Question

A person covers one third of total distance at x m/s and one fourth of total distance at y m/s .find the average speed of person?

Answer 1

Let total distance be z.

Distance covered at x m/s = z/3
Time taken = z/3 / x = z/3x

Distance covered at y m/s = z/4
Time taken = z/4 / y = z/4y

Total distance covered = z/3+z/4 = 7z/12
Total time taken = z/3x + z/4y = (4zy + 3xz)/12xy

Average speed = total distance/ total time
                         = (7z/12) / [(4zy + 3xz)/12xy]
                         = (7z * 12xy) / [12 * z(4x+3y)]
Cancelling z and 12 from numerator and denominator.

Answer 7xy/4y+3x.

Answer 2

Given :
Let the total distance travelled by the person be = s meters
speed for the one third distance = x m/s
Speed for the one fourth distance = y m/s
Time taken to cover one third distance = Distance/ speed =s/3x
Time taken to cover one fourth distance = Distance/speed=s/4y
Total time= s/3x +s/4y
=s[4y+3x]/12xy
Total distance = s/3 + s/4
=4s+3s/12
=7s/12
Average speed= total distance / total time
=7s*12xy/12*s*(4y+3x)
=7xy/(4y+3x)

∴ The average speed of person is 7xy/(4y+3x)