INTRODUCTION:

The problems based on time & distances are based on speed, time & distance. The general process to find out distance is as follows:

DISTANCE= SPEED*TIME.

Where S is the speed of the moving object, T is the time period & D is the distance covered.

Process to convert km/h in to m/s or vice versa is as follows:

1km/ h = 5/18 m/s

1 m/s = 18/5 km/h

Relative speed: The different between the speeds of two moving object with respect to their direction is known as relative speed. The relative speed according to the two different directions is as follows:

Case I: In case of same direction it is the difference between the speeds of the two moving objects.

Case II: In case of opposite direction it is the sum of the speeds of the two moving objects.

Example: Two objects are moving @ 30 km/h & 50 km/h. then the relative speed between them in two different cases are as follows:

Case I: In the same direction it is (50-30) i.e. 20 km/h

Case II: In case of opposite direction it is (50+30) i.e.80 km/h.

QUESTION:

A boy goes to school @ 10 km/h & returns to his home @ 12 km/h respectively. If he takes 5 ½ hrs in all, find the distance between school & his home.

A.20km.

B.25 km.

C.30 km .

D.30 km.

Sol:

The L.C.M of two different speeds of 10 &12 is 60.

If 60 km is the distance, then at the rate of two different speeds the times are 6 & 5 hrs respectively.

When the total time is 6+5 i.e. 11hrs, the distance between the two points is 60 km.

So when the total time is 5 ½ hrs, the distance between school & home is 30 km.

QUESTION:

A person covers half of his journey @ 21 km/h, & the rest half @ 24 km/h respectively. If he takes 15 hrs in all, find the distance covered by the person.

A.168 km.

B.224 km.

C.336 km.

D.280 km.

Sol:

The L.C.M of 21 &24 i.e. 168.

If 168 km is the distance, then the person covers ½ of it i.e. 84 km @ 21 km/h in 4 hrs and the rest half @ 24 km/h in 3 ½ hrs.

So when the total time is 7 ½ hrs, the distance is 168 km.

And if the total time is 15 hrs, the distance is 336 km.

QUESTION:

A person goes a certain distance @ 48 km/h & returns to the starting point @ 96 km/h respectively. Find his average speed during the whole journey.

A.60 km/h.

B.64 km/h

C.68 km/h

D.72 km/h

Sol:

The L.C.M of 48 @ 96 is 96.

If the distance is 96 km, then @ two different speeds the required time are 2 &1 hr respectively.

Average speed= Total distance/Total time

So the average speed i.e.(96+96)/2+1=64 km/h.

QUESTION:

A person covers a certain distance between his house & office on scooter. Having an average speed of 45 km/h, he is late by 15 minutes. How ever, with a speed of 60 km/h, he reaches his office 7 ½ min earlier. Find the distance between his house & office.

A.67.5 km.

B.60 km.

C.75 km.

D.120 km.

Sol: Let the L.C.M of the two different speeds i.e. the L.C.M of 45 & 60 =180 is the distance.

So @ 45km/h, it will take him 4 hrs & @ 60 km/h, it will take him 3 hours.

In this case the difference between the two times is 1hr for a distance of 180 km.

So when difference between the two time i.e. [15-(-7 ½)], the distance is 67.5 km.

QUESTION:

A boy walking at a speed of 15 km/h reaches his school 20 minutes late. Next time he increases his speed 5km/h, but still he is late by 10 minutes. Find the distance of his school from his house.

A.10 km.

B.12 km.

C.15 km.

D.20 km.

Sol: In this case the two different speeds are 15 & 20 km/h of which the L.C.M is 60.

If the distance is 60 km, then @ 15km/h it will take him 4 hours where as @ 20 km/h it will take him 3 hours. So the difference between two times is 1 hr.

When the difference is 1 hr, the distance is 60 km.

So when the difference is (20-10) i.e. 10 minutes or 1/6 hr, the distance is 10 km.

QUESTION:

: The distance between two stations is 900 km. A train starts from A and moves towards B at an average speed of 30 km/h. Another train starts from B, 20 minutes earlier than the train A, and moves towards A at an average speed of 40 km/h. How far from A will the two trains meet?

Sol: If the train B starts 20minutes earlier than the train A, it must have covered (1/3*40) i.e.40/3 km. So at the time of the starting of A, the difference between the distances between them is (900-40/3) i.e.2660/3 km.

The ratio between the speeds of the two trains is 30: 40 i.e.3:4.So at the same time the difference between the two train is 3+4 i.e. 7 parts which is 2660/3

The distance from A at which they will meet i.e. 3 parts =380 km.

QUESTION:

Walking ¾ of his usual speed, a person is 15 minutes late to reach his office. Find his usual time to cover the distance.

A.30 min.

B.45 min.

C.60 min.

D.75 min.

Sol:

If the usual speed is 1 part, then the new speed is ¾.

