Aptitude Day 273

Problems on Time and Distance:

1. Walking 4/9^th of his usual speed, a man is 5 minutes too late. The usual time taken by him to cover that distance is:

Solution:

                 New speed = 4/9 of the usual speed(U.T).

                 New time = 9/4 of the usual time.   

    

                 Note: Speed and Time are inversely proportional to each other.

                 9/4 U.T – U.T = 5/60

                 U.T (9-1)/4 = 1/12

                 By solving the above equation, we get

                 U.T = 5/2 

                        = 2(1/2) hours.

2. Starting from his house one day, a student walks at a speed of 1(1/2) km/hr and reaches his school 9 minutes late. Next day he increases his speed by 1 km/hr and reaches the school 9 minutes early. How far is the school from his house?

Solution:

                  Let the distance be x.

                  Difference in timings = 18 minutes = 18/60 = 3/10 hours.

                  2x/3 – 2x/5 = 3/10

                  By solving the above equation, we get

                  x = 45/40 

                     = 1(1/8) km.