◆ QA · Arithmetic

Time, Speed & Distance, formulas + CAT PYQs

The single most tested arithmetic topic in CAT: relative speed, trains, boats & streams, races, circular tracks and clocks. Master the proportionality tricks and most questions collapse into one line of working.

16formulas

57CAT PYQs

★★★priority

Formula Sheet CAT PYQs

Formula & Concept Sheet

A-to-Z. Everything you need for this chapter, distilled from the Oswaal Revision Notes.

1The core relation

Speed = Distance ÷ Time. The whole topic is rearrangements of this.
Distance = Speed × Time; Time = Distance ÷ Speed.

Speed = Distance / Time

2Proportionality (the real shortcut)

Time constant → Speed ∝ Distance: S₁/S₂ = D₁/D₂.
Distance constant → Speed ∝ 1/Time: S₁/S₂ = T₂/T₁.
Most CAT problems are solved by spotting which quantity is constant.

D const ⇒ S₁/S₂ = T₂/T₁

3Unit conversion

km/hr → m/s: multiply by 5/18.
m/s → km/hr: multiply by 18/5.

1 km/hr = 5/18 m/s · 1 m/s = 18/5 km/hr

4Average speed

Always = Total distance ÷ Total time. Never just the mean of speeds.
Equal times at two speeds → arithmetic mean: (S₁+S₂)/2.
Equal distances at two speeds → harmonic mean.

Avg (equal dist) = 2·S₁·S₂ / (S₁ + S₂)

5Relative speed

Same direction → subtract: |S₁ − S₂|.
Opposite directions / towards each other → add: S₁ + S₂.
Time to meet/catch = gap distance ÷ relative speed.

Opposite: S₁ + S₂ · Same: S₁ − S₂

6Trains, distance to cover

Crossing a pole/man (length ≈ 0): distance = length of train.
Crossing a platform/another train: distance = sum of both lengths.
Use relative speed when the object is also moving.

Cross object: D = L(train) + L(object)

7Two trains crossing

Opposite directions: time = (L₁ + L₂) / (S₁ + S₂).
Same direction: time = (L₁ + L₂) / (S₁ − S₂).

t = (L₁ + L₂) / (S₁ ± S₂)

8Boats & streams, basics

B = boat in still water, R = stream speed.
Downstream D = B + R; Upstream U = B − R.

D = B + R · U = B − R

9Boats, recover B and R

From the two speeds: add and halve, subtract and halve.
If equal distances up & down: time-up/time-down = D/U (inverse of speeds).

B = (D + U)/2 · R = (D − U)/2

10Races, terminology

"A beats B by y m": when A finishes, B is y m short → speeds ∝ distances in equal time.
"A gives B a start of y m": B begins y m ahead.
"Dead heat": all reach the finish at the same instant.

Equal time ⇒ S(A)/S(B) = D(A)/D(B)

11Chained beats

If A beats B in ratio and B beats C in ratio, multiply the ratios.
e.g. V(A):V(B) = 10:9 and V(B):V(C) = 10:9 ⇒ V(A):V(C) = 100:81.

V(A):V(C) = V(A):V(B) × V(B):V(C)

12Circular tracks, relative speed

Same direction: relative speed = V₁ − V₂.
Opposite direction: relative speed = V₁ + V₂.
Time to meet (anywhere) = Circumference / relative speed.

First meeting = Circumference / Rel. speed

13Circular, meet at start

First meeting at the starting point = LCM of each body's lap time.
Lap time of a body = Circumference / its own speed.

Meet at start = LCM(C/V₁, C/V₂, …)

14Clocks, hand angle

H = hour, M = minutes. Angle between the hands:
Minute hand moves 6°/min; hour hand 0.5°/min.

Angle = |30H − 5.5M| degrees

15Escalators & moving walkways

Effective speed = person's stepping speed ± escalator speed.
If the escalator helps, fewer of your own steps are needed; total steps (off) = your steps + escalator's steps in that time.

Steps(off) = Person steps + Escalator steps

16Equal-distance meeting (one-line)

Two bodies start towards each other; after meeting they take t₁ and t₂ to finish.
Then S₁/S₂ = √(t₂/t₁), speeds are in the inverse square-root of remaining times.

S₁ / S₂ = √(t₂ / t₁)

Practice questions generated · up to 100

Original easy-hard warm-up drills (not CAT PYQs). Pick the levels, generate a set, reveal answers.

CAT Previous-Year Questions

Real CAT questions with worked solutions from the Oswaal book. Difficulty: Easy Moderate Hard. Click any to reveal the solution.

CAT 1999

ModerateCAT 1999

1. Navjivan Express from Ahmedabad to Chennai leaves Ahmedabad at 6.30 a.m. and travels at 50 kmph towards Baroda situated 100 km away. At 7.00 a.m. Howrah-Ahmedabad Express leaves Baroda towards Ahmedabad and travels at 40 kmph. At 7.30 a.m. Mr Shah, the traffic controller at Baroda realizes that both the trains are running on the same track. How much time does he have to avert a head-on collision between the two trains?

(1) 15 min
(2) 20 min
(3) 25 min
(4) 30 min

Show solution

(2) 20 min. At 7:30, Navjivan is 50 km from A and Howrah-Ahmedabad is 20 km from B, so the gap between them is 100 − 50 − 20 = 30 km. Relative speed (towards each other) = 50 + 40 = 90 kmph. Time = 30/90 hr = 20 min.

CAT 2001

HardCAT 2001

2. Shyama and Vyom walk up an escalator (moving stairway). The escalator moves at a constant speed. Shyama takes three steps for every two of Vyom's steps. Shyama gets to the top of the escalator after having taken 25 steps. While Vyom (because his slower pace lets the escalator do a little more of the work) takes only 20 steps to reach the top. If the escalator were turned off, how many steps would they have to take to walk up?

