Questions on Time, Spped and Distance

Question1: A man leaves his house at H₁:X  ( at a time X minutes past H₁ hours) and returns home after a few hours at H₂:Y ( at Y minutes past H₂ hours). He finds that the position of the minutes hands and hour hands had interchanged during this period. In other words the angle between the hands was the same just when he left the house and when he just returned home. Find the value of X and Y in terms of H₁ and H_2.

Solution:   We know the speed of minute hand = 6⁰ per minute, and

Speed of Hour hand = 30⁰ per hour or ½⁰ per minute

Angle made by the Minute hand with vertical line ( 12-6 line in the clock) when he left the home at H₁:X= 

Angle made by the Hour hand with vertical line when he returned home.

Therefore,                 6X = 30 H₂+ Y/2……………. Equation 1

Similarly;

Angle made by the Hour hand with vertical line ( 12-6 line in the clock) when he left the home at H₁:X= 

Angle made by the Minute hand with vertical line when he returned home.

Therefore,

30H₁+X/2 = 6Y…………. Equation 2

Solving the above two equations we get

X= (720H₂ +60H₁)/143

Y= (720H₁ +60H₂)/143

From the above two values of X and Y, it can be found for how much  time was the man away from his home. Apart from the above, we can also derive a few more short cuts as mentioned below:

  

 The angle between the hands=(H2-H1)360/13

Time Away=(H2-H1)+(60/13)(H1-H2)