Time, Speed, Distance, Boats and Trains Previous Year Questions (PYQs) for CAT: 17+ Solved Questions with Step-by-Step Solutions

Solve 17+ Time, Speed, Distance, Boats and Trains previous year questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Time, Speed, Distance, Boats and Trains

↔

Chapter Journey

Time, Speed, Distance, Boats and Trains

Step 1 • 11 CAT PYQs • Importance 0.76

🚶 Linear Motion and Relative Speed

Master distance-speed-time, meetings, chases, speed changes, stoppages, and multi-person motion.

Step 2 • 2 CAT PYQs • Importance 0.30

🚣 Boats, Streams and Water Travel

Apply upstream-downstream logic using still-water speed and stream speed.

Step 3 • 3 CAT PYQs • Importance 0.35

🚆 Trains, Crossings and Track Problems

Handle train lengths, poles, platforms, and opposite-direction crossings.

Step 4 • 3 CAT PYQs • Importance 0.35

🕒 Clocks, Circular Motion and Escalators

Convert movement into relative angular speed or effective speed on moving platforms.

By the end of this chapter: you should be able to convert every motion story into distance covered, time taken, and speed used.

Linear Motion and Relative Speed

Arithmetic → Time, Speed, Distance, Boats and Trains → Topic 1

Linear Motion and Relative Speed

Motion questions become easy when you track who covers which distance in whose time.

D=ST

Basic distance-speed-time

Meeting and chasing

Speed-change equations

CAT relative-speed patterns

Time, Speed, Distance, Boats and Trains: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Quantitative Ability NAT

Arvind travels from town A to town B, and Surbhi from town B to town A, both starting at the same time along the same route. After meeting each other, Arvind takes 6 hours to reach town B while Surbhi takes 24 hours to reach town A. If Arvind travelled at a speed of 54 km/h, then the distance, in km, between town A and town B is

Correct Answer:

972

Step-by-Step Solution

This is an after-meeting relative-speed problem — recognisable because two travellers start at the same time from opposite ends, meet somewhere in between, and each is given the time taken to finish the remaining distance after the meeting.

Step 1: Understand what "after meeting" means.

Before meeting, Arvind covers the distance from A to the meeting point, and Surbhi covers the distance from B to the meeting point, both in the same time $T$ . After meeting, Arvind covers the remaining distance (B to meeting point, which Surbhi had already covered) in 6 hours, and Surbhi covers the remaining distance (A to meeting point, which Arvind had already covered) in 24 hours.

Step 2: Write the two key equations.

Let $S_{a} = 54$ km/h be Arvind's speed and $S_{b}$ be Surbhi's speed, and $T$ the time until they meet.

$S_{a} \times T = distance B to meeting point = S_{b} \times 24 ...(i)$

$S_{b} \times T = distance A to meeting point = S_{a} \times 6 ...(ii)$

Step 3: Divide (i) by (ii) to find the speed ratio.

$\frac{S _{a} T}{S _{b} T} = \frac{S _{b} \times 24}{S _{a} \times 6} ⟹ \frac{S _{a}}{S _{b}} = \frac{24 S _{b}}{6 S _{a}} ⟹ S_{a}^{2} = 4 S_{b}^{2} ⟹ \frac{S _{a}}{S _{b}} = 2$

So $S_{b} = \frac{S _{a}}{2} = \frac{54}{2} = 27$ km/h.

Step 4: Find the meeting time $T$ .

There is a useful shortcut here: $T = t_{1} \times t_{2}$ where $t_{1}, t_{2}$ are the two after-meeting times.

$T = 6 \times 24 = 144 = 12 hours$

(You can also verify this directly from equation (ii): $S_{b} \times T = S_{a} \times 6 ⟹ 27 T = 324 ⟹ T = 12$ .)

Step 5: Find the total distance.

$D = (S_{a} + S_{b}) \times T = (54 + 27) \times 12 = 81 \times 12 = 972$

Answer: 972 km.