(1) 40
(2) 50
(3) 60
(4) 80

Show solution

(2) 50. Ratio of Shyama:Vyom stepping rate = 3:2; steps taken 25:20 = 5:4, so their times are in ratio 25/3 : 20/2 = 5:6. If the escalator has x steps, it carries (x − 25) while Shyama steps and (x − 20) while Vyom steps, and these are in the same 5:6 ratio of times ⇒ (x − 25)/(x − 20) = 5/6 ⇒ x = 50 steps.

ModerateCAT 2001

3. A train X departs from station A at 11.00 a.m. for station B, which is 180 km away. Another train Y departs from station B at 11.00 a.m. for station A. Train X travels at an average speed of 70 km/hr and does not stop anywhere until it arrives at station B. Train Y travels at an average speed of 50 km/hr, but has to stop for 15 minutes at station C, which is 60 km away from station B enroute to station A. Ignoring the lengths of the trains, what is the distance, to the nearest km, from station A to point where the trains cross each other?

(1) 112
(2) 118
(3) 120
(4) None of these

Show solution

(1) 112 km. Y reaches C after 60/50 = 1.2 hr = 72 min, then halts 15 min, so at 87 min Y has done 60 km while X has done 87/60×70 = 101.5 km, they haven't met yet. After the halt, with Y again moving, let total t hr from start: 70t + 50(t − 1/4) = 180 ⇒ 120t = 192.5 ⇒ t = 192.5/120 hr. Distance from A = 70t ≈ 112 km.

CAT 2002

HardCAT 2002

4. Only a single rail track exists between station A and B on a railway line. One hour after the north bound superfast train N leaves station A for station B, a south passenger train S reaches station A from station B. The speed of the superfast train is twice that of a normal express train E, while the speed of a passenger train S is half that of E. On a particular day N leaves for station B from station A, 20 minutes behind the normal schedule. In order to maintain the schedule both N and S increased their speed. If the superfast train doubles its speed, what should be the ratio (approximately) of the speed of passenger train to that of the superfast train so that passenger train S reaches exactly at the scheduled time at the station A on that day.

(1) 1 : 3
(2) 1 : 4
(3) 1 : 5
(4) 1 : 6

Show solution

(4) 1 : 6. Normally d/s + d/p = 60 min with s = 4p ⇒ d = 4p/5. After the 20-min delay only 40 min remain; superfast runs at 2s = 8p and passenger at new speed xp: d/(8p) + d/(xp) = 40/60. Substituting d gives x = 24/17. So passenger:superfast = xp : 2s = (24/17) : 8 = 3 : 17 ≈ 1 : 6.

HardCAT 2002

5. There is a tunnel connecting city A & B. There is a CAT which is standing at 3/8 the length of the tunnel from A. It listens a whistle of the train and starts running towards the entrance where, the train and the CAT meet. In another case, the CAT started running towards the exit and the train again met the CAT at the exit. What is the ratio of their speeds?

(1) 4 : 1
(2) 1 : 2
(3) 8 : 1
(4) None of these

Show solution

(1) 4 : 1. Towards entrance the cat covers 3 parts; towards exit it covers 5 parts (3/8 vs 5/8 of the tunnel). The difference 5 − 3 = 2 parts of cat-distance corresponds to the extra 8 parts the train covers in that time. So speed ratio train:cat = 8 : 2 = 4 : 1.

CAT 2003

ModerateCAT 2003

6. In a 4000 meter race around a circular stadium having a circumference of 1000 meters, the fastest runner and the slowest runner reach the same point at the end of the 5th minute, for the first time after the start of the race. All the runners have the same starting point and each runner maintains a uniform speed throughout the race. If the fastest runner runs at twice the speed of the slowest runner, what is the time taken by the fastest runner to finish the race?

(1) 20 min
(2) 15 min
(3) 10 min
(4) 5 min

Show solution

(3) 10 min. Let slowest = x m/min, fastest = 2x. First meeting (same direction) after Circumference/(2x − x) = 1000/x = 5 min ⇒ x = 200, fastest = 400 m/min. Time to finish 4000 m = 4000/400 = 10 min.

HardCAT 2003

7. Two straight roads R₁ and R₂ diverge from a point A at an angle of 120°. Ram starts walking from point A along R₁ at a uniform speed of 3 km/hr. Shyam starts walking at the same time from A along R₂ at a uniform speed of 2 km/hr. They continue walking for 4 hours along their respective roads and reach points B and C on R₁ and R₂, respectively. There is a straight line path connecting B and C. Then Ram returns to point A after walking along the line segments BC and CA. Shyam also returns to A after walking along line segments CB and BA. Their speeds remain unchanged. The time interval (in hours) between Ram's and Shyam's return to the point A is:

(1) (10√19 + 26)/3
(2) (2√19 + 10)/3
(3) (√19 + 26)/3
(4) (√19 + 10)/3

Show solution

(2) (2√19 + 10)/3. AB = 4×3 = 12, AC = 4×2 = 8. By cosine rule, BC² = 8² + 12² − 2·8·12·cos120° = 64 + 144 + 96 = 304 ⇒ BC = 4√19. Ram's return time (BC at 3, then CA = 12 at 3) and Shyam's return time (CB at 2, then BA = 8 at 2) differ by (4√19 + 12)/2 − (4√19 + 8)/3 = (2√19 + 10)/3 hr.

CAT 2004

HardCAT 2004

8. A sprinter starts running on a circular path of radius r metres. Her average speed (in metres/minute) is πr during the first 30 seconds, πr/2 during next one minute, πr/4 during next 2 minutes, πr/8 during next 4 minutes, and so on. What is the ratio of the time taken for the nth round to that for the previous round?

(1) 4
(2) 8
(3) 16
(4) 32

Show solution

(3) 16. Each segment covers a fixed distance (one round = 2πr) with halving speeds, so the segment times go 1/2, 1, 2, 4, 8, 16, 32, 64,… One round = 4 such segments. Time for round (segments ½,1,2,4) = 15/2; time for next round (8,16,32,64) = 120. Ratio = 120 ÷ (15/2) = 16.

