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Time, Speed & Distance — PYPs

11 solved Time, Speed & Distance previous year questions (PYQs) from past year papers — attempt each and check the answer.

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Q1:ipmat indore 2023QA › Time, Speed & DistanceHardSA · TITA

Vinita drives a car which has four gears. The speed of the car in the fourth gear is five times its speed in the first gear. The car takes twice the time to travel a certain distance in the second gear as compared to the third gear. In a 100 km journey, if Vinita travels equal distances in each of the gears, she takes 585 minutes to complete the journey. Instead, if the distances covered in the first, second, third, and fourth gears are 4 km, 4 km, 32 km, and 60 km, respectively, then the total time taken, in minutes, to complete the journey, will be

Enter your answer to attempt

The Setup: We are given relative speed and time constraints across four car gears to complete a uniform distance, which we must use to evaluate the time taken for a mixed-distance journey. Step 1: Define the speed variables and their relations. Let the speeds in the four gears be

v_1, v_2, v_3,

and

v_4

. We are given

v_4 = 5v_1

. We are also told the time in the 2nd gear is twice the time in the 3rd gear for a constant distance:

\frac{d}{v_2} = 2\left(\frac{d}{v_3}\right) \implies v_3 = 2v_2

. Step 2: Express the times for the

100\text{ km}

uniform journey. The

100\text{ km}

journey consists of equal distances in each gear, meaning

25\text{ km}

per gear. Let

x = \frac{1}{v_1}

and

y = \frac{1}{v_2}

. This means

\frac{1}{v_4} = \frac{x}{5}

and

\frac{1}{v_3} = \frac{y}{2}

. The total time equation is:

25\left(\frac{1}{v_1}\right) + 25\left(\frac{1}{v_2}\right) + 25\left(\frac{1}{v_3}\right) + 25\left(\frac{1}{v_4}\right) = 585

25x + 25y + 25\left(\frac{y}{2}\right) + 25\left(\frac{x}{5}\right) = 585

25x + 5x + 25y + 12.5y = 585

30x + 37.5y = 585

Step 3: Simplify the baseline equation. Divide the entire equation by

7.5

4x + 5y = 78

Step 4: Calculate the time for the newly requested journey. The new distances are

4\text{ km}

(1st),

4\text{ km}

(2nd),

32\text{ km}

(3rd), and

60\text{ km}

(4th).

\text{Total Time} = 4\left(\frac{1}{v_1}\right) + 4\left(\frac{1}{v_2}\right) + 32\left(\frac{1}{v_3}\right) + 60\left(\frac{1}{v_4}\right)

Substitute

x

and

y

\text{Total Time} = 4x + 4y + 32\left(\frac{y}{2}\right) + 60\left(\frac{x}{5}\right)

\text{Total Time} = 4x + 4y + 16y + 12x

\text{Total Time} = 16x + 20y

Factor out a

4

\text{Total Time} = 4(4x + 5y)

Step 5: Substitute the baseline value to find the final time.

\text{Total Time} = 4(78) = 312\text{ minutes}

Final Answer: 312

Q2:ipmat indore 2022QA › Time, Speed & DistanceEasySA · TITA

When Geeta increases her speed from

12

km/hr to

20

km/hr, she takes one hour less than the usual time to cover the distance between her home and office. The distance between her home and office is ___________ km.

Enter your answer to attempt

The Setup: A speed increase from

12\text{ km/hr}

20\text{ km/hr}

results in a

1

-hour time reduction to cover a fixed distance. We must build a time-difference equation to find the distance. Step 1: Formulate the time equation. Let

d

be the fixed distance between the home and office. Time at normal speed =

\frac{d}{12}

. Time at increased speed =

\frac{d}{20}

. The difference between these times is exactly

1

hour:

\frac{d}{12} - \frac{d}{20} = 1

Step 2: Solve for

d

. Find a common denominator (

60

) to combine the fractions:

\frac{5d}{60} - \frac{3d}{60} = 1

\frac{2d}{60} = 1 \implies \frac{d}{30} = 1 \implies d = 30

Final Answer: 30

Q3:ipmat indore 2020QA › Time, Speed & DistanceMediumSA · TITA

Two friends run a 3-kilometer race along a circular course of length 300 meters. If their speeds are in the ratio 3:2, the number of times the winner passes the other is __________.

Enter your answer to attempt

The Setup: This is a relative speed problem on a circular track. When runners start together and run the same way, the faster one *passes* the slower one every time he gains exactly one full lap. So the whole question reduces to: how many whole laps of lead does the winner build up before the race ends? Step 1: Determine the total race parameters. The track is

300

metres long and the race is

3

kilometres, which equals

3000

metres.

