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

The best way to prepare for Arithmetic - Time, Speed & Distance is by going through the previous year **Arithmetic - Time, Speed & Distance XAT questions**.
Here we bring you all previous year Arithmetic - Time, Speed & Distance XAT questions along with detailed solutions.

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**XAT 2024 QADI | Arithmetic - Time, Speed & Distance XAT Question**

A flight, traveling to a destination 11,200 kms away, was supposed to take off at 6:30 AM. Due to bad weather, the departure of the flight got delayed by three hours. The pilot increased the average speed of the airplane by 100 km/hr from the initially planned average speed, to reduce the overall delay to one hour. Had the pilot increased the average speed by 350 km/hr from the initially planned average speed, when would have the flight reached its destination?

- (a)
11:30 PM

- (b)
7:50 PM

- (c)
5:10 PM

- (d)
8:10 PM

- (e)
10:36 PM

Answer: Option D

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**Text Explanation** :

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**XAT 2023 QADI | Arithmetic - Time, Speed & Distance XAT Question**

**The problem below consists of a question and two statements numbered 1 & 2. You have to decide whether the data provided in the statements are sufficient to answer the question.**

Rahim is riding upstream on a boat, from point A to B, at a constant speed. The distance from A to B is 30 km. One minute after Rahim leaves from point A, a speedboat starts from point A to go to point B. It crosses Rahim’s boat after 4 minutes. If the speed of the speedboat is constant from A to B, what is Rahim’s speed in still water?

1. The speed of the speedboat in still water is 30 km/hour.

2. Rahim takes three hours to reach point B from point A.

- (a)
Statement 1 alone is sufficient to answer the question, but statement 2 alone is not sufficient

- (b)
Statement 2 alone is sufficient to answer the question, but statement 1 alone is not sufficient

- (c)
Each statement alone is sufficient

- (d)
Both statements together are sufficient, but neither of them alone is sufficient

- (e)
Statements 1 & 2 together are not sufficient

Answer: Option D

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**Text Explanation** :

Let the speed of river be 'r' kmph.

Rahim and Speedboat start from A and meet at a point M (say).

Going upstream, Rahim takes 5 minutes to cover this distance while Speedboat takes 4 minutes to cover the same distance.

∴ $\frac{{\mathrm{S}}_{\mathrm{Rahim}}-\mathrm{r}}{{\mathrm{S}}_{\mathrm{Speedboat}}-\mathrm{r}}$ = $\frac{4}{5}$ ...(1)

We need to find S_{Rahim}

**Statement (1)**: S_{Speedboat} = 30

If we put S_{Speedboat} = 30 in (1), this alone would not be sufficient to calculate S_{Rahim}.

∴ Statement (1) alone is not sufficient to answer the question.

**Statement (2)**: Rahim takes 3 hours to reach B.

⇒ S_{Rahim} - r = 30/3 = 10 kmph.

Since we don't know the value of r, we cannot determine the S_{Rahim}.

∴ Statement (2) alone is not sufficient to answer the question.

Using both statements together we have, S_{Speedboat} = 30 kmph and S_{Rahim} - r = 10 kmph.

Putting these values in (1), we get

$\frac{10}{30-\mathrm{r}}$ = $\frac{4}{5}$

⇒ r = 17.5 kmph.

⇒ S_{Rahim} - 17.5 = 10

⇒ S_{Rahim} - 17.5 = 27.5 kmph

∴ Using both statements we can get the answer.

Hence, option (d).

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**XAT 2023 QADI | Arithmetic - Time, Speed & Distance XAT Question**

**The problem below consists of a question and two statements numbered 1 & 2. You have to decide whether the data provided in the statements are sufficient to answer the question.**

In a cricket match, three slip fielders are positioned on a straight line. The distance between 1st slip and 2nd slip is the same as the distance between 2nd slip and the 3rd slip. The player X, who is not on the same line of slip fielders, throws a ball to the 3rd slip and the ball takes 5 seconds to reach the player at the 3rd slip. If he had thrown the ball at the same speed to the 1st slip or to the 2nd slip, it would have taken 3 seconds or 4 seconds, respectively. What is the distance between the 2nd slip and the player X?