ModerateCAT 2004

9. Karan and Arjun run a 100-metre race, where Karan beats Arjun by 10 metres. To do a favour to Arjun, Karan starts 10 metres behind the starting line in a second 100-metre race. They both run at their earlier speeds. Which of the following is true in connection with the second race?

(1) Karan and Arjun reach the finishing line simultaneously.
(2) Arjun beats Karan by 1 metre.
(3) Arjun beats Karan by 11 metres.
(4) Karan beats Arjun by 1 metre.

Show solution

(4) Karan beats Arjun by 1 m. Speeds are in ratio 100:90 = 10:9. In race 2 Karan must run 110 m; in that time Arjun runs 110×9/10 = 99 m. So Karan finishes when Arjun is at 99 m, Karan wins by 1 m.

CAT 2005

ModerateCAT 2005

Directions (Q. 10 - 11): Answer the questions on the basis of the information given below. Ram and Shyam run a race between points A and B, 5 km apart. Ram starts at 9 a.m. from A at a speed of 5 km/hr, reaches B, and returns to A at the same speed. Shyam starts at 9:45 a.m. from A at a speed of 10 km/hr, reaches B and comes back to A at the same speed.

10. At what time do Ram and Shyam first meet each other?

(1) 10:00 a.m.
(2) 10:10 a.m.
(3) 10:20 a.m.
(4) 10:30 a.m.

Show solution

(2) 10:10 a.m. Ram takes 1 hr to reach B (at 10:00), Shyam 0.5 hr (reaches B at 10:15). When Ram reaches B at 10:00, Shyam is 2.5 km from A. They then close the remaining gap at combined 10 + 5 = 15 km/hr ⇒ 2.5/15 hr = 10 min ⇒ they meet at 10:10 a.m.

ModerateCAT 2005

11. At what time does Shyam overtake Ram?

(1) 10:20 a.m.
(2) 10:30 a.m.
(3) 10:40 a.m.
(4) 10:50 a.m.

Show solution

(2) 10:30 a.m. Shyam reaches B at 10:15; at that moment Ram is (15/60)×5 = 1.25 km from B (heading back to A). Now both move towards A; Shyam overtakes in 1.25/(10 − 5) = 0.25 hr = 15 min after 10:15 ⇒ 10:30 a.m.

CAT 2006

ModerateCAT 2006

12. Arun, Barun and Kiranmala start from the same place and travel in the same direction at speeds of 30, 40 and 60 km per hour respectively. Barun starts two hours after Arun. If Barun and Kiranmala overtake Arun at the same instant, how many hours after Arun did Kiranmala start?

(1) 3
(2) 3.5
(3) 4
(4) 4.5
(5) 5

Show solution

(3) 4 hours. When Barun starts, Arun has gone 60 km. Barun overtakes Arun in 60/(40 − 30) = 6 hr, by which point Barun (= overtaking point) is 6×40 = 240 km from start. Kiranmala covers 240 km in 240/60 = 4 hr; Arun reaches there in 240/30 = 8 hr. To arrive together Kiranmala starts 8 − 4 = 4 hr after Arun.

CAT 2008

HardCAT 2008

13. Rahim plans to drive from city A to station C, at the speed of 70 km per hour, to catch a train arriving there from B. He must reach C at least 15 minutes before the arrival of the train. The train leaves B, located 500 km south of A, at 8:00 a.m. and travels at a speed of 50 km per hour. It is known that C is located between west and northwest of B, with BC at 60° to AB. Also, C is located between south and southwest of A with AC at 30° to AB. The latest time by which Rahim must leave A and still catch the train is closest to

(1) 6:15 a.m.
(2) 6:30 a.m.
(3) 6:45 a.m.
(4) 7:00 a.m.

Show solution

(2) 6:30 a.m. Triangle ABC has B = 60°, A = 30°, so C = 90°. BC = AB·cos60° = 500×½ = 250 km; AC = AB·sin60° = 250√3. Train covers BC at 50 km/hr in 5 hr, reaching C at 13:00; Rahim must be there by 12:45. AC time = 250√3/70 ≈ 6 hr 11 min. So latest departure ≈ 12:45 − 6:11 = 6:34 ≈ 6:30 a.m.

CAT 2017

EasyCAT 2017TITA

14. A man leaves his home and walks at a speed of 12 km per hour, reaching the railway station 10 minutes after the train had departed. If instead he had walked at a speed of 15 km per hour, he would have reached the station 10 minutes before the train's departure. The distance (in km) from his home to the railway station is

Show solution

20 km. Time difference between the two trips = 10 + 10 = 20 min = 1/3 hr. So D/12 − D/15 = 1/3 ⇒ D/60 = 1/3 ⇒ D = 20 km.

HardCAT 2017

15. A man travels by a motor boat down a river to his office and back. With the speed of the river unchanged, if he doubles the speed of his motor boat, then his total travel time gets reduced by 75%. The ratio of the original speed of the motor boat to the speed of the river is

(1) √6 : √2
(2) √7 : 2
(3) 2√5 : 3
(4) 3 : 2

Show solution

(2) √7 : 2. Let stream = x, boat = y, one-way distance = d. Round-trip time t = d/(y+x) + d/(y−x) = 2yd/(y²−x²). Doubling the boat to 2y gives t/4 = 2(2y)d/(4y²−x²). Setting (t/4) equal: 16yd/(4y²−x²) = 2yd/(y²−x²) ⇒ 8(y²−x²) = 4y²−x² ⇒ 4y² = 7x² ⇒ y/x = √7/2.

EasyCAT 2017TITA

16. In a 10 km race, A, B and C, each running at uniform speed, get the gold, silver, and bronze medals, respectively. If A beats B by 1 km and B beats C by 1 km, then by how many metres does A beat C?

