Question #59ae1

1 Answer
Apr 20, 2016

Answer:

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"#