So the ratio between the speed of usual & new is 4:3.

There fore the ratio between usual time & new time is 3:4. (As speed α 1/Time)

Due to 1 part more time in case of new time it takes 15 minutes more.

In this case the usual time is 3 parts i.e. 45 minutes.

QUESTION:

Walking 5/3 of his usual speed, a person is 20 minutes earlier to reach his office. Find his usual time to cover the distance.

A.25 min.

B.50 min.

C.60 min.

D.100min.

Sol:

The ratio between the usual & new speed is 3:5, so that the ratio between the time between usual & new is 5:3.

New time is less than the usual time by 2 parts, so it takes 20 minutes earlier.

Usual time is 5 parts i.e. 50 minutes.

QUESTION:

A train leaves from kolkata at 7.30 am and travels @ 40 km/h; another train leaves kolkata at noon and travels @ 64 km/h, when and where the second train overtakes the first?

A.400 km.

B.460 km.

C.480 km.

D.500 km.

Sol:

The first train starts its journey 4 hrs 30 minutes before the second train in which it must have covered 180 km.

When the second train starts its journey, the difference between the distances from the first train is 180 km.

So the second train meets the first train after (180/24) i.e.7 hrs 30 minutes. (As required time to meet= difference between the distance/ relative speed)

And the distance after which the second train meets i.e. 64*7 ½ =480 km.

QUESTION:

One man takes 150 steps a minute, each 4 decimeters long, another walks 4 km/h, if they start together, how soon will one of them be 60 meters ahead to the other?

A.3 min.

B.6 min.

C.9 min.

D.10min.

Sol:

The speed of the first man is 150* 4/10 m i.e.60 meters/min & the second man it is 200/3 meters/min.

The difference between them is 20/3 meter in 1minute.

So one will be ahead of 60 meters to the other in 9 minutes.

QUESTION:

Two men A & B start from a place P walking @ 5.5 & 6.5 km/h respectively. How many km will they be apart at the end of 5 hrs, if they walk in the opposite direction?

A.20km.

B.30 km.

C.40km.

D.60 km.

Sol: In 1 hr, they are 12 km apart.

So in 5 hrs, they are 60 km apart.

QUESTION:

Two men A & B start their journey from P to Q, a distance of 36 km, at 4 & 5 km/h respectively. B reaches Q and returns immediately and meets A at the point R.

(i) Find the distance from P to R.

(ii) And also find the distance covered by B.

(iii) the difference between the distances covered by B & A.

Sol: The ratio between the speeds of A & B is 4:5, so that the ratio between the distance covered by A & B is also 4: 5. (As speed α Distance)

So that the total distance covered by A & B by both i.e. 9 parts. And also the total distance covered by A & B by both i.e. twice of 36km

So 9parts = 72 km.

The distance from P to R is 4 parts = 32 km.

Similarly, the distance traveled by B is 5 parts i.e.40 km.

And the difference between the distances covered by B & A is 1 part i.e. 8 km.

QUESTION:

A man sets out to cycle from Mumbai to kolkata, and at the same time another man starts from Kolkata to Mumbai. After passing each other they complete their journey in 4 & 9 hrs respectively. At what rate does the second man cycle if the first cycle at 9 km/h?

A.4 km/h.

B.5 km/h.

C.6 km/h.

D.8 km/h.

Sol: According to the method

A’s speed : B’s speed=√B’s time :√A’s time.

Speed of second man= 6 km/h.

QUESTION:

Two bullets were fired at a place at an interval of 38 minutes. A person approaching the firing point in his car hears the two sounds at an interval of 36 minutes. If the speed of the sound is 330 m/sec, what is the speed of the car?

A.60km/h.

B. 66km/h.

C.72km/h.

D.76km/h.

Sol: The speed of car in 36 min could be traveled by sound in (38-36) i.e. in 2 min.

The speed of the car in 36 min=330*120 m/sec.

So the speed of car per hour =66 km/h.

QUESTION:

Two persons do the same journey by traveling respectively @ 9 & 10 km/h. Find the length of the journey when one takes 32 minutes longer than the other.

A.36km.

B.40km.

C.48km.

D.50km.

Sol: If the total distance is the L.C.M of 9 & 10i.e.90 km, then @ 9 km/h, it will take 10 hrs & @ 10 km/h, it will take 9 hrs.

When the difference between two timing is 1 hr, the distance is 90km.

So when the difference between two timing is 32min, the distance is 48km.

QUESTION:

A carriage driving in a fog passed a man who was walking @ 5km/h in the same direction. He could see the carriage for 6 minutes and it was visible to him up to a distance of 120metres. What was the speed of the carriage?

A.5km/h.

B.6 km/h.

C.6.2km/h.

D.7.2 km/h.

Sol: The speed of carriage in 6 min = The speed of man in 6 min @ 5km/h + 120metres.

So, the speed of carriage per hour=6.2 km/h.