1. The ball travels at a speed of 3.6 km/hour.

2. The distance between the 1st slip and the 3rd slip is 2 meters.

- (a)
Statement 1 alone is sufficient to answer the question, but statement 2 alone is not sufficient

- (b)
Statement 2 alone is sufficient to answer the question, but statement 1 alone is not sufficient

- (c)
Each statement alone is sufficient

- (d)
Both statements together are sufficient, but neither of them alone is sufficient

- (e)
Statements 1 & 2 together are not sufficient

Answer: Option A

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**Text Explanation** :

Discrepancy was found in this question and full marks were awarded to all candidates.

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**XAT 2021 QADI | Arithmetic - Time, Speed & Distance XAT Question**

Rajesh, a courier delivery agent, starts at point A and makes a delivery each at points B, C and D, in that order. He travels in a straight line between any two consecutive points. The following are known:

(i) AB and CD intersect at a right angle at E, and

(ii) BC, CE and ED are respectively 1.3 km, 0.5 km and 2.5 km long.

If AD is parallel to BC, then what is the total distance (in km) that Rajesh covers in travelling from A to D?

- (a)
10.2

- (b)
12

- (c)
11.5

- (d)
5.5

- (e)
18

Answer: Option C

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**Text Explanation** :

Given, CE = 0.5, BC = 1.3 and ED = 2.5

Triangle CEB is a right-angled triangle ⇒ EB = 1.2

Triangles ECB is similar to triangle EDA

EB/EC = AE/ED ⇒ AE = 6

Hence total distance travelled = AB + BC + CD = 7.2 + 1.3 + 3.5 = 11.5km

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**XAT 2021 QADI | Arithmetic - Time, Speed & Distance XAT Question**

Swati can row a boat on still water at a speed of 5 km/hr. However, on a given river, it takes her 1 hour more to row the boat 12 km upstream than downstream. One day, Swati rows the boat on this river from X to Y, which is N km upstream from X. Then she rows back to X immediately. If she takes at least 2 hours to complete this round trip, what is the minimum possible value of N?

- (a)
3

- (b)
4.8

- (c)
2

- (d)
3.6

- (e)
2.1

Answer: Option B

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**Text Explanation** :

Let the speed of the stream be x

Let the speed of the stream be

$\frac{12}{5-X}$ = $\frac{12}{5+X}$ + 1

The value of x satisfying the above equation is 1

Now,

$\frac{N}{5+1}$ + $\frac{N}{5-1}$ ≥ 2

$\frac{2N+3N}{12}$ ≥ 2

⇒ N ≥ 4.8

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**XAT 2021 QADI | Arithmetic - Time, Speed & Distance XAT Question**

On the bank of the pristine Tunga river, a deer and a tiger are joyfully playing with each other. The deer notices that it is 40 steps away from the tiger and starts running towards it. At the same time, the tiger starts running away from the deer. Both run on the same straight line. For every five steps the deer takes, the tiger takes six. However, the deer takes only two steps to cover the distance that the tiger covers in three. In how many steps can the deer catch the tiger?

- (a)
200

- (b)
To solve this, the length of a deer’s step must also be given

- (c)
120

- (d)
360

- (e)
320

Answer: Option A

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**Text Explanation** :

Let speed of deer = 5steps/second and speed of tiger = 6 steps/sec

Let deer cover 1 m in a step ⇒ tiger covers 2/3 m in a step

Hence speed of deer = 5m/s and spped of tiger = 6 x 2/3 m/s = 4m/s

Hence time taken by a deer to catch tiger = 40 seconds

Distance travelled by deer in 40 seconds = 5 x 40 = 200 steps

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**XAT 2021 DM | Arithmetic - Time, Speed & Distance XAT Question**

Four friends, Ashish, Brian, Chaitra, and Dorothy, decide to jog for 30 minutes inside a stadium with a circular running track that is 200 metres long. The friends run at different speeds. Ashish completes a lap exactly every 60 seconds. Likewise, Brian, Chaitra and Dorothy complete a lap exactly every 1 minute 30 seconds, 40 seconds and 1 minute 20 seconds respectively. The friends begin together at the start line exactly at 4 p.m. What is the total of the numbers of laps the friends would have completed when they next cross the start line together ?

