A boy walks to his school at a distance of 6 km with constant speed of 2.5 km h-1 and walks back with a constant speed of 4 km h-1. His average speed for round trip expressed in km h-1, is

Question

A boy walks to his school at a distance of 6 km with constant speed of 2.5 km h-1 and walks back with a constant speed of 4 km h-1. His average speed for round trip expressed in km h-1, is (A)  (24/13) (B)  (40/13) (C) 3 (D)  (1/2)

Tardigrade Admin · Accepted Answer

Total times t = (6/2.5) + (6/4) = (12/5) + (3/2) = (39/10) h Average speed, v = (6+6/39/10) = (120/39) = (40/13) km h-1

Q. A boy walks to his school at a distance of 6 km with constant speed of $2.5 km h^{- 1}$ and walks back with a constant speed of $4 km h^{- 1}$ . His average speed for round trip expressed in $km h^{- 1}$ , is

$\frac{24}{13}$

$\frac{40}{13}$

3

$\frac{1}{2}$

Total times $t = \frac{6}{2.5} + \frac{6}{4} = \frac{12}{5} + \frac{3}{2} = \frac{39}{10} h$
Average speed, $v = \frac{6 + 6}{39/10} = \frac{120}{39} = \frac{40}{13} km h^{- 1}$

Q. A boy walks to his school at a distance of 6 km with constant speed of 2.5kmh−1 and walks back with a constant speed of 4kmh−1. His average speed for round trip expressed in kmh−1, is

1324​

1340​

3

21​

Total times t=2.56​+46​=512​+23​=1039​h Average speed, v=39/106+6​=39120​=1340​kmh−1

Q. A boy walks to his school at a distance of 6 km with constant speed of $2.5 km h^{- 1}$ and walks back with a constant speed of $4 km h^{- 1}$ . His average speed for round trip expressed in $km h^{- 1}$ , is

$\frac{24}{13}$

$\frac{40}{13}$

$\frac{1}{2}$

Total times $t = \frac{6}{2.5} + \frac{6}{4} = \frac{12}{5} + \frac{3}{2} = \frac{39}{10} h$
Average speed, $v = \frac{6 + 6}{39/10} = \frac{120}{39} = \frac{40}{13} km h^{- 1}$