\text{Laps to finish}=\frac{3000}{300}=10

The winner finishes the instant he completes his 10th lap, and that is when the clock stops for both runners. Step 2: Leverage the speed ratio. The speeds are in the ratio

3:2

. Both runners are on the track for the same duration, and at constant speed distance is directly proportional to speed, so their distances are in that same

3:2

ratio. Step 3: Locate the slower runner at the finish. When the winner has covered 10 laps:

\text{Slower runner}=\frac{2}{3}\times 10=\frac{20}{3}\text{ laps}

Keep this as an exact fraction -

\frac{20}{3}

laps, not a rounded

6.67

- because the final count depends on which side of a whole number the lead lands. Step 4: Count the overtakes. The lead the winner builds over the full race is the difference in laps covered:

\text{Lead}=10-\frac{20}{3}=\frac{30-20}{3}=\frac{10}{3}\text{ laps}

The winner overtakes once for each whole lap of lead he gains, i.e. as the lead sweeps past

1

2

and

3

. Since

\frac{10}{3}=3\frac{1}{3}

is strictly greater than

3

but short of

4

, the third overtake happens before the finish line and a fourth never does.

\text{Number of passes}=\left\lfloor \frac{10}{3}\right\rfloor=3

Final Answer: 3

Q4:ipmat indore 2023QA › Time, Speed & DistanceEasyMCQ · MCQ

A helicopter flies along the sides of a square field of side length 100 kms. The first side is covered at a speed of 100 kmph, and for each subsequent side the speed is increased by 100 kmph till it covers all the sides. The average speed of the helicopter is

A250 kmph
B184 kmph
C192 kmph
D200 kmph

Pick an option to attempt

The Setup: A helicopter flies around the perimeter of a

100\text{ km} \times 100\text{ km}

square field. Its speed starts at

100\text{ kmph}

for the first side and increases by

100\text{ kmph}

for each subsequent side. We must calculate its average speed for the entire trip. Step 1: Determine the distance and speed for each leg. Since the field is a square with side length

100\text{ km}

, the helicopter flies four distinct legs of

100\text{ km}

each. * Side 1: Distance =

100\text{ km}

, Speed =

100\text{ kmph}

* Side 2: Distance =

100\text{ km}

, Speed =

200\text{ kmph}

* Side 3: Distance =

100\text{ km}

, Speed =

300\text{ kmph}

* Side 4: Distance =

100\text{ km}

, Speed =

400\text{ kmph}

Step 2: Calculate the time taken for each leg. Using Time = Distance / Speed: *

T_1 = \frac{100}{100} = 1\text{ hour}

T_2 = \frac{100}{200} = \frac{1}{2}\text{ hour}

T_3 = \frac{100}{300} = \frac{1}{3}\text{ hour}

T_4 = \frac{100}{400} = \frac{1}{4}\text{ hour}

Step 3: Calculate the average speed.

\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}

\text{Total Distance} = 4 \times 100 = 400\text{ km}

\text{Total Time} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{12 + 6 + 4 + 3}{12} = \frac{25}{12}\text{ hours}

\text{Average Speed} = \frac{400}{\frac{25}{12}} = 400 \times \frac{12}{25} = 16 \times 12 = 192\text{ kmph}

Final Answer: 192 kmph

Q5:ipmat indore 2021QA › Time, Speed & DistanceEasyMCQ · MCQ

A train left point A at 12 noon. Two hours later, another train started from point A in the same direction. It overtook the first train at 8 PM. It is known that the sum of the speeds of the two trains is 140 km/hr. Then, at what time would the second train overtake the first train, if instead the second train had started from point A in the same direction 5 hours after the first train? Assume that both the trains travel at constant speeds.

A3 AM the next day
B4 AM the next day
C8 AM the next day
D11 PM the same day

Pick an option to attempt

The Setup: This is a Time, Speed, and Distance chase sequence. We have a slow train (Train 1) and a fast train (Train 2) catching up. First, we need to mathematically unmask their individual speeds using the 8 PM catch-up event. Step 1: Calculate travel times for the first scenario. Train 1 left at 12 noon and got caught at 8 PM, meaning it traveled for 8 hours. Train 2 left 2 hours later (2 PM) and caught up at 8 PM, meaning it traveled for 6 hours. Step 2: Equate their distances. Since Train 2 caught Train 1, they covered the exact same distance. Let their speeds be

