# 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                              ?

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