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

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

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

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

**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).

Workspace:

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

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

**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).

Workspace:

**IIFT 2019 QA | Arithmetic - Time, Speed & Distance**

A motorboat takes the passengers from Rishikesh to Haridwar and back. Both the cities, Rishikesh and Haridwar are located on the banks of River Ganga. During Kumbh Mela, to earn more money, the owner of the motorboat decided to have more trips from Rishikesh to Haridwar and back, so he increased the speed of the motorboat in still water, by 50%. By increasing the speed, he was able to cut down the travel time from Rishikesh to Haridwar and back, by 60%. What is the ratio of the speed of motorboat in still water to that of the speed of river Ganga?

- A.
√(11/6)

- B.
√11/6

- C.
√(3/2)

- D.
√3/2

Answer: Option A

**Explanation** :

Let the speed of boat in still water be ‘b’ and speed of river be ‘r’ and the one-way distance between the two cities be ‘d’.

Time taken initially to go and come back = $\frac{d}{b+r}$ + $\frac{d}{b-r}$

When the speed of boat increases by 50%, time taken to go and come back = $\frac{d}{1.5b+r}$ + $\frac{d}{1.5b-r}$

New time taken is 40% of the initial time taken

⇒ $\frac{d}{1.5b+r}+\frac{d}{1.5b-r}$ = $\frac{2}{5}\left[\frac{d}{b+r}+\frac{d}{b-r}\right]$

⇒ $\frac{3b}{2.25{b}^{2}-{r}^{2}}$ = $\frac{2}{5}\left[\frac{2b}{{b}^{2}-{r}^{2}}\right]$

⇒ 15b^{2} – 15r^{2 }= 9b^{2} – 4r^{2}

⇒ 6b^{2} = 11r^{2}

⇒ $\frac{b}{r}$ = $\sqrt{\frac{11}{6}}$

Hence option (a).

Workspace:

**IIFT 2019 QA | Arithmetic - Time, Speed & Distance**

You travel by Delhi Metro everyday from Botanical Garden, Noida to Hauz Khas, Delhi. At Hauz Khas metro station, you use an escalator to get out from the station. The escalator takes 80 seconds. One day, escalator was not working and you walk up the escalator in 50 seconds. How many minutes does it approximately take you to walk down the working escalator?

- A.
1.5 minutes

- B.
2.2 minutes

- C.
2.8 minutes

- D.
2.6 minutes

Answer: Option B

**Explanation** :

Let, the number of steps on the escalator be LCM(80, 50) = 400

∴ Speed of escalator = 400/80 = 5 steps/ second, and

Speed of the person = 400/50 = 8 steps/ second

When the person tries to go up a moving down escalator, the effective speed of the person = (8 − 5) steps /second = 3 steps/second

∴ Time taken = 400/3 = 133.33 steps/ second = 2.2 minutes (approx.)

Hence, option (b).

Workspace:

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

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

**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).

Workspace:

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

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 B

**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

Hence, option (b).

Workspace:

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

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

**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).

Workspace:

**IIFT 2017 QA | Arithmetic - Time, Speed & Distance**

Ramesh and Sohan start walking away from each other from a point P at an angle of 120°. Ramesh walks at a speed of 3 kmph while Sohan walks at a speed of 4 kmph. What is the distance between them after 90 minutes?

- A.
9.89 km

- B.
10.56 km

- C.
9.12 km

- D.
12.42 km

Answer: Option C

**Explanation** :

Let Ramesh and Sohan reach points R and S respectively after 90 minutes.

∴ PR = distance travelled by Ramesh in 90 minutes = 1.5 × 3 = 4.5 km

Similarly, PS = 1.5 × 4 = 6 km

Now, PRS forms a triangle, where two sides (PR and PS) as well as the included angle (RPS = 120°) are known. Hence, the third side can be found using the cosine rule.

∴ Cos 120 = [(4.5)^{2} + (6)^{2} − (RS)^{2}] / (2 × 4.5 × 6)

∴ −0.5 = [20.25 + 36 − (RS)^{2}] / 54

∴ 56.25 − (RS)^{2 }= −27

∴ (RS)^{2 }= 83.25 i.e. RS = 9.12

Hence, they are 9.12 km apart.

Hence, option 3.

Workspace:

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

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

**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).

Workspace:

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

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

**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:

**IIFT 2016 QA | Arithmetic - Time, Speed & Distance**

Shruti and Krishna left Delhi for Noida at the same time. While Shruti was driving her car, Krishna, an environmentalist by profession, was traveling on his bicycle. Having reached Noida, Shruti turned back and met Krishna an hour after they started. Krishna continued his journey to Noida after the meeting, while Shruti turned back and also headed for Noida. Having reached Noida, Shruti again turned back and met Krishna 30 minutes after their first meeting. The time taken by Krishna to cover the distance between Delhi and Noida is

- A.
2 hours

- B.
2.5 hours

- C.
3 hours

- D.
None of the above

Answer: Option A

**Explanation** :

Shruti and Krishna leave for the first time. Their first meeting is after one hour, the second is after an additional half an hour.

So, the interval between subsequent meetings keeps halving every time.

∴ Total time = 60 + 30 + 15 + ….

This is an infinite G.P. with a = 60 and r = 1/2

∴ Sum of terms = a / (1 – r) = 60 /[1 – (1/2)] = 120 minutes = 2 hours

Hence, option 1.

Workspace:

**IIFT 2015 QA | Arithmetic - Time, Speed & Distance**

A chartered bus carrying office employees travels everyday in two shifts- morning and evening. In the evening, the bus travels at an average speed which is 50% greater than the morning average speed; but takes 50% more time than the amount of time it takes in the morning. The average speed of the chartered bus for the entire journey is greater/less than its average speed in the morning by:

- A.
18% less

- B.
30% greater

- C.
37.5% greater

- D.
50% less

Answer: Option B

**Explanation** :

Let the bus travels at speed ‘V’km/hr for time ‘T’ hours in the morning.

∴ Distance travelled in the morning = VT km

∴ In the evening, it travels at speed ‘1.5V’km/hr for ‘1.5T’ hours respectively.

∴ Distance travelled in the evening = 1.5V × 1.5T = 2.25VT

∴ Average speed = $\frac{TotalDis\mathrm{tan}ce}{TotalTime}$

$=\frac{(VT+2.25VT)}{(T+1.5T)}$

$\frac{}{}\phantom{\rule{0ex}{0ex}}=\frac{3.25V}{2.5}=1.3V$

∴ The average speed of the chartered bus for the entire journey is greater than its average speed in the morning by 30%.

Hence, option 2.

Workspace:

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

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

**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 4.

Workspace:

## Feedback

**Help us build a Free and Comprehensive CAT/MBA Preparation portal by providing
us your valuable feedback about Apti4All and how it can be improved.**