# Arithmetic - Time, Speed & Distance - Previous Year CAT/MBA Questions

You can practice all previous year CAT questions from the topic Arithmetic - Time, Speed & Distance. This will help you understand the type of questions asked in CAT. It would be best if you clear your concepts before you practice previous year CAT questions.

**CAT 2022 QA Slot 1 | Arithmetic - Time, Speed & Distance**

Trains A and B start traveling 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

- A.
12

- B.
15

- C.
10

- D.
6

Answer: Option B

**Explanation** :

Let the time taken to meet after starting be t mins.

Time taken by A to reach Y after meeting = (10 – t) mins

Time taken by B to reach X after meeting = 9 mins

⇒ t^{2} = (10 - t) × 9

⇒ t^{2} +9t – 90 = 0

⇒ (t + 15)(t - 6) = 0

⇒ t = 6 (-15 is rejected)

Total time taken by B to reach Y = 6 + 9 = 15 mins.

Hence, option (b).

Workspace:

**CAT 2022 QA Slot 2 | Arithmetic - Time, Speed & Distance**

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

- A.
18

- B.
24

- C.
12

- D.
20

Answer: Option A

**Explanation** :

Let the speed of the slower ship be x km/hr.

Distance travelled by it in 2 hours = 2x kms

Speed of faster ship = (x + 6) km/hr.

Distance travelled by it in 2 hours = 2(x + 6) kms

⇒ (2x)^{2} + (2(x + 6))^{2} = (60)2

⇒ 4x^{2} + 4(x^{2} 12x + 36) = 3600

⇒ 8x^{2} + 48x + 144 = 3600

⇒ 8x^{2} + 48x - 3456 = 0

⇒ x^{2} + 6x - 432 = 0

⇒ (x + 24)(x - 18) = 0

⇒ x = 18 (-24 is rejected)

∴ Speed of slower person is 18 km/hr

Hence, option (a).

Workspace:

**CAT 2022 QA Slot 3 | Arithmetic - Time, Speed & Distance**

Moody takes 30 seconds to finish riding an escalator if he walks on it at his normal speed in the same direction. He takes 20 seconds to finish riding the escalator if he walks at twice his normal speed in the same direction. If Moody decides to stand still on the escalator, then the time, in seconds, needed to finish riding the escalator is

Answer: 60

**Explanation** :

Let speed of Moody and escalator be ‘m’ and ‘e’ steps/sec and total number of steps in the escalator be ‘N’.

We will assume that Moody is going in the same direction as the escalator. If that is not the case value of e will come out to be negative.

Moody takes 30 seconds at normal speed ⇒ N = 30m + 30 ×e …(1)

Moody takes 20 seconds at twice the speed ⇒ N = 20 × 2m + 20 ×e …(2)

4 × (1) – 3 × (2)

N = 60e

⇒ N/e = 60

∴ If someone was standing on the escalator, he/she will take 60 seconds to climb the escalator.

Hence, 60.

Workspace:

**CAT 2022 QA Slot 3 | Arithmetic - Time, Speed & Distance**

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 traveling towards each other, is

- A.
90

- B.
120

- C.
100

- D.
150

Answer: Option A

**Explanation** :

When travelling towards each other, the two cars take 1.5 hours to meet.

⇒ The distance travelled by slower car in this case till they meet = 1.5 × 60 = 90 kms.

Hence, option (a).

Workspace:

**CAT 2021 QA Slot 1 | Arithmetic - Time, Speed & Distance**

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

- A.
180

- B.
184

- C.
190

- D.
192

Answer: Option C

**Explanation** :

Let the length of the smaller train be L meters.

Also, let the speeds of faster and slower trains be ‘f’ and ‘s’ m/s.

Given, f - s = 6 × 5/18 = 5/3 m/s

⇒ f = s + 5/3 …(1)

Faster train crosses a lamp post in 12 seconds.

∴ 160/f = 12

⇒ f = 40/3

∴ s = f – 5/3 = 35/3

The two trains cross each other in 14 seconds.

∴ (160 + L)/(f + s) = 14

⇒ L = 14(f + s) – 160

⇒ L = 14 × 25 – 160

⇒ L = 350 – 160 = 190

Hence, option (c).

