# Time, Speed and Distance - Solved Questions - Part 2

• In this article we wil try to cover various types of questions from the topic Time, Speed and Distance.

We all know the basic formula, viz Speed = Distance / Time. There is no much theory in this topic as it all boils down to applying this equation in various scenarios. We will solve some questions.

Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first met after Brenda has run 100 meters. They next meet after Sally has run 150 meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters ?

Let length is x.

At first meeting point, Brenda runs 100 meters and sally runs (x/2) – 100

At second meeting point, Sally runs 150 meters and Brenda runs x-150 meters

100/x-150 = (x/2) -100/150

15000 = x^2/2-100x -75x+15000

X^2/2 – 175x = 0

X^2-350x = 0

x = 350 meters.

A and B start from same point on the circular track in same direction. A completes a track in 7 minutes and B takes 3 minutes. When they started running it was 12.00 PM by clock. When they meet 4th time what is time by clock?

If A and B takes m and n minutes respectively to complete one circular track then A and B would meet for the first time after

T = (m * n) /|m-n|

They will keep meeting at the interval T.

m = 7 n= 3

T = 21/4 minutes

4th time it will take 4T = > 4 * 21/4 = 21 minutes

So time 12:21

Vijay left P for Q at 10:00 AM. At the same time Ajay left Q for P. After their meeting on a point on the way Vijay took 24 minute to reach Q and Ajay took 54 minute to reach P.At what time they met?

Speed of Vijay/Speed of Ajay = root (54/24) = root (9/4) = 3/2

Distance = 3x*24/60 + 2x*54/60 = 180x/60 = 3x

Time taken to meet = 3x/5x = 3/5 = 36/60

= > after 36 minutes …that is 10:36

A fires two bullets from same place at an interval of 6 minutes but B sitting in a car approaching the place of firing hears the second fire 5 minutes 32 seconds after the first firing. What is the speed of the car if speed of sound is 332m/s?

It took 28 seconds for him to travel from the place where he heard sound first to current place.

Distance (sound) = 28 *332 m

Same distance he travelled in 5 minutes 32 seconds.

Speed of car = 28 *332/332 = 28 m/s

What is the time interval between two successive meets of minute hand and hour hand?

Once the minute hand and hour hand are together, minute hand starts increasing the gap between the hour hand itself by 11/2 degree every minute. Therefore when it has increased the gap by 360 degree it again meets the hour hand. Time taken to increase the gap by 360 degree it again meets the hour hand.

Time taken to increase the gap by 360 degree = 360 / (11/2) = 720/11 = 65 5/11 minutes.

Therefore minute hand and hour hand meets in every 65 5/11 minutes.

Rahul starts a trip when the hands of clock are together between 8am and 9am.He arrives his destination between 2pm and 3pm when the hands are exactly 180 degree apart. How long did the trip take?

At 8am distance between minute hand and hour hand is 240 degree.

It takes minute hand 240/(11/2) = 480/11 = 43 7/11 minutes to catch up with hour hand. Therefore at 43 7/11 minutes past 8.00 the two hands are together.

At 2 pm the distance between minute hand and hour hand is 60 degree. The total relative distance minute hand needs to travel is 240 degree with respect to hour hand.

Therefore time taken = 240/ (11/2) = 480/11 = 43 7/11 minutes.

Therefore at 43 7/11 minutes past 2.00 the two hands are 180 degree apart.

Total time taken  = 6 hours.

Rahul left his house between 2 o clock and 3 o clock in the afternoon. He returned home between 5 o clock and  6 o clock in the evening and noticed that the hour hand and minute hand interchanged their position with what there when he went out. At what time did he return?

Let angle between minute hand and hour hand be x.

So hour hand moves x degree, minute hand moves 1080-x degree

When hour hand moves x degree minute hand moves 12x degree

12x = 1080-x

13x = 1080 = > x = 1080/13

Now we need to find when angle will be 1080/13 between 5 and 6

30*5 – 11M/2 = 1080/13

M = 1740/143

So time is 1740/143 minutes past 5.

