Q A man starts walking at 3 pm ha walks at a speed of 4 km hr on level ground

avg speed=2*3*6/(3+6)=4
so speed of man=4km/hr
distance=4*6=24km
since we want one way distance it will be
24/2 =12
11 years agoHelpfull: Yes(18) No(3)

let distance of ground level is g and for hill level is h .
now, (s/4)+(h/3)+(s/4)+(h/6)=(9-3)
=> s+h=12 ....so ans is 12km
10 years agoHelpfull: Yes(18) No(3)
Data inadequate
11 years agoHelpfull: Yes(7) No(6)
My answer :-
Given man has started at 3pm and reached home at 9 pm : diff is 6 hrs

4km/hr on level ground :- (4km - 1hr)
3km/hr on uphill [ given,downhill 6km/hr,therefore] : [6km - 2hr]
6km/hr downhill : [6km : 1hr]
remaining 2 hrs
4km/hr on level ground : [ 2hr : 8km]

Therefore Ans is 24
10 years agoHelpfull: Yes(5) No(2)
I think LCM of 4,3,6,4= 12 km/hr..
12*6= 72km...
10 years agoHelpfull: Yes(3) No(4)
let the distance be equal=d for each part, then
d/4/4+d/4/6+d/4/3+d/4/4=9-3
d/8+d/24/+d/12=6
6*d=6*24
d=24(for whole journey)
for one way (24/2=12Kms)
10 years agoHelpfull: Yes(3) No(0)
Data inadequate
Bcoz Let he Spend Time as Following
0 0 0 6 in hrs
6 0 0 6
2 2 2 0
1 1 1 4
no solution or say infinite solution
10 years agoHelpfull: Yes(2) No(0)
@rakesh..................why r u considering downhill also while calculating???
and given one way in the quesstion i think its.. 12
10 years agoHelpfull: Yes(1) No(0)
avg speed=2*3*6/(3+6)=4
so speed of man=4km/hr
distance=4*6=24km
11 years agoHelpfull: Yes(0) No(4)
short data
10 years agoHelpfull: Yes(0) No(0)

Q. A man starts walking at 3 pm . ha walks at a speed of 4 km/hr on level ground and at a speed of 3 km/hr on uphill , 6 km/hr downhill and then 4 km/hr on level ground to reach home at 9 pm. What is the distance covered on one way