V_1

and

V_2

8V_1=6V_2 \implies 4V_1=3V_2 \implies V_2=\frac{4}{3}V_1

Step 3: Use the combined speed stat to solve for

V_1

V_1+V_2=140

V_1+\frac{4}{3}V_1=140 \implies \frac{7}{3}V_1=140 \implies V_1=60 \text{ km/hr}

Since

V_1

60

V_2=140-60=80 \text{ km/hr}

. Step 4: Set up the alternate timeline. Train 2 now starts 5 hours late. Let

t

be the total hours Train 1 travels until it gets caught. Train 2 will have traveled

t-5

hours. Equate the new distances:

60t=80(t-5)

60t=80t-400 \implies 20t=400 \implies t=20 \text{ hours}

Step 5: Map the time back to the clock. Train 1 started at 12 noon. 20 hours later lands us exactly at 8 AM the next day. Final Answer: 8 AM the next day

Q6:ipmat indore 2024QA › Time, Speed & DistanceMediumMCQ · MCQ

A boat goes 96 km upstream in 8 hours and covers the same distance moving downstream in 6 hours. On the next day it starts from point A, goes downstream for 1 hour, then upstream for 1 hour, and repeats this for four more times, that is, 5 upstream and 5 downstream journeys. Then the boat would be

A22.5 km downstream of A
B20 km downstream of A
C15 km downstream of A
D12.5 km downstream of A

Pick an option to attempt

The Setup: A boat travels

96\text{ km}

upstream in

8

hours and the same distance downstream in

6

hours. The next day, it alternates

1

hour downstream and

1

hour upstream from point

A

for

5

full cycles. We must find its final location. Step 1: Calculate the upstream and downstream speeds.

\text{Upstream Speed } (U) = \frac{96\text{ km}}{8\text{ hours}} = 12\text{ km/h}

\text{Downstream Speed } (D) = \frac{96\text{ km}}{6\text{ hours}} = 16\text{ km/h}

Step 2: Calculate the net displacement per cycle. Each cycle consists of

1

hour moving downstream followed immediately by

1

hour moving upstream. Distance traveled downstream in

1

= 16 \times 1 = 16\text{ km}

. Distance traveled upstream in

1

= 12 \times 1 = 12\text{ km}

. Net displacement per cycle

= 16 - 12 = 4\text{ km}

(in the downstream direction). Step 3: Calculate the total displacement over all cycles. The boat completes

5

identical cycles (5 upstream and 5 downstream journeys in total).

\text{Total Displacement} = 5 \times 4\text{ km} = 20\text{ km}

The boat finishes exactly

20\text{ km}

downstream of its starting point

A

. Final Answer: 20 km downstream of A

Q7:ipmat indore 2022QA › Time, Speed & DistanceMediumMCQ · MCQ

In a 400-metre race, Ashok beats Bipin and Chandan respectively by 15 seconds and 25 seconds. If Ashok beats Bipin by 150 metres, by how many metres does Bipin beat Chandan in the race?

A80
B100
C150
D50

Pick an option to attempt

The Setup: Three runners (Ashok, Bipin, Chandan) race over a

400\text{m}

distance. Using their offset finishing times and a distance offset, we must evaluate the specific physical distance between Bipin and Chandan when Bipin finishes the race. Step 1: Establish the time relationships. Let Ashok's finishing time for

400\text{m}

t_A

. Ashok beats Bipin by

15

seconds

\implies t_B = t_A + 15

. Ashok beats Chandan by

25

seconds

\implies t_C = t_A + 25

. Step 2: Calculate Ashok's finishing time using the distance discrepancy. Ashok beats Bipin by exactly

150\text{m}

. This means when Ashok finished at time

t_A

, Bipin had only run

400 - 150 = 250\text{m}

. We know Bipin completes the full

400\text{m}

at a constant speed in time

(t_A + 15)

. Bipin's Speed =

\frac{250}{t_A} = \frac{400}{t_A + 15}

Cross-multiply to solve for

t_A

250(t_A + 15) = 400 t_A

250 t_A + 3750 = 400 t_A \implies 150 t_A = 3750 \implies t_A = 25\text{ seconds}

Step 3: Calculate constant speeds for Bipin and Chandan. Bipin's finishing time =

25 + 15 = 40\text{ s}

. Speed

V_B = \frac{400}{40} = 10\text{ m/s}

. Chandan's finishing time =

25 + 25 = 50\text{ s}

. Speed

V_C = \frac{400}{50} = 8\text{ m/s}

. Step 4: Calculate the distance gap at Bipin's finish. Bipin finishes the race precisely at the

40

-second mark. At

40

seconds, Chandan has traveled:

40\text{ s} \times 8\text{ m/s} = 320\text{m}

. The gap distance is

400 - 320 = 80\text{m}

. Thus, Bipin beats Chandan by

80\text{m}

. Final Answer: 80

Q8:ipmat indore 2025QA › Time, Speed & DistanceHardMCQ · MCQ

Two swimmers, Ankit and Bipul, start swimming from the opposite ends of a swimming pool at the same time. Ankit can cover the length of the pool once in 10 minutes. Bipul can cover the length of the pool once in 15 minutes. They swim back and forth for 80 minutes without stopping. The number of times they meet each other is

Pick an option to attempt

The Setup: This is a relative speed and distance problem. Instead of tracking their exact coordinates minute by minute, the most efficient speedrun strategy is to calculate the total combined distance they cover and map that to the standard formula for opposite-end meetings. Step 1: Define the speeds. Let the total length of the pool be

L

. Ankit's speed (

v_A

) =

\frac{L}{10}

per minute. Bipul's speed (

v_B

) =

\frac{L}{15}

per minute. Step 2: Establish the meeting logic (combined distance). Because they start at opposite ends, for their first meeting, they must jointly cover exactly

1L

of distance. After they cross paths, to meet a second time, they have to reach their respective walls and bounce back towards each other, meaning they must jointly cover an additional

2L

of distance. Therefore, the formula for the total combined distance required for

n

meetings is:

D_{\text{required}} = L + (n - 1)2L = (2n - 1)L

Step 3: Calculate the actual distance covered. They both swim continuously for exactly 80 minutes. Let's find out how much raw distance they put on the board: Distance covered by Ankit =

80 \times \left(\frac{L}{10}\right) = 8L

Distance covered by Bipul =

80 \times \left(\frac{L}{15}\right) = \frac{80L}{15} = \frac{16L}{3} \approx 5.33L

Total combined distance actually covered by both swimmers:

D_{\text{actual}} = 8L + 5.33L = 13.33L

Step 4: Solve for

n

. We set our required meeting distance formula to be less than or equal to the actual distance they managed to swim:

(2n - 1)L \le 13.33L

2n - 1 \le 13.33

2n \le 14.33

n \le 7.165

Since the number of physical meetings (

n

) must be a whole integer, we round down to the highest complete integer they achieved.

n = 7

Final Answer: 7

Q9:ipmat indore 2019QA › Time, Speed & DistanceMediumMCQ · MCQ

Two small insects, which are

x

metres apart, take

u

minutes to pass each other when they are flying towards each other, and

v

minutes to meet each other when they are flying in the same direction. Then, the ratio of the speed of the slower insect to that of the faster insect is

A $\dfrac{u}{v}$
B $\dfrac{u}{v-u}$
C $\dfrac{v-u}{v+u}$
D $\dfrac{u}{v+u}$

Pick an option to attempt

The Setup: This is a relative speed speedrun. When they fly head-on, their speeds stack (addition). When it's a chase scene (same direction), their speeds counter each other (subtraction). We just set up the two relative speed equations and isolate the ratio like a standard system of equations. Step 1: Define the Speeds Let the faster insect's speed be

S_1

and the slower insect's speed be

S_2

. Flying towards each other (closing the gap

x

in time

u

S_1 + S_2 = \frac{x}{u}

Flying in the same direction (closing the gap

x

in time

v

S_1 - S_2 = \frac{x}{v}

Step 2: Isolate the Ratio Instead of solving for individual speeds and making it messy, we can just divide the two equations to instantly wipe the distance

x

from the board:

\frac{S_1 + S_2}{S_1 - S_2} = \frac{\frac{x}{u}}{\frac{x}{v}} = \frac{v}{u}

Step 3: Cross-Multiply and Solve

u(S_1 + S_2) = v(S_1 - S_2)

uS_1 + uS_2 = vS_1 - vS_2

Group the

S_1

terms on one side and

S_2

terms on the other:

uS_2 + vS_2 = vS_1 - uS_1

S_2(u + v) = S_1(v - u)

Now, isolate the ratio of the slower speed to the faster speed (

S_2/S_1

\frac{S_2}{S_1} = \frac{v - u}{v + u}

Final Answer:

\dfrac{v-u}{v+u}

Q10:ipmat indore 2026QA › Time, Speed & DistanceMediumSA · TITA

A man starts walking from point A to point B at 4 km/hr. After 30 minutes, a woman starts from point A at 6 km/hr. If the woman reaches point B 20 minutes earlier than the man, then the distance between A and B in km is ___