Local trains leave a station at an interval of 15 minutes at a speed of 16KMPH.A man moving from opposite direction meets the train at an interval of 12 minutes. What is the speed of man?

Let speed of Man is S

Relative speed = 16+S

15 minutes = 1/4 hour

Trains are 16/4 = 4 km apart

So 4 = (16+S) *1/5

S = 20-16 = 4KMPH

Ravi takes 40 seconds to walk up on an escalator which is moving upwards but he takes 60 seconds to walk up on an escalator which is moving downwards. How much time will he take to walk up if the escalator is not moving?

Let speed of Ravi r and speed of escalator is x

When both are moving up, effective speed (r+x)

When escalator moving down, Ravi moving up effective speed r-x

When escalator not moving his speed r

Speeds (r-x), r and (r+x) are in A P

so time taken when escalator is not moving will be HM of other two time.

Time taken when escalator is not moving = 2*40*60/100 = 48 seconds.

Ravi takes 60 seconds on an escalator which is moving down when he walks down but takes 40 seconds when he runs down. He takes 20 steps when he is walking whereas he takes 30 steps when he is running. What is the total number of steps in escalator ?

Let  speed of escalator x steps/second

Distance covered by Ravi is same whether he walks or runs

20+60x = 30+40x

20x = 10

x = ½ steps/second

total number of steps = 20+60*1/2 = 50 steps

Ravi and Rakesh are climbing on a moving escalator that is going up. Ravi takes 10 seconds to reach the top but Rakesh takes 8 seconds to reach the top. Rakesh takes 4 steps whereas Ravi can take only 3 steps in one second. What is the total number of steps in Escalator?

Speed of escalator  X steps/second

Distance covered by both is same

10(3+x) = 8(4+x)

x = 1 step/second

Number of steps = 10*4 =40 steps.

Two swimmers started simultaneously from the beach to the south and other to the east. Two hours later the distance between them turned out to be 100 KM. Find the speed of faster swimmer knowing that speed of one of them was 75% of speed of other?

Let speed s. Distance travelled in 2 hours 2s

speed 3s/4 = > distance travelled in 2 hours = 3s/2

it will form a right angled triangle.

(3s/2)^2 + (2s)^2 = 100^2

9s^2/4 + 4s^2 = 100^2

S = 40KMPH

Two men A and B run a 4 km race on a circular course of ¼ KM. If their speeds are in the ratio of 5:4. How often does the winner pass the other?

When a completes 5 rounds B completes 4 rounds

So A will overtake B when he completes 5 round = > 5*1/4 = 1 ¼ km

So there are three 1 ¼ km in 4 KM

So A will pass B in 3 times.

The Howrah –puri express can move at 45 km/hr without its rake, and the speed is diminished by a constant that varies as the square root of the number of wagons attatched.If its known with 9 wagons the speed is 30 KM/HR  what’s greatest number of wagons with which the train can just move?

S1/S2 = root ((W1/W2))

where S is change in speed and w is number of wagons

S1 = 45-30 = 15

W1 = 9

S2 = 45-0 = 45

15/45 = 3/ROOT(W2)

W2 = 81

So if 81 wagons are there then train wont move

A watch which gains uniformly is 2 minutes slow at noon on Sunday and is 4 minutes 48 seconds fast at 2PM on the following Sunday. When was it correct?

From Sunday noon to following Sunday 2 PM = 170 hours

The watch gains 2+4+48/60 = 6+48/60 minutes or 6 4/5 minutes in 170 hours.

Watch will show correct time when it has gained 2 minutes.

The watch gains 2 minutes in 2*170/6 4/5 = 50 hours = > 2 days and 2 hours = > 2PM Tuesday

We will solve more questions from this topic in the subsequent articles. Please share your doubts as comments.

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