Workspace:

**CAT 2021 QA Slot 2 | Arithmetic - Time, Speed & Distance**

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

- A.
2 : 1

- B.
3 : 2

- C.
2 : 3

- D.
5 : 3

Answer: Option C

**Explanation** :

Let the lengths of trains A and B be ‘A’ and ‘B’ and their speeds be ‘a’ and ‘b’ respectively.

Time taken for front end of A to cross rear end of B = 46 seconds

⇒ 46 = $\frac{B}{a+b}$

⇒ B = 46a + 46b …(1)

Total time taken for rear ends of the two train to cross each other = (46 + 69) = 115 seconds

⇒ 115 = $\frac{A+B}{a+b}$

⇒ A + B = 115a + 115b …(2)

(2) - (1)

⇒ A = 69a + 69b …(3)

Now, $\frac{A}{B}$ = $\frac{69(a+b)}{46(a+b)}$ = $\frac{69}{46}$ = $\frac{3}{2}$

Hence, option (c).

Workspace:

**CAT 2021 QA Slot 3 | Arithmetic - Time, Speed & Distance**

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

Answer: 8

**Explanation** :

Let the length of the track be ‘L’ meter and speeds of Mira and Amal be ‘m’ and ‘a’ meters/minute respectively.

When they walk in opposite direction, they meet in 3 minutes.

∴ 3(m + a) = L

⇒ 3m + 3a = L …(1)

When they walk in same direction, Amal completes 3 rounds more than Mira. Hence, Amal completes 3L meters more than Mira.

∴ 45a – 45m = 3L …(2)

(1) × 15 - (2)

⇒ 90m = 12L

⇒ L/m = 7.5

Hence, it takes Mira 7.5 minutes to complete one round.

Hence, in 60 minutes she will cover = 60/7.5 = 8 rounds.

Hence, 8.

Workspace:

**CAT 2020 QA Slot 1 | Arithmetic - Time, Speed & Distance**

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

- A.
75

- B.
82

- C.
87

- D.
78

Answer: Option B

**Explanation** :

Let the speed and length of train be ‘s’ m/s and ‘t’ m respectively.

Speeds of two persons is 2 kmph and 4 kmph i.e., 5/9 m/s and 10/9 m/s

Train passes the first person in 90 seconds.

∴ $\frac{t}{s-\frac{5}{9}}$ = 90

⇒ t = 90s – 50 …(1)

Train passes the second person in 100 seconds.

∴ $\frac{t}{s-\frac{10}{9}}$ = 100

⇒ t = 100s – 1000/9 …(2)

(1) = (2)

⇒ 90s – 50 = 100s – 1000/9

⇒ s = 55/9 m/s and t = 500 m

∴ Time taken to cross an electric pole = $\frac{500}{55/9}$ = $\frac{4500}{55}$ = $\frac{900}{11}$ = 81.81 ≈ 82 seconds.

Hence, option (b).

Workspace:

**CAT 2020 QA Slot 1 | Arithmetic - Time, Speed & Distance**

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

- A.
100

- B.
90

- C.
70

- D.
80

Answer: Option B

**Explanation** :

Let the time taken for them to meet after starting from opposite ends be t mins.

Also, let speeds of Car 2 be b kmph.

Distance AM = 60 × t = b × 20 …(1)

Distance BM = b × t = 60 × 45 …(2)

(1) × (2)

⇒ t^{2} = 900

⇒ t = 30 minutes

From (2) we get,

60 × 30 = b × 20

⇒ b = 90 kmph

Hence, option (b).

Workspace:

**CAT 2020 QA Slot 1 | Arithmetic - Time, Speed & Distance**

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?

- A.
13

- B.
14

- C.
11

- D.
12

Answer: Option D

**Explanation** :

Let the distance between home and office be d kms.

∴ $\frac{d}{8}-\frac{d}{15}=\frac{35}{60}=\frac{7}{12}$

⇒ d = 10 kms

If he leaves home at 9:10 a.m. and reaches office at 10 a.m., i.e., time taken = 50 minutes = 5/6 hours

∴ Required speed = $\frac{10}{5/6}$ = 12 kmph.

Hence, option (d).

Workspace:

**CAT 2020 QA Slot 1 | Arithmetic - Time, Speed & Distance**

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

- A.
58

- B.
61

- C.
67

- D.
50

Answer: Option C

**Explanation** :

Since the train travelled at 1/3rd of its speed, it will take thrice the usual time.

Hence, if the normal time taken for train is t minutes, now it will take 3t minutes.

∴ 3t – t = 30

⇒ t = 15 minutes.

