Time Speed and Distance Questions for IIFT PDF

Time Speed and Distance Questions for IIFT PDF
Time Speed and Distance Questions for IIFT PDF

Time Speed and Distance Questions for IIFT PDF

Download important IIFT Time Speed and Distance sense Questions PDF based on previously asked questions in IIFT and other MBA exams. Practice Time Speed and Distance questions and answers for IIFT exam.

Download Time Speed and Distance Questions for IIFT PDF

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Question 1: 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 am 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

a) 6 : 15 am

b) 6 : 30 am

c) 6 :45 am

d) 7 : 00 am

e) 7 : 15 am


DIRECTIONS for the following three questions: Answer the questions on the basis of the information given below.

A city has two perfectly circular and concentric ring roads, the outer ring road (OR) being twice as long as the inner ring road (IR). There are also four (straight line) chord roads from E1, the east end point of OR to N2, the north end point of IR; from N1, the north end point of OR to W2, the west end point of IR; from W1, the west end point of OR, to S2, the south end point of IR; and from S1 the south end point of OR to E2, the east end point of IR. Traffic moves at a constant speed of $30\pi$ km/hr on the OR road, 20$\pi$ km/hr on the IR road, and 15$\sqrt5$ km/hr on all the chord roads.

Question 2: Amit wants to reach E2 from N1 using first the chord N1 – W2 and then the inner ring road. What will be his travel time in minutes on the basis of information given in the above question?

a) 60

b) 45.

c) 90

d) 105

Question 3: A train running at the speed of 66 kmph crosses a signal pole in 18 seconds. What is the length of the train?

a) 330 meters

b) 300 metres

c) 360 metres

d) 320 metres

e) None of these

Question 4: To reach point B from point A, at 4pm, Sara will have to travel at an average speed of 18 kmph. She will reach point B at 3 pm if she travels at an average speed of 24 kmph. At what average speed should Sara travel to reach point B at 2 pm ?

a) 36 kmph

b) 28 kmph

c) 25 kmph

d) 30 kmph

e) 32 kmph

Question 5: At his usual rowing rate, Rahul can travel 12 miles downstream in a certain river in 6 hr less than it takes him to travel the same distance upstream. But if he could double his usual rowing rate for this 24 miles round trip, the downstream 12 miles would then take only 1 hr less than the upstream 12 miles. What is the speed of the current in miles per hour?

a) $\frac{7}{3}$

b) $\frac{4}{3}$

c) $\frac{5}{3}$

d) $\frac{8}{3}$

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Question 6: On a 20 km tunnel, connecting two cities A and B, there are three gutters (1, 2 and 3). The distance between gutters 1 and 2 is half the distance between gutters 2 and 3. The distance from city A to its nearest gutter, gutter 1, is equal to the distance of city B from gutter 3. On a particular day, the hospital in city A receives information that an accident has happened at gutter 3. The victim can be saved only if an operation is started within 40 min. An ambulance started from city A at 30 km/hr and crossed gutter 1 after 5 min. If the driver had doubled the speed after that, what is the maximum amount of time would the doctor get to attend the patient at the hospital. Assume 1 min is elapsed for taking the patient into and out of the ambulance?

a) 4 min

b) 2.5 min

c) 1.5 min

d) The patient died before reaching the hospital

Question 7: A truck covers a distance of 330 kms at the speed of 30 km. /hr. What is the average speed of a car which travels a distance of 110 kms more than the truck in the same time ?

a) 42 km./hr,

b) 48 km./ hr.

c) 39 km. /hr.

d) 36 km. / hr.

e) None of these


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.

Question 8: At what time do Ram and Shyam first meet each other?

a) 10 a.m.

b) 10:10 a.m.

c) 10:20 a.m.

d) 10:30 a.m.

Question 9: Two boats, traveling at 5 and 10 kms per hour, head directly towards each other. They begin at a distance of 20 kms from each other. How far apart are they (in kms) one minute before they collide.

a) 1/12

b) 1/6

c) 1/4

d) 1/3

Question 10: 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

a) 6

b) 4

c) 3

d) 5

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Answers & Solutions:

1) Answer (B)

According to given conditions angle between AC and AB is 30 degrees and between AB and BC is 60 degrees. So the triangle formed is a 30-60-90 triangle.

