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Solved Example

Example-1
A person covers a certain distance at 72 km/h. How many meters does he cover in 2 minutes.
Solution
Speed = 72 km/h = 72 × 5/18 = 20 m/s
Distance covered in 2min = 20 × 2 × 60 = 2400 m
 
 
Example-2

If a man runs at 3 m/s. How many km does he run in 1 hr 40 minutes?

Solution

Speed of the man = 3 × 18/5 km/h = 54/5 km/h
Distance covered in 5/3 hrs = 54/5 × 5/3 = 18k m
 

 
Example-3
Walking at the rate of 4 kmph a man covers certain distance in 2 hr 45 min. running at a speed of 16.5 km/h the man will cover the same distance in.
Solution
Distance = Speed × time
4 × 11/4 = 11 km
New speed = 16.5 km/h
Therefore Time = D/S = 11/16.5 = 40 min
 
 
 
Example-4

A train covers a distance in 50 min, if it runs at a speed of 48 km/h on an average. The speed at which the train must run to reduce the time of journey to 40 min will be.

Solution

Time = 50/60 hr = 5/6 hr
Speed = 48mph
Distance = S × T = 48 × 5 /6 = 40 km
Time = 40/60 hr = 2/3 hr
New speed = 40 × 3/2 km/h = 60 km/h
 

 
Example-5
Vikas can cover a distance in 1 hr 24 min by covering 2/3 of the distance at 4 km/h and the rest at 5 km/h. The total distance is?
Solution
Let total distance be S
Total time = 1 hr 24 min
A to T:
Speed = 4 km/h Distance = 2/3S
T to S:
Speed = 5 km/h Distance = 1 - 2/3 S = 1/3 S
 = > 21/15 hr = 2/3 S/4 + 1/3S/5
 = > 84 = 14/3S × 3
S = 84 × 3/14 × 3 = 6 km
 
 
 
Example-6

Walking at ¾ of his usual speed, a man is late by 2½ hr. The usual time is?

Solution

Usual speed = S Usual time = T Distance = D
New Speed is 3/4 S
New time is 4/3 T
4/3 T – T = 5/2
T = 15/2 = 7 ½
 

 
Example-7
A man covers a distance on scooter. Had he moved 3 km/h faster he would have taken 40 min less. If he had moved 2 km/h slower he would have taken 40 min more. The distance is?
Solution
Let distance = x m Usual rate = y km/h
x/y – x/y + 3 = 40/60 hr
2y (y + 3) = 9x [Eqn 1]
x/y - 2 – x/y = 40/60 hr y(y - 2) = 3x [Eqn 2]
Divide 1 & 2 and solving we get
x = 40
 
 
 
Example-8

Excluding stoppages, the speed of the bus is 54 km/h and including stoppages, it is 45 km/h. For how many min does the bus stop per hr?

Solution

Due to stoppages, it covers 9 km less.
Time taken to cover 9 km is [9/54 × 60] min = 10 min
 

 
Example-9
Two boys starting from the same place walk at a rate of 5 km/h and 5.5 km/h respectively. What time will they take to be 8.5 km apart, if they walk in the same direction?
Solution
The relative speed of the boys = 5.5 km/h – 5 km/h = 0.5 km/h
Distance between them is 8.5 km
Time = 8.5/0.5 = 17 hrs
 
 
Example-10

Two trains starting at the same time from 2 stations 200 km apart and going in opposite direction cross each other at a distance of 110 km from one of the stations. What is the ratio of their speeds?

Solution

In same time, they cover 110 km & 90 km respectively
So ratio of their speed = 110:90 = 11:9
 

 
Example-11
Two trains start from A & B and travel towards each other at speed of 50 km/h and 60 km/h resp. At the time of the meeting the second train has travelled 120 km more than the first. The distance between them.
Solution
Let the distance travelled by the first train be x km
Then distance covered by the second train is x + 120 km
x/50 = x + 120 / 60
x = 600
So the distance between A & B is x + x + 120 = 1320 km
 
 
 
Example-12

A thief steals a car at 2.30 pm and drives it at 60 km/h. The theft is discovered at 3pm and the owner sets off in another car at 75 km/h when he will overtake the thief.

