# 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:

**IIFT 2014 QA | Arithmetic - Time, Speed & Distance**

A fery carries passengers to Rock of Vivekananda and back from Kanyakumari. The distance of Rock of Vivekananda from Kanyakumari is 100 km. One day, the ferry started for Rock of Vivekananda with passengers on board, at a speed of 20 km per hour. After 90 minutes, the crew realized that there is a hole in the ferry and 15 gallons of sea water had already entered the ferry. Sea water is entering the ferry at the rate of 10 gallons per hour. It requires 60 gallons of water to sink the ferry. At what speed should the driver now drive the ferry so that it can reach the Rock of Vivekananda and return back to Kanyakumari just in time before the ferry sinks?

(Current of the sea water from Rock of Vivekananda to Kanyakumari is 2 km per hour.)

- A.
40 km/hr towards the Rock & 39 km/hr while returning to Kanyakumari

- B.
41 km/hr towards the Rock & 38 km/hr while returning to Kanyakumari

- C.
42 km/hr towards the Rock & 36 km/hr while returning to Kanyakumari

- D.
35 km/hr towards the Rock & 39 km/hr while returning to Kanyakumari

Answer: Option C

**Explanation** :

Since the current is from the Rock towards the Kanyakumari, the boat travels the first part against the stream and the second part with the stream.

Thus, in 90 minutes the boat has travelled 27 kms.

Since 15 gallons of water is already filled, there are 45 more gallons required to sink the ship which will take 4.5 hours at 10 gallons/hr.

Thus, the boat needs to cover 73 kms of first part and 100 kms of the second part within 4.5 hrs.

Let the speed of the boat be S_{B}

Since the first part is upstream, the time required will be = 73/(S_{B} − 2)

The time required for the second part will be = 100/(S_{B} + 2)

Thus,

Time taken by the ship ≤ 4.5 i.e.

$\frac{73}{{S}_{B}-2}$ + $\frac{100}{{S}_{B}+2}$ ≤ 4.5

We now put the values of SB from the options and calculate the time taken by the ship.

Thus, the calculated times from options (1), (2), (3), (4) would be 4.36, 4.37, 4.455, 4.65 hrs respectively.

Please note that in the question, the ferry needs to complete the journey just in time and hence the answer would be the option closest to 4.5 hrs i.e. 4.455 hrs.

Hence, option 3.

Workspace:

**IIFT 2014 QA | Arithmetic - Time, Speed & Distance**

A bouncing tennis ball is dropped from a height of 32 metre. The ball rebounds each time to a height equal to half the height of the previous bounce. The approximate distance travelled by the ball when it hits the ground for the eleventh time, is:

- A.
64 metre

- B.
96 metre

- C.
128 metre

- D.
150 metre

Answer: Option B

**Explanation** :

Distance covered when the ball hits the ground the first time = 32 m

The ball covers 16 + 16 = 32 m more when it hits the ground 2^{nd} time.

The ball covers 8 + 8 = 16 m more when it hits the ground 3^{rd} time.

And so on i.e., at the nth hit, the ball covers 32/2^{(n – 2)} more distance.

Thus, at the 11^{th} hit, the ball covers 32/29 m more.

Thus, the total distance the ball covers when it hits the ground 11^{th} time

= 32 + 32(1 + ½ + ¼ + … 1/2^{9})

= 32 + 32(1-1/2^{10})/(1-1/2) ≈ 32 + 64

= 96 m

Hence, option 2.

Workspace:

**IIFT 2013 QA | Arithmetic - Time, Speed & Distance**

A tennis ball is initially dropped from a height of 180 m. After striking the ground, it rebounds (3/5)^{th} of the height from which it has fallen. The total distance that the ball travels before it comes to rest is

- A.
540 m

- B.
600 m

- C.
720 m

- D.
900 m

Answer: Option C

**Explanation** :

Initial distance travelled = 180 m

Distance travelled after 1st rebound

Upward = $\frac{3}{5}$ × 180 = 108 m

Downward = 108 m

Total = 108 + 108 = 216 m

Distance travelled after 2nd rebound (upward and downward)

= $\frac{3}{5}$ × 108 × 2 = $\frac{3}{5}$ × 216 m

This gives an infinite G.P. with

a = 216 and r = $\frac{3}{5}$.

