Loading...

Snap 2026 Notification Out. Check out what changed this year! | Solve previous year questions in quiz format to check your preparation | CAT Free Mock Test Is Live on the Website! Click here to test your preparation.
Login Register

Account

Log In Register

Contact Info

Follow Us

Trending Searches
CAT Exams CAT & OMETs Mock test Banking Mock test
Shopping cart

Your Cart is empty

Go to Shop
Subtotal:
₹ 0.00
View Cart Checkout
End Practice
Sections
TIME, SPEED & DISTANCE (LEVEL - 3)
TIME, SPEED & DISTANCE (LEVEL - 3)
Q1–10 of 50
1

With an average speed of 50 km/hr, a train reaches its destination in time. If it goes with an average speed of 40 km/hr, it is late by 24 minutes. The total journey is

Correct Answer: D. 80 km
Explanation:


Let total distance be $D\text{ km}$.

$$\text{Time difference} = \frac{D}{40} - \frac{D}{50} = \frac{24}{60}\text{ hours}$$

$$\frac{5D - 4D}{200} = \frac{2}{5} \implies \frac{D}{200}
= \frac{2}{5} \implies D = 80\text{ km}$$

Answer: (d) 80 km

2

Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. If he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.?

Correct Answer: C. 12 kmph
Explanation:


Let distance be $D$ and planned time to reach at 1 P.M. be $T$ hours.

$$\frac{D}{10} = T + 1, \quad \frac{D}{15} = T - 1$$

Subtracting the equations:

$$\frac{D}{10} - \frac{D}{15} = 2 \implies \frac{3D - 2D}{30} = 2 \implies D = 60\text{ km}$$
$$\text{Since } \frac{60}{10} = T + 1 \implies 6 = T + 1 \implies T = 5\text{ hours}$$

To reach at 1 P.M. (5 hours journey time):

$$\text{Speed} = \frac{60}{5} = 12\text{ kmph}$$

Answer: (c) 12 kmph

3

Ravi walks to and fro to a shopping mall. He spends 30 minutes shopping. If he walks at a speed of 10 km an hour, he returns home at 19.00 hours. If he walks at 15 km an hour, he returns home at 18.30 hours. How far must he walk in order to return home at 18.15 hours?

Correct Answer: D. None of these
Explanation:


Let walking distance one-way be $D$ (total walking distance $= 2D$). Shopping time $= 30\text{ min} = 0.5\text{ hrs}$. Difference in total time $= 19:00 - 18:30 = 30\text{ min} = 0.5\text{ hrs}$.

$$\frac{2D}{10} - \frac{2D}{15} = 0.5 \implies \frac{6D - 4D}{30} = 0.5 \implies 2D = 15\text{ km}$$

So $D = 7.5\text{ km}$. Walking time at $10\text{ km/hr} = \frac{15}{10} = 1.5\text{ hrs}$. Total trip $= 1.5 + 0.5 = 2\text{ hrs}$ (started at 17.00). To return at 18.15 (total trip duration $= 1.25\text{ hrs} = 75\text{ min}$): Walking time required $= 75 - 30 = 45\text{ min} = 0.75\text{ hrs}$.

$$\text{Speed} = \frac{2D}{0.75} = \frac{15}{0.75} = 20\text{ km/hr}$$

Since $20\text{ km/hr}$ is not in the options:Answer: (d) None of these

4

A person travels 285 km in 6 hours in two stages. In the first part of the journey, he travels by bus at the speed of 40 km/hr. In the second part of the journey, he travels by train at the speed of 55 km/hr. How much distance does he travel by train?