Solution

Let the thief is overtaken x hrs after 2.30pm
Distance covered by the thief in x hrs = distance covered by the owner in x - 1/2 hr
60x = 75 (x - ½)
x = 5/2 hr
Thief is overtaken at 2.30 pm + 2 ½ hr = 5 pm
 

 
Example-13
In covering distance, the speed of A & B is in the ratio of 3 : 4. A takes 30 min more than B to reach the destination. The time taken by A to reach the destination is.
Solution
Ratio of speed = 3:4 Ratio of time = 4:3
Let A takes 4x hrs, B takes 3x hrs
Then 4x - 3x = 30/60 hr
x = ½ hr
Time taken by A to reach the destination is 4x = 4 x ½ = 2 hr
 
 
 
Example-14

A motorist covers a distance of 39 km in 45 min by moving at a speed of x km/h for the first 15 min. Then moving at double the speed for the next 20 min and then again moving at his original speed for the rest of the journey. Then x = ?

Solution

Total distance = 39 km Total time = 45 min
D = S x T
X × 15/60 + 2x × 20/60 + x × 10/60 = 39 km
x = 36 km/h
 

 
Example-15
A & B are two towns. Mr. Fara covers the di stance from A to Bon cycle at 17 km/h and returns to A by a Tonga running at a uniform speed of 8 km/h. His average speed during the whole journey is.
Solution
When same distance is covered with different speeds, then the average speed
 = 2xy / x + y = 10.88 km/h
 
 
 
Example-16

A car covers 4 successive 3 km stretches at speed of 10 km/h, 20 km/h, 30 km/h & 60 km/h respectively. Its average speed is.

Solution

Average speed = total distance / total time
Total distance = 4 × 3 = 12 km
Total time = 3/10 + 3/20 + 3/30 + 3/60 = 36/60 hr
Speed = 12/36 × 60 = 20 km/h
 

 
Example-17
The ratio between the speeds of the A & B is 2:3 and therefore A takes 10 min more than the time taken by B to reach the destination. If A had walked at double the speed, he would have covered the distance in?
Solution
Ratio of speed = 2 : 3 Ratio of time = 3 : 2
A takes 10 min more
3x - 2x = 10 min
A's time = 30 min Thus, A covers the distance in 30 min, if its speed is x
He will cover the same distance in 15 min, if its speed doubles (i.e 2x)
 
 
 
Example-18

A is twice as fast as B and B is thrice as fast as C is. The journey covered by B in?

Solution

Speed’s ratio a : b = 2 : 1 b : c = 3 : 1
Time's ratio b : c = 1 : 3 b : c = 18 : 54
(if c covers in 54 min i.e. twice to 18 min)
 

 
Example-19
A train M leaves Meerut at 5 am and reaches Delhi at 9 am. Another train N leaves Delhi at 7 am and reaches Meerut at 10:30 am. At what time do the 2 trains cross one another?
Solution
Let the distance between Meerut & Delhi be x
They meet after y hr after 7am
M covers x in 4hr
N covers x in 3 ½ i.e. 7/2 hr
Speed of M = x/4
Speed of N = 2x/7
Distance covered by M in y + 2 hr + Distance covered by N in y hr is x
x/4 (y + 2) + 2x/7(y) = x
y = 14/15 hr or 56 min
 
 
Example-20

A man takes 5 hr 45 min in walking to certain place and riding back. He would have gained 2 hrs by riding both ways. The time he would take to walk both ways is?

Solution

Let x be the speed of walked
Let y be the speed of ride
Let D be the distance
Then D/x + D/y = 23/4 hr - - - - - - - 1
D/y + D/y = 23/4 – 2 hr
D/y = 15/8 - - - - - - - - 2
Substitute 2 in 1
D/x + 15/8 = 23/4 - → D/x = 23/4 - 15/8 = 46 - 15/8 = 31/8
Time taken for walk one way is 31/8 hr
Time taken to walk to and fro is 2 × 31/8 = 31/4 hr → = 7 hr 45 min
 





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