For an infinite G.P. with r <>

s = $\frac{a}{1-r}$ = $\frac{216}{1-{\displaystyle \frac{3}{5}}}$ = 540 m

Since initial distance was 180 m,

Total distance = (180 + 540) m = 720 m

Hence, option 3.

Workspace:

**IIFT 2013 QA | Arithmetic - Time, Speed & Distance**

It was a rainy morning in Delhi when Rohit drove his mother to a dentist in his Maruti Alto. They started at 8.30 AM from home and Rohit maintained the speed of the vehicle at 30 Km/hr. However, while returning from the doctor’s chamber, rain intensified and the vehicle could not move due to severe water logging. With no other alternative, Rohit kept the vehicle outside the doctor’s chamber and returned home along with his mother in a rickshaw at a speed of 12 Km/hr. They reached home at 1.30 PM. If they stayed at the doctor’s chamber for the dental check-up for 48 minutes, the distance of the doctor’s chamber from Rohit’s house is

- A.
15 km

- B.
30 km

- C.
36 km

- D.
45 km

Answer: Option C

**Explanation** :

Let the distance between the doctor’s chamber and Rohit’s house be x km.

Total time spent = 5 hours

Time spent at the doctor’s chamber

= $\frac{48}{60}$ = $\frac{4}{5}$

From the given conditions,

$\frac{x}{30}$ + $\frac{x}{12}$ + $\frac{4}{5}$ = 5

∴ x = 36

Hence, option 3.

Workspace:

**IIFT 2013 QA | Arithmetic - Time, Speed & Distance**

The duration of the journey from your home to the College in the local train varies directly as the distance and inversely as the velocity. The velocity varies directly as the square root of the diesel used per km., and inversely as the number of carriages in the train. If, in a journey of 70 km. in 45 minutes with 15 carriages, 10 litres of diesel is required, then the diesel that will be consumed in a journey of 50 km. in half an hour with 18 carriages is

- A.
2.9 litres

- B.
11.8 litres

- C.
15.7 litres

- D.
None of the above

Answer: Option B

**Explanation** :

T = $\frac{K\times D\times C}{\sqrt{A}}$

Where T is the time taken, D is the distance, C is the number of carriages, A is the diesel used per km and K is the proportionality constant.

Based on the data given:

45 = $\frac{K\times 70\times 15}{\sqrt{{\displaystyle \frac{1}{7}}}}$

K = $\frac{3}{70\sqrt{7}}$

T' = $\frac{3\times D\text{'}\times C\text{'}}{70\sqrt{7}\times \sqrt{A\text{'}}}$

30 = $\frac{3\times 50\times 18}{70\sqrt{7}\times \sqrt{A\text{'}}}$

Solving, A’ = 11.8 liters

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:

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

It takes 15 seconds for a train travelling at 60 km/hour to cross entirely another train half its length and travelling in opposite direction at 48 km/hour. It also passes a bridge in 51 seconds. The length of the bridge is

- A.
550 m

- B.
450 m

- C.
500 m

- D.
600 m

Answer: Option A

**Explanation** :

Relative speed of the longer train with respect to the

shorter one = (60 + 48) × $\frac{5}{18}$ = 30 m/s

Let the length of the longer train be x m.

Now the distance travelled by the longer train to cross the shorter train = length of the longer train + length of the shorter train = x + x/2 = 3x/2

The trains cross each other with a speed of (60 + 48) km/h

(60 + 48) km/h = (60 + 48) × $\frac{5}{18}$ = 30 m/s

Therefore,

$\frac{3x}{2}$ = 30 × 15

∴ x = 300 mts

Let the length of the bridge be L.

Now total distance covered in crossing the bridge

= length of the longer train + length of the bridge

= 300 + L

∴ 300 + L = 60 × $\left(\frac{5}{18}\right)$ × 51 = 850

∴ L = 550 mts

Hence, option 1.

Workspace:

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

The Howrah-Puri express can move at 45 km/hour without its rake, and the speed is diminished by a constant that varies as the square root of the number of wagons attached. If it is known that with 9 wagons, the speed is 30 km/hour, what is the greatest number of wagons with which the train can just move?

- A.
63

- B.
64

- C.
80

- D.
81

Answer: Option C

**Explanation** :

Let the reduced speed of the train be denoted by S and the number of wagons attached to it be denoted by N.

