a person walks from home to office 5km hr he late by office 3 mins if the person walks 6km

x=dist b/w home and office
t=time taken to reach the office

x=5*(t+3)--->eq1
x=6*(t-7)--->eq2

5*(t+3)=6*(t-7)
by solving
t=57
putting the value in any eq
x=300km
10 years agoHelpfull: Yes(4) No(3)
sorry....change min into hr
so it will be 57/60
so ans will be 5 km
10 years agoHelpfull: Yes(2) No(0)
let required distance be x km
difference in times taken at two speed=10 min=1/6
x/5-x/6=1/6
6x-5x=5
x=5km
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5,diff. in time=10 min=1/6 hour.So x/5-x/6=1/6=> x=5.
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d1=5*518*3*60=250
d2=6*518*7*60=700
total distance=700+250=950m
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let distance be 'x'
(x/5)-(x/6)=10
x=300
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a small change in my solution
we have to convert min into hrs
so 300/60=5kms
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as the distance is same so(v1/v2)=(t2/t1)
where t2=t+7 and t1=t-7 where t=exact time required for any ideal speed!
hence t=57;therefor dis=5*(57+3)/60=5
10 years agoHelpfull: Yes(0) No(0)

a person walks from home to office 5km/hr he late by office 3 mins if the person walks 6km/hr he will reach 7 mins earlier to office. what is the distance between home to office