Show solution

1900 m. When A does 10 km, B does 9 km ⇒ V(A):V(B) = 10:9. Similarly V(B):V(C) = 10:9. So V(A):V(C) = 100:81. When A finishes 10 km, C has done 8.1 km, A beats C by 1.9 km = 1900 m.

ModerateCAT 2017

17. Arun drove from home to his hostel at 60 miles per hour. While returning home he drove half way along the same route at a speed of 25 miles per hour and then took a bypass road which increased his driving distance by 5 miles, but allowed him to drive at 50 miles per hour along this bypass road. If his return journey took 30 minutes more than his onward journey, then the total distance traveled by him is

(1) 55 miles
(2) 60 miles
(3) 65 miles
(4) 70 miles

Show solution

(3) 65 miles. Let one-way (direct) distance = D. Onward time = D/60. Return = (D/2)/25 + (D/2 + 5)/50, which exceeds onward by 0.5 hr. Solving (D/2)/25 + (D/2+5)/50 − D/60 = 0.5 ⇒ D = 30. Total = onward D + return (D + 5) = 30 + 35 = 65 miles.

ModerateCAT 2017

18. A motorbike leaves point A at 1 p.m. and moves towards point B at a uniform speed. A car leaves point B at 2 p.m. and moves towards point A at a uniform speed which is double that of the motorbike. They meet at 3:40 p.m. at a point which is 168 km away from A. What is the distance, in km, between A and B?

(1) 364
(2) 378
(3) 380
(4) 388

Show solution

(2) 378 km. The bike travels 168 km in 2 hr 40 min ⇒ bike speed = 63 km/hr, so car = 126 km/hr. The car runs from 2:00 to 3:40 = 1 hr 40 min, covering 126×(5/3) = 210 km. Total AB = 168 + 210 = 378 km.

CAT 2018

ModerateCAT 2018TITA

19. Point P lies between points A and B such that the length of BP is thrice that of AP. Car 1 starts from A and moves towards B. Simultaneously, car 2 starts from B and moves towards A. Car 2 reaches P one hour after car 1 reaches P. If the speed of car 2 is half that of car 1, then the time, in minutes, taken by car 1 in reaching P from A is

Show solution

12 minutes. Let AB = 4d, so AP = d, BP = 3d. Car 1 speed s, Car 2 speed 0.5s. Car 1 reaches P in d/s; Car 2 reaches P (covering 3d) in 3d/0.5s = 6d/s. Given 6d/s = d/s + 60 ⇒ 5d/s = 60 ⇒ d/s = 12 min.

ModerateCAT 2018

20. The distance from A to B is 60 km. Partha and Narayan start from A at the same time and move towards B. Partha takes four hours more than Narayan to reach B. Moreover, Partha reaches the mid-point of A and B two hours before Narayan reaches B. The speed of Partha, in km per hour, is

(1) 6
(2) 4
(3) 3
(4) 5

Show solution

(4) 5 km/hr. Let Partha's time = t, Narayan's = t − 4. Partha reaches midpoint (30 km) in t/2. Given t/2 = (t − 4) − 2 ⇒ t/2 = t − 6 ⇒ t = 12. Partha's speed = 60/12 = 5 km/hr.

HardCAT 2018

21. Points A, P, Q and B lie on the same line such that P, Q and B are, respectively, 100 km, 200 km and 300 km away from A. Cars 1 and 2 leave A at the same time and move towards B. Simultaneously, car 3 leaves B and moves towards A. Car 3 meets car 1 at Q, and car 2 at P. If each car is moving in uniform speed then the ratio of the speed of car 2 to that of car 1 is

(1) 2 : 7
(2) 2 : 9
(3) 1 : 2
(4) 1 : 4

Show solution

(4) 1 : 4. Car 3 meets Car 1 at Q (200 km from A): in equal time Car 1 did 200, Car 3 did 100 ⇒ V₁:V₃ = 2:1. Car 3 meets Car 2 at P (100 from A): Car 2 did 100, Car 3 did 200 ⇒ V₂:V₃ = 1:2. So V₁:V₂:V₃ = 4:1:2 ⇒ V₂:V₁ = 1:4.

ModerateCAT 2018TITA

22. Points A and B are 150 km apart. Cars 1 and 2 travel from A to B, but car 2 starts from A when car 1 is already 20 km away from A. Each car travels at a speed of 100 kmph for the first 50 km, at 50 kmph for the next 50 km, and at 25 kmph for the last 50 km. The distance, in km, between car 2 and B when car 1 reaches B is

Show solution

5 km. Car 1's total time A→B = 50/100 + 50/50 + 50/25 = 0.5 + 1 + 2 = 3.5 hr. Car 1 covers the first 20 km at 100 kmph in 12 min = 0.2 hr; that is when Car 2 starts, so Car 2 runs for 3.5 − 0.2 = 3.3 hr. In 3.3 hr Car 2 does 50 km (0.5 hr) + 50 km (1 hr), using 1.5 hr; the remaining 1.8 hr at 25 kmph covers 45 km. So Car 2 has gone 50 + 50 + 45 = 145 km, leaving 150 − 145 = 5 km to B.

ModerateCAT 2018TITA

23. 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 hrs. The difference between their speeds, in km per hour, is

Show solution

50 km/hr. Let speeds be v (from A) and u (from B). Same direction (east): the faster catches up over 350 km in 7 hr ⇒ |u − v| = 350/7 = 50 km/hr.

CAT 2019

ModerateCAT 2019

24. Two cars travel the same distance starting at 10:00 a.m. and 11:00 a.m., respectively, on the same day. They reach their common destination at the same point of time. If the first car travelled for at least 6 hours, then the highest possible value of the percentage by which the speed of the second car could exceed that of the first car is

(1) 20
(2) 30
(3) 25
(4) 10

Show solution

(1) 20%. For maximum difference, take the minimum: car 1 runs 6 hr, car 2 runs 5 hr (it starts 1 hr later, same finish). Same distance ⇒ speeds inverse to times ⇒ V₂:V₁ = 6:5. Excess = (6 − 5)/5 × 100 = 20%.