Common trap: stopping after finding only one traveller's pre-meeting distance (like $54 \times 12 = 648$ ) — that's only half the journey. The full distance needs both travellers' pre-meeting distances added together, i.e. $(S_{a} + S_{b}) \times T$ .

Question 2 · Quantitative Ability MCQ

Two places A and B are 45 kms apart and connected by a straight road. Anil goes from A to B while Sunil goes from B to A. Starting at the same time, they cross each other in exactly 1 hour 30 minutes. If Anil reaches B exactly 1 hour 15 minutes after Sunil reaches A, the speed of Anil, in km per hour, is

A.
18
B.
16
C.
14
D.
12

Correct Answer:

Step-by-Step Solution

This is a meeting-then-continuing relative speed question. Recognise it because two people start together from opposite ends, cross each other once, and then each continues to the other's starting point — and we are told the time gap between their two arrivals.

Step 1 — use the meeting condition.

Let Anil's speed be $a$ km/h and Sunil's speed be $s$ km/h. They start together and meet after 1.5 hours, together covering the full 45 km:

$1.5 a + 1.5 s = 45 ⟹ a + s = 30$

Step 2 — find each person's remaining distance after meeting.

By the time they meet, Anil has covered $1.5 a$ km, so his remaining distance to B is what Sunil already covered: $1.5 s$ km. Similarly, Sunil's remaining distance to A is $1.5 a$ km.

Step 3 — find the extra time each needs after meeting.

$Anil’s extra time = \frac{1.5 s}{a}, Sunil’s extra time = \frac{1.5 a}{s}$

Step 4 — use the arrival-time gap.

Anil reaches B exactly $1 h 15 min = 1.25$ h after Sunil reaches A. Since both traveled the same 1.5 h before meeting, this gap is exactly the difference in their extra times:

$\frac{1.5 s}{a} - \frac{1.5 a}{s} = 1.25$

Factor the left side:

$1.5 \times \frac{s ^{2} - a ^{2}}{a s} = 1.25$

Since $s^{2} - a^{2} = (s - a) (s + a) = (s - a) (30)$ :

$1.5 \times \frac{30 ( s - a )}{a s} = 1.25$

$\frac{45 ( s - a )}{a s} = 1.25 ⟹ 36 (s - a) = a s$

Step 5 — solve using $a + s = 30$ .

Substitute $s = 30 - a$ :

$36 (30 - 2 a) = a (30 - a)$

$1080 - 72 a = 30 a - a^{2}$

$a^{2} - 102 a + 1080 = 0$

Using the quadratic formula, discriminant $= 10 2^{2} - 4 (1080) = 10404 - 4320 = 6084 = 7 8^{2}$ :

$a = \frac{102 \pm 78}{2} = 90 or 12$

Since $a + s = 30$ , $a = 90$ is impossible (would make $s$ negative). So $a = 12$ .

Step 6 — verify.

$a = 12$ , $s = 18$ . Anil's remaining distance after meeting $= 1.5 (18) = 27$ km, extra time $= 27/12 = 2.25$ h, total time $= 3.75$ h. Sunil's remaining distance $= 1.5 (12) = 18$ km, extra time $= 18/18 = 1$ h, total time $= 2.5$ h. Gap $= 3.75 - 2.5 = 1.25$ h $= 1$ h 15min ✓.

Trap: students often assume Anil's remaining distance is $45 - 1.5 a$ computed independently, forgetting this equals $1.5 s$ exactly because both people started together — using this equivalence is what makes the algebra solvable without knowing $a$ and $s$ separately upfront.

The final answer is 12 km/h, option D.

Question 3 · Quantitative Ability MCQ

A bus starts at 9 am and follows a fixed route every day. One day, it traveled at a constant speed of 60 km per hour and reached its destination 3.5 hours later than its scheduled arrival time. Next day, it traveled two-thirds of its route in one-third of its total scheduled travel time, and the remaining part of the route at 40 km per hour to reach just on time. The scheduled arrival time of the bus is

A.
7 : 30 pm
B.
7 : 00 pm
C.
9 : 00 pm
D.
10 : 30 pm

Correct Answer:

Step-by-Step Solution

This is a "scheduled travel vs actual travel" pattern - recognisable because the question compares two different days' journeys against one unknown scheduled arrival time. The method is to express both days' travel times using the same unknowns (distance D and scheduled time T), then solve the resulting equations.

