Question

Question #59ae1

Answer 1

Approximately `47.4 "km"/"hr"`

Explanation:

Let the distance up (and down) the hill be `d` km.

Going up the hill at `37` km/hr would take
`color(white)("XXX")d cancel(" km")xx "hr"/(37 cancel(" km"))= d/ 37 " hr."`

Similarly going down the hill at `66` km/hr would take
`d/66" hr".`

`color(red)("Total distance traveled: " 2d" km")`

`color(blue)("Time time taken:")`
`color(white)("XXX")d/37+d/66" hr"`

`color(white)("XXX")=(66d+37d)/(66xx37)" hr."`

`color(white)("XXX")=color(blue)((103d)/(2442) " hr.")`

Average speed `=color(red)("total distance")/color(blue)("total time")`

`color(white)("XXX")=(color(red)(2cancel(d)))/(color(blue)((103cancel(d))/(2442))) "km"/"hr"`

`color(white)("XXX")=4884/103 "km"/"hr"`

`color(white)("XXX")~~47.4 "km"/"hr"`