ModerateCAT 2019

25. 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

(1) 18π
(2) 14π
(3) 16π
(4) 12π

Show solution

(3) 16π. Per revolution A covers 60π cm, B covers 80π cm, so revolutions are in ratio 4:3. 4x − 3x = 5000 ⇒ x = 5000, B's revolutions = 3x = 15000. Distance = 15000×80π cm = 12π km. Speed = 12π × 60/45 = 16π km/hr.

ModerateCAT 2019

26. One can use three different transports which move at 10, 20, and 30 kmph, respectively. To reach from A to B, Amal took each mode of transport 1/3 of his total journey time, while Bimal took each mode of transport 1/3 of the total distance. The percentage by which Bimal's travel time exceeds Amal's travel time is nearest to

(1) 22
(2) 20
(3) 19
(4) 21

Show solution

(1) 22%. Amal (equal times) → arithmetic mean speed = (10+20+30)/3 = 20 kmph. Bimal (equal distances) → harmonic mean = 3/(1/10+1/20+1/30) = 180/11 kmph. Time ∝ 1/speed ⇒ time ratio Bimal:Amal = 20 : 180/11 = 11:9. Excess = 2/9 × 100 ≈ 22%.

ModerateCAT 2019

27. Two ants A and B start from a point P on a circle at the same time, with A moving clockwise and B moving anti-clockwise. They meet for the first time at 10:00 am when A has covered 60% of the track. If A returns to P at 10:12 am, then B returns to P at

(1) 10:25 am
(2) 10:45 am
(3) 10:18 am
(4) 10:27 am

Show solution

(4) 10:27 a.m. At first meeting A did 60%, B did 40%. A covers the remaining 40% in 12 min ⇒ A's 60% took 18 min (so meeting was at 10:00, start at 9:42). B covers 40% in the same 18 min, and must still cover 60%, taking 18×60/40 = 27 min from 10:00 ⇒ 10:27 a.m.

HardCAT 2019

28. A cyclist leaves A at 10 am and reaches B at 11 am. Starting from 10:01 am, every minute a motorcycle leaves A and moves towards B. Forty-five such motorcycles reach B by 11 am. All motorcycles have the same speed. If the cyclist had doubled his speed, how many motorcycles would have reached B by the time the cyclist reached B?

(1) 22
(2) 23
(3) 15
(4) 20

Show solution

(3) 15. The 45th motorcycle left at 10:45 and reached B at 11:00, so each motorcycle takes 15 min A→B. If the cyclist doubles speed he reaches B at 10:30. Motorcycles reaching B by 10:30 are those that left by 10:15, i.e. those leaving 10:01 to 10:15 = 15 motorcycles.

HardCAT 2019TITA

29. John jogs on track A at 6 kmph and Mary jogs on track B at 7.5 kmph. The total length of tracks A and B is 325 metres. While John makes 9 rounds of track A, Mary makes 5 rounds of track B. In how many seconds will Mary make one round of track A?

Show solution

48 seconds. John = 5/3 m/s, Mary = 25/12 m/s. In equal time 9A = (5/3)t and 5B = (25/12)t ⇒ 9A = 4B and A + B = 325 ⇒ B = 9A/4 ⇒ A + 9A/4 = 325 ⇒ A = 100 m. Mary on track A: 100 ÷ (25/12) = 48 s.

CAT 2020

HardCAT 2020

30. Two persons are walking beside a railway track at respective speeds of 2 and 4 km per hour in the same direction. A train came from behind them and crossed them in 90 and 100 seconds, respectively. The time, in seconds, taken by the train to cross an electric post is nearest to:

(1) 78
(2) 87
(3) 75
(4) 82

Show solution

(4) 82. Let train length l, speed s (km/hr). Then l = (s − 2)·90 and l = (s − 4)·100 (in consistent units) ⇒ (s − 2)/(s − 4) = 100/90 ⇒ s = 22 kmph. Length = (22 − 4)×100×(5/18) = 500 m. Time to cross a post = 500 ÷ (22×5/18) ≈ 82 s.

ModerateCAT 2020

31. 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:

(1) 90
(2) 70
(3) 100
(4) 80

Show solution

(1) 90. For two bodies meeting then finishing, S₁/S₂ = √(t₂/t₁) = √(20/45) = 2/3. So 60/S₂ = 2/3 ⇒ S₂ = 90 km/hr.

HardCAT 2020

32. A train travelled at one-thirds of its usual speed, and hence reached the destination 30 minutes after the scheduled time. On its return journey, the train initially travelled at its usual speed for 5 minutes but then stopped for 4 minutes for an emergency. The percentage by which the train must now increase its usual speed so as to reach the destination at the scheduled time, is nearest to

(1) 58
(2) 67
(3) 50
(4) 61

Show solution

(2) 67%. Let usual speed = s, usual time = t. At 1/3 the speed, time becomes 3t, and 3t − t = 30 min ⇒ t = 15 min. On the return, the train runs 5 min at usual speed then stops 4 min, so 9 min are gone and only 15 − 9 = 6 min remain to cover the same distance that usually needs the remaining 10 min. Since distance is fixed, speed ∝ 1/time, so the ratio of times 10 : 6 = 5 : 3 means speed must rise in ratio 3 : 5. Percentage increase = (5 − 3)/3 × 100 = 2/3 × 100 = 66.67% ≈ 67%.

ModerateCAT 2020

33. Leaving home at the same time, Amal reaches office at 10:15 am if he travels at 8 km/hr, and at 9:40 am if he travels at 15 km/hr. Leaving home at 9:10 am, at what speed, in km/hr, must he travel so as to reach office exactly at 10 am?

(1) 14
(2) 12
(3) 13
(4) 11

Show solution

(2) 12. Time difference 35 min = 35/60 hr; d/8 − d/15 = 35/60 ⇒ d(7/120) = 7/120 ⇒ d = 10 km. Departing 9:10 to arrive 10:00 gives 50 min = 5/6 hr ⇒ speed = 10 ÷ (5/6) = 12 km/hr.

