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A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?
Speed = Distance/ Time
Speed = (600 / 5 x 60) = 2 m/s
i.e. Speed = 2 x (18/5) km/hr = 7.2 km/hr
Q. A man takes 5 hours 45 min in walking to a certain place and riding back. He would have gained 2 hours by riding both ways. The time he would take to walk both ways, is:
Given that time taken for riding both ways will be 2 hours lesser than the time needed for walking one way and riding back.
From this, we can understand that
Time needed for riding one way = time needed for walking one way  2 hours
Given that time taken in walking one way and riding back = 5 hours 45 min
Hence, the time he would take to walk both ways = 5 hours 45 min + 2 hours = 7 hours 45 min
Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?
Due to stoppages, it covers 9 km less in an hour.
Time taken to cover 9 km =(9/54 )*60 min = 10 min.
A man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.
Let time taken to travel the first half = x hr
⇒ Time taken to travel the second half = (10  x) hr
∵ Distance = Time * Speed
Distance covered in the first half = 21x
Distance covered in the the second half = 24(10  x)
∵ Distance covered in the the first half = Distance covered in the the second half
⇒ 21x = 24(10  x)
⇒ 45x = 240
⇒ x = 16/3
Total Distance = 2 * 21(16/3) = 224 Km [∵ multiplied by 2 as 21x was the distance of half way]
A car traveling with 5/7 of its actual speed covers 42 km in 1 hr 40 min 48 sec. What is the actual speed of the car?
Time taken = 1 hr 40 min 48 sec = 1 hr (40 + 4/5) min = 1 + 51/75 hrs = 126/75 hrs
Let the actual speed be x km/hr
Then, (5/7) * x * (126/75) = 42
⇒ x = 42 * 7 * (75 / 5) * 126
⇒ x = 35 km/hr
A man covered a certain distance at some speed. If he had moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. What is the the distance in km?
According to given condition:
On solving we will get:
t = 10/3 hrs and v = 12 km/hr
So, distance = (10/3) * 12 = 40 km
A and B walk around a circular track. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. If they start at 8 a.m. from the same point in opposite directions, how many times shall they cross each other before 9.30 a.m.?
Relative speed = Speed of A + Speed of B = 2 + 3 = 5 rounds per hour
(∴ they walk in opposite directions)
⇒ They cross each other 5 times in 1 hour and 2 times in 1/2 hour
∵ Time duration from 8 a.m. to 9.30 a.m. = 1.5 hour
∴ They cross each other 7 times before 9.30 a.m
Two boys starts from the same place walking at the rate of 5 kmph and 5.5 kmph respectively in the same direction. What time will they take to be 8.5 km apart?
In this type of question, we need to get the relative speed between them.
The relative speed of the boys = 5.5 – 5 = 0.5 km/h
The distance between them is 8.5 km.
Time = Distance/Speed
Time = 8.5/0.5 = 17 hrs
In covering a distance of 30 km, Arun takes 2 hours more than Anil. If Arun doubles his speed, then he would take 1 hour less than Anil. What is Arun's speed?
Let Anil takes x hrs.
⇒ Arun takes x+2
If Arun doubles the speed, he will take x1 hrs
⇒ He needs 3 hours less.
∵ Double speed means half time.
∴ Half of the time required by Arun to cover 30 km = 3 hours
⇒ Time required by Arun to cover 30 km = 6 hour
Thus, Arun's speed = 30/6 = 5 km/h
A car travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. What is the average speed for the first 320 km of the tour?
Total time taken = (160/64) + (160/80) = 9/2 hrs
∴ Average speed = 320 x (2/9) = 71.11 km/hr.
A jeep travels a distance of 100 km at a uniform speed. If the speed of the jeep is 5 kmph more, then it takes 1 hour less to cover the same distance. The original speed of the jeep is:
Let the original speed of the jeep be x kmph.
⇒100/x−100/(x+5)=1
Solving this, we get x = 20 kmph.
Two athletes cover the same distance at the rate of 10 and 15 kmph respectively. Find the distance travelled when one takes 15 minutes longer than the other.
Let the distance be D km.
⇒ D/10−D/15 = 15/60
⇒ D/30 = 1/4
⇒ D = 7.5 km
If Sita walks at 5 kmph, she misses her train by 10 minutes. If she walks at 7 kmph, she reaches the station 10 minutes early. How much distance does she walk to the station?
Let the distance be D.
D/5−D/7=(10+10)/60
⇒ D = 35/6 = 5.8km
A friend is spotted by Lalloo at a distance of 200 m. When Lalloo starts to approach him, the friend also starts moving in the same direction as Lalloo. If the speed of his friend is 15 kmph, and that of Lalloo is 20 kmph, then how far will the friend have to walk before Lalloo meets him?
Lalloo is unfortunate that the friend is moving away from him.
(Because the friend moves in same direction as Lalloo).
relative speed= 20 15= 5,kmph. distance= 200 m.
Thus, Lalloo will meet his friend when he gains 200 m over him.
=> time required = distance / speed = 0.2/5 = 1/25 hrs.
=> Distance travelled by the friend in 1/25 hrs. (when Lalloo catches up him)
=> Time x Speed = 1/25 x 15 = 3/5 km = 600 m
A person going from Pondicherry to Ootacamond travels 120 km by steamer, 450 km by rail and 60 km by horse transit. The journey occupies 13 hours 30 minutes, and the speed of the train is three times that of the horsetransit and 1(1/2) times that of the steamer. Find the speed of the train.
If speed of horse is X, then speed of steamer = 2X and
Speed of train = 3X => (120/2x)+ (450/3x)+(60/x) = 13.5
=> X = 20 km/hr => 3X = 60 km/hr.
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