While coming back train travels at usual speed for 5 minutes and then takes 4 minutes break.

∴ If the train has to come back at scheduled time, it needs to cover the remaining distance in (15 – 5 – 4 =) 6 minutes. Train usually takes 10 minutes to travel this distance.

∴ The time reduced to 6/10 = 3/5th hence speed should become 5/3 times.

∴ % increase = $\frac{\frac{5}{3}-1}{1}$ × 100 = 66.66% ≈ 67%

Hence, option (c).

Workspace:

**CAT 2020 QA Slot 2 | Arithmetic - Time, Speed & Distance**

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

- A.
550

- B.
450

- C.
475

- D.
500

Answer: Option B

**Explanation** :

Let their speeds be a, b and c respectively and the length of the race be D.

Car A beats car B by 45 kms

∴ $\frac{a}{b}=\frac{D}{D-45}$ …(1)

Car B beats car C by 50 kms

∴ $\frac{b}{c}=\frac{D}{D-50}$ …(2)

Car A beats car C by 90 kms

∴ $\frac{a}{c}=\frac{D}{D-90}$ …(3)

(1) × (2) = (3)

⇒ $\frac{D}{D-45}\times \frac{D}{D-50}=\frac{D}{D-90}$

⇒ $\frac{D}{(D-45)(D-50)}=\frac{1}{D-90}$

⇒ D^{2} – 95D + 2250 = D^{2} – 90D

⇒ 5D = 2250

⇒ D = 450

Hence, option (b).

Workspace:

**CAT 2020 QA Slot 2 | Arithmetic - Time, Speed & Distance**

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

- A.
√2

- B.
2√2

- C.
1/2

- D.
2

Answer: Option D

**Explanation** :

Let the time taken for them to meet after starting from opposite ends be t mins.

Also, let speeds of Ram and Rahim be a and b respectively.

Distance AM = a × t = b × 4 …(1)

Distance BM = b × t = a × 1 …(2)

(1) × (2)

⇒ t^{2} = 4

⇒ t = 2

From (1) we get,

a × 2 = b × 4

⇒ a : b = 4 : 2 = 2 : 1

Hence, option (d).

Workspace:

**CAT 2020 QA Slot 2 | Arithmetic - Time, Speed & Distance**

Two circular tracks T_{1} and T_{2} 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 T_{1} and track T_{2} 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

- A.
3

- B.
5

- C.
4

- D.
2

Answer: Option C

**Explanation** :

Time taken by Ram to reach the starting point A = $\frac{2\times \pi \times 0.1}{15}=\frac{\pi}{75}$

Time taken by Rahim to reach the starting point A = $\frac{2\times \pi \times 0.02}{5}=\frac{\pi}{125}$

Both of them will meet only at A since they are running in different circles which meet at A.

Both of them will reach the starting point after LCM$\left[\frac{\pi}{75},\frac{\pi}{125}\right]$ = $\frac{\pi}{25}$

∴ Number of rounds completed by Ram when both of them meet again = $\frac{\frac{\pi}{25}}{\frac{\pi}{75}}$ = 3

Hence, 3.

Workspace:

**CAT 2020 QA Slot 2 | Arithmetic - Time, Speed & Distance**

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 travelling from A to C is

- A.
4 : 1

- B.
7 : 5

- C.
5 : 7

- D.
1 : 4

Answer: Option C

**Explanation** :

Let the distance from A to B = d and that from B to C = 3d.

Speed of train T_{1} = s from A to B and 2s from B to C.

Total time taken by T_{1} = $\frac{d}{s}+\frac{3d}{2s}=\frac{5d}{2s}$

Speed of train T_{2} = 2s from A to B and s from B to C.

Total time taken by T_{2} = $\frac{d}{2s}+\frac{3d}{s}=\frac{7d}{2s}$

∴ Ratio of time taken by T_{1} and T_{2} = $\frac{5d}{2s}:\frac{7d}{2s}$ = 5 : 7

Hence, option (c).

Workspace:

**CAT 2020 QA Slot 3 | Arithmetic - Time, Speed & Distance**

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

- A.
11:45 a.m.

- B.
11 a.m.

- C.
10:45 a.m.

- D.
11:20 a.m.

Answer: Option B

**Explanation** :

Train A starts at 9 a.m. whereas Train B starts at 10:30 a.m.

Distance travelled by A till 10:30 a.m. = 40 × 1.5 = 60 kms.