- (a)
43

- (b)
36

- (c)
They will never be at the start line together again before 4:30 p.m.

- (d)
47

- (e)
28

Answer: Option D

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**Text Explanation** :

All the four friends will meet at the starting point after LCM(60,90,40,80) = 720 seconds.

Number of laps by A in 720 seconds = 12

Number of laps by B in 720 seconds = 8

Number of laps by C in 720 seconds = 18

Number of laps by D in 720 seconds = 9

Together they complete = 47 laps

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**XAT 2020 QADI | Arithmetic - Time, Speed & Distance XAT Question**

Two friends, Ram and Shyam, start at the same point, at the same time. Ram travels straight north at a speed of 10km/hr, while Shyam travels straight east at twice the speed of Ram. After 15 minutes, Shyam messages Ram that he is just passing by a large telephone tower and after another 15 minutes Ram messages Shyam that he is just passing by an old banyan tree. After some more time has elapsed, Ram and Shyam stop. They stop at the same point of time. If the straight-line distance between Ram and Shyam now is 50 km, how far is Shyam from the banyan tree (in km)? (Assume that Ram and Shyam travel on a flat surface.)

- (a)
20√5 + 5

- (b)
20√5 - 5

- (c)
5√21

- (d)
45

- (e)
115/3

Answer: Option D

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**Text Explanation** :

Distance travelled by Shyam till Telephone tower = 20 × 1/4 = 5 kms

Distance travelled by Ram till Banyan tree = 10 × ½ = 5 kms

Let the two of them stop after t hours from starting.

∴ (50)^{2} = (10t)^{2} + (20t)^{2}

⇒ 2500 = 100t^{2} + 400t^{2}

⇒ 5 = t^{2}

Now, (BS)^{2} = 52 + (20t)^{2}

⇒ (BS)^{2} = 25 + 400t^{2}

⇒ (BS)^{2} = 25 + 400 × 5 = 2025

⇒ BS = √2025 = 45

Hence, option (d).

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**XAT 2020 QADI | Arithmetic - Time, Speed & Distance XAT Question**

A hare and a tortoise run between points O and P located exactly 6 km from each other on a straight line. They start together at O, go straight to P and then return to O along the same line. They run at constant speeds of 12 km/hr and 1 km/hr respectively. Since the tortoise is slower than the hare, the hare shuttles between O and P until the tortoise goes once to P and returns to O. During the run, how many times are the hare and the tortoise separated by an exact distance of 1 km from each other?

- (a)
40

- (b)
24

- (c)
48

- (d)
42

- (e)
22

Answer: Option A

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**Text Explanation** :

Tortoise will take 12 hours to complete 1 round. During this, hare will make 12 rounds of OP.

In the first round, both started from point O. After some time, distance between them will be 1 km.

After some more time, when hare is returning from P to O, before and after crossing tortoise, hare will be two more times 1 km apart from tortoise. So, in first round, there are three such occurrences.

In the second round, when hare starts from point O, while going and returning, there will be four occurrences when before and after crossing the tortoise, hare will be exactly 1 km apart. But the first occurrence of round 2 is already counted in round 1. So, in second round as well, there will be total 3 occurrences.

In each of the third, fourth and fifth rounds, there will be 4 such occurrences.

In the sixth round, because tortoise will be at point P, there will be only 2 cases.

Now, till round 6 there are 20 such occurrences.

And from round 7 to 12, it will exactly same but in reverse order.

∴ Total such occurrences = 20 × 2 = 40.

Hence, option (a).