ModerateCAT 2020

34. In a car race, car A beats car B by 45 km, car B beats car C by 50 km, and car A beats car C by 90 km. The distance (in km) over which the race has been conducted is:

(1) 475
(2) 550
(3) 450
(4) 500

Show solution

(3) 450. Let race = x. When A finishes, B is at x − 45; B beats C by 50 means when B does x, C does x − 50, so when B is at x − 45, C is at (x − 45)(x − 50)/x = x − 90. Solving x²−90x = x²−95x+2250 ⇒ x = 450 km.

HardCAT 2020

35. The distance from B to C is thrice that from A to B. Two trains travel from A to C via B. The speed of train 2 is double that of train 1 while traveling from A to B and their speeds are interchanged while traveling from B to C. The ratio of the time taken by train 1 to that taken by train 2 in traveling from A to C is:

(1) 1 : 4
(2) 5 : 7
(3) 4 : 1
(4) 7 : 5

Show solution

(2) 5 : 7. Let AB = x, BC = 3x; Train 1 speed = s (A→B) and 2s (B→C); Train 2 = 2s then s. T₁ = x/s + 3x/2s = 5x/2s; T₂ = x/2s + 3x/s = 7x/2s. Ratio T₁:T₂ = 5:7.

HardCAT 2020

36. Two circular tracks T1 and T2 of radii 100 m and 20 m, respectively touch at a point A. Starting from A at the same time, Ram and Rahim are walking on track T1 and track T2 at speeds 15 km/hr and 5 km/hr respectively. The number of full rounds that Ram will make before he meets Rahim again for the first time is:

(1) 4
(2) 2
(3) 5
(4) 3

Show solution

(4) 3. Lap time = circumference/speed ⇒ T₁: 2π·100/15, T₂: 2π·20/5. Ratio of one-round times = (2π·100/15) : (2π·20/5) = 5:3. They reunite at A after the LCM of laps; Ram completes 3 rounds in that interval.

ModerateCAT 2020

37. A and B are two points on a straight line. Ram runs from A to B while Rahim runs from B to A. After crossing each other, Ram and Rahim reach their destinations in one minute and four minutes, respectively. If they start at the same time, then the ratio of Ram's speed to Rahim's speed is:

(1) 1 : 2
(2) √2 : 1
(3) 2 : 1
(4) 2√2 : 1

Show solution

(3) 2 : 1. Two bodies crossing then finishing: S(Ram)/S(Rahim) = √(t(Rahim)/t(Ram)) = √(4/1) = 2 ⇒ ratio 2 : 1.

ModerateCAT 2020

38. Vimla starts for office every day at 9 a.m. and reaches exactly on time if she drives at her usual speed of 40 km/hr. She is late by 6 minutes if she drives at 35 km/hr. One day, she covers two-thirds of her distance to office in one-thirds of her usual time to reach office, and then stops for 8 minutes. The speed, in km/hr, at which she should drive the remaining distance to reach office exactly on time is:

(1) 29
(2) 27
(3) 26
(4) 28

Show solution

(4) 28. Same distance ⇒ 40/35 = (t+6)/t ⇒ t = 42 min usual time; distance = 35×48/60 = 28 km. She covers 2/3 (≈18.67 km) in 14 min, stops 8 min, leaving 42 − 22 = 20 min for the remaining 28/3 km ⇒ speed = (28/3)/(20/60) = 28 km/hr.

EasyCAT 2020

39. A and B are two railway stations 90 km apart. A train leaves A at 9:00 a.m., heading towards B at a speed of 40 km/hr. Another train leaves B at 10:30 a.m., heading towards A at a speed of 20 km/hr. The trains meet each other at

(1) 11:20 a.m.
(2) 11:45 a.m.
(3) 11:00 a.m.
(4) 10:45 a.m.

Show solution

(3) 11:00 a.m. By 10:30 the first train has gone 40×1.5 = 60 km, leaving a 30 km gap. Both now approach at 40 + 20 = 60 kmph ⇒ 30/60 = 0.5 hr ⇒ they meet at 11:00 a.m.

HardCAT 2020

40. Anil, Sunil, and Ravi run along a circular path of length 3 km, starting from the same point at the same time, and going in the clockwise direction. If they run at speeds of 15 km/hr, 10 km/hr, and 8 km/hr, respectively, how much distance in km will Ravi have run when Anil and Sunil meet again for the first time at the starting point?

(1) 4.2
(2) 4.8
(3) 5.2
(4) 4.6

Show solution

(2) 4.8. Lap times: Anil 3/15, Sunil 3/10 hr. Anil & Sunil meet at start after LCM(3/15, 3/10) = 3/5 hr. Ravi at 8 km/hr covers 8 × 3/5 = 4.8 km.

CAT 2021

ModerateCAT 2021 · Slot 1

41. 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

(1) 184
(2) 192
(3) 190
(4) 180

Show solution

(3) 190. Faster train speed v₁ = 160/12 = 40/3 m/s. Slower v₂ = 40/3 − 6×5/18 = 35/3 m/s. Crossing (opposite) ⇒ (160 + l₂)/(v₁+v₂) = 14 ⇒ (160 + l₂)/(40/3 + 35/3) = 14 ⇒ 160 + l₂ = 25×14 = 350 ⇒ l₂ = 190 m.

HardCAT 2021 · Slot 2

42. 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

(1) 2 : 3
(2) 5 : 3
(3) 2 : 1
(4) 3 : 2

Show solution

(4) 3 : 2. Relative speed = 8x. In the first 46 s the relative motion covers l(B); in the next 69 s it covers l(A). Since speed is constant, l(A)/l(B) = 69/46 = 3/2.

