## Time and Distance Part 6

Problem 1: The ratio between the rates of traveling of A and  B IS 2:3 and therefore A takes 10minutes more than the time taken by B to reach the destination. If A had walked at double the speed the time taken by him to reach the destination would have been?

Solution:

Note: Speed and time are inversely proportion

A    :    B

S          2     :       3

T         3      :       2

1———->10

3———->?$\frac{10\times&space;3}{1}$=30

2———->?$\frac{10\times&space;2}{1}$=20

Problem 2: A man is walking at a speed of 10km/hr. After every km, he takes rest for 4min. How much time will he take to cover a distance of 10km?

Solution:

1h———->10km

60min—–>10km

10-1=9

9*4=36

Total time=1 hour 36 min

Problem 3: A boy walks a certain distance and rides back taking a total time of 37min. I could walk both ways in 55min. How long would it take me to ride both ways?

Solution:

w+r=37

2w=55

(w+r)*2=37*2

## Time and Distance Part 7

Problem 1: If a man walks 5km/hr, he misses a train by 7 min. However, if he walked at the rate of 6km/hr, he reaches the station 5 minutes before the arrival of the train. Find the distance covered by him to reach the station?

Solution:

5km/hr ————–>+7

6km/hr—————>-5

7-(-5)=12

Distance=$\frac{S_{1}\times&space;S_{2}}{S_{1}\sim&space;S_{2}}\times&space;\Delta&space;t$

$\frac{5\times&space;6}{1}\times&space;\frac{12}{60}=6km$

Problem 2: If a train runs at the rate of 40km/hr it reaches its destination late by 11min but if it reaches 50km/hr it is late by 5 minutes early. the correct time for the train to complete its journey is?

Solution:

40km/hr———–>+11

50km/hr————->+5

11-5=6

Distance=$\frac{S_{1}\times&space;S_{2}}{S_{1}\sim&space;S_{2}}\times&space;\Delta&space;t$

=$\frac{40\times&space;50}{10}\times&space;\frac{6}{60}=20$

$Speed=\frac{Distance}{time}$

$40=\frac{20}{t}$

$t=\frac{1}{2}=30min$

Original time=30-11=19min

Problem 3: A and B ate 285 km apart. A train starts from A at 9:30 am and travels towards B at 60km/hr another train starts from B at 10:30 AM  and travels towards A at 40km/hr. At what time they will meet?

Solution: Time gap between Train A and B is 1 hour

285 – 60=225

Time = $\frac{Distance}{Speed}$

=$\frac{225}{40+60}=\frac{225}{100}=2.25&space;hrs$

At 12:55 both trains will meet

## Time and Distance Part 8

Problem 1: Walking at $\frac{3}{4}$th of his usual speed a man covers a certain distance in 2 hours more than the time taken to cover the distance at his usual speed. The time taken by him to cover the same distance with usual speed is?

Solution:

New     :     Usual

S             3         :          4

t              4         :          3

4-3=1

1——->2

4———>?$(\frac{4\times&space;2}{1})=8$hours

3———->?$(\frac{2\times&space;3}{1})=6$hours

The time is taken by him to cover the same distance with usual speed is 6 hours.

Problem 2: A car traveling with  $\frac{5}{7}$ th of his usual speed covers 42km in 1hr 40 min 40 sec. What is the usual speed of the car?

Solution:

Given time =1hr 40 min 40 sec

$1+\frac{40}{60}+\frac{48}{3600}$

$1+\frac{2}{3}+\frac{1}{75}$

$\frac{75+50+1}{75}=\frac{126}{75}$

$S=7X,S^{1}=5X$

5X——–>42KM

$Speed=\frac{Distance}{Time}$

$5x=\frac{42}{126}\times&space;75$

$x=5$

S=7*5=35

Problem 3: A thief is reported to a policeman. The thief starts running and the policeman chases him when the policeman starts chasing the thief the thief was at the distance of 250km. The thief and the policeman run at the speed of 8km/hr and 9km/hr respectively. Find the time that policeman will take to catch the thief?

Solution:

1 hr———–>1km

1000meters————->60min

2.5*100meters————>6min*2.5

250meters———>15min

## Time and Distance Part 9

Problem 1: From a point on a circular track 5km long A, B, C  started running in the same direction at the same time will speed of $2\frac{1}{2}$km/hr, 3km/hr,2km/hr then on the starting point all three will meet again?

Solution :

L.C.M of time A,B,C

T = $\frac{5\times&space;2}{5},\frac{5}{3},\frac{5}{2}$

$T=2,\frac{5}{3},\frac{5}{2}$

$\frac{10}{1}=10hrs$

Problem 2: A man covered a certain distance at some speed, if he had move 3km/h faster, he would have taken 40min less. If he moved 2km/h slower, he would have taken 40min more. Find the distance in km?

Solution:

+3           -40

-2            +40

Cross multiply +3 * +40 = 120

-2 * -40 = 80

120:80  = 3:2

1———> 5

3——–>?$(\frac{5\times&space;3}{1})=15$

2———->?$(\frac{5\times&space;2}{1})=10$

$Distance=\frac{S_{1}\times&space;S_{2}}{S_{1}\sim&space;S_{2}}\times&space;\Delta&space;t$

$\frac{15\times&space;10}{5}\times&space;\frac{80}{60}=40km$

Problem 3: “A” leaves Mumbai at 6 AM and reaches Bangalore at 10 AM. “B” leaves Bangalore at 8 AM and reaches Mumbai at 11:30 AM. At what time do they cross each other?

Solution:

The time gap between 6 AM-10 AM=4 hours

The time gap between 8 AM and 11:30 AM=3.5 hours

A                       B

T            4                      3.5

S            3.5                     4

Distance = Speed * Time = 4*3.5 = 14

In two hours A travel 7 km (2*3.5=7km)

$Time=\frac{Distance}{Speed}=\frac{7}{7.5}\times&space;60=56min$

At 8:56 both trains will meet.

## Time and Distance Part 10

Problem 1: A person leaves a place A to place B at 9 AM and reaches place B at 2 pm. Another person starts from point B at 11 AM and reaches point A at 2 PM. Find the ratio of distance covered by them at the time they meet each other?

Solution:

The time gap between 9AM-2PM=5hours

The time gap between 11AM-2PM=3hours

A                    B

T             5                      3

S             3                      5

Distance=Speed*Time=5*3=15

In two hours the distance traveled by A is 6km

15-6=9

$D_{A}=6+9\times&space;\frac{3}{8}=\frac{48+27}{8}=\frac{75}{8}$

$D_{B}=9\times&space;\frac{5}{8}=\frac{45}{8}$

$\frac{75}{8}:\frac{45}{8}$

$75:45$

Problem 2: Two guns were fired from the same place at an interval of 10min and  30 sec but a person on the train approaching the hear the second shot 10 min after the first. The speed of the train (in km/hr) supposing that the speed of sound travels at 330m/s is?

Solution:

10*60*S=30*330

$S=\frac{33}{2}m/s$

$S=\frac{33}{2}\times&space;\frac{18}{5}=\frac{97}{5}=59.4km/h$

Shortcut:

less time                 time difference

10min                           30sec

T=600sec                           30sec

S=330m/s                              ?

$\frac{330\times&space;30}{600}=\frac{33}{2}\times&space;\frac{18}{5}=59.4km/h$