[Manco]

A car completes a journey at an average speed of 40km/h. At what speed must it travel in the return journey so that the average speed of the complete journey (out and back) is 60km/h?

The answer given in the book is 120km/h.

[Celine91]

let D be the distance of the path (same out and back )
Vout is the average speed of the journey out
and Vback is the average speed of the journey back
Tout = D / Vout
and Tback = D / Vback

T_total = Tout + Tback =>
average speed of whole journey (out and back ) = 2D / T_total
the average speed = 2D / (Vout + Vback)
Vout = D / (40 km/hr)
V back = D / (Vback)

60 km/hr = 2D / (D/40 + D/Vback)... we can divide the whole right side fraction with D so we get:

60 = 2 / ( 1/40 + 1/Vback)... simplify the calculation

1/40 + 1/Vback = 2/60

1/Vback = 1/120
Vback = 120 km/hr

good luck!

[Manco]

Thanks... (idk how to comment : p )

[Celine91]

let D be the distance of the path (same out and back )
Vout is the average speed of the journey out
and Vback is the average speed of the journey back
Tout = D / Vout
and Tback = D / Vback

T_total = Tout + Tback =>
average speed of whole journey (out and back ) = 2D / T_total
the average speed = 2D / (Tout + Tback)
Tout = D / (40 km/hr)
T back = D / (Vback)

60 km/hr = 2D / (D/40 + D/Vback)... we can divide the whole right side fraction with D so we get:

60 = 2 / ( 1/40 + 1/Vback)... simplify the calculation

1/40 + 1/Vback = 2/60

1/Vback = 1/120
Vback = 120 km/hr

good luck!

(mixed up with the variables! :p corrected now.. )