HardCAT 2021 · Slot 3TITA

43. Mira and Amal walk along a circular track, starting from the same point at the same time. If they walk in the same direction, then in 45 minutes, Amal completes exactly 3 more rounds than Mira. If they walk in opposite directions, then they meet for the first time exactly after 3 minutes. The number of rounds Mira walks in one hour is

Show solution

8. Let track length = x. Same direction: v₁ − v₂ = 3x/45 = x/15. Opposite: v₁ + v₂ = x/3. Adding/subtracting gives Mira's speed v₂ = 2x/15 per min, so in 15 min Mira does 2 rounds ⇒ in 60 min, 8 rounds.

ModerateCAT 2022 · Slot 1

45. Trains A and B start travelling at the same time towards each other with constant speeds from stations X and Y, respectively. Train A reaches station Y in 10 minutes while train B takes 9 minutes to reach station X after meeting train A. Then, the total time taken, in minutes, by train B to travel from station Y to station X is:

(1) 12
(2) 6
(3) 15
(4) 10

Show solution

(3) 15. Let the meeting take t minutes after start. After meeting, A takes (10 − t) min to cover B's pre-meet distance, and B takes 9 min to cover A's pre-meet distance. For two bodies meeting then finishing, t² = (10 − t)·9, i.e. the meeting time is the geometric mean of the two post-meeting times. So t² = 9(10 − t) ⇒ t² + 9t − 90 = 0 ⇒ t = 6. Then B's total time = t + 9 = 6 + 9 = 15 minutes.

CAT 2022

HardCAT 2022 · Slot 2

46. Two ships meet mid-ocean, and then, one ship goes south and the other ship goes west, both travelling at constant speeds. Two hours later, they are 60 km apart. If the speed of one of the ships is 6 km per hour more than the other one, then the speed, in km per hour, of the slower ship is:

(1) 24
(2) 18
(3) 20
(4) 12

Show solution

(2) 18. Let slower = x, faster = x + 6. In 2 hr they cover 2x and 2(x+6); these are perpendicular legs. (2x)² + (2x+12)² = 60² ⇒ x² + 6x − 432 = 0 ⇒ (x − 18)(x + 24) = 0 ⇒ x = 18 km/hr.

HardCAT 2022 · Slot 3

47. Two ships are approaching a port along straight routes at constant speeds. Initially, the two ships and the port formed an equilateral triangle with sides of length 24 km. When the slower ship travelled 8 km, the triangle formed by the new positions of the two ships and the port became right-angled. When the faster ship reaches the port, the distance, in km, between the other ship and the port will be:

(1) 4
(2) 6
(3) 12
(4) 8

Show solution

(3) 12. Slower ship now 24 − 8 = 16 km from port (AS = 16), angle A = 60°. cos60° = AF/AS ⇒ AF = 8, so faster has travelled 24 − 8 = 16 km when slower travelled 8 km ⇒ speed ratio faster:slower = 2:1. When faster covers the full 24 km to port, slower covers 12 km, leaving 24 − 12 = 12 km from the port.

ModerateCAT 2022 · Slot 3

48. Two cars travel from different locations at constant speeds. To meet each other after starting at the same time, they take 1.5 hours if they travel towards each other, but 10.5 hours if they travel in the same direction. If the speed of the slower car is 60 km/hr, then the distance travelled, in km, by the slower car when it meets the other car while travelling towards each other, is:

(1) 100
(2) 90
(3) 150
(4) 120

Show solution

(2) 90. Distance covered by the slower car before meeting (towards each other) = 60 × 1.5 = 90 km.

CAT 2023

ModerateCAT 2023 · Slot 1

49. Brishti went on an 8-hour trip in a car. Before the trip, the car had travelled a total of x km till then, where x is a whole number and is palindromic, i.e., x remains unchanged when its digits are reversed. At the end of the trip, the car had travelled a total of 26862 km till then, this number again being palindromic. If Brishti never drove at more than 110 km/h, then the greatest possible average speed at which she drove during the trip, in km/h, was

(1) 90
(2) 80
(3) 100
(4) 110

Show solution

(3) 100. Max distance in 8 hr at ≤110 is 880 ⇒ x = 26862 − 880 = 25982 (not a palindrome). Try avg 100: distance 800 ⇒ x = 26062, which is palindromic. So the greatest feasible average speed is 100 km/h.

ModerateCAT 2023 · Slot 1TITA

50. 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

Show solution

972 km. Meet-then-finish: S(Arvind)/S(Surbhi) = √(t(Surbhi)/t(Arvind)) = √(24/6) = 2 ⇒ Surbhi = 27 km/h. Distance AB = (pre-meet by Arvind) + (pre-meet by Surbhi). Surbhi's pre-meet time = Arvind's finish time = 6 hr ⇒ AB = 27×24 + 6×54 = 648 + 324 = 972 km.

HardCAT 2023 · Slot 2

51. Ravi is driving at a speed of 40 km/h on a road. Vijay is 54 metres behind Ravi and driving in the same direction as Ravi. Ashok is driving along the same road from the opposite direction at a speed of 50 km/h and is 225 metres away from Ravi. The speed, in km/h, at which Vijay should drive so that all the three cross each other at the same time, is

(1) 67.2
(2) 58.8
(3) 61.6
(4) 64.4

Show solution

(3) 61.6. Ravi 40 km/h = 100/9 m/s; Ashok 50 km/h = 125/9 m/s. Ravi and Ashok (opposite) close 225 m at (100/9 + 125/9) = 25 m/s ⇒ they meet in 9 s. Vijay must also close his 54 m gap on Ravi in 9 s: (V − 100/9)×9 = 54 ⇒ 9V − 100 = 54 ⇒ V = 154/9 m/s = 61.6 km/h.