Step 1 - Set up Day 1.

On Day 1, the bus travels the whole distance D at a constant 60 km/h, taking 3.5 hours MORE than the scheduled travel time T.

$\frac{D}{60} = T + 3.5 ...(i)$

Step 2 - Set up Day 2.

On Day 2, the bus covers two-thirds of the distance in one-third of the scheduled travel time T (this part is not linked to any given speed, it's just a time fact).

The remaining one-third of the distance is covered at 40 km/h, taking $\frac{D /3}{40} = \frac{D}{120}$ hours.

Since Day 2 ends exactly on time, the total time used must equal T:

$\frac{T}{3} + \frac{D}{120} = T$

Step 3 - Simplify Day 2's equation.

$\frac{D}{120} = T - \frac{T}{3} = \frac{2 T}{3}$

$D = 120 \times \frac{2 T}{3} = 80 T ...(ii)$

Step 4 - Solve equations (i) and (ii) together.

From (i): $D = 60 T + 210$ .

Substitute into (ii): $60 T + 210 = 80 T$

$20 T = 210 \Rightarrow T = 10.5 hours$

Step 5 - Find the scheduled arrival time.

The bus starts at 9:00 am, and the scheduled travel time is 10.5 hours.

$9 : 00 am + 10.5 hours = 7 : 30 pm$

Trap to avoid: don't confuse "one-third of the total scheduled travel time" with a distance fraction - it is a TIME fraction of T, separate from the distance fraction of D. Mixing the two up leads to an unsolvable or wrong equation.

The scheduled arrival time is 7:30 pm, option A.

Question 4 · Quantitative Ability MCQ

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

A.
12
B.
15
C.
18
D.
20

Correct Answer:

Step-by-Step Solution

Key idea: This is a stoppage compensation question, recognisable because a traveller stops partway, then increases speed to still arrive on time, and two different stop/speed scenarios are given.

Why this method: The total distance is fixed. The stop reduces available travel time, so the increased speed must cover the remaining distance in less time. Two scenarios give two equations.

Step-by-step working:

Set up the basic journey:

Start 5 pm, arrive 11 pm: total time = 6 hours
Let initial speed = $v$ km/h
Total distance = $6 v$ km

Scenario 1 (20 min stop, speed +3):

Travels for time $t$ at speed $v$ , covering $v t$ km
Stops for $\frac{1}{3}$ hour
Remaining distance = $6 v - v t = v (6 - t)$
Remaining time = $6 - t - \frac{1}{3}$
New speed = $v + 3$
Equation: $v (6 - t) = (v + 3) (6 - t - \frac{1}{3})$ ... (1)

Scenario 2 (30 min stop, speed +5):

Same stopping point, so same $t$ and same $v t$ covered before stop
$v (6 - t) = (v + 5) (6 - t - \frac{1}{2})$ ... (2)

Simplify equation (1):

$v (6 - t) = (v + 3) (\frac{17}{3} - t)$
$6 v - v t = \frac{17 v}{3} + 17 - v t - 3 t$
$6 v = \frac{17 v}{3} + 17 - 3 t$
$\frac{18 v - 17 v}{3} = 17 - 3 t$
$\frac{v}{3} = 17 - 3 t$
$v = 51 - 9 t$ ... (3)

Simplify equation (2):

$v (6 - t) = (v + 5) (\frac{11}{2} - t)$
$6 v - v t = \frac{11 v}{2} + \frac{55}{2} - v t - 5 t$
$6 v = \frac{11 v}{2} + \frac{55}{2} - 5 t$
$\frac{12 v - 11 v}{2} = \frac{55}{2} - 5 t$
$\frac{v}{2} = \frac{55}{2} - 5 t$
$v = 55 - 10 t$ ... (4)