At 10:30 a.m. remaining distance between A and B = 90 – 60 = 30 kms.

Time taken for both of them to meet now = 30/(40 + 20) = 0.5 hours i.e., 30 mins.

∴ Both trains meet half after 10:30 a.m. i.e., at 11 a.m.

Hence, option (b).

Workspace:

**CAT 2020 QA Slot 3 | Arithmetic - Time, Speed & Distance**

Vimla starts for office every day at 9 am 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

- A.
26

- B.
28

- C.
29

- D.
27

Answer: Option B

**Explanation** :

Let the distance from home to office be d kms.

∴ $\frac{d}{35}-\frac{d}{40}=\frac{6}{60}$

⇒ d = 28 kms.

∴ Normal time taken = $\frac{28}{40}$ = 0.7 hour = 42 minutes.

Now, she covers 2/3^{rd} of this distance in 1/3rd time i.e., in 14 minutes.

She now takes rest for 8 minutes.

∴ Time elapsed = 14 + 8 = 22 minutes.

Time remaining = 20 minutes.

In 20 minutes she needs to travel 1/3^{rd} of the distance i.e., 28/3 kms.

∴ Time taken = $\frac{28/3}{20/60}$ = 28 kmph.

Hence, option (b).

Workspace:

**CAT 2020 QA Slot 3 | Arithmetic - Time, Speed & Distance**

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 15km/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?

- A.
4.6

- B.
4.2

- C.
5.2

- D.
4.8

Answer: Option D

**Explanation** :

Speeds of Anil, Sunil and Ravi, which are 15 kmph, 10 kmph and 8 kmph respectively.

Length of the track is 3 km.

Time taken by Anil to complete one round = 3/15 × 60 = 12 minutes.

Time taken by Sunil to complete one round = 3/10 × 60 = 18 minutes.

Both Anil and Sunil will reach at the starting points after LCM[12, 18] = 36 minutes.

The distance covered by Ravi in 36 minutes (= 0.6 hours) = 8 × 0.6 = 4.8 km.

Hence, option (d).

Workspace:

**CAT 2019 QA Slot 1 | Arithmetic - Time, Speed & Distance**

Two cars travel the same distance starting at 10:00 am and 11:00 am, 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

- A.
25

- B.
10

- C.
30

- D.
20

Answer: Option D

**Explanation** :

First car starts at 10 am : Let the speed and time taken be a and t respectively.

Second car starts at 11 am : Let the speed be b. Time taken = t − 1.

Required percentage = [(b − a)/a] × 100 = [(b/a) − 1] × 100

Required percentage will be highest when b/a is highest.

Now, since distances covered by both cars are same, so D = at = b(t − 1)

∴ b/a = t/(t − 1) = 1/[1 − (1/t)]

(b/a) is maximum when t is minimum.

*t*_{min} = 6 (given)

∴ b/a = 6/5.

Maximum required percentage = [(6/5) − 1] × 100 = 20%.

Hence, option 4.

Workspace:

**CAT 2019 QA Slot 1 | Arithmetic - Time, Speed & Distance**

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 for 1/3 of his total journey time, while Bimal took each mode of transport for 1/3 of the total distance. The percentage by which Bimal’s travel time exceeds Amal’s travel time is nearest to

- A.
22

- B.
21

- C.
20

- D.
19

Answer: Option A

**Explanation** :

Let time taken by Amal be 3t, so time taken with each speed = 3t/3 = t.

Let total distance travelled be 3d. Hence, Bimal travels d at each of the 3 different speeds.

For Amal : 3d = 10t + 20t + 30t = 60t.

∴ d = 20t.

Time taken by Bimal = (d/10) + (d/20) + (d/30) = (11d/60) = 11t/3.

Required percentage = [{(11t/3)/3t} − 1] × 100 = (2/9) × 100 = 22.22 ≈ 22%.

Hence option 1.

Workspace:

**CAT 2019 QA Slot 1 | Arithmetic - Time, Speed & Distance**

In a race of three horses, the first beat the second by 11 metres and the third by 90 metres. If the second beat the third by 80 metres, what was the length, in metres, of the racecourse?

Answer: 880

**Explanation** :

Let x be the length of the racecourse.

The first horse beat the second by 11 metres and the third by 90 metres.

∴ Distances travelled by the first, second and third horse are x, x − 11 and x − 90 respectively.

The second horse beat the third by 80 metres.