∴ S ∝ $\sqrt{N}$

∴ $\frac{{S}_{1}}{{S}_{2}}$ = $\sqrt{\frac{{N}_{1}}{{N}_{2}}}$

The speed reduces from 45 km/hr to 30 km/hr with 9 wagons.

∴ The reduction in speed = *S*_{1} = 45 – 30 = 15 km/hr

Now, let *N*_{2} number of wagons attached when the train halts completely.

Hence, *S*_{2} = reduction in speed at this point = 45 – 0

= 45 km/hr

∴ $\frac{15}{45}$ = $\frac{\sqrt{9}}{\sqrt{{N}_{2}}}$

∴ *N _{2}* = 81 wagons

Hence, when 81 wagons are attached, the train halts completely. For the train to just move, the number of wagons attached should be 1 less than 81 i.e. 80.

Hence, option 3.

Workspace:

**IIFT 2011 QA | Arithmetic - Time, Speed & Distance**

Mandeep and Jagdeep had gone to visit Ranpur, which is a seaside town and also known for the presence of the historical ruins of an ancient kingdom. They stayed in a hotel which is exactly 250 meters away from the railway station. At the hotel, Mandeep and Jagdeep learnt from a tourist information booklet that the distance between the sea-beach and the gate of the historical ruins is exactly 1 km. Next morning they visited the sea-beach to witness sunrise and afterwards decided to have a race from the beach to the gate of the ruins. Jagdeep defeated Mandeep in the race by 60 meters or 12 seconds. The following morning they had another round of race from the railway station to the hotel. How long did Jagdeep take to cover the distance on the second day?

- A.
53 seconds

- B.
47 seconds

- C.
51 seconds

- D.
45 seconds

Answer: Option B

**Explanation** :

Jagdeep defeated Mandeep in a 1000 m race by 60 m or 12 s.

This means that Mandeep would travel 60 m in 12 s.

∴ Mandeep’s speed = 5 m/s

∴ Mandeep covered 1000 m in 1000/5 = 200 s

∴ Jagdeep covered 1000 m in 200 – 12 = 188 s

∴ Jagdeep covered 250 m in 47 s.

Hence, option 2.

Workspace:

**IIFT 2010 QA | Arithmetic - Time, Speed & Distance**

Mukesh, Suresh and Dinesh travel from Delhi to Mathura to attend Janmasthmi Utsav. They have a bike which can carry only two riders at a time as per traffic rules. Bike can be driven only by Mukesh. Mathura is 300Km from Delhi. All of them can walk at 15Km/Hrs. All of them start their journey from Delhi simultaneously and are required to reach Mathura at the same time. If the speed of bike is 60Km/Hrs then what is the shortest possible time in which all three can reach Mathura at the same time.

- A.
8$\frac{2}{7}$ Hrs

- B.
9$\frac{2}{7}$ Hrs

- C.
10 Hrs

- D.
None of these

Answer: Option B

**Explanation** :

Let Mukesh take Suresh on his bike till B and leave him there to walk till C(Mathura). In the meanwhile, Dinesh keeps walking to reach D, Mukesh comes back picks Dinesh and then both ride to Mathura.

When Mukesh comes back, let us say he meets Dinesh at E.

Let AB = x, then BC = 300 – x

Since Dinesh walks at 15 kmph and bike’s speed is 60 kmph, we have AD = x/4.

∴ BD = $\frac{3x}{4}$

∴ BE = $\left(\frac{60}{75}\right)$$\left(\frac{3x}{4}\right)$ = $\frac{3x}{5}$

Hence, $\frac{300-x}{15}$ = $\frac{{\displaystyle \frac{3x}{5}}+{\displaystyle \frac{3x}{5}}+300-x}{60}$

∴ 4(300 – x) = 300 + x/5

∴ 900 = 4x + $\frac{x}{5}$

$\frac{4500}{21}$ = x

x = $\frac{1500}{7}$ km

Hence, minimum time

= $\frac{x}{60}$ + $\frac{300-x}{15}$

= $\frac{1500}{60\times 7}$ + 20 - $\frac{1500}{15\times 7}$

= $\frac{25}{7}$ + 20 - $\frac{100}{7}$

= 20 - $\frac{75}{7}$

= $\frac{65}{7}$ = 9$\frac{2}{7}$

Hence, option 2.