Solve the system:

From (3) and (4): $51 - 9 t = 55 - 10 t$
$t = 4$
$v = 51 - 36 = 15$

Verify:

Distance = $6 \times 15 = 90$ km
Scenario 1: Travel 4h at 15 = 60 km. Stop 20 min. Remaining 30 km in $\frac{5}{3}$ h at 18 km/h = 30 km ✓
Scenario 2: Travel 4h at 15 = 60 km. Stop 30 min. Remaining 30 km in $\frac{3}{2}$ h at 20 km/h = 30 km ✓

Answer: Rahul's initial speed was 15 km/h.

Question 5 · Quantitative Ability MCQ

Shruti travels a distance of 224 km in four parts for a total travel time of 3 hours. Her speeds in these four parts follow an arithmetic progression, and the corresponding time taken to cover these four parts follow another arithmetic progression. If she travels at a speed of 960 meters per minute for 30 minutes to cover the first part, then the distance, in meters, she travels in the fourth part is

A.
76800
B.
112000
C.
96000
D.
86400

Correct Answer:

Step-by-Step Solution

This is an AP-speeds-with-AP-times question, recognisable because both the speeds AND the times of the four parts form arithmetic progressions, and we must find the common differences of both APs using the given totals.

Step 1: Convert given data to consistent units (meters and minutes).

Total distance = 224 km = 224000 m. Total time = 3 hours = 180 minutes.

Step 2: Find the first part's time.

First part: speed = 960 m/min for 30 minutes, so $t_{1} = 30$ min.

Step 3: Set up the AP of times.

Let the times be $t_{1}, t_{2}, t_{3}, t_{4}$ in AP with common difference $e$ :

$t_{1} = 30, t_{2} = 30 + e, t_{3} = 30 + 2 e, t_{4} = 30 + 3 e$

Since total time is 180 minutes:

$4 (30) + (0 + 1 + 2 + 3) e = 180$

$120 + 6 e = 180 ⟹ e = 10$

So the times are 30, 40, 50, 60 minutes.

Step 4: Set up the AP of speeds.

Let the speeds be $s_{1}, s_{2}, s_{3}, s_{4}$ in AP with common difference $d$ , and $s_{1} = 960$ :

$s_{1} = 960, s_{2} = 960 + d, s_{3} = 960 + 2 d, s_{4} = 960 + 3 d$

Step 5: Use total distance = sum of (speed × time) for each part.

$s_{1} t_{1} + s_{2} t_{2} + s_{3} t_{3} + s_{4} t_{4} = 224000$

Compute each term:

$960 \times 30 = 28800$

$(960 + d) \times 40 = 38400 + 40 d$

$(960 + 2 d) \times 50 = 48000 + 100 d$

$(960 + 3 d) \times 60 = 57600 + 180 d$

Step 6: Add all terms and solve for $d$ .

$28800 + 38400 + 48000 + 57600 + (40 + 100 + 180) d = 224000$

$172800 + 320 d = 224000$

$320 d = 51200 ⟹ d = 160$

Step 7: Find the fourth part's speed and distance.

$s_{4} = 960 + 3 (160) = 960 + 480 = 1440 m/min$

$Distance_{4} = s_{4} \times t_{4} = 1440 \times 60 = 86400 m$

Why the trap exists: it's easy to assume the times AP has the SAME common difference as speeds, or to forget converting km to meters (since the first part's speed is given in m/min), both of which throw off the final numeric answer.

The answer is 86400 meters.

Time, Speed, Distance, Boats and Trains Previous Year Questions (PYQs) for CAT: 17+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Time, Speed, Distance, Boats and Trains

Time, Speed, Distance, Boats and Trains

🚶 Linear Motion and Relative Speed

🚣 Boats, Streams and Water Travel

🚆 Trains, Crossings and Track Problems

🕒 Clocks, Circular Motion and Escalators

Linear Motion and Relative Speed

Linear Motion and Relative Speed

Time, Speed, Distance, Boats and Trains: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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