QUESTION:

A man takes 6 hrs 30 min in walking to a certain place and riding back. He would have gained 2hrs 10 min by riding both ways. How long would he take to walk both ways? And also find how long would he take to ride both ways?

Sol: Required time to walk both ways= Required time to walk + riding back +Time of Gain.

=6 hrs 30min+2 hrs 10 min=8 hrs 40 min.

Required time to ride both ways= Required time to walk + riding back -Time of Gain.

=6 hrs 30min-2 hrs 10 min=4 hrs 20 min.

QUESTION:

A man leaves a point P and reaches the point Q in 4 hrs. Another man leaves the point Q, 2 hrs later and reaches the point P in 4 hours. Find the time in which first man meets to the second man.

QUESTION:

A person covers a certain distance in 24 minutes if he runs at a speed of 27 km/h on an average. Find the speed at which he must run to reduce the time of journey to 18minutes.

A.36 km/h.

B.40.5km/h.

C.44km/h.

D.48km/h.

Sol: The ratio between two time =24:18 i.e.4:3.

The ratio between two speeds =3:4.

( As time α1/speed)

So if 3 parts of speed = 27 km/h.

Then 4 parts of speed = 36 km/h.

QUESTION:

Without any stoppage a person travels a certain distance at an average speed of 15 km/h, and with stoppage he covers the same distance at an average speed of 12 km/h. How many minutes per hour does he stop?

A.10min.

B.12min.

C.15min.

D.20min.

Sol: The ratio between speed without stoppage & with stoppage =15:12 i.e.5:4.

So, the ratio between time without stoppage & with stoppage =4:5.

Therefore, in 5 parts there is a rest of 1 part .

So, in 60 parts or 60 min, there is a rest of 12 part or12min..

QUESTION:

A person has to cover a distance of 100km in 10 hrs. If he covers half of the journey in 3/5 of the time, what should be his speed to cover the remaining distance in the time left?

QUESTION:

A man travels 480 km in 6 hrs, partly by air and partly by train. If he had traveled all the way by air, he would have saved3/4 of the time he was in train and would have arrived at his destination 3hrs early .Find the distance traveled by the train.

QUESTION:

One aeroplane started 1 hr later than the scheduled time from a place 3000 km away from its destination. To reach the destination at the scheduled time the pilot had to increase the speed by500 km/h. What was the speed of the aeroplane per hour during the journey?

QUESTION:

A train leaves the station 1 hour before the scheduled time. The driver decreases its speed by 50 km/h. At the next station 300 km away, the train reached on time .Find the original speed of the train.

QUESTION:

When a person travels equal distance at speeds V1 and V2 km/h, his average speed is 8 km/h. But when he travels at these speeds for equal times his average speed is 9 km/h. find the difference of the two speeds.

QUESTION:

A person covers a certain distance on scooter. Had he moved 6 km/h faster, he would have taken 30 minutes less. If he had moved 4 km/hr slower, he would have taken 1 hr 30 min more.Find the original speed.

QUESTION:

A train does a journey without stopping in 9 hrs. If it had traveled 8 km/h faster, it would have done the journey in 6 hours. What is its original speed?

QUESTION:

A car travels a distance of 80 km in 3 hrs partly at a speed of 45 km/h & partly at 20 km/h. Find the distance traveled at a speed of 45 km/h.

.

QUESTION:

A distance is covered by a person in 6 hrs. He covers ¾ of it at 12 km/h and the remaining at 16 km/h. Find the total distance

QUESTION:

The ratio between the speeds of Ajay & Bijay is 6:7. If Ajay takes 30 minutes more than Bijay to cover a distance, then find the actual time taken by Ajay and Bijay.

QUESTION:

A person covers 2/3 rd of his journey at 30 km/h and remaining journey at 60 km/h. If the total journey is of 270 km, what is his average speed for the whole journey?

QUESTION:

A hare sees a dog 200 meters away from her and scuds off in the opposite direction at a speed of 24 km/h. Two minutes later the dog perceives her and gives chase at a speed of 32 km/h. How soon will the dog overtake the hare, and at what distance from the spot whence the hare took flight?

QUESTION:

Train A traveling at 60 km/hr leaves Mumbai for Delhi at 6 P.M. Train B traveling at 90 km/hr also leaves Mumbai for Delhi at 9 P.M. Train C leaves Delhi for Mumbai at 9 P.M. If all three trains meet at the same time between Mumbai and Delhi, what is the speed of Train C if the distance between Delhi and Mumbai is 1260kms?

A.60 km/hr

B.90 km/hr

C.120 km/hr

D.135 km/hr

QUESTION:

A man moves from A to B at the rate of 4 km/hr. Had he moved at the rate of 3.67 km/hr, he would have taken 3 hours more to reach the destination. What is the distance between A and B?

33 kms

132 kms

36 kms

144 kms

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