Distances travelled by the second and third horse are x and x − 80 respectively.

Ratio of speeds of second and third horse is constant which is equal to the ratio of the distances travelled by the second and third horse.

∴ (x − 11)/(x − 90) = x/(x − 80)

⇒ x^{2} - 91x + 880 = x^{2} - 90x

⇒ x = 880

Hence, 880.

Workspace:

**CAT 2019 QA Slot 1 | Arithmetic - Time, Speed & Distance**

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

- A.
14π

- B.
18π

- C.
16π

- D.
12π

Answer: Option C

**Explanation** :

Let the radius of A and B be a and b respectively. (a = 30 and b = 40)

Distance travelled by any wheel D = (Circumference C) × (Number of revolutions N)

∴ D ∝ R × N (∵ C ∝ Radius R)

∴ N ∝ 1/R

∴ N_{a}/N_{b} = b/a = 40/30 = 4/3 ...(1)

It is given that N_{a} = N_{b} + 5000 ...(2)

Solving (1) and (2), we get; N_{a} = 20000 and N_{b} = 15000.

D = C_{a} × N_{a} = 2πa × N_{a} = v_{b} × (3/4) [where v_{b} is the velocity of B]

∴ 2π × (30 × 10^{-5}) × 20000 = v_{b} × (3/4)

∴ v_{b} = 16π.

Hence option 3.

Workspace:

**CAT 2019 QA Slot 2 | Arithmetic - Time, Speed & Distance**

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?

- A.
22

- B.
20

- C.
15

- D.
23

Answer: Option C

**Explanation** :

It is given that the cyclist starts at 10:00 am from A and reaches B at 11:00 am

Now, Motorcyclists start every minute from 10:01 am, and 45 such motorcyclists reach B before 11:00 am

If they leave one by one every minute, the 45^{th} motorcyclist would have left by 10:45 am to reach B at 11:00 am.

Thus, time taken by one motorcyclist to reach B from A = 15 minutes.

Now, the cyclist doubles his speed. This means, he reaches B at 10:30 am

So, the last motorcyclist should have left A by 10:15 am

Thus, 15 motorcyclists would have reached B by the time the cyclist reaches B

Hence, option (3).

Workspace:

**CAT 2019 QA Slot 2 | Arithmetic - Time, Speed & Distance**

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?

Answer: 48

**Explanation** :

Let the track length for John be 'a' and for Mary be 'b'

So, Distance travelled by John = 9a

Distance travelled by Mary = 5b

Now, Time taken by John = 9a/6 hours

Time taken by Mary = 5b/7.5 hours

We know that Time taken by John = Time taken by Mary

⇒ 9a/6 = 5b/7.5

⇒ a = 4b/9

Total track length = 325 meters

So, 4b/9 + b = 325 meters

⇒ 13b/9 = 325 meters

⇒ b = 225 meters

⇒ a = 100 meters

Mary jogs at 7.5 Kmph = 7.5 × 5/18

So, time taken = $\frac{100}{(7.5\times {\displaystyle \frac{5}{18}})}$ = 48 seconds.

Hence, 48.

Workspace:

**CAT 2019 QA Slot 2 | Arithmetic - Time, Speed & Distance**

Two ants A and B start from a point P on a circle at the same time, with A moving clock-wise 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

- A.
10 : 27 am

- B.
10 : 25 am

- C.
10 : 45 am

- D.
10 : 18 am

Answer: Option A

**Explanation** :

We are told that, by the time A and B meet for the first time, A covers 60% of the distance, while B covers 40% of the distance.

So, the speeds of A and B are in the ratio 60:40 or 3:2

Hence, the time they take to cover a particular distance will be in the ratio 2:3

We know that A covers 60% of the distance at 10:00 AM and covers 100% of the distance at 10:12 AM.

That means A takes 12 minutes to cover 40% of the track. So to cover the entire track he must have taken 12 + 12 + 6 = 30 minutes. (because 40% + 40% + 20% = 100%)

Since the time taken by A and B to complete the track are in the ratio 2:3, the time taken by B to complete the track will be 45 minutes.

At 10:00 AM, B has covered 40% of the track. If we can find out what time does B take to complete the remaining 60% of the track, we can find the finish time of B.

Time required to complete 60% of the track = 60% of 45 = 27 minutes.

Hence, B complete one single round at 10:27 AM.

Hence, option (1).

Workspace:

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