = 20kmph
Time taken for the car to meet bus = 120/20 = 6 hours
Distance travelled by the bike in 6 hours = 100 * 6 = 600 km

In this question, it is given that during the initial journey from A to B => It takes 1 hour travelling at 30 kmph => distance between A and B = d = 30 kms.

Let us assume the speed of the train during the return journey is 's' kmph.

We know that the average speed is the ratio of total distance through out the journey to the total time taken.

Waiting time of the time during the return journey = 4 * 3 mins = 12 mins = $$\dfrac{12}{60}=\dfrac{1}{5}$$ hr

Therefore, Average Speed = Total distance/ Total time

=> $$40=\dfrac{\left(30+30\right)}{1+\dfrac{1}{5}+\dfrac{30}{s}}$$

=> $$\dfrac{6}{5}+\dfrac{30}{s}=\dfrac{3}{2}$$

=> $$\dfrac{30}{s}=\dfrac{3}{2}-\dfrac{6}{5}=\dfrac{3}{10}$$ => s = 100 kmph.

Let the length of the escalator be $$l$$.

Assuming the speed of the first escalator to be $$4x$$ and that of the second escalator will be $$5x$$.

Let us assume that the speed of Shyam be $$y$$.

As the question $$\left(4x+y\right)t_1=l$$ .....(I)

When going down the equation will be $$\left(5x+y\right)t_2=l$$ ....(II)

Equating both the equation

$$\left(5x+y\right)t_2=\left(4x+y\right)t_1$$

$$\frac{5x+y}{4x+y}=\frac{t_1}{t_2}$$

$$\frac{5x+y}{4x+y}=\frac{6}{5}$$

Upon rearranging we get

$$25x+5y=24x+6y$$

or $$y=x$$

Required ratio  = $$y: 4x$$ = $$x:4x$$ = 1:4

Let us assume that the time taken by Abhay and Mukesh to reach point B, which is 150 km, is "t" minutes.
It is given that Abhay stops after every 5 km. Therefore, he will stop 29 times before covering a distance of 150 km.

It is also given that Abhay stops for 3 minutes at every break, so the time consumed by Abhay during breaks = 29*3 = 87 minutes.
Thus, we can infer that Abhay actually drove for t - 87 minutes for travelling from point A to point B.
Therefore, the speed of Abhay = $$\dfrac{150}{t-87}$$ km/minute.

Similarly, Mukesh stops at the end of 3 km, so he will stop 49 times before covering a distance of 150 km.
Mukesh stops for 2 minutes at end of every 3 km, thus the time consumed by Mukesh during breaks = 49*2 = 98 minutes.
Hence, we can say that Mukesh drove for t-98 minutes
Therefore, the speed of Mukesh is $$\dfrac{150}{t-98}$$ km/minute.

It is given that the speed of Abhay is V and the speed of Mukesh is 3V/2.

Hence, comparing both speeds

=> $$\dfrac{150}{t - 87} = \dfrac{2}{3}\times \dfrac{150}{t - 98}$$

=> 3(t-98) = 2(t-87)

=> 3t-294 = 2t - 174

=> t = 120 minutes

Therefore, Abhay and Mukesh will reach point B in 120 minutes.

The ratio of speeds of A and B is 3 : 5. So, by the time bus A travels 375 km, bus B would have travelled 625 km. So, the first time that the two buses meet when travelling in opposite directions is at a distance of 125 km from city Y. Time taken for this meeting from the start of the journey = 6.25 hours.
From this moment onwards, the two buses will meet each other while travelling in opposite directions every 6.25 hours. The second meeting occurs at the midpoint of the two cities. The third meeting occurs at a point that is 125 km from city X. This is the final meeting between the two buses. Bus B comes to rest before the next meeting between the buses can take place. So, the answer is 3.

Converting given speed in metres per second convention:

Krushna’s speed = $$\frac{5}{18} \times 54 = 15$$ metres per second
Sudesh’s speed = $$\frac{5}{18}\times72=20$$ metres per second

So the relative speed of Krushna and Sudesh = 15 +20 = 35 metres per second

Since they are approaching towards at 35 mps hence the distance between them 30 seconds before they meet = Distance covered by them in 30 seconds with the relative speed
$$\Rightarrow$$ 30 * 35 = 1050 metres.

Hence, they will be 1050 metres apart 30 seconds prior to the meeting.

When Car 1 completes 2 rounds, Car 2 completes 3 rounds whereas when Car 2 completes 1 round, Car 3 completes 2 rounds.

Car1 -- 2rounds....... Car2 -- 3 rounds

Car2 -- 1round....... Car3 -- 2 rounds

By the time car1 completes 2 rounds, car2 completes 3 rounds. And subsequently, Car 3 will complete 6 rounds.

At this point, they will meet for the first time at the starting point.

The number of times they will meet = 50/6= 8.3

They will meet 8 times as we take an integral part of 8.3 as the number has to be an integer.

Option A