HardCAT 2023 · Slot 3

52. A boat takes 2 hours to travel downstream a river from part A to part B, and 3 hours to return to part A. Another boat takes a total of 6 hours to travel from part B to part A and return to part B. If the speeds of the boats and the river and constant, then the time, in hours, taken by the slower boat to travel from part A to part B is

(1) 3(3 + √5)
(2) 3(3 − √5)
(3) 3(√5 − 1)
(4) 12(√5 − 2)

Show solution

(2) 3(3 − √5). Let boat speed x and river speed y. First boat (downstream 2 hr, upstream 3 hr): 2(x + y) = 3(x − y) ⇒ x = 5y; take y = 1, x = 5 ⇒ distance d = 2(x + y) = 12 km. Since the first boat's downstream time (2 hr) is less than the second's, the second boat is slower; let its speed be S: 12/(S − 1) + 12/(S + 1) = 6 ⇒ 2[(S − 1) + (S + 1)]/(S² − 1) = 1 ⇒ 4S = S² − 1 ⇒ S² − 4S − 1 = 0 ⇒ S = (4 + √20)/2 = 2 + √5. Time A→B (downstream) = 12/(S + 1) = 12/(3 + √5) = 12(3 − √5)/4 = 3(3 − √5) hours.

CAT 2024 & 2025, recent

Fresh questions distributed from the real CAT 2024 & CAT 2025 papers into this chapter.

HardCAT 2024 · Slot 1

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

Show solution

(D) 12. For two bodies meeting then finishing, the meeting time is the geometric mean of the post-meeting times: t² = t₁·t₂. They cross at 90 min. Let Sunil take x min after meeting to reach A; Anil takes x + 75 min to reach B. So 90² = x(x + 75) ⇒ x² + 75x − 8100 = 0 ⇒ (x − 60)(x + 135) = 0 ⇒ x = 60. Anil's full travel time = 90 + (x + 75) = 90 + 135 = 225 min = 15/4 hr. Anil's speed = 45 ÷ (15/4) = 12 km/hr.

ModerateCAT 2024 · Slot 3

A train travelled a certain distance at a uniform speed. Had the speed been 6 km per hour more, it would have needed 4 hours less. Had the speed been 6 km per hour less, it would have needed 6 hours more. The distance, in km, travelled by the train is

(A) 720
(B) 800
(C) 780
(D) 640

Show solution

(A) 720. Let speed = s, time = t, distance d = st. Speed +6 ⇒ (s+6)(t−4) = st ⇒ −2s + 3t − 12 = 0. Speed −6 ⇒ (s−6)(t+6) = st ⇒ s − t − 6 = 0. Solving: s = 30, t = 24 ⇒ d = 30 × 24 = 720 km.

HardCAT 2025 · Slot 1

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

Show solution

(D) 86400. Total time = 3 hr = 180 min; the four times are in AP with first term 30 min. AP summing to 180 with first term 30 ⇒ 30, 40, 50, 60 min (common difference 10). First speed = 960 m/min. Speeds in AP: 960, 960+d, 960+2d, 960+3d. Total distance = 224 km = 224000 m = 960·30 + (960+d)·40 + (960+2d)·50 + (960+3d)·60. This gives 960·180 + d(40+100+180) = 224000 ⇒ 172800 + 320d = 224000 ⇒ d = 160 m/min. Fourth speed = 960 + 480 = 1440 m/min; distance = 1440 × 60 = 86400 m.

HardCAT 2025 · Slot 2 TITA

Rita and Sneha can row a boat at 5 km/h and 6 km/h in still water, respectively. In a river flowing with a constant velocity, Sneha takes 48 minutes more to row 14 km upstream than to row the same distance downstream. If Rita starts from a certain location in the river, and returns downstream to the same location, taking a total of 100 minutes, then the total distance, in km, Rita will cover is

Show solution

8. Let stream speed = v. Sneha: 14/(6−v) − 14/(6+v) = 48/60 = 4/5 hr ⇒ 14·2v/(36−v²) = 4/5 ⇒ 140v = 4(36−v²) ⇒ v² + 35v − 36 = 0 ⇒ (v+36)(v−1) = 0 ⇒ v = 1 km/h. Rita: upstream speed = 5−1 = 4 km/h, downstream = 5+1 = 6 km/h. Let one-way distance = d. Total time 100 min = 5/3 hr: d/4 + d/6 = 5/3 ⇒ 5d/12 = 5/3 ⇒ d = 4 km. Total distance Rita covers = 2d = 8 km.

HardCAT 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

(A) 12
(B) 20
(C) 18
(D) 15

Show solution

(D) 15. Usual time = 6 hr (5 pm to 11 pm), so total distance = 6x, where x = initial speed. Let k = distance covered at speed x before the stop. Scenario 1: stop 20 min = 1/3 hr, so remaining (6x − k) is done at (x+3) and total driving time = 17/3 hr: k/x + (6x − k)/(x+3) = 17/3 ⇒ x² = 51x − 9k …(1). Scenario 2: stop 30 min = 1/2 hr at (x+5), driving time = 11/2 hr: k/x + (6x − k)/(x+5) = 11/2 ⇒ x² = 55x − 10k …(2). Subtracting: 4x = k. Substituting into (1): x² = 51x − 36x = 15x ⇒ x = 15 km/h.

HardCAT 2025 · Slot 3 TITA

Ankita walks from A to C through B, and runs back through the same route at a speed that is 40% more than her walking speed. She takes exactly 3 hours 30 minutes to walk from B to C as well as to run from B to A. The total time, in minutes, she would take to walk from A to B and run from B to C, is

Show solution

444. Take walk speed = 5 and run speed = 1.4 × 5 = 7 (ratio 5:7). Let AB = a, BC = b. "Takes exactly 3.5 hr to walk B→C as well as to run B→A" means each equals 3.5 hr: walk B→C = b/5 = 3.5 hr, and run B→A = a/7 = 3.5 hr. Required = walk A→B + run B→C = a/5 + b/7. From a/7 = 3.5 ⇒ a/5 = 3.5 × 7/5 = 4.9 hr. From b/5 = 3.5 ⇒ b/7 = 3.5 × 5/7 = 2.5 hr. Total = 4.9 + 2.5 = 7.4 hr = 444 minutes.