So, total time taken by train is 5 hrs, hence the train reaches at 1 pm. Accordingly, Rahim has to reach C fifteen minutes before i.e. at 12:45 PM.

Time taken by Rahim to travel by car is around 6.2 hrs. So, the latest time by which Rahim must leave A and still be able to catch the train is 6:30 am.

2) Answer (D)

Let the radii of 2 circles be R and r respectively such that R=2*r. Triangle O$N_2$$E_1$and all the other 3 similar triangles form a right angle at the centre . So using pythagoras theorem the value of chords come out to be $\sqrt5$ * R/2 . Hence the total distance traveled is $\sqrt5$ * R/2 + 0.5*R*pi. Total time required can be calculated by distance / speed which comes out to be 3.5*R. Among options only 105 is integral multiple of 3.5.

3) Answer (A)

66 kmph = 66 *5/18 m/s

It takes 18 secs to cross the post. So, length = 66 * 18* 5/18 = 330 metres.

4) Answer (A)

average speed = (total distance ) /(total time)

Let her time taken to finish race with 18km/hr average speed be y hours

And distance covered in both cases is same


D = 18 × y…..(1)

D = 24 x (y-1)….(2)

From equation 1 and 2

y = 4 hours

And D = 72 km

Now if she need to reach at 2pm i.e in 2 hours then the average speed = 72/2 = 36 km/hr

5) Answer (D)

$12/(R – S) = T$
$12/(R + S) = T – 6$
$12/(2R – S) = t$
$12/(2R + S) = t – 1$
=> $12/(R – S) – 12/(R + S) = 6$ and $12/(2R – S) – 12/(2R + S) = 1$
=> $12R + 12S – 12R + 12S = 6R^2 – 6S^2$ and $24R + 12S – 24R + 12S = 4R^2 – S^2$
=> $24S = 6R^2 – 6S^2 and 24S = 4R^2 – S^2$
=> $6R^2 – 6S^2 = 4R^2 – S^2$
=> $2R^2 = 5S^2$
=> $24S = 10S^2 – S^2 = 9S^2$
=> $S = 24/9 = 8/3$

6) Answer (C)

Let the distance between gutter 1 and A be x and between gutter 1 and 2 be y.

Hence, x + y + 2y + x = 20 => 2x+3y=20

Also x = 30kmph * 5/60 = 2.5km

Hence, y = 5km

After the ambulance doubles its speed it goes at 60kmph i.e. 1km per min. Hence, time taken for the rest of the journey = 15*2 + 2.5 = 32.5

It takes 1 min to load and unload the patient.

Hence, total time = 5 + 32.5 + 1 = 38.5 mins

So, the doctor would get 1.5 min to attend to the patient.

7) Answer (E)

Distance tarvelled by car = 330+110 = 440

Time taken by truck = 330km/(30km/hr)

= 11 hours

speed of car = 440km/11 hours

= 40kmph

8) Answer (B)

Let the time at which they meet be t minutes past 10.

So, distance run by Ram + distance run by Shyam = 10 km

=> (60+t)5/60 [t+60 because he would have traveled for 9 am to 10 am and t minutes more before meeting Shyam]+ (15+t)*10/60 [15+t because he would have traveled from 9:45 to 10:00 and t minutes more]= 10

=> 300+5t+150+10t = 600 => t = 10

So, they meet at 10.10 am

9) Answer (C)

The relative speed is 15 km/hr = 15 km/60 min = 0.25 km/min = 250 m/min.
Therefore, one minute before they collide, they are at a distance of 250m.

10) Answer (D)

Let the time taken by Partha to cover 60 km be x hours.
Narayan will cover 60 km in x-4 hours.

Speed of Partha = $\frac{60}{x}$
Speed of Narayan = $\frac{60}{x-4}$

Partha reaches the mid-point of A and B two hours before Narayan reaches B.

=> $\dfrac{30}{\frac{60}{x}} + 2 = \dfrac{60}{\frac{60}{(x-4)}}$

$\frac{x}{2} + 2 = x-4$

Partha will take 12 hours to cross 60 km.
=> Speed of Partha = $\frac{60}{12}=5$ Kmph.
Therefore, option D is the right answer.

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