Correct Answer: B. 165 km
Explanation:


Let time spent on train be $t$ hours. Time on bus $= 6 - t$ hours.

$$40(6 - t) + 55t = 285 \implies 240 + 15t = 285 \implies 15t = 45 \implies t = 3\text{ hrs}$$

$$\text{Distance by train} = 55 \times 3 = 165\text{ km}$$

Answer: (b) 165 km

5

A train covered a certain distance at a uniform speed. If the train had been 6 km/hr faster, then it would have taken 4 hours less than the scheduled time. And, if the train were slower by 6 km/hr, then the train would have taken 6 hours more than the scheduled time. The length of the journey is

Correct Answer: B. 720 km
Explanation:


Using the speed-distance-time formula $D = \frac{S(S + \Delta S)}{\Delta S} \times \Delta T$:

$$D = \frac{S(S + 6)}{6} \times 4 \quad \text{and} \quad D = \frac{S(S - 6)}{6} \times 6$$

Equating both:

$$4(S + 6) = 6(S - 6) \implies 4S + 24 = 6S - 36 \implies 2S = 60 \implies S = 30\text{ km/hr}$$
$$D = \frac{30 \times 36}{6} \times 4 = 720\text{ km}$$

Answer: (b) 720 km

6

A car travels from P to Q at a constant speed. If its speed were increased by 10 km/hr, it would have taken one hour lesser to cover the distance. It would have taken further 45 minutes lesser if the speed was further increased by 10 km/hr. What is the distance between the two cities?

Correct Answer: A. 420 km
Explanation:


Let original speed be $S$ and original time be $T$.

$$(S + 10)(T - 1) = ST \implies ST - S + 10T - 10 = ST \implies 10T - S = 10$$

For further speed increase of $10\text{ km/hr}$ (total increase $= 20\text{ km/hr}$, time decrease $= 1\text{ hr } 45\text{ min} = \frac{7}{4}\text{ hrs}$):

$$(S + 20)\left(T - \frac{7}{4}\right) = ST \implies 20T - \frac{7}{4}S = 35 \implies 80T - 7S = 140$$

Solving the simultaneous equations gives $S = 60\text{ km/hr}$ and $T = 7\text{ hours}$.

$$\text{Distance} = 60 \times 7 = 420\text{ km}$$

Answer: (a) 420 km

7

A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is

Correct Answer: C. 120 kmph
Explanation:


Let speed of car be $S_c$. Speed of train $S_t = 1.5 S_c$. Time taken by car $= \frac{75}{S_c}$. Time taken by train $= \frac{75}{1.5 S_c} + \frac{12.5}{60} = \frac{50}{S_c} + \frac{5}{24}$.

$$\frac{75}{S_c} - \frac{50}{S_c} = \frac{5}{24} \implies \frac{25}{S_c} = \frac{5}{24} \implies S_c = 120\text{ kmph}$$

Answer: (c) 120 kmph

8

Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?

Correct Answer: B. 10
Explanation:


$$\text{Stoppage time per hour} = \frac{\text{Speed without stoppages} - \text{Speed with stoppages}}{\text{Speed without stoppages}} \times 60\text{ min}$$

$$= \frac{54 - 45}{54} \times 60 = \frac{9}{54} \times 60 = 10\text{ minutes}$$

Answer: (b) 10

9

A bus covered a certain distance from village A to village B at the speed of 60 km/hr. However on its return journey it got stuck in traffic and covered the same distance at the speed of 40 km/hr and took 2 hours more to reach its destination. What is the distance covered between villages A and B?

Correct Answer: B. 240 km
Explanation:


Let distance between A and B be $D$.

$$\frac{D}{40} - \frac{D}{60} = 2 \implies \frac{3D - 2D}{120} = 2 \implies D = 240\text{ km}$$

Answer: (b) 240 km

10

A train covers a distance between two stations A and B in 45 minutes. If the speed of the train is reduced by 5 km/hr, then it covers the distance in 48 minutes. The distance between the stations A and B is

Correct Answer: B. 60 km
Explanation:


Let distance be $D$.

$$\text{Usual speed} = \frac{D}{45/60} = \frac{4D}{3}$$

$$\text{Reduced speed} = \frac{4D}{3} - 5$$

$$\left(\frac{4D}{3} - 5\right) \times \frac{48}{60}
= D \implies \left(\frac{4D}{3} - 5\right) \times \frac{4}{5} = D$$

$$\frac{16D}{15} - 4 = D \implies \frac{16D}{15} - D = 4 \implies \frac{D}{15} = 4 \implies D = 60\text{ km}$$

Answer: (b) 60 km

1 2 3 4 5
Next
Page 1 of 5
Home Courses PYQs Exams Login