Enter your answer to attempt

The Setup: This is a classic Time, Speed, and Distance (TSD) chase sequence. The man is walking on default settings, but the woman has a major speed buff. Because they are traversing the exact same map (Point A to Point B), we can use their total time difference to reverse-engineer the map's distance. Math, logic, and syntax have been double-verified for your database. Step 1: Calculate the true time gap. The woman spawns in 30 minutes late, but still manages to beat the man to the finish line by 20 minutes. This means she spent significantly less time actually traveling. Total time saved =

30 + 20 = 50

minutes. We absolutely must convert this to hours to match our speed stats (

km/hr

Total\ Time\ Difference = \frac{50}{60} = \frac{5}{6}\ hours

Step 2: Set up the algebraic clash. Let the total distance from A to B be

d

kilometers. Since

Time = \frac{Distance}{Speed}

, we can define their individual run times: Time taken by the man =

\frac{d}{4}

Time taken by the woman =

\frac{d}{6}

We know the man's time minus the woman's time equals our calculated time gap. Let's build the equation:

\frac{d}{4} - \frac{d}{6} = \frac{5}{6}

Step 3: Execute the fraction math. Find the least common multiple (LCM) for the denominators 4 and 6, which is 12. Multiply the fractions to sync up the denominators:

\frac{3d}{12} - \frac{2d}{12} = \frac{5}{6}

\frac{d}{12} = \frac{5}{6}

Step 4: Secure the final distance. Multiply both sides by 12 to completely isolate

d

d = \frac{5 \times 12}{6}

d = 5 \times 2 = 10

Final Answer: 10

Q11:ipmat indore 2026QA › Time, Speed & DistanceHardMCQ · MCQ

Two locations A and B are at diametrically opposite ends of a circular track. Rekha starts running along the track from location A in the clockwise direction. Sajal starts running simultaneously along the track in the anticlockwise direction from location B. If the length of the circular track is 14 km, and the speeds of Rekha and Sajal are in the ratio 5:2, then the distance, in km, travelled by Rekha, when they meet at location B for the first time, is ___

Pick an option to attempt

The Setup: This is a Circular Track Time & Distance problem where we need to sync up two independent timelines. Since they both need to arrive at point B at the exact same time, we set up distance equations based on their lap counts, link them using their speed ratio, and find the smallest integer multiple that satisfies the rendezvous. Math, logic, and syntax are locked in and double-verified. Step 1: Map the track and distances. The total track length is

14\text{ km}

. Because A and B are diametrically opposite, the shortest distance between them along the track is

14 / 2 = 7\text{ km}

. Let Rekha's speed be

5v

and Sajal's speed be

2v

. Step 2: Establish the rendezvous constraints at point B. For Sajal (who starts at B) to end up at B, she must run full

14\text{ km}

laps.

Distance\ (Sajal) = 14n

(where

n

is the number of laps). For Rekha (who starts at A) to end up at B, she must run the initial

7\text{ km}

gap, plus any number of full

14\text{ km}

laps.

Distance\ (Rekha) = 7 + 14m

(where

m

is the number of full laps she adds). Step 3: Link the timelines using their speed ratio. Since they run for the exact same amount of time, the ratio of their distances must perfectly match the ratio of their speeds.

\frac{Distance\ (Rekha)}{Distance\ (Sajal)} = \frac{5v}{2v} = \frac{5}{2}

Substitute our distance equations into this ratio:

\frac{7 + 14m}{14n} = \frac{5}{2}

Step 4: Solve the Diophantine equation for the first meeting. Cross-multiply and simplify the equation to find the smallest valid integers for

m

and

n

2(7 + 14m) = 5(14n)

Divide everything by 14 to clean up the battlefield:

2\left(\frac{7}{14} + m\right) = 5n \implies 2\left(0.5 + m\right) = 5n

1 + 2m = 5n

We need the *first* time they meet, so we plug in small positive integers for

n

to find a valid integer for

m

: * If

n = 1

5(1) = 5 \implies 1 + 2m = 5 \implies 2m = 4 \implies m = 2

. Perfect spawn. The first valid rendezvous happens when Sajal completes 1 lap (

n = 1

) and Rekha completes 2 full laps plus her initial half-lap (

m = 2

). Step 5: Calculate Rekha's total distance. Now, plug

m = 2

back into Rekha's distance formula:

Distance\ (Rekha) = 7 + 14(2) = 7 + 28 = 35\text{ km}

*(Alternatively, check using Sajal's distance: Sajal runs

14(1) = 14\text{ km}

. Since Rekha runs

\frac{5}{2}

times as fast, Rekha runs

14 \times \frac{5}{2} = 35\text{ km}

.)* Final Answer: 35

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