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**XAT 2018 QADI | Arithmetic - Time, Speed & Distance XAT Question**

Every day a person walks at a constant speed, V_{1} for 30 minutes. On a particular day, after walking for 10 minutes at V_{1}, he rested for 5 minutes. He finished the remaining distance of his regular walk at a constant speed, V_{2}, in another 30 minutes. On that day, find the ratio of V_{2} and his average speed (i.e., total distance covered /total time taken including resting time).

- (a)
1 : 1

- (b)
1 : 2

- (c)
2 : 3

- (d)
2 : 1

- (e)
None of the above

Answer: Option A

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**Text Explanation** :

The person usually walks at V_{1} for 30 minutes.

∴ Distance travelled = 30V_{1}.

Now the person travels for 10 minutes. Distance remaining after this = 30V_{1} – 10V_{1} = 20V_{1}

This remaining distance is covered by him at V_{2} in 30 minutes.

⇒ V_{2} = 20V_{1}/30 = 2V_{1}/3

∴ His average speed for the entire journey = 30V_{1}/45 = 2V_{1}/3

Now,

V_{2} : Average speed = 2V_{1}/3 : 2V_{1}/3 = 1 : 1

Hence, option (a).

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**XAT 2018 QADI | Arithmetic - Time, Speed & Distance XAT Question**

A girl travels along a straight line, from point A to B at a constant speed, V_{1} meters/sec for T seconds. Next, she travels from point B to C along a straight line, at a constant speed of V_{2} meters/sec for another T seconds. BC makes an angle 105° with AB. If CA makes an angle 30° with BC, how much time will she take to travel back from point C to A at a constant speed of V_{2} meters/sec, if she travels along a straight line from C to A?

- (a)
0.5(√3 - 1)T

- (b)
T

- (c)
0.5(√3 + 1)T

- (d)
√3T

- (e)
None of the above

Answer: Option C

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**Text Explanation** :

Refer to the diagram below.

Distance AB = V_{1}T and distance BC = V_{2}T

Also, ∠CAB = 180 – 30 – 105 = 45°

Let BD be drawn perpendicular to AC.

In ΔBDA,

BD = AB sin 45° = V_{1}T/√2

Also, AD = AB cos 45° = V_{1}T/√2

In ΔBDC,

BD = BC sin 30° = V_{2}T/2

Also, DC = BC cos 30° = √3V_{2}T/2

Now, V_{1}T/√2 = V_{2}T/2

⇒ V_{1} = V_{2}/√2

AC = AD + DC = V_{1}T/√2 + √3V_{2}T/2

= V_{2}T/2 + √3V_{2}T/2

= V_{2}T(1 + √3)/2

Time taken to travel AC at speed V_{2} = [V_{2}T(1 + √3)/2] ÷ V_{2} = T(1 + √3)/2 = 0.5(1 + √3)T

Hence, option (c).

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**XAT 2017 QADI | Arithmetic - Time, Speed & Distance XAT Question**

Arup and Swarup leave point A at 8 AM to point B. To reach B, they have to walk the first 2 km, then travel 4 km by boat and complete the final 20 km by car. Arup and Swarup walk at a constant speed of 4 km/hr and 5 km/hr respectively. Each rows his boat for 30 minutes. Arup drives his car at a constant speed of 50 km/hr while Swarup drives at 40 km/hr. If no time is wasted in transit, when will they meet again?

- (a)
at 9:15 am

- (b)
at 9:18 am

- (c)
at 9:21 am

- (d)
at 9:24 am

- (e)
at 9:30 am

Answer: Option D

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**Text Explanation** :

At 4 km/hr, Arup walks 2 km in ½ hour = 30 minutes and at 50 km/hr, he covers 20 km in ⅖ hours = 24 minutes.

At 5 km/hr, Swarup walks 2 km in ⅖ hours = 40 minutes and at 40 km/hr, he covers 20 km in ½ hour = 30 minutes.

Thus, time taken by each of them to reach B is equal.

∴ Initially Swarup will go ahead of Arup and both of them will reach the finish line together.

Time = 30 + 30 + 24 = 84 minutes.

So, they meet again at 9:24 AM.

Hence, option (d).

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**XAT 2016 QADI | Arithmetic - Time, Speed & Distance XAT Question**

Pradeep could either walk or drive to office. The time taken to walk to the office is 8 times the driving time. One day, his wife took the car making him walk to office. After walking 1 km, he reached a temple when his wife call to say that he can now take the car. Pradeep figure that continuing to walk to the office will take as long as walking back home and then driving to the office. Calculate the distance between the temple and the office.

- (a)
1

- (b)
7/3

- (c)
9/7

- (d)
16/7

- (e)
16/9

Answer: Option C

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**Text Explanation** :

Time taken to walk to the office is 8 times the driving time. It means that speed while driving is 8 times the speed while walking.

Let the walking speed be ‘s’ kmph

∴ Driving speed = 8s kmph

Let the distance between home and office be D kms.

Now, time taken to (D – 1) kms is same as time taken to walk 1 km and then drive D kms.

∴ $\frac{D-1}{s}$ = $\frac{1}{s}$ + $\frac{D}{8s}$

⇒ 8(D – 1) = 8 + D

⇒ 7D = 16

⇒ D = 16/7

∴ Distance between temple and office = D – 1 = 16/7 - 1 = 9/7

Hence, option (c).

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**XAT 2016 QADI | Arithmetic - Time, Speed & Distance XAT Question**

Study the figure below and answer the question:

Four persons walk from Point A to Point D following different routes. The one following ABCD takes 70 minutes. Another person takes 45 minutes following ABD. The third person takes 30 minutes following route ACD. The last person takes 65 minutes following route ACBD. IF all were to walk at the same speed, how long will it take to go from point B to point C?

- (a)
10 mins

- (b)
20 mins

- (c)
30 mins

- (d)
40 mins

- (e)
Cannot be answered as the angles are unknown.

Answer: Option C

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**Text Explanation** :

Let AB = a , BC = b , BD = c , AC = d and CD = e and let the speed of the persons be V.

From given information :

a + b + e = 70V …(1)

a +c = 45V … (2)

d + e = 30V … (3)

d + b + c = 65V … (4)

Adding (1) and (4) we get (a + c) + (d + e) + 2b = 135V

From (2) and (3) ; 45V + 30V + 2b = 135V

b = 30V.

b/V = 30.

Hence , it takes 30 min to go from B to C

Hence, option (c).

Workspace:

**XAT 2012 QA | Arithmetic - Time, Speed & Distance XAT Question**

City Bus Corporation runs two buses from terminus A to terminus B, each bus making 5 round trips in a day. There are no stops in between. These buses ply back and forth on the same route at different but uniform speeds. Each morning the buses start at 7 AM from the respective terminuses. They meet for the first time at a distance of 7 km from terminus A. Their next meeting is at a distance of 4 km from terminus B, while travelling in opposite directions. Assuming that the time taken by the buses at the terminuses is negligibly small, and the cost of running a bus is Rs. 20 per km, find the daily cost of running the buses (in Rs.).

- (a)
3200

- (b)
4000

- (c)
6400

- (d)
6800

- (e)
None of the above

Answer: Option D

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**Text Explanation** :

Let the distance between the two terminuses be x km.

Now, relative distance travel by the two buses before they meet for the first time = x km

Similarly, relative distance travelled by the two buses after first meet and before second meet = 2x

Now, bus originating from terminus A travels 7 km before the first meet.

Hence, it should travel 2 × 7 = 14 km after first meet and before second meet.

Hence, total distance travelled by that bus before second meet = 7 + 14 = 21 km

Now, second meet occurs at 4 km from the terminus B.

Hence, total distance travelled by the bus starting from terminus A (from the beginning till they meet for the second time) = x + 4 km

Hence, x + 4 = 21

Hence, x = 17 km

Hence, one bus travels 34 × 5 = 170 km a day.

Hence, cost of running one bus = 170 × 20 = Rs. 3400

Hence, cost of running two buses = 3400 × 2 = Rs. 6800

Hence, option (d).

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