Workspace:

**IIFT 2009 QA | Arithmetic - Time, Speed & Distance**

Fortuner, the latest SUV by Toyota Motors, consumes diesel at the rate of $\frac{1}{400}$$\left\{\left(\frac{1000}{x}\right)+x\right\}$ litres per km, when driven at the speed of x km per hour. If the cost of diesel is Rs.35 per litre and the driver is paid at the rate of Rs.125 per hour then find the approximate optimal speed (in km per hour) of Fortuner that will minimize the total cost of the round trip of 800 kms.

- A.
49

- B.
55

- C.
50

- D.
53

Answer: Option A

**Explanation** :

Rate of consumption of diesel when speed = x kmph

= $\frac{1}{400}$$\left\{\left(\frac{1000}{x}\right)+x\right\}$ litres/km

Diesel cost = Rs. 35 per litre

Driver is paid Rs. 125 per hour

Distance of 800 km

∴ Total cost = $\left(\frac{800}{x}\right)$ 125 + $\frac{1}{400}$$\left\{\left(\frac{1000}{x}\right)+x\right\}$ × 35 × 800

= 800$\left[\frac{125}{x}+\frac{1000+{x}^{2}}{400x}\times 35\right]$

= $\frac{170000+70{x}^{2}}{x}$

Substituting the options in this, we find that x = 49 gives the minimum cost of Rs. 6899.38

Hence, option 1.

Workspace:

**IIFT 2009 QA | Arithmetic - Time, Speed & Distance**

Two motorists Anil and Sunil are practicing with two different sports cars: Ferrari and Maclarun, on the circular racing track, for the car racing tournament to be held next month. Both Anil and Sunil start from the same point on the circular track .Anil completes one round of the track in 1 minute and Sunil takes 2 minutes to compete a round. While Anil maintains same speed for all the rounds, Sunil halves his speed after the completion of each round. How many times Anil and Sunil will meet between the 6th round and 9th round of Sunil (6th and 9th round is excluded)? Assume that the speed of Sunil remains steady throughout each round and changes only after the completion of that round.

- A.
260

- B.
347

- C.
382

- D.
None of the above

Answer: Option C

**Explanation** :

Anil and Sunil meet every time Anil gains a round over Sunil.

When Sunil completes one round of the track of length, say d, in 2 minutes, Anil meets Sunil every $\frac{d}{d-{\displaystyle \frac{d}{2}}}$ = 2 minutes

∴ Anil meets Sunil 2/2 = 1 time in his first round.

When Sunil completes one round of the track of length, say d, in 2* ^{n}* minutes, Anil meets Sunil every $\frac{d}{d-{\displaystyle \frac{d}{{2}^{n}}}}$ = $\frac{{2}^{n}}{{2}^{n}-1}$ minutes

∴ Anil meets Sunil $\frac{{2}^{n}}{{\displaystyle \frac{{2}^{n}}{{2}^{n}-1}}}$ = 2* ^{n}* - 1 times in his nth round.

∴Anil meets Sunil 2^{7} – 1 = 127 times in the 7^{th} round and 2^{8} – 1 = 255 times in the 8^{th} round.

∴ They will meet 127 + 255 = 382 times between the 6^{th} and 9^{th} rounds.

Hence, option 3.

Workspace:

**IIFT 2009 QA | Arithmetic - Time, Speed & Distance**

Sukriti and Saloni are athletes. Sukriti covers a distance of 1 km in 5 minutes and 50 seconds, while Saloni covers the same distance in 6 minutes and 4 seconds. If both of them start together and run at uniform speed, by what distance will Sukriti win a 5 km mini marathon:

- A.
150 m

- B.
200 m

- C.
175 m

- D.
225 m

Answer: Option B

**Explanation** :

Sukriti runs 1 km in 5 minutes and 50 seconds = 350 seconds.

Saloni runs 1 km in 6 minutes and 4 seconds = 364 seconds.

∴ Sukriti runs 5 km in 5 × 350 = 1750 seconds

Saloni runs a distance of $\frac{1750}{364}$ ≈ 4.8 km in this time.

∴ Sukriti will win the mini marathon by 5 – 4.8 = 0.2 km = 200 m